Student Solutions Manual for SINGLE VARIABLE CALCULUS

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SEVENTH EDITION

DANIEL ANDERSON
University of Iowa

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JEFFERY A. COLE
Anoka-Ramsey Community College

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DANIEL DRUCKER
Wayne State University

Australia . Brazil . Japan . Korea . Mexico . Singapore . Spain . United Kingdom . United States

ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States
Copyright Act, without the prior written permission of the publisher except as may be permitted by the license terms below. For product information and technology assistance, contact us at
Cengage Learning Customer & Sales Support,

ISBN-13: 978-0-8400-4949-0
ISBN-10: 0-8400-4949-8
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© 2012 Brooks/Cole, Cengage Learning

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Printed in the United States of America
1 2 3 4 5 6 7 8

15 14 13 12 11

PREFACE

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This Student Solutions Manual contains strategies for solving and solutions to selected exercises in the text Single Variable Calculus, Seventh Edition, by James Stewart. It contains solutions to the odd-numbered exercises in each section, the review sections, the True-False Quizzes, and the
Problem Solving sections, as well as solutions to all the exercises in the Concept Checks.
This manual is a text supplement and should be read along with the text. You should read all exercise solutions in this manual because many concept explanations are given and then used in subsequent solutions. All concepts necessary to solve a particular problem are not reviewed for every exercise. If you are having difficulty with a previously covered concept, refer back to the section where it was covered for more complete help.
A significant number of today’s students are involved in various outside activities, and find it difficult, if not impossible, to attend all class sessions; this manual should help meet the needs of these students. In addition, it is our hope that this manual’s solutions will enhance the understanding of all readers of the material and provide insights to solving other exercises.
We use some nonstandard notation in order to save space. If you see a symbol that you don’t recognize, refer to the Table of Abbreviations and Symbols on page v.
We appreciate feedback concerning errors, solution correctness or style, and manual style. Any comments may be sent directly to jeff.cole@anokaramsey.edu, or in care of the publisher:
Brooks/Cole, Cengage Learning, 20 Davis Drive, Belmont CA 94002-3098.
We would like to thank Stephanie Kuhns and Kathi Townes, of TECHarts, for their production services; and Elizabeth Neustaetter of Brooks/Cole, Cengage Learning, for her patience and support. All of these people have provided invaluable help in creating this manual.

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Jeffery A. Cole
Anoka-Ramsey Community College
James Stewart
McMaster University and University of Toronto
Daniel Drucker
Wayne State University
Daniel Anderson
University of Iowa

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ABBREVIATIONS AND SYMBOLS
CD

concave downward

CU

concave upward

D

the domain of f

FDT

First Derivative Test

HA

horizontal asymptote(s)

I

interval of convergence
Increasing/Decreasing Test

IP

inflection point(s)

R

radius of convergence

VA

vertical asymptote(s)

CAS

=

indicates the use of a computer algebra system.

H

indicates the use of l’Hospital’s Rule.

j

indicates the use of Formula j in the Table of Integrals in the back endpapers.

s

indicates the use of the substitution {u = sin x, du = cos x dx}.

=
=
c

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indicates the use of the substitution {u = cos x, du = − sin x dx}.

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I/D

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CONTENTS
■

1

DIAGNOSTIC TESTS

■

1

FUNCTIONS AND LIMITS

9

1.1

Four Ways to Represent a Function

1.2

Mathematical Models: A Catalog of Essential Functions

1.3

New Functions from Old Functions

1.4

The Tangent and Velocity Problems

1.5

The Limit of a Function

1.6

Calculating Limits Using the Limit Laws

1.7

The Precise Definition of a Limit

1.8

Continuity

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24

29

34

38

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51

53

DERIVATIVES

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18

43

Principles of Problem Solving

2

2.1

Derivatives and Rates of Change

2.2

The Derivative as a Function

2.3

Differentiation Formulas

2.4

Derivatives of Trigonometric Functions

2.5

The Chain Rule

2.6

Implicit Differentiation

2.7

Rates of Change in the Natural and Social Sciences

2.8

Related Rates

2.9

Linear Approximations and Differentials
Review

Problems Plus

14

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9

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Review

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53

58

64
71

74
79
85

89
94

97

105

vii

CONTENTS

3

■

111

APPLICATIONS OF DIFFERENTIATION
3.1

Maximum and Minimum Values

3.2

The Mean Value Theorem

3.3

How Derivatives Affect the Shape of a Graph

3.4

Limits at Infinity; Horizontal Asymptotes

3.5

Summary of Curve Sketching

3.6

Graphing with Calculus and Calculators

3.7

Optimization Problems

3.8

Newton’s Method

3.9

Antiderivatives

128

135
144

152

162
167

INTEGRALS

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183

4.1

Areas and Distances

189

4.2

The Definite Integral

194

4.3

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4.4
4.5

The Fundamental Theorem of Calculus

199

Indefinite Integrals and the Net Change Theorem
The Substitution Rule
Review

Problems Plus

5

■

118

172

Problems Plus

4

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Review

111

208

212

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217

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viii

APPLICATIONS OF INTEGRATION
5.1

Areas Between Curves

5.2

Volumes

5.3

Volumes by Cylindrical Shells

5.4

Work

5.5

Average Value of a Function
Review

Problems Plus

219

226
234

238
242

247

241

219

205

CONTENTS

6

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INVERSE FUNCTIONS:

6.1

Inverse Functions

6.2

Exponential Functions and
Their Derivatives 254
Logarithmic
Functions 261
Derivatives of Logarithmic
Functions 264

6.2*

6.5

Exponential Growth and Decay

286

6.6

Inverse Trigonometric Functions

288

6.7

Hyperbolic Functions

6.8

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Exponential, Logarithmic, and Inverse Trigonometric Functions

Indeterminate Forms and L’Hospital’s Rule
Review

Problems Plus

■

6.3*
6.4*

294

306

315

7.1
7.2
7.3

Integration by Parts

319

Trigonometric Integrals

Trigonometric Substitution

325
329

Integration of Rational Functions by Partial Fractions

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7.4

298

319

TECHNIQUES OF INTEGRATION

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7

The Natural Logarithmic
Function 270
The Natural Exponential
Function 277
General Logarithmic and
Exponential Functions 283

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6.4

251

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6.3

334

Strategy for Integration

7.6

Integration Using Tables and Computer Algebra Systems

7.7

Approximate Integration

7.8

Improper Integrals

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7.5

Review

343
349

353

361

368

Problems Plus 375

8

■

251

FURTHER APPLICATIONS OF INTEGRATION
8.1

Arc Length

8.2

Area of a Surface of Revolution

8.3

Applications to Physics and Engineering

379
382
386

379

■

ix

CONTENTS

8.4

Applications to Economics and Biology

8.5

Probability
Review

Problems Plus

■

394

401

405

DIFFERENTIAL EQUATIONS
9.1

Modeling with Differential Equations

9.2

Direction Fields and Euler’s Method

9.3

Separable Equations

9.4

Models for Population Growth

9.5

Linear Equations

9.6

Predator-Prey Systems

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425

433

PARAMETRIC EQUATIONS AND POLAR COORDINATES

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417

421

427

Problems Plus

10

406

411

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Review

405

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9

393

397

10.1

Curves Defined by Parametric Equations

10.2

Calculus with Parametric Curves

10.3

Polar Coordinates

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443

Areas and Lengths in Polar Coordinates

10.5

Conic Sections

10.6

Conic Sections in Polar Coordinates
Review

■

456

462
468

471

Problems Plus

11

437

449

10.4

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479

INFINITE SEQUENCES AND SERIES
11.1

Sequences

11.2

Series

11.3

The Integral Test and Estimates of Sums

11.4

The Comparison Tests

481

481

487
498

495

437

CONTENTS

11.5

Alternating Series

11.6

Absolute Convergence and the Ratio and Root Tests

11.7

Strategy for Testing Series

11.8

Power Series

11.9

Representations of Functions as Power Series

11.10

Taylor and Maclaurin Series

11.11

Applications of Taylor Polynomials

510
519
526

541

547

APPENDIXES

Numbers, Inequalities, and Absolute Values

B

Coordinate Geometry and Lines

C

Graphs of Second-Degree Equations

D

Trigonometry

E

Sigma Notation

G

Graphing Calculators and Computers

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A

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552

554

558

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Complex Numbers

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H

514

533

Problems Plus

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504

508

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Review

501

564

561

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DIAGNOSTIC TESTS
Test A Algebra
1. (a) (−3)4 = (−3)(−3)(−3)(−3) = 81

(c) 3−4 =
(e)

 2 −2
3

1
1
=
34
81
 2
= 3 =
2

2. (a) Note that

(b) −34 = −(3)(3)(3)(3) = −81
523
= 523−21 = 52 = 25
521
1
1
1
1
(f ) 16−34 = 34 =  √ 3 = 3 =
4
2
8
16
16

(d)
9
4

√
√
√
√
√
√
√
√
√
√
√
200 = 100 · 2 = 10 2 and 32 = 16 · 2 = 4 2. Thus 200 − 32 = 10 2 − 4 2 = 6 2.

332  3
2  −12

−2

=



2  −12
332  3

2

=


(2  −12 )2
4  −1
4
=
= 3 6 = 7
93  6
9  
9
(332  3 )2

3. (a) 3( + 6) + 4(2 − 5) = 3 + 18 + 8 − 20 = 11 − 2

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(c)



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(b) (33 3 )(42 )2 = 33 3 162 4 = 485 7

(c)

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(b) ( + 3)(4 − 5) = 42 − 5 + 12 − 15 = 42 + 7 − 15

√
√ 2
√ √
√  √
√  √ 2 √ √
+ 
−  =
 −  +  −
 =−

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Or: Use the formula for the difference of two squares to see that

√
√ √
√  √ 2 √ 2
+ 
−  =
 −
 =  − .

(d) (2 + 3)2 = (2 + 3)(2 + 3) = 42 + 6 + 6 + 9 = 42 + 12 + 9.

Note: A quicker way to expand this binomial is to use the formula ( + )2 = 2 + 2 + 2 with  = 2 and  = 3:
(2 + 3)2 = (2)2 + 2(2)(3) + 32 = 42 + 12 + 9

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(e) See Reference Page 1 for the binomial formula ( + )3 = 3 + 32  + 32 + 3 . Using it, we get
( + 2)3 = 3 + 32 (2) + 3(22 ) + 23 = 3 + 62 + 12 + 8.

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4. (a) Using the difference of two squares formula, 2 − 2 = ( + )( − ), we have

42 − 25 = (2)2 − 52 = (2 + 5)(2 − 5).
(b) Factoring by trial and error, we get 22 + 5 − 12 = (2 − 3)( + 4).
(c) Using factoring by grouping and the difference of two squares formula, we have
3 − 32 − 4 + 12 = 2 ( − 3) − 4( − 3) = (2 − 4)( − 3) = ( − 2)( + 2)( − 3).
(d) 4 + 27 = (3 + 27) = ( + 3)(2 − 3 + 9)
This last expression was obtained using the sum of two cubes formula, 3 + 3 = ( + )(2 −  + 2 ) with  = 

and  = 3. [See Reference Page 1 in the textbook.]

(e) The smallest exponent on  is − 1 , so we will factor out −12 .
2
332 − 912 + 6−12 = 3−12 (2 − 3 + 2) = 3−12 ( − 1)( − 2)
(f ) 3  − 4 = (2 − 4) = ( − 2)( + 2) c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

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5. (a)

(b)

(c)

DIAGNOSTIC TESTS

( + 1)( + 2)
+2
2 + 3 + 2
=
=
2 −  − 2
( + 1)( − 2)
−2
(2 + 1)( − 1)  + 3
−1
22 −  − 1  + 3
·
=
·
=
2 − 9
2 + 1
( − 3)( + 3) 2 + 1
−3
2
+1
2
+1
2
+1 −2
2 − ( + 1)( − 2)
−
=
−
=
−
·
=
−4
+2
( − 2)( + 2)
+2
( − 2)( + 2)
+2 −2
( − 2)( + 2)

2

=

2 − (2 −  − 2)
+2
1
=
=
( + 2)( − 2)
( + 2)( − 2)
−2





−
−
 2 − 2
( − )( + )
+



 
=
·
=
=
=
= −( + )
(d)
1
1
1
1 
−
−( − )
−1
−
−




√
√
√
√
√
√
√
√
√
10
10
5+2
50 + 2 10
5 2 + 2 10
= 5 2 + 2 10
=
= √
·√
= √ 2
2
5−4
5−2
5−2
5+2
5 −2

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6. (a) √

1
4



1
4

+1−

2

= + 1 +
2

3
4

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

7. (a) 2 +  + 1 = 2 +  +

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√
√
√
4+−2
4+−2
4++2
4+−4

1
 = √
 = √
= √
=
·√
(b)


4++2
 4++2
 4++2
4++2
(b) 22 − 12 + 11 = 2(2 − 6) + 11 = 2(2 − 6 + 9 − 9) + 11 = 2(2 − 6 + 9) − 18 + 11 = 2( − 3)2 − 7

(b)

2
2 − 1
=
+1


⇔  + 1  = 14 − 5 ⇔
2

3

2

=9 ⇔ =

2
3

·9 ⇔ =6

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8. (a)  + 5 = 14 − 1 
2

⇒ 22 = (2 − 1)( + 1) ⇔ 22 = 22 +  − 1 ⇔  = 1

(c) 2 −  − 12 = 0 ⇔ ( + 3)( − 4) = 0 ⇔  + 3 = 0 or  − 4 = 0 ⇔  = −3 or  = 4

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(d) By the quadratic formula, 22 + 4 + 1 = 0 ⇔
√ 


√
√
√
2 −2 ± 2
−4 ± 42 − 4(2)(1)
−4 ± 8
−4 ± 2 2
−2 ± 2
=
=
=
=
= −1 ±
=
2(2)
4
4
4
2

1
2

√
2.

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2

(e) 4 − 32 + 2 = 0 ⇔ (2 − 1)(2 − 2) = 0 ⇔ 2 − 1 = 0 or 2 − 2 = 0 ⇔ 2 = 1 or 2 = 2 ⇔
√
 = ±1 or  = ± 2
(f ) 3 | − 4| = 10 ⇔ | − 4| =

10
3

⇔  − 4 = − 10 or  − 4 =
3

(g) Multiplying through 2(4 − )−12 − 3

⇔ =

2
3

or  =

22
3

√
4 −  = 0 by (4 − )12 gives 2 − 3(4 − ) = 0 ⇔

2 − 12 + 3 = 0 ⇔ 5 − 12 = 0 ⇔ 5 = 12 ⇔  =
9. (a) −4  5 − 3 ≤ 17

10
3

12
.
5

⇔ −9  −3 ≤ 12 ⇔ 3   ≥ −4 or −4 ≤   3.

In interval notation, the answer is [−4 3).

(b) 2  2 + 8 ⇔ 2 − 2 − 8  0 ⇔ ( + 2)( − 4)  0. Now, ( + 2)( − 4) will change sign at the critical values  = −2 and  = 4. Thus the possible intervals of solution are (−∞ −2), (−2 4), and (4 ∞). By choosing a single test value from each interval, we see that (−2 4) is the only interval that satisfies the inequality.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

TEST B ANALYTIC GEOMETRY

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(c) The inequality ( − 1)( + 2)  0 has critical values of −2 0 and 1. The corresponding possible intervals of solution are (−∞ −2), (−2 0), (0 1) and (1 ∞). By choosing a single test value from each interval, we see that both intervals
(−2 0) and (1 ∞) satisfy the inequality. Thus, the solution is the union of these two intervals: (−2 0) ∪ (1 ∞).
(d) | − 4|  3 ⇔ −3   − 4  3 ⇔ 1    7. In interval notation, the answer is (1 7).
2 − 3
≤1 ⇔
+1

2 − 3
−1 ≤0 ⇔
+1

Now, the expression

−4 may change signs at the critical values  = −1 and  = 4, so the possible intervals of solution
+1

(e)

2 − 3
+1
−
≤0 ⇔
+1
+1

2 − 3 −  − 1
≤0 ⇔
+1

−4
≤ 0.
+1

are (−∞ −1), (−1 4], and [4 ∞). By choosing a single test value from each interval, we see that (−1 4] is the only

interval that satisfies the inequality.

10. (a) False. In order for the statement to be true, it must hold for all real numbers, so, to show that the statement is false, pick

e

 = 1 and  = 2 and observe that (1 + 2)2 6= 12 + 22 . In general, ( + )2 = 2 + 2 +  2 .

√
12 + 22 6= 1 + 2.

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(c) False. To see this, let  = 1 and  = 2, then

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(b) True as long as  and  are nonnegative real numbers. To see this, think in terms of the laws of exponents:
√ √
√
 = ()12 = 12 12 =  .

(d) False. To see this, let  = 1 and  = 2, then

(f ) True since

1
1
1
6= − .
2−3
2
3

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(e) False. To see this, let  = 2 and  = 3, then

1 + 1(2)
6= 1 + 1.
2


1
1
· =
, as long as  6= 0 and  −  6= 0.
 −  
−

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Test B Analytic Geometry

1. (a) Using the point (2 −5) and  = −3 in the point-slope equation of a line,  − 1 = ( − 1 ), we get

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 − (−5) = −3( − 2) ⇒  + 5 = −3 + 6 ⇒  = −3 + 1.
(b) A line parallel to the -axis must be horizontal and thus have a slope of 0. Since the line passes through the point (2 −5), the -coordinate of every point on the line is −5, so the equation is  = −5.

(c) A line parallel to the -axis is vertical with undefined slope. So the -coordinate of every point on the line is 2 and so the equation is  = 2.
(d) Note that 2 − 4 = 3 ⇒ −4 = −2 + 3 ⇒  = 1  − 3 . Thus the slope of the given line is  = 1 . Hence, the
2
4
2
slope of the line we’re looking for is also
So the equation of the line is  − (−5) =

1
2

(since the line we’re looking for is required to be parallel to the given line).

1
2 (

− 2) ⇒  + 5 = 1  − 1 ⇒  = 1  − 6.
2
2

2. First we’ll find the distance between the two given points in order to obtain the radius, , of the circle:

=



√
[3 − (−1)]2 + (−2 − 4)2 = 42 + (−6)2 = 52. Next use the standard equation of a circle,

( − )2 + ( − )2 = 2 , where ( ) is the center, to get ( + 1)2 + ( − 4)2 = 52.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

DIAGNOSTIC TESTS

3. We must rewrite the equation in standard form in order to identify the center and radius. Note that

2 +  2 − 6 + 10 + 9 = 0 ⇒ 2 − 6 + 9 +  2 + 10 = 0. For the left-hand side of the latter equation, we factor the first three terms and complete the square on the last two terms as follows: 2 − 6 + 9 +  2 + 10 = 0 ⇒
( − 3)2 +  2 + 10 + 25 = 25 ⇒ ( − 3)2 + ( + 5)2 = 25. Thus, the center of the circle is (3 −5) and the radius is 5.
4. (a) (−7 4) and (5 −12)

⇒  =

−12 − 4
−16
4
=
=−
5 − (−7)
12
3

(b)  − 4 = − 4 [ − (−7)] ⇒  − 4 = − 4  −
3
3

⇒ 3 − 12 = −4 − 28 ⇒ 4 + 3 + 16 = 0. Putting  = 0,

28
3

we get 4 + 16 = 0, so the -intercept is −4, and substituting 0 for  results in a -intercept of − 16 .
3
(c) The midpoint is obtained by averaging the corresponding coordinates of both points:

−7+5 4+(−12)
2 
2



√
√
[5 − (−7)]2 + (−12 − 4)2 = 122 + (−16)2 = 144 + 256 = 400 = 20



= (−1 −4).

e

(d)  =



perpendicular bisector passes through (−1 −4) and has slope

3
4

al

(e) The perpendicular bisector is the line that intersects the line segment  at a right angle through its midpoint. Thus the
[the slope is obtained by taking the negative reciprocal of

rS

the answer from part (a)]. So the perpendicular bisector is given by  + 4 = 3 [ − (−1)] or 3 − 4 = 13.
4
(f ) The center of the required circle is the midpoint of , and the radius is half the length of , which is 10. Thus, the equation is ( + 1)2 + ( + 4)2 = 100.

Fo

5. (a) Graph the corresponding horizontal lines (given by the equations  = −1 and

 = 3) as solid lines. The inequality  ≥ −1 describes the points ( ) that lie on or above the line  = −1. The inequality  ≤ 3 describes the points ( ) that lie on or below the line  = 3. So the pair of inequalities −1 ≤  ≤ 3

ot

describes the points that lie on or between the lines  = −1 and  = 3.
(b) Note that the given inequalities can be written as −4    4 and −2    2,

N

4

respectively. So the region lies between the vertical lines  = −4 and  = 4 and between the horizontal lines  = −2 and  = 2. As shown in the graph, the region common to both graphs is a rectangle (minus its edges) centered at the origin. (c) We first graph  = 1 − 1  as a dotted line. Since   1 − 1 , the points in the
2
2 region lie below this line.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

TEST C FUNCTIONS

¤

5

(d) We first graph the parabola  = 2 − 1 using a solid curve. Since  ≥ 2 − 1, the points in the region lie on or above the parabola.


(e) We graph the circle 2 +  2 = 4 using a dotted curve. Since 2 +  2  2, the region consists of points whose distance from the origin is less than 2, that is, the points that lie inside the circle.

(f ) The equation 92 + 16 2 = 144 is an ellipse centered at (0 0). We put it in
2
2
+
= 1. The -intercepts are
16
9

e

standard form by dividing by 144 and get

rS

al

√ located at a distance of 16 = 4 from the center while the -intercepts are a
√
distance of 9 = 3 from the center (see the graph).

Test C Functions

1. (a) Locate −1 on the -axis and then go down to the point on the graph with an -coordinate of −1. The corresponding

Fo

-coordinate is the value of the function at  = −1, which is −2. So,  (−1) = −2.

(b) Using the same technique as in part (a), we get  (2) ≈ 28.
(c) Locate 2 on the -axis and then go left and right to find all points on the graph with a -coordinate of 2. The corresponding

ot

-coordinates are the -values we are searching for. So  = −3 and  = 1.
(d) Using the same technique as in part (c), we get  ≈ −25 and  ≈ 03.

N

(e) The domain is all the -values for which the graph exists, and the range is all the -values for which the graph exists.
Thus, the domain is [−3 3], and the range is [−2 3].
2. Note that  (2 + ) = (2 + )3 and (2) = 23 = 8. So the difference quotient becomes

(2 + )3 − 8
8 + 12 + 62 + 3 − 8
12 + 62 + 3
(12 + 6 + 2 )
 (2 + ) −  (2)
=
=
=
=
= 12 + 6 + 2 .





3. (a) Set the denominator equal to 0 and solve to find restrictions on the domain: 2 +  − 2 = 0

⇒

( − 1)( + 2) = 0 ⇒  = 1 or  = −2. Thus, the domain is all real numbers except 1 or −2 or, in interval notation, (−∞ −2) ∪ (−2 1) ∪ (1 ∞).

(b) Note that the denominator is always greater than or equal to 1, and the numerator is defined for all real numbers. Thus, the domain is (−∞ ∞).
(c) Note that the function  is the sum of two root functions. So  is defined on the intersection of the domains of these two root functions. The domain of a square root function is found by setting its radicand greater than or equal to 0. Now, c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

DIAGNOSTIC TESTS

4 −  ≥ 0 ⇒  ≤ 4 and 2 − 1 ≥ 0 ⇒ ( − 1)( + 1) ≥ 0 ⇒  ≤ −1 or  ≥ 1. Thus, the domain of
 is (−∞ −1] ∪ [1 4].
4. (a) Reflect the graph of  about the -axis.

(b) Stretch the graph of  vertically by a factor of 2, then shift 1 unit downward.
(c) Shift the graph of  right 3 units, then up 2 units.
5. (a) Make a table and then connect the points with a smooth curve:



−2

−1

0

1

2



−8

−1

0

1

8

rS

al

e

(b) Shift the graph from part (a) left 1 unit.

Fo

(c) Shift the graph from part (a) right 2 units and up 3 units.

(d) First plot  = 2 . Next, to get the graph of  () = 4 − 2 ,

ot

reflect  about the x-axis and then shift it upward 4 units.

N

6

(e) Make a table and then connect the points with a smooth curve:


0

1

4

9



0

1

2

3

(f ) Stretch the graph from part (e) vertically by a factor of two.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

TEST D TRIGONOMETRY

¤

(g) First plot  = 2 . Next, get the graph of  = −2 by reflecting the graph of
 = 2 about the x-axis.

(h) Note that  = 1 + −1 = 1 + 1. So first plot  = 1 and then shift it upward 1 unit.

6. (a)  (−2) = 1 − (−2)2 = −3 and  (1) = 2(1) + 1 = 3

e

(b) For  ≤ 0 plot  () = 1 − 2 and, on the same plane, for   0 plot the graph

rS

al

of  () = 2 + 1.

7. (a) ( ◦ )() =  (()) = (2 − 3) = (2 − 3)2 + 2(2 − 3) − 1 = 42 − 12 + 9 + 4 − 6 − 1 = 42 − 8 + 2

Fo

(b) ( ◦ )() = ( ()) = (2 + 2 − 1) = 2(2 + 2 − 1) − 3 = 22 + 4 − 2 − 3 = 22 + 4 − 5
(c) ( ◦  ◦ )() = ((())) = ((2 − 3)) = (2(2 − 3) − 3) = (4 − 9) = 2(4 − 9) − 3
= 8 − 18 − 3 = 8 − 21

ot

Test D Trigonometry

   300
5
=
=
180◦
180
3


◦
5 180
5
=
= 150◦
2. (a)
6
6


  
18

=−
=−
180◦
180
10


360◦
180◦
=
≈ 1146◦
(b) 2 = 2



1. (a) 300◦ = 300◦

N

(b) −18◦ = −18◦

3. We will use the arc length formula,  = , where  is arc length,  is the radius of the circle, and  is the measure of the

central angle in radians. First, note that 30◦ = 30◦
4. (a) tan(3) =

   
 
= . So  = (12)
= 2 cm.
◦
180
6
6

√ 
√ 
3 You can read the value from a right triangle with sides 1, 2, and 3.

(b) Note that 76 can be thought of as an angle in the third quadrant with reference angle 6. Thus, sin(76) = − 1 ,
2
since the sine function is negative in the third quadrant.
(c) Note that 53 can be thought of as an angle in the fourth quadrant with reference angle 3. Thus, sec(53) =

1
1
=
= 2, since the cosine function is positive in the fourth quadrant. cos(53) 12

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

7

¤

DIAGNOSTIC TESTS

5. sin  = 24
6. sin  =

1
3

⇒  = 24 sin 

and cos  = 24 ⇒  = 24 cos 

and sin2  + cos2  = 1 ⇒ cos  =

So, using the sum identity for the sine, we have


1−

1
9

=

2

√
2
. Also, cos  =
3

4
5

⇒ sin  =


1−

16
25

√
√
√ 
2 2 3
4+6 2
1
1 4
· =
=
4+6 2 sin( + ) = sin  cos  + cos  sin  = · +
3 5
3
5
15
15
7. (a) tan  sin  + cos  =

2 sin (cos ) sin 
2 tan 
=
=2 cos2  = 2 sin  cos  = sin 2
1 + tan2  sec2  cos 

8. sin 2 = sin 

⇔ 2 sin  cos  = sin  ⇔ 2 sin  cos  − sin  = 0 ⇔ sin  (2 cos  − 1) = 0 ⇔

sin  = 0 or cos  =

1
2

⇒  = 0,


,
3

,

5
,
3

2.

e

(b)

sin  sin2  cos2 
1
sin  + cos  =
+
=
= sec  cos  cos  cos  cos 

al

9. We first graph  = sin 2 (by compressing the graph of sin 

ot

Fo

rS

by a factor of 2) and then shift it upward 1 unit.

N

8

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

= 3.
5

1

FUNCTIONS AND LIMITS

1.1 Four Ways to Represent a Function
1. The functions  () =  +

and  are equal.

√
√
2 −  and () =  + 2 −  give exactly the same output values for every input value, so 

3. (a) The point (1 3) is on the graph of , so  (1) = 3.

(b) When  = −1,  is about −02, so  (−1) ≈ −02.
(c)  () = 1 is equivalent to  = 1 When  = 1, we have  = 0 and  = 3.

e

(d) A reasonable estimate for  when  = 0 is  = −08.
(e) The domain of  consists of all -values on the graph of  . For this function, the domain is −2 ≤  ≤ 4, or [−2 4].

al

The range of  consists of all -values on the graph of  . For this function, the range is −1 ≤  ≤ 3, or [−1 3].
(f ) As  increases from −2 to 1,  increases from −1 to 3. Thus,  is increasing on the interval [−2 1].

rS

5. From Figure 1 in the text, the lowest point occurs at about ( ) = (12 −85). The highest point occurs at about (17 115).

Thus, the range of the vertical ground acceleration is −85 ≤  ≤ 115. Written in interval notation, we get [−85 115]. the Vertical Line Test.

Fo

7. No, the curve is not the graph of a function because a vertical line intersects the curve more than once. Hence, the curve fails

9. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is [−3 2] and the range

is [−3 −2) ∪ [−1 3].

ot

11. The person’s weight increased to about 160 pounds at age 20 and stayed fairly steady for 10 years. The person’s weight

dropped to about 120 pounds for the next 5 years, then increased rapidly to about 170 pounds. The next 30 years saw a gradual

N

increase to 190 pounds. Possible reasons for the drop in weight at 30 years of age: diet, exercise, health problems.
13. The water will cool down almost to freezing as the ice melts. Then, when

the ice has melted, the water will slowly warm up to room temperature.

15. (a) The power consumption at 6 AM is 500 MW which is obtained by reading the value of power  when  = 6 from the

graph. At 6 PM we read the value of  when  = 18 obtaining approximately 730 MW
(b) The minimum power consumption is determined by finding the time for the lowest point on the graph,  = 4 or 4 AM. The maximum power consumption corresponds to the highest point on the graph, which occurs just before  = 12 or right before noon. These times are reasonable, considering the power consumption schedules of most individuals and businesses. c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

9

10

¤

CHAPTER 1

FUNCTIONS AND LIMITS

17. Of course, this graph depends strongly on the

19. As the price increases, the amount sold

geographical location!

decreases.

21.

(b) From the graph, we estimate the number of US cell-phone

e

23. (a)

al

subscribers to be about 126 million in 2001 and 207 million

25.  () = 32 −  + 2

Fo

rS

in 2005.

ot

 (2) = 3(2)2 − 2 + 2 = 12 − 2 + 2 = 12

 (−2) = 3(−2)2 − (−2) + 2 = 12 + 2 + 2 = 16

N

 () = 32 −  + 2

 (−) = 3(−)2 − (−) + 2 = 32 +  + 2
 ( + 1) = 3( + 1)2 − ( + 1) + 2 = 3(2 + 2 + 1) −  − 1 + 2 = 32 + 6 + 3 −  + 1 = 32 + 5 + 4
2 () = 2 ·  () = 2(32 −  + 2) = 62 − 2 + 4
 (2) = 3(2)2 − (2) + 2 = 3(42 ) − 2 + 2 = 122 − 2 + 2
 (2 ) = 3(2 )2 − (2 ) + 2 = 3(4 ) − 2 + 2 = 34 − 2 + 2

2 


[ ()]2 = 32 −  + 2 = 32 −  + 2 32 −  + 2

= 94 − 33 + 62 − 33 + 2 − 2 + 62 − 2 + 4 = 94 − 63 + 132 − 4 + 4

 ( + ) = 3( + )2 − ( + ) + 2 = 3(2 + 2 + 2 ) −  −  + 2 = 32 + 6 + 32 −  −  + 2
27.  () = 4 + 3 − 2 , so  (3 + ) = 4 + 3(3 + ) − (3 + )2 = 4 + 9 + 3 − (9 + 6 + 2 ) = 4 − 3 − 2 ,

and

(4 − 3 − 2 ) − 4
(−3 − )
 (3 + ) −  (3)
=
=
= −3 − .



c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.1

FOUR WAYS TO REPRESENT A FUNCTION

¤

11

1
1
−
−
 () −  ()

 =  =  −  = −1( − ) = − 1
29.
=
−
−
−
( − )
( − )

31.  () = ( + 4)(2 − 9) is defined for all  except when 0 = 2 − 9

⇔ 0 = ( + 3)( − 3) ⇔  = −3 or 3, so the

domain is { ∈ R |  6= −3 3} = (−∞ −3) ∪ (−3 3) ∪ (3 ∞).
33.  () =


√
3
2 − 1 is defined for all real numbers. In fact 3 (), where () is a polynomial, is defined for all real numbers.

Thus, the domain is R or (−∞ ∞).
35. () = 1

√
4
2 − 5 is defined when 2 − 5  0

( − 5)  0. Note that 2 − 5 6= 0 since that would result in

⇔

division by zero. The expression ( − 5) is positive if   0 or   5. (See Appendix A for methods for solving inequalities.) Thus, the domain is (−∞ 0) ∪ (5 ∞).

√
√
√
√
√
2 −  is defined when  ≥ 0 and 2 −  ≥ 0. Since 2 −  ≥ 0 ⇒ 2 ≥  ⇒
≤2 ⇒

e

37.  () =

al

0 ≤  ≤ 4, the domain is [0 4].

39.  () = 2 − 04 is defined for all real numbers, so the domain is R,

rS

or (−∞ ∞) The graph of  is a line with slope −04 and -intercept 2.

Fo

41.  () = 2 + 2 is defined for all real numbers, so the domain is R, or

(−∞ ∞). The graph of  is a parabola opening upward since the coefficient of 2 is positive. To find the -intercepts, let  = 0 and solve for . 0 = 2 + 2 = (2 + ) ⇒  = 0 or  = −2. The -coordinate of

ot

the vertex is halfway between the -intercepts, that is, at  = −1. Since
 (−1) = 2(−1) + (−1)2 = −2 + 1 = −1, the vertex is (−1 −1).

N

√
 − 5 is defined when  − 5 ≥ 0 or  ≥ 5, so the domain is [5 ∞).
√
Since  =  − 5 ⇒  2 =  − 5 ⇒  =  2 + 5, we see that  is the

43. () =

top half of a parabola.

3 + ||
. Since || =
45. () =

 3 + 



() =
 3 − 






−

if  ≥ 0 if   0

 4



=
 2

if   0

if   0

, we have

if   0
=
if   0



4

if   0

2

if   0

Note that  is not defined for  = 0. The domain is (−∞ 0) ∪ (0 ∞).

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

12

¤

CHAPTER 1

47.  () =



FUNCTIONS AND LIMITS

+2

if   0

1−

if  ≥ 0

The domain is R.

49.  () =


 + 2 if  ≤ −1

if   −1

2

Note that for  = −1, both  + 2 and 2 are equal to 1.
The domain is R.

2 − 1 and an equation of the line
2 − 1

e

51. Recall that the slope  of a line between the two points (1  1 ) and (2  2 ) is  =

al

connecting those two points is  − 1 = ( − 1 ). The slope of the line segment joining the points (1 −3) and (5 7) is

53. We need to solve the given equation for .

11
,
2

1 ≤  ≤ 5.

rS

7 − (−3)
5
= , so an equation is  − (−3) = 5 ( − 1). The function is  () = 5  −
2
2
5−1
2

√
 + ( − 1)2 = 0 ⇔ ( − 1)2 = − ⇔  − 1 = ± − ⇔

√
−. The expression with the positive radical represents the top half of the parabola, and the one with the negative
√
radical represents the bottom half. Hence, we want  () = 1 − −. Note that the domain is  ≤ 0.

Fo

 =1±

55. For 0 ≤  ≤ 3, the graph is the line with slope −1 and -intercept 3, that is,  = − + 3. For 3   ≤ 5, the graph is the line

ot

with slope 2 passing through (3 0); that is,  − 0 = 2( − 3), or  = 2 − 6. So the function is

− + 3 if 0 ≤  ≤ 3
 () =
2 − 6 if 3   ≤ 5

N

57. Let the length and width of the rectangle be  and  . Then the perimeter is 2 + 2 = 20 and the area is  =  .

Solving the first equation for  in terms of  gives  =

20 − 2
= 10 − . Thus, () = (10 − ) = 10 − 2 . Since
2

lengths are positive, the domain of  is 0    10. If we further restrict  to be larger than  , then 5    10 would be the domain.
59. Let the length of a side of the equilateral triangle be . Then by the Pythagorean Theorem, the height  of the triangle satisfies
√
 1 2
 = 2 , so that  2 = 2 − 1 2 = 3 2 and  = 23 . Using the formula for the area  of a triangle,
2
4
4
√  √
 = 1 (base)(height), we obtain () = 1 () 23  = 43 2 , with domain   0.
2
2

2 +

61. Let each side of the base of the box have length , and let the height of the box be . Since the volume is 2, we know that

2 = 2 , so that  = 22 , and the surface area is  = 2 + 4. Thus, () = 2 + 4(22 ) = 2 + (8), with domain   0.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.1

FOUR WAYS TO REPRESENT A FUNCTION

¤

13

63. The height of the box is  and the length and width are  = 20 − 2,  = 12 − 2. Then  =   and so

 () = (20 − 2)(12 − 2)() = 4(10 − )(6 − )() = 4(60 − 16 + 2 ) = 43 − 642 + 240.
The sides ,  , and  must be positive. Thus,   0 ⇔ 20 − 2  0 ⇔   10;

  0 ⇔ 12 − 2  0 ⇔   6; and   0. Combining these restrictions gives us the domain 0    6.
65. We can summarize the amount of the fine with a

piecewise defined function.

 15(40 − )

 () = 0


15( − 65)

if 0 ≤   40

if 40 ≤  ≤ 65

if   65

67. (a)

(b) On $14,000, tax is assessed on $4000, and 10%($4000) = $400.

e

On $26,000, tax is assessed on $16,000, and

al

10%($10,000) + 15%($6000) = $1000 + $900 = $1900.

rS

(c) As in part (b), there is $1000 tax assessed on $20,000 of income, so

the graph of  is a line segment from (10,000 0) to (20,000 1000).
The tax on $30,000 is $2500, so the graph of  for   20,000 is

(30,000 2500).

Fo

the ray with initial point (20,000 1000) that passes through

69.  is an odd function because its graph is symmetric about the origin.  is an even function because its graph is symmetric with

ot

respect to the -axis.
71. (a) Because an even function is symmetric with respect to the -axis, and the point (5 3) is on the graph of this even function,

N

the point (−5 3) must also be on its graph.
(b) Because an odd function is symmetric with respect to the origin, and the point (5 3) is on the graph of this odd function, the point (−5 −3) must also be on its graph.
73.  () =


.
2 + 1

 (−) =

−

−
= 2
=− 2
= −().
(−)2 + 1
 +1
 +1

75.  () =

−


, so (−) =
=
.
+1
− + 1
−1

Since this is neither  () nor − (), the function  is neither even nor odd.

So  is an odd function.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

14

¤

CHAPTER 1 FUNCTIONS AND LIMITS

77.  () = 1 + 32 − 4 .

 (−) = 1+3(−)2 −(−)4 = 1+32 −4 =  ().
So  is an even function.

79. (i) If  and  are both even functions, then  (−) =  () and (−) = (). Now

( + )(−) = (−) + (−) =  () + () = ( + )(), so  +  is an even function.
(ii) If  and  are both odd functions, then  (−) = − () and (−) = −(). Now
( + )(−) = (−) + (−) = − () + [−()] = −[ () + ()] = −( + )(), so  +  is an odd function.

e

(iii) If  is an even function and  is an odd function, then ( + )(−) =  (−) + (−) =  () + [−()] =  () − (), which is not ( + )() nor −( + )(), so  +  is neither even nor odd. (Exception: if  is the zero function, then

1. (a)  () = log2  is a logarithmic function.

rS

1.2 Mathematical Models: A Catalog of Essential Functions

al

 +  will be odd. If  is the zero function, then  +  will be even.)

√
4
 is a root function with  = 4.

(c) () =

23 is a rational function because it is a ratio of polynomials.
1 − 2

Fo

(b) () =

(d) () = 1 − 11 + 2542 is a polynomial of degree 2 (also called a quadratic function).

ot

(e) () = 5 is an exponential function.

N

(f ) () = sin  cos2  is a trigonometric function.

3. We notice from the figure that  and  are even functions (symmetric with respect to the -axis) and that  is an odd function



(symmetric with respect to the origin). So (b)  = 5 must be  . Since  is flatter than  near the origin, we must have




(c)  = 8 matched with  and (a)  = 2 matched with .

5. (a) An equation for the family of linear functions with slope 2

is  =  () = 2 + , where  is the -intercept.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS

¤

(b)  (2) = 1 means that the point (2 1) is on the graph of  . We can use the point-slope form of a line to obtain an equation for the family of linear functions through the point (2 1).  − 1 = ( − 2), which is equivalent to  =  + (1 − 2) in slope-intercept form.

(c) To belong to both families, an equation must have slope  = 2, so the equation in part (b),  =  + (1 − 2), becomes  = 2 − 3. It is the only function that belongs to both families.

7. All members of the family of linear functions () =  −  have graphs

rS

al

e

that are lines with slope −1. The -intercept is .

9. Since  (−1) =  (0) = (2) = 0,  has zeros of −1, 0, and 2, so an equation for  is  () = [ − (−1)]( − 0)( − 2),

or  () = ( + 1)( − 2). Because  (1) = 6, we’ll substitute 1 for  and 6 for  ().

Fo

6 = (1)(2)(−1) ⇒ −2 = 6 ⇒  = −3, so an equation for  is  () = −3( + 1)( − 2).
11. (a)  = 200, so  = 00417( + 1) = 00417(200)( + 1) = 834 + 834. The slope is 834, which represents the

change in mg of the dosage for a child for each change of 1 year in age.

N

13. (a)

ot

(b) For a newborn,  = 0, so  = 834 mg.

(b) The slope of

9
5

means that  increases

9
5

degrees for each increase

of 1◦ C. (Equivalently,  increases by 9 when  increases by 5 and  decreases by 9 when  decreases by 5.) The  -intercept of
32 is the Fahrenheit temperature corresponding to a Celsius temperature of 0.

15. (a) Using  in place of  and  in place of , we find the slope to be

equation is  − 80 = 1 ( − 173) ⇔  − 80 = 1  −
6
6
(b) The slope of

1
6

173
6

10
1
2 − 1
80 − 70
=
= . So a linear
=
2 − 1
173 − 113
60
6


⇔  = 1  + 307 307 = 5116 .
6
6
6

means that the temperature in Fahrenheit degrees increases one-sixth as rapidly as the number of cricket

chirps per minute. Said differently, each increase of 6 cricket chirps per minute corresponds to an increase of 1◦ F.
(c) When  = 150, the temperature is given approximately by  = 1 (150) +
6

307
6

= 7616 ◦ F ≈ 76 ◦ F.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

15

16

¤

CHAPTER 1 FUNCTIONS AND LIMITS

17. (a) We are given

434 change in pressure
=
= 0434. Using  for pressure and  for depth with the point
10 feet change in depth
10

(  ) = (0 15), we have the slope-intercept form of the line,  = 0434 + 15.
(b) When  = 100, then 100 = 0434 + 15 ⇔ 0434 = 85 ⇔  =

85
0434

≈ 19585 feet. Thus, the pressure is

100 lbin2 at a depth of approximately 196 feet.
19. (a) The data appear to be periodic and a sine or cosine function would make the best model. A model of the form

 () =  cos() +  seems appropriate.
(b) The data appear to be decreasing in a linear fashion. A model of the form  () =  +  seems appropriate.
Exercises 21 – 24: Some values are given to many decimal places. These are the results given by several computer algebra systems — rounding is left to the reader.

e

(b) Using the points (4000 141) and (60,000 82), we obtain

21. (a)

rS

Fo

A linear model does seem appropriate.

al

82 − 141
( − 4000) or, equivalently,
60,000 − 4000
 ≈ −0000105357 + 14521429.
 − 141 =

(c) Using a computing device, we obtain the least squares regression line  = −00000997855 + 13950764.
The following commands and screens illustrate how to find the least squares regression line on a TI-84 Plus. to enter the editor.

N

ot

Enter the data into list one (L1) and list two (L2). Press

Find the regession line and store it in Y1 . Press

.

Note from the last figure that the regression line has been stored in Y1 and that Plot1 has been turned on (Plot1 is

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS

highlighted). You can turn on Plot1 from the Y= menu by placing the cursor on Plot1 and pressing

¤

17

or by

.

pressing

Now press

to produce a graph of the data and the regression

line. Note that choice 9 of the ZOOM menu automatically selects a window

(d) When  = 25,000,  ≈ 11456; or about 115 per 100 population.
(e) When  = 80,000,  ≈ 5968; or about a 6% chance.
23. (a) A linear model seems appropriate over the time interval

ot

Fo

rS

considered.

al

(f ) When  = 200,000,  is negative, so the model does not apply.

e

that displays all of the data.

(b) Using a computing device, we obtain the regression line  ≈ 00265 − 468759. It is plotted in the graph in part (a).

N

(c) For  = 2008, the linear model predicts a winning height of 6.27 m, considerably higher than the actual winning height of 5.96 m.

(d) It is not reasonable to use the model to predict the winning height at the 2100 Olympics since 2100 is too far from the
1896–2004 range on which the model is based.
25. If  is the original distance from the source, then the illumination is  () = −2 = 2 . Moving halfway to the lamp gives

us an illumination of 

 −2
1 
 =  1
= (2)2 = 4(2 ), so the light is 4 times as bright.
2
2

27. (a) Using a computing device, we obtain a power function  =  , where  ≈ 31046 and  ≈ 0308.

(b) If  = 291, then  =  ≈ 178, so you would expect to find 18 species of reptiles and amphibians on Dominica.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

18

¤

CHAPTER 1 FUNCTIONS AND LIMITS

1.3 New Functions from Old Functions
1. (a) If the graph of  is shifted 3 units upward, its equation becomes  =  () + 3.

(b) If the graph of  is shifted 3 units downward, its equation becomes  =  () − 3.
(c) If the graph of  is shifted 3 units to the right, its equation becomes  =  ( − 3).
(d) If the graph of  is shifted 3 units to the left, its equation becomes  =  ( + 3).
(e) If the graph of  is reflected about the -axis, its equation becomes  = − ().
(f ) If the graph of  is reflected about the -axis, its equation becomes  =  (−).
(g) If the graph of  is stretched vertically by a factor of 3, its equation becomes  = 3 ().
(h) If the graph of  is shrunk vertically by a factor of 3, its equation becomes  = 1  ().
3

e

3. (a) (graph 3) The graph of  is shifted 4 units to the right and has equation  =  ( − 4).

al

(b) (graph 1) The graph of  is shifted 3 units upward and has equation  =  () + 3.

(c) (graph 4) The graph of  is shrunk vertically by a factor of 3 and has equation  = 1 ().
3

rS

(d) (graph 5) The graph of  is shifted 4 units to the left and reflected about the -axis. Its equation is  = − ( + 4).
(e) (graph 2) The graph of  is shifted 6 units to the left and stretched vertically by a factor of 2. Its equation is
 = 2 ( + 6).

ot

horizontally by a factor of 2.

N

The point (4 −1) on the graph of  corresponds to the

 point 1 · 4 −1 = (2 −1).
2

(c) To graph  =  (−) we reflect the graph of  about the -axis.

The point (4 −1) on the graph of  corresponds to the point (−1 · 4 −1) = (−4 −1).

(b) To graph  = 

Fo

5. (a) To graph  =  (2) we shrink the graph of 

1 
 we stretch the graph of 
2

horizontally by a factor of 2.

The point (4 −1) on the graph of  corresponds to the

point (2 · 4 −1) = (8 −1).

(d) To graph  = − (−) we reflect the graph of  about the -axis, then about the -axis.

The point (4 −1) on the graph of  corresponds to the

point (−1 · 4 −1 · −1) = (−4 1).

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS

7. The graph of  =  () =

¤

√
3 − 2 has been shifted 4 units to the left, reflected about the -axis, and shifted downward

1 unit. Thus, a function describing the graph is
=

−1 ·
   reflect about -axis

 ( + 4)
   shift 4 units left

− 1
   shift 1 unit left

This function can be written as


√
 = − ( + 4) − 1 = − 3( + 4) − ( + 4)2 − 1 = − 3 + 12 − (2 + 8 + 16) − 1 = − −2 − 5 − 4 − 1
1
: Start with the graph of the reciprocal function  = 1 and shift 2 units to the left.
+2

√

√
3
 and reflect about the -axis.

√
√
 − 2 − 1: Start with the graph of  = , shift 2 units to the right, and then shift 1 unit downward.

N

13.  =

ot

Fo

11.  = − 3 : Start with the graph of  =

rS

al

e

9.  =

15.  = sin(2): Start with the graph of  = sin  and stretch horizontally by a factor of 2.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

19

20

¤

17.  =

CHAPTER 1 FUNCTIONS AND LIMITS
1
2 (1

− cos ): Start with the graph of  = cos , reflect about the -axis, shift 1 unit upward, and then shrink vertically by

a factor of 2.

19.  = 1 − 2 − 2 = −(2 + 2) + 1 = −(2 + 2 + 1) + 2 = −( + 1)2 + 2: Start with the graph of  = 2 , reflect about

rS

al

e

the -axis, shift 1 unit to the left, and then shift 2 units upward.

√

ot

Fo

21.  = | − 2|: Start with the graph of  = || and shift 2 units to the right.

N

23.  = |  − 1|: Start with the graph of  =

√
, shift it 1 unit downward, and then reflect the portion of the graph below the

-axis about the -axis.

25. This is just like the solution to Example 4 except the amplitude of the curve (the 30◦ N curve in Figure 9 on June 21) is


 2
14 − 12 = 2. So the function is () = 12 + 2 sin 365 ( − 80) . March 31 is the 90th day of the year, so the model gives

(90) ≈ 1234 h. The daylight time (5:51 AM to 6:18 PM) is 12 hours and 27 minutes, or 1245 h. The model value differs from the actual value by

1245−1234
1245

≈ 0009, less than 1%.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS

¤

27. (a) To obtain  =  (||), the portion of the graph of  =  () to the right of the -axis is reflected about the -axis.

(c)  =

(b)  = sin ||

29.  () = 3 + 22 ; () = 32 − 1.


||

 = R for both  and .

(a) ( + )() = (3 + 22 ) + (32 − 1) = 3 + 52 − 1,  = R.
(b) ( − )() = (3 + 22 ) − (32 − 1) = 3 − 2 + 1,  = R.

e

(c) ( )() = (3 + 22 )(32 − 1) = 35 + 64 − 3 − 22 ,  = R.

31.  () = 2 − 1,  = R;

() = 2 + 1,  = R.

al



 
3 + 22
1
 since 32 − 1 6= 0.
() =
,  =  |  6= ± √
(d)

32 − 1
3

rS

(a) ( ◦ )() =  (()) = (2 + 1) = (2 + 1)2 − 1 = (42 + 4 + 1) − 1 = 42 + 4,  = R.
(b) ( ◦ )() = ( ()) = (2 − 1) = 2(2 − 1) + 1 = (22 − 2) + 1 = 22 − 1,  = R.

Fo

(c) ( ◦  )() = ( ()) =  (2 − 1) = (2 − 1)2 − 1 = (4 − 22 + 1) − 1 = 4 − 22 ,  = R.
(d) ( ◦ )() = (()) = (2 + 1) = 2(2 + 1) + 1 = (4 + 2) + 1 = 4 + 3,  = R.
33.  () = 1 − 3; () = cos .

 = R for both  and , and hence for their composites.

ot

(a) ( ◦ )() =  (()) = (cos ) = 1 − 3 cos .

(b) ( ◦ )() = ( ()) = (1 − 3) = cos(1 − 3).

N

(c) ( ◦  )() = ( ()) =  (1 − 3) = 1 − 3(1 − 3) = 1 − 3 + 9 = 9 − 2.
(d) ( ◦ )() = (()) = (cos ) = cos(cos ) [Note that this is not cos  · cos .]
+1
1
,  = { |  6= 0}; () =
,  = { |  6= −2}

+2


+1
+1
1
+2
+1
=
=
+
+
(a) ( ◦ )() =  (()) = 
+1
+2
+2
+2
+1
+2
 2
 

 + 2 + 1 + 2 + 4 + 4
22 + 6 + 5
( + 1)( + 1) + ( + 2)( + 2)
=
=
=
( + 2)( + 1)
( + 2)( + 1)
( + 2)( + 1)

35.  () =  +

Since () is not defined for  = −2 and  (()) is not defined for  = −2 and  = −1, the domain of ( ◦ )() is  = { |  6= −2 −1}.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

21

22

¤

CHAPTER 1 FUNCTIONS AND LIMITS



1
2 + 1 + 


+
+1

2 +  + 1
2 +  + 1
1

= 
=
= 2
(b) ( ◦ )() = ( ()) =   +
= 2 

 + 2 + 1
( + 1)2
1
 + 1 + 2
+
+2


Since () is not defined for  = 0 and ( ()) is not defined for  = −1, the domain of ( ◦  )() is  = { |  6= −1 0}.

 


1
1
1

1
1
1
= +
+
+ 2
=+ + 2
(c) ( ◦  )() =  ( ()) =   +
1 = +
 +1




 +1
+ 

 2

 2

4
2
2
()  + 1 + 1  + 1 + ()
 +  +  + 1 + 2
=
=
2 + 1)
(
(2 + 1)
 = { |  6= 0}



+1
(d) ( ◦ )() = (()) = 
+2



e

4 + 32 + 1

(2 + 1)

+1
 + 1 + 1( + 2)
+1
2 + 3
+1++2
+2
+2
=
=
=
=
+1
 + 1 + 2( + 2)
 + 1 + 2 + 4
3 + 5
+2
+2
+2

al

=

rS

Since () is not defined for  = −2 and (()) is not defined for  = − 5 ,
3


the domain of ( ◦ )() is  =  |  6= −2 − 5 .
3

Fo

37. ( ◦  ◦ )() =  ((())) =  ((2 )) =  (sin(2 )) = 3 sin(2 ) − 2
39. ( ◦  ◦ )() =  ((())) =  ((3 + 2)) =  [(3 + 2)2 ]

=  (6 + 43 + 4) =


√
(6 + 43 + 4) − 3 = 6 + 43 + 1

√
3
 and  () =

√
3
√


√ =  ().
. Then ( ◦ )() =  (()) =  ( 3  ) =
1+
1+ 3

N

43. Let () =

ot

41. Let () = 2 + 2 and  () = 4 . Then ( ◦ )() = (()) =  (2 + 2 ) = (2 + 2 )4 =  ().

45. Let () = 2 and  () = sec  tan . Then ( ◦ )() =  (()) =  (2 ) = sec(2 ) tan(2 ) = ().
47. Let () =

√
√
, () =  − 1, and () = . Then

√
√
√
( ◦  ◦ )() =  ((())) =  (( )) =  (  − 1) =
 − 1 = ().

49. Let () =

√
, () = sec , and () = 4 . Then

√
√
√ 4
√
( ◦  ◦ )() =  ((())) =  ((  )) =  (sec  ) = (sec  ) = sec4 (  ) = ().
51. (a) (2) = 5, because the point (2 5) is on the graph of . Thus, ((2)) =  (5) = 4, because the point (5 4) is on the

graph of  .
(b) ((0)) = (0) = 3
(c) ( ◦ )(0) =  ((0)) =  (3) = 0 c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS

¤

(d) ( ◦  )(6) = ( (6)) = (6). This value is not defined, because there is no point on the graph of  that has
-coordinate 6.
(e) ( ◦ )(−2) = ((−2)) = (1) = 4
(f ) ( ◦  )(4) =  ( (4)) =  (2) = −2
53. (a) Using the relationship distance = rate · time with the radius  as the distance, we have () = 60.

(b)  = 2

⇒ ( ◦ )() = (()) = (60)2 = 36002 . This formula gives us the extent of the rippled area

(in cm2 ) at any time .
55. (a) From the figure, we have a right triangle with legs 6 and , and hypotenuse .

By the Pythagorean Theorem, 2 + 62 = 2

⇒  =  () =

√
 2 + 36.

(b) Using  = , we get  = (30 kmh)( hours) = 30 (in km). Thus,


√
(30)2 + 36 = 9002 + 36. This function represents the distance between the

al

(c) ( ◦ )() =  (()) =  (30) =

e

 = () = 30.

lighthouse and the ship as a function of the time elapsed since noon.
(b)

0

if   0

1

if  ≥ 0

Fo

() =



rS

57. (a)

 () =


0

if   0

120 if  ≥ 0

so  () = 120().

Starting with the formula in part (b), we replace 120 with 240 to reflect the

(c)

different voltage. Also, because we are starting 5 units to the right of  = 0,

ot

we replace  with  − 5. Thus, the formula is  () = 240( − 5).

N

59. If () = 1  + 1 and () = 2  + 2 , then

( ◦ )() =  (()) =  (2  + 2 ) = 1 (2  + 2 ) + 1 = 1 2  + 1 2 + 1 .

So  ◦  is a linear function with slope 1 2 .

61. (a) By examining the variable terms in  and , we deduce that we must square  to get the terms 42 and 4 in . If we let

 () = 2 + , then ( ◦ )() =  (()) =  (2 + 1) = (2 + 1)2 +  = 42 + 4 + (1 + ). Since
() = 42 + 4 + 7, we must have 1 +  = 7. So  = 6 and  () = 2 + 6.
(b) We need a function  so that  (()) = 3(()) + 5 = (). But
() = 32 + 3 + 2 = 3(2 + ) + 2 = 3(2 +  − 1) + 5, so we see that () = 2 +  − 1.
63. We need to examine (−).

(−) = ( ◦ )(−) = ((−)) =  (()) [because  is even]

= ()

Because (−) = (),  is an even function. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

23

24

¤

CHAPTER 1

FUNCTIONS AND LIMITS

1.4 The Tangent and Velocity Problems
1. (a) Using  (15 250), we construct the following table:

(b) Using the values of  that correspond to the points closest to  ( = 10 and  = 20), we have

slope =  





5

(5 694)

694−250
5−15

= − 444 = −444
10

10

(10 444)

444−250
10−15

= − 194 = −388
5

20

(20 111)

111−250
20−15

= − 139 = −278
5

25

(25 28)

28−250
25−15

30

(30 0)

0−250
30−15

−388 + (−278)
= −333
2

= − 222 = −222
10
= − 250 = −166
15

−300
9

= −333.

1
,  (2 −1)
1−

(i)

15

(ii)

19

(iii)

199

(iv)

1999

(v)

25

(b) The slope appears to be 1.
(c) Using  = 1, an equation of the tangent line to the

( 1(1 − ))

 

(15 −2)

2

(199 −1010 101)

(19 −1111 111)

1111 111

(1999 −1001 001)

21

201

(21 −0909 091)

(viii)

2001

0909 091

(2001 −0999 001)

 =  − 3.

1001 001

0999 001

N

(vi)
(vii)

curve at  (2 −1) is  − (−1) = 1( − 2), or

1010 101

ot

3. (a)  =

Fo

rS

al

tangent line at  to be

e

(c) From the graph, we can estimate the slope of the

(25 −0666 667)

0666 667

(201 −0990 099)

0990 099

5. (a)  = () = 40 − 162 . At  = 2,  = 40(2) − 16(2)2 = 16. The average velocity between times 2 and 2 +  is

ave



40(2 + ) − 16(2 + )2 − 16
(2 + ) − (2)
−24 − 162
=
=
=
= −24 − 16, if  6= 0.
(2 + ) − 2



(i) [2 25]:  = 05, ave = −32 fts
(iii) [2 205]:  = 005, ave = −248 fts

(ii) [2 21]:  = 01, ave = −256 fts
(iv) [2 201]:  = 001, ave = −2416 fts

(b) The instantaneous velocity when  = 2 ( approaches 0) is −24 fts. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.4

7. (a) (i) On the interval [1 3], ave =

THE TANGENT AND VELOCITY PROBLEMS

¤

(3) − (1)
107 − 14
93
=
=
= 465 ms.
3−1
2
2

(ii) On the interval [2 3], ave =

(3) − (2)
107 − 51
=
= 56 ms.
3−2
1

(iii) On the interval [3 5], ave =

258 − 107
151
(5) − (3)
=
=
= 755 ms.
5−3
2
2

(iv) On the interval [3 4], ave =

177 − 107
(4) − (3)
=
= 7 ms.
4−3
1

(b)

Using the points (2 4) and (5 23) from the approximate tangent
23 − 4
≈ 63 ms.
5−2

rS

al

e

line, the instantaneous velocity at  = 3 is about

9. (a) For the curve  = sin(10) and the point  (1 0):




(2 0)

15

(15 08660)

14

(14 −04339)

13
12

(13 −08230)
(12 08660)

(11 −02817)



0

05

(05 0)

17321

06

(06 08660)

−10847

07

(07 07818)

08

(08 1)

43301

09

(09 −03420)

−27433
−28173

ot

11



Fo

2

 

 
0
−21651
−26061
−5

34202

(b)

N

As  approaches 1, the slopes do not appear to be approaching any particular value.
We see that problems with estimation are caused by the frequent oscillations of the graph. The tangent is so steep at  that we need to take -values much closer to 1 in order to get accurate estimates of its slope.

(c) If we choose  = 1001, then the point  is (1001 −00314) and   ≈ −313794. If  = 0999, then  is
(0999 00314) and   = −314422. The average of these slopes is −314108. So we estimate that the slope of the tangent line at  is about −314.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

25

26

¤

CHAPTER 1

FUNCTIONS AND LIMITS

1.5 The Limit of a Function
1. As  approaches 2,  () approaches 5. [Or, the values of  () can be made as close to 5 as we like by taking  sufficiently

close to 2 (but  6= 2).] Yes, the graph could have a hole at (2 5) and be defined such that  (2) = 3.
3. (a) lim  () = ∞ means that the values of  () can be made arbitrarily large (as large as we please) by taking 
→−3

sufficiently close to −3 (but not equal to −3).
(b) lim () = −∞ means that the values of  () can be made arbitrarily large negative by taking  sufficiently close to 4
→4+

through values larger than 4.
5. (a) As  approaches 1, the values of  () approach 2, so lim  () = 2.
→1

(b) As  approaches 3 from the left, the values of () approach 1, so lim  () = 1.
→3−

→3+

e

(c) As  approaches 3 from the right, the values of  () approach 4, so lim () = 4.
(d) lim  () does not exist since the left-hand limit does not equal the right-hand limit.

al

→3

(e) When  = 3,  = 3, so  (3) = 3.

(b) lim () = −2

rS

7. (a) lim () = −1
→0−

→0+

(c) lim () does not exist because the limits in part (a) and part (b) are not equal.
→0

(d) lim () = 2

(e) lim () = 0

→2−

→2+

Fo

(f ) lim () does not exist because the limits in part (d) and part (e) are not equal.
→2

(g) (2) = 1

(h) lim () = 3
→4

9. (a) lim  () = −∞

(b) lim  () = ∞

(d) lim  () = −∞
→6−

→−3

(c) lim  () = ∞
→0

(e) lim  () = ∞

ot

→−7

→6+

N

(f ) The equations of the vertical asymptotes are  = −7,  = −3,  = 0, and  = 6.
11. From the graph of


1 +  if   −1

if −1 ≤   1 ,
 () = 2


2 −  if  ≥ 1

we see that lim () exists for all  except  = −1. Notice that the
→

right and left limits are different at  = −1.
13. (a) lim  () = 1
→0−

(b) lim () = 0
→0+

(c) lim  () does not exist because the limits in
→0

part (a) and part (b) are not equal. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.5

15. lim  () = −1,
→0−

17. lim  () = 4,

lim () = 2,  (0) = 1

→0+

THE LIMIT OF A FUNCTION

lim  () = 2, lim () = 2,
→−2

→3−

→3+

 (3) = 3,  (−2) = 1



2 − 2
:
−−2

21. For  () =

2

 ()



 ()

19

21

0677419

195

0661017

205

0672131

199

0665552

201

0667774

1995

0666110

2005
2001

0667221
0666778

1999



0655172

0666556

2

→2



√
+4−2
:

 ()



0236068

05

0242641

01

0248457

005

0249224

001

0249844

N

1

It appears that lim

→0

−1

 ()
0329033

±05

0458209

±01

0498333

±001

0499983

±02

→0

25. For  () =

 ()

0493331

±005

It appears that lim

ot

23. For  () =

 − 2
= 0¯ = 2 .
6 3
2 −  − 2

Fo

It appears that lim

±1

al

0714286

rS

25

sin 
:
 + tan 

e

19. For  () =

0499583

sin 
1
= 05 = .
 + tan 
2

6 − 1
:
10 − 1



 ()



 ()

0267949

05

0985337

15

0183369

−05

0258343

09

0719397

11

0484119

−01

0251582

095

0660186

105

0540783

−005

0250786

099

0612018

101

0588022

−001

0250156

0999

0601200

1001

0598800

√
+4−2
= 025 = 1 .
4


27. (a) From the graphs, it seems that lim

→0

cos 2 − cos 
= −15.
2

It appears that lim

→1

6 − 1
= 06 = 3 .
5
10 − 1

(b)


 ()

±01

−1493759

±0001

−1499999

±001

±00001

−1499938
−1500000

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

27

28

29.

¤

CHAPTER 1

lim

→−3+

31. lim

→1

33.
35.

FUNCTIONS AND LIMITS

+2
= −∞ since the numerator is negative and the denominator approaches 0 from the positive side as  → −3+ .
+3

2−
= ∞ since the numerator is positive and the denominator approaches 0 through positive values as  → 1.
( − 1)2

lim

→−2+

−1
−1
= −∞ since ( + 2) → 0 as  → −2+ and 2
 0 for −2    0.
2 ( + 2)
 ( + 2)

lim  csc  = lim

→2 −

→2 −


= −∞ since the numerator is positive and the denominator approaches 0 through negative sin 

values as  → 2− .
37. lim

→2+

2 − 2 − 8
( − 4)( + 2)
= lim
= ∞ since the numerator is negative and the denominator approaches 0 through
2 − 5 + 6 →2+ ( − 3)( − 2)
1
.
3 − 1


05

From these calculations, it seems that
→1−

→1+

−114

09
099

−369



0999

09999

099999

 ()

15

042

11

302

−337

101

330

−3337

1001

3330

−33337

10001

33330

−33,3337

100001

33,3333

rS

lim  () = −∞ and lim  () = ∞.

 ()

al

39. (a)  () =

e

negative values as  → 2+ .

Fo

(b) If  is slightly smaller than 1, then 3 − 1 will be a negative number close to 0, and the reciprocal of 3 − 1, that is,  (), will be a negative number with large absolute value. So lim () = −∞.
→1−

If  is slightly larger than 1, then  − 1 will be a small positive number, and its reciprocal,  (), will be a large positive
3

→1+

ot

number. So lim  () = ∞.

(c) It appears from the graph of  that

N

lim  () = −∞ and lim  () = ∞.

→1−

→1+

41. For  () = 2 − (21000):

(a)

(b)



 ()



 ()

1
08
06
04
02
01

0998000
0638259
0358484
0158680
0038851
0008928

004
002
001
0005
0003

0000572
−0000614
−0000907
−0000978
−0000993

005

0001465

It appears that lim  () = 0.

0001

−0001000

It appears that lim  () = −0001.
→0

→0

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.6

CALCULATING LIMITS USING THE LIMIT LAWS

¤

29

43. No matter how many times we zoom in toward the origin, the graphs of  () = sin() appear to consist of almost-vertical

al

e

lines. This indicates more and more frequent oscillations as  → 0.

There appear to be vertical asymptotes of the curve  = tan(2 sin ) at  ≈ ±090

45.

rS

and  ≈ ±224. To find the exact equations of these asymptotes, we note that the graph of the tangent function has vertical asymptotes at  = must have 2 sin  =


2

+ , or equivalently, sin  =


4


2

+ . Thus, we

+  . Since
2

4

(corresponding

Fo

−1 ≤ sin  ≤ 1, we must have sin  = ±  and so  = ± sin−1
4

to  ≈ ±090). Just as 150◦ is the reference angle for 30◦ ,  − sin−1  is the
4


−1 
−1  reference angle for sin 4 . So  = ±  − sin 4 are also equations of

ot

vertical asymptotes (corresponding to  ≈ ±224).

N

1.6 Calculating Limits Using the Limit Laws
1. (a) lim [ () + 5()] = lim  () + lim [5()]
→2

→2

→2

= lim  () + 5 lim ()
→2

→2

[Limit Law 1]

(b) lim [()]3 =
→2

[Limit Law 3]



3 lim ()

→2

[Limit Law 6]

= ( −2)3 = −8

= 4 + 5(−2) = −6
(c) lim

→2



 () = lim  ()
→2

=

[Limit Law 11]

(d) lim

→2

lim [3 ()]
3 ()
→2
=
()
lim ()

[Limit Law 5]

→2

√
4=2

3 lim  ()
=

→2

lim ()

[Limit Law 3]

→2

=

3(4)
= −6
−2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

30

¤

CHAPTER 1

FUNCTIONS AND LIMITS

(e) Because the limit of the denominator is 0, we can’t use Limit Law 5. The given limit, lim

→2

()
, does not exist because the
()

denominator approaches 0 while the numerator approaches a nonzero number.
(f) lim

→2

lim [() ()]
() ()
→2
=
 () lim  ()

[Limit Law 5]

→2

=

lim () · lim ()

→2

→2

[Limit Law 4]

lim  ()

→2

=

−2 · 0
=0
4
[Limit Laws 2 and 1]

3. lim (53 − 32 +  − 6) = lim (53 ) − lim (32 ) + lim  − lim 6
→3

→3

→3

→3

→3

= 5 lim 3 − 3 lim 2 + lim  − lim 6

[3]

= 5(33 ) − 3(32 ) + 3 − 6

[9, 8, and 7]

→3

→3

→3

e

→3

al

= 105 lim (4 − 2)
4 − 2
→−2
=
5. lim
→−2 22 − 3 + 2 lim (22 − 3 + 2)
=

lim 4 − lim 2

→−2
2 lim 2 −
→−2

→−2

3 lim  + lim 2
→−2

=

→8

14
7
=
16
8

[9, 7, and 8]

√
√
3
 ) (2 − 62 + 3 ) = lim (1 + 3  ) · lim (2 − 62 + 3 )

ot

7. lim (1 +

16 − 2
2(4) − 3(−2) + 2

[1, 2, and 3]

Fo

=

→−2

rS

[Limit Law 5]

→−2

→8

=



lim 1 + lim

→8

[Limit Law 4]

→8

→8


√  
3
 · lim 2 − 6 lim 2 + lim 3
→8

N

√  


= 1 + 3 8 · 2 − 6 · 82 + 83

→8

→8

[1, 2, and 3]
[7, 10, 9]

= (3)(130) = 390

9. lim

→2



22 + 1
=
3 − 2



22 + 1
→2 3 − 2

 lim (22 + 1)
 →2
=
lim (3 − 2)

[Limit Law 11]

lim

[5]

→2


 2 lim 2 + lim 1
 →2
→2
=
3 lim  − lim 2

[1, 2, and 3]

2(2)2 + 1
=
3(2) − 2

[9, 8, and 7]

→2

=



→2



9
3
=
4
2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.6

CALCULATING LIMITS USING THE LIMIT LAWS

11. lim

2 − 6 + 5
( − 5)( − 1)
= lim
= lim ( − 1) = 5 − 1 = 4
→5
→5
−5
−5

13. lim

¤

2 − 5 + 6 does not exist since  − 5 → 0, but 2 − 5 + 6 → 6 as  → 5.
−5

→5

→5

15. lim

→−3

2 − 9
( + 3)( − 3)
−3
−3 − 3
−6
6
= lim
= lim
=
=
=
22 + 7 + 3 →−3 (2 + 1)( + 3) →−3 2 + 1
2(−3) + 1
−5
5

(−5 + )2 − 25
(25 − 10 + 2 ) − 25
−10 + 2
(−10 + )
= lim
= lim
= lim
= lim (−10 + ) = −10
→0
→0
→0
→0





17. lim

→0

19. By the formula for the sum of cubes, we have

lim

→−2

+2
+2
1
1
1
= lim
= lim
=
=
.
3 + 8 →−2 ( + 2)(2 − 2 + 4) →−2 2 − 2 + 4
4+4+4
12

→0




1
1
1
√
 = lim √
=
=
→0
3+3
6
9++3
9++3

al

= lim

e

√
2
√
√
√
9 +  − 32
9+−3
9+−3
9++3
(9 + ) − 9
 = lim √

21. lim
= lim √
= lim
·√
→0
→0
→0 


9 +  + 3 →0  9 +  + 3
9++3

rS

1
+4
1
+
+4
1
1
1
4
 = lim 4 = lim
23. lim
= lim
=
=−
→−4 4 + 
→−4 4 + 
→−4 4(4 + )
→−4 4
4(−4)
16

Fo

√
2 √
2
√
√
√
√
√
√
1+ −
1−
1+− 1−
1+− 1−
1++ 1−
√

√
√
= lim
25. lim
= lim
·√
→0
→0


1 +  + 1 −  →0  1 +  + 1 − 

(1 + ) − (1 − )
2
2
 = lim √
 = lim √
√
√
√
= lim √
→0 
→0 
→0
1++ 1−
1++ 1−
1++ 1−

ot

2
2
√ = =1
= √
2
1+ 1

√
√
√
4− 
(4 −  )(4 +  )
16 − 
√
√
= lim
= lim
→16 16 − 2
→16 (16 − 2 )(4 +
 ) →16 (16 − )(4 +  )

N

27. lim

= lim

→16

29. lim

→0



1
1
√
−

 1+



1
1
1
1
√  =
√
=
=

16(8)
128
(4 +  )
16 4 + 16

= lim

→0




√
√
√
1− 1+ 1+ 1+
1− 1+
−

 = lim √


√
√
√
√
= lim
→0
→0 
 1+
 +1 1+ 1+
1+ 1+ 1+

−1
−1
1

 = √

 =−
√
√
= lim √
→0
2
1+ 1+ 1+
1+0 1+ 1+0

31. lim

→0

( + )3 − 3
(3 + 32  + 32 + 3 ) − 3
32  + 32 + 3
= lim
= lim
→0
→0



= lim

→0

(32 + 3 + 2 )
= lim (32 + 3 + 2 ) = 32
→0


c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

31

32

¤

CHAPTER 1

FUNCTIONS AND LIMITS

(b)

33. (a)


−0001
−00001
−000001
−0000001
0000001
000001
00001
0001


2
≈ lim √
3
1 + 3 − 1

→0

(c) lim

→0



 ()
06661663
06666167
06666617
06666662
06666672
06666717
06667167
06671663

The limit appears to be

2
.
3

√

√

√

 1 + 3 + 1
 1 + 3 + 1
1 + 3 + 1

√
= lim
·√
= lim
→0
→0
(1 + 3) − 1
3
1 + 3 − 1
1 + 3 + 1
√

1
1 + 3 + 1 lim 3 →0


1  lim (1 + 3) + lim 1
=
→0
→0
3


1 
=
lim 1 + 3 lim  + 1
→0
→0
3

[Limit Law 3]

e

=

al

[1 and 11]


1 √
1+3·0+1
3

[7 and 8]

1
2
(1 + 1) =
3
3

Fo

=

rS

=

[1, 3, and 7]

35. Let () = −2 , () = 2 cos 20 and () = 2 . Then

−1 ≤ cos 20 ≤ 1 ⇒ −2 ≤ 2 cos 20 ≤ 2

⇒  () ≤ () ≤ ().

So since lim  () = lim () = 0, by the Squeeze Theorem we have
→0

→0

ot

lim () = 0.

→0





37. We have lim (4 − 9) = 4(4) − 9 = 7 and lim 2 − 4 + 7 = 42 − 4(4) + 7 = 7. Since 4 − 9 ≤  () ≤ 2 − 4 + 7
→4

N

→4

for  ≥ 0, lim  () = 7 by the Squeeze Theorem.
→4

39. −1 ≤ cos(2) ≤ 1



⇒ −4 ≤ 4 cos(2) ≤ 4 . Since lim −4 = 0 and lim 4 = 0, we have
→0

→0



lim 4 cos(2) = 0 by the Squeeze Theorem.

→0

41. | − 3| =



if  − 3 ≥ 0

−3

if  − 3  0

−( − 3)

=



−3
3−

if  ≥ 3

if   3

Thus, lim (2 + | − 3|) = lim (2 +  − 3) = lim (3 − 3) = 3(3) − 3 = 6 and
→3+

→3+

→3+

lim (2 + | − 3|) = lim (2 + 3 − ) = lim ( + 3) = 3 + 3 = 6. Since the left and right limits are equal,

→3−

→3−

→3−

lim (2 + | − 3|) = 6.

→3

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.6













43. 23 − 2  = 2 (2 − 1) = 2  · |2 − 1| = 2 |2 − 1|

|2 − 1| =



2 − 1

−(2 − 1)

if 2 − 1 ≥ 0

if 2 − 1  0

=



2 − 1

CALCULATING LIMITS USING THE LIMIT LAWS

¤

if  ≥ 05

if   05

−(2 − 1)



So 23 − 2  = 2 [−(2 − 1)] for   05.
Thus,

lim

→05−

2 − 1
2 − 1
−1
−1
−1
= lim
= lim
= −4.
=
=
|23 − 2 | →05− 2 [−(2 − 1)] →05− 2
(05)2
025

45. Since || = − for   0, we have lim

→0−



1
1
−

||



= lim

→0−



1
1
−

−



= lim

→0−

2
, which does not exist since the


denominator approaches 0 and the numerator does not.
47. (a)

(b) (i) Since sgn  = 1 for   0, lim sgn  = lim 1 = 1.

e

→0+

→0+

(ii) Since sgn  = −1 for   0, lim sgn  = lim −1 = −1.

al

→0−

→0−

(iii) Since lim sgn  6= lim sgn , lim sgn  does not exist.
→0−

→0

→0+

(iv) Since |sgn | = 1 for  6= 0, lim |sgn | = lim 1 = 1.

49. (a) (i) lim () = lim
→2+

→2+

→0

rS

→0

2 +  − 6
( + 3)( − 2)
= lim
| − 2|
| − 2|
→2+
= lim

( + 3)( − 2)
−2

Fo

→2+

[since  − 2  0 if  → 2+ ]

= lim ( + 3) = 5
→2+

(ii) The solution is similar to the solution in part (i), but now | − 2| = 2 −  since  − 2  0 if  → 2− .
Thus, lim () = lim −( + 3) = −5.
→2−

ot

→2−

(b) Since the right-hand and left-hand limits of  at  = 2

(c)

N

are not equal, lim () does not exist.
→2

51. (a) (i) [[]] = −2 for −2 ≤   −1, so

→−2+

(ii) [[]] = −3 for −3 ≤   −2, so

→−2−

lim [[]] = lim [[]] =

lim (−2) = −2

→−2+

lim (−3) = −3.

→−2−

The right and left limits are different, so lim [[]] does not exist.
→−2

(iii) [[]] = −3 for −3 ≤   −2, so

lim [[]] =

→−24

lim (−3) = −3.

→−24

(b) (i) [[]] =  − 1 for  − 1 ≤   , so lim [[]] = lim ( − 1) =  − 1.
→−

→−

(ii) [[]] =  for  ≤    + 1, so lim [[]] = lim  = .
→+

→+

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

33

34

¤

CHAPTER 1

FUNCTIONS AND LIMITS

(c) lim [[]] exists ⇔  is not an integer.
→

53. The graph of  () = [[]] + [[−]] is the same as the graph of () = −1 with holes at each integer, since  () = 0 for any

integer . Thus, lim  () = −1 and lim  () = −1, so lim  () = −1. However,
→2−

→2

→2+

 (2) = [[2]] + [[−2]] = 2 + (−2) = 0, so lim  () 6=  (2).
→2

55. Since () is a polynomial, () = 0 + 1  + 2 2 + · · · +   . Thus, by the Limit Laws,



lim () = lim 0 + 1  + 2 2 + · · · +   = 0 + 1 lim  + 2 lim 2 + · · · +  lim 

→

→

→

→

→

= 0 + 1  + 2 2 + · · · +   = ()

Thus, for any polynomial , lim () = ().
→

→1

→1




 () − 8
() − 8
· ( − 1) = lim
· lim ( − 1) = 10 · 0 = 0.
→1
→1
−1
−1

e

57. lim [() − 8] = lim

Thus, lim  () = lim {[ () − 8] + 8} = lim [ () − 8] + lim 8 = 0 + 8 = 8.
→1

Note: The value of lim

→1

→1

 () − 8 does not affect the answer since it’s multiplied by 0. What’s important is that
−1

 () − 8 exists. −1

rS

lim

→1

→1

al

→1

59. Observe that 0 ≤  () ≤ 2 for all , and lim 0 = 0 = lim 2 . So, by the Squeeze Theorem, lim  () = 0.
→0

→0

→0

Fo

61. Let () = () and () = 1 − (), where  is the Heaviside function defined in Exercise 1.3.57.

Thus, either  or  is 0 for any value of . Then lim  () and lim () do not exist, but lim [ ()()] = lim 0 = 0.
→0

→0

→0

→0

63. Since the denominator approaches 0 as  → −2, the limit will exist only if the numerator also approaches

ot

0 as  → −2. In order for this to happen, we need lim

→−2

 2

3 +  +  + 3 = 0 ⇔

lim

→−2

N

3(−2)2 + (−2) +  + 3 = 0 ⇔ 12 − 2 +  + 3 = 0 ⇔  = 15. With  = 15, the limit becomes
32 + 15 + 18
3( + 2)( + 3)
3( + 3)
3(−2 + 3)
3
= lim
= lim
=
=
= −1.
→−2 ( − 1)( + 2)
→−2  − 1
2 +  − 2
−2 − 1
−3

1.7 The Precise Definition of a Limit
1. If |() − 1|  02, then −02   () − 1  02

⇒ 08   ()  12. From the graph, we see that the last inequality is

true if 07    11, so we can choose  = min {1 − 07 11 − 1} = min {03 01} = 01 (or any smaller positive

number).

3. The leftmost question mark is the solution of

√
√
 = 16 and the rightmost,  = 24. So the values are 162 = 256 and

242 = 576. On the left side, we need | − 4|  |256 − 4| = 144. On the right side, we need | − 4|  |576 − 4| = 176.
To satisfy both conditions, we need the more restrictive condition to hold — namely, | − 4|  144. Thus, we can choose
 = 144, or any smaller positive number.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.7

THE PRECISE DEFINITION OF A LIMIT

¤

35

From the graph, we find that  = tan  = 08 when  ≈ 0675, so

5.


4

−  1 ≈ 0675 ⇒  1 ≈

when  ≈ 0876, so


4


4

− 0675 ≈ 01106. Also,  = tan  = 12

+  2 ≈ 0876 ⇒  2 = 0876 −


4

≈ 00906.

Thus, we choose  = 00906 (or any smaller positive number) since this is the smaller of  1 and  2 .
From the graph with  = 02, we find that  = 3 − 3 + 4 = 58 when

7.

 ≈ 19774, so 2 − 1 ≈ 19774 ⇒ 1 ≈ 00226. Also,
 = 3 − 3 + 4 = 62 when  ≈ 2022, so 2 +  2 ≈ 20219 ⇒
 2 ≈ 00219. Thus, we choose  = 00219 (or any smaller positive number) since this is the smaller of  1 and  2 .

e

For  = 01, we get  1 ≈ 00112 and  2 ≈ 00110, so we choose

al

 = 0011 (or any smaller positive number).
9. (a)

From the graph, we find that  = tan2  = 1000 when  ≈ 1539 and

.
2

Thus, we get  ≈ 1602 −

rS

 ≈ 1602 for  near


2

≈ 0031 for

 = 1000.

From the graph, we find that  = tan2  = 10,000 when  ≈ 1561 and

Fo

(b)

 ≈ 1581 for  near


2.

Thus, we get  ≈ 1581 −


2

≈ 0010 for

ot

 = 10,000.

N

11. (a)  = 2 and  = 1000 cm2

⇒ 2 = 1000 ⇒ 2 =

1000


⇒ =



1000


(  0)

≈ 178412 cm.

(b) | − 1000| ≤ 5 ⇒ −5 ≤ 2 − 1000 ≤ 5 ⇒ 1000 − 5 ≤ 2 ≤ 1000 + 5 ⇒






995
1000
≤  ≤ 1005 ⇒ 177966 ≤  ≤ 178858.
− 995 ≈ 004466 and 1005 − 1000 ≈ 004455. So





 if the machinist gets the radius within 00445 cm of 178412, the area will be within 5 cm2 of 1000.

(c)  is the radius,  () is the area,  is the target radius given in part (a),  is the target area (1000),  is the tolerance in the area (5), and  is the tolerance in the radius given in part (b).
13. (a) |4 − 8| = 4 | − 2|  01

⇔ | − 2| 

01
01
, so  =
= 0025.
4
4

(b) |4 − 8| = 4 | − 2|  001 ⇔ | − 2| 

001
001
, so  =
= 00025.
4
4

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

36

¤

CHAPTER 1

FUNCTIONS AND LIMITS

15. Given   0, we need   0 such that if 0  | − 3|  , then







(1 + 1 ) − 2  . But (1 + 1 ) − 2   ⇔  1  − 1   ⇔
3
3
3
1
  | − 3|   ⇔ | − 3|  3. So if we choose  = 3, then
3


0  | − 3|   ⇒ (1 + 1 ) − 2  . Thus, lim (1 + 1 ) = 2 by
3
3
→3

the definition of a limit.

17. Given   0, we need   0 such that if 0  | − (−3)|  , then

|(1 − 4) − 13|  . But |(1 − 4) − 13|   ⇔
|−4 − 12|   ⇔ |−4| | + 3|   ⇔ | − (−3)|  4. So if we choose  = 4, then 0  | − (−3)|  

⇒ |(1 − 4) − 13|  .

Thus, lim (1 − 4) = 13 by the definition of a limit.

e

→−3

al


 2 + 4

19. Given   0, we need   0 such that if 0  | − 1|  , then 


x






 2 + 4
− 2  . But 
− 2   ⇔

 3


Fo

rS

3


 4 − 4 
4
3
3


 
 3    ⇔ 3 | − 1|   ⇔ | − 1|  4 . So if we choose  = 4 , then 0  | − 1|  


 2 + 4

2 + 4

− 2  . Thus, lim
= 2 by the definition of a limit.
 3

→1
3
 2
 +  − 6

21. Given   0, we need   0 such that if 0  | − 2|  , then 




 ( + 3)( − 2)


− 5  


−2



− 5   ⇔


| + 3 − 5|   [ 6= 2] ⇔ | − 2|  . So choose  = .



 ( + 3)( − 2)

⇒ | − 2|   ⇒ | + 3 − 5|   ⇒ 
− 5   [ 6= 2] ⇒


−2

ot

Then 0  | − 2|  

⇔

−2

⇒

N


 2

 +  − 6
2 +  − 6

− 5  . By the definition of a limit, lim
= 5.

 −2
→2
−2

23. Given   0, we need   0 such that if 0  | − |  , then | − |  . So  =  will work.









25. Given   0, we need   0 such that if 0  | − 0|  , then 2 − 0  

Then 0  | − 0|  

⇔ 2  



⇒ 2 − 0  . Thus, lim 2 = 0 by the definition of a limit.

⇔ || 

√
√
. Take  = .

→0





27. Given   0, we need   0 such that if 0  | − 0|  , then || − 0  . But || = ||. So this is true if we pick  = .

Thus, lim || = 0 by the definition of a limit.
→0





⇔ 2 − 4 + 4   ⇔


√
⇔ | − 2|   ⇔ ( − 2)2   . Thus,




29. Given   0, we need   0 such that if 0  | − 2|  , then  2 − 4 + 5 − 1  



( − 2)2   . So take  = √. Then 0  | − 2|  

 lim 2 − 4 + 5 = 1 by the definition of a limit.
→2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.7





THE PRECISE DEFINITION OF A LIMIT

¤

37



31. Given   0, we need   0 such that if 0  | − (−2)|  , then  2 − 1 − 3   or upon simplifying we need

 2

 − 4   whenever 0  | + 2|  . Notice that if | + 2|  1, then −1   + 2  1 ⇒ −5   − 2  −3 ⇒

| − 2|  5. So take  = min {5 1}. Then 0  | + 2|   ⇒ | − 2|  5 and | + 2|  5, so
 2


  − 1 − 3 = |( + 2)( − 2)| = | + 2| | − 2|  (5)(5) = . Thus, by the definition of a limit, lim (2 − 1) = 3.
→−2


. If 0  | − 3|  , then | − 3|  2 ⇒ −2   − 3  2 ⇒




4   + 3  8 ⇒ | + 3|  8. Also | − 3|  8 , so 2 − 9 = | + 3| | − 3|  8 · 8 = . Thus, lim 2 = 9.


33. Given   0, we let  = min 2


8

→3

35. (a) The points of intersection in the graph are (1  26) and (2  34)

with 1 ≈ 0891 and 2 ≈ 1093. Thus, we can take  to be the

e

smaller of 1 − 1 and 2 − 1. So  = 2 − 1 ≈ 0093.

rS

al

(b) Solving 3 +  + 1 = 3 +  gives us two nonreal complex roots and
√
23

− 12
216 + 108 + 12 336 + 324 + 812 one real root, which is () =
√
13 . Thus,  = () − 1.

2
6 216 + 108 + 12 336 + 324 + 81

(c) If  = 04, then () ≈ 1093 272 342 and  = () − 1 ≈ 0093, which agrees with our answer in part (a).

37. 1. Guessing a value for 

√
√
Given   0, we must find   0 such that |  − |   whenever 0  | − |  . But

Fo

√
√
√
√
| − |
√   (from the hint). Now if we can find a positive constant  such that  +    then
|  − | = √
+ 




√
√ 
1
 is a suitable choice for the constant. So | − | 
 +  . This suggests that we let
2

1

2

+



1

2



√  
  .

N

=

ot

| − |
| − |
√
√ 
 , and we take | − |  . We can find this number by restricting  to lie in some interval

+ 

√
√
√ centered at . If | − |  1 , then − 1    −   1  ⇒ 1     3  ⇒
 +   1  + , and so
2
2
2
2
2
2

 = min

1

2

+

2. Showing that  works
| − |  1  ⇒
2

Given   0, we let  = min



1

2



1

2

+

√  
  . If 0  | − |  , then



√
√
√
√ 
1
 +   1  +  (as in part 1). Also | − | 
 +  , so
2
2


√ 
2 +  
√
√
√
√
| − |
√  
|  − | = √ lim √  = . Therefore, →  =  by the definition of a limit.
+ 
2 + 

39. Suppose that lim  () = . Given  =
→0

1
,
2

there exists   0 such that 0  ||  

⇒ |() − |  1 . Take any rational
2

number  with 0  ||  . Then  () = 0, so |0 − |  1 , so  ≤ ||  1 . Now take any irrational number  with
2
2
0  ||  . Then  () = 1, so |1 − |  1 . Hence, 1 −   1 , so   1 . This contradicts   1 , so lim () does not
2
2
2
2
→0

exist. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

38

¤

41.

1
1
 10,000 ⇔ ( + 3)4 
( + 3)4
10,000

CHAPTER 1

FUNCTIONS AND LIMITS

1
⇔ | + 3|  √
4
10,000

⇔

| − (−3)| 

1
10


5
5
5
3
43. Let   0 be given. Then, for   −1, we have
 ⇔
 ( + 1) ⇔ 3
  + 1.
( + 1)3




5
5
5
5
. Then −1 −     −1 ⇒ 3
+10 ⇒
 , so lim
= −∞.
Let  = − 3


( + 1)3
→−1− ( + 1)3

1.8 Continuity
1. From Definition 1, lim  () =  (4).
→4

3. (a)  is discontinuous at −4 since  (−4) is not defined and at −2, 2, and 4 since the limit does not exist (the left and right

lim  () =  (−2).  is continuous from the right at 2 and 4 since

→−2−

al

(b)  is continuous from the left at −2 since

e

limits are not the same).

lim () =  (2) and lim  () =  (4). It is continuous from neither side at −4 since  (−4) is undefined.
→4+

5. The graph of  =  () must have a discontinuity at

 = 2 and must show that lim  () =  (2).
→2+

7. The graph of  =  () must have a removable

rS

→2+

discontinuity (a hole) at  = 3 and a jump discontinuity

ot

Fo

at  = 5.

N

9. (a) The toll is $7 between 7:00 AM and 10:00 AM and between 4:00 PM and 7:00 PM .

(b) The function  has jump discontinuities at  = 7, 10, 16, and 19. Their significance to someone who uses the road is that, because of the sudden jumps in the toll, they may want to avoid the higher rates between  = 7 and  = 10 and between  = 16 and  = 19 if feasible.
11. If  and  are continuous and (2) = 6, then lim [3 () +  () ()] = 36
→2

⇒

3 lim  () + lim  () · lim () = 36 ⇒ 3 (2) + (2) · 6 = 36 ⇒ 9 (2) = 36 ⇒  (2) = 4.
→2

→2

13. lim  () = lim
→−1

→−1

→2


4
 + 23 =



lim  + 2 lim 3

→−1

→−1

By the definition of continuity,  is continuous at  = −1.

4

4

= −1 + 2(−1)3 = (−3)4 = 81 = (−1).

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.8

CONTINUITY

¤

15. For   2, we have

lim (2 + 3)
2 + 3
→
=
→  − 2 lim ( − 2)

lim  () = lim

→

[Limit Law 5]

→

2 lim  + lim 3
=

→

→

=

→

[1, 2, and 3]

lim  − lim 2
→

2 + 3
−2

[7 and 8]

=  ()

Thus,  is continuous at  =  for every  in (2 ∞); that is,  is continuous on (2 ∞).
1
is discontinuous at  = −2 because  (−2) is undefined.
+2

19.  () =



1

if   1

rS

1 − 2

al

e

17.  () =

if  ≥ 1

The left-hand limit of  at  = 1 is
→1−

Fo

lim  () = lim (1 − 2 ) = 0 The right-hand limit of  at  = 1 is

→1−

lim  () = lim (1) = 1 Since these limits are not equal, lim  ()

→1+

→1+

→1

does not exist and  is discontinuous at 1. if   0

ot


 cos 

21.  () = 0


1 − 2

if  = 0

N

if   0

lim  () = 1, but  (0) = 0 6= 1, so  is discontinuous at 0.

→0

23.  () =

( − 2)( + 1)
2 −  − 2
=
=  + 1 for  6= 2. Since lim  () = 2 + 1 = 3, define  (2) = 3. Then  is
→2
−2
−2

continuous at 2.
25.  () =

22 −  − 1 is a rational function, so it is continuous on its domain, (−∞ ∞), by Theorem 5(b).
2 + 1

27. 3 − 2 = 0

⇒ 3 = 2 ⇒  =

√
3
2, so () =

continuous everywhere by Theorem 5(a) and

√
3
√  √


−2
has domain −∞ 3 2 ∪ 3 2 ∞ . Now 3 − 2 is
3 −2


√
3
 − 2 is continuous everywhere by Theorems 5(a), 7, and 9. Thus,  is

continuous on its domain by part 5 of Theorem 4.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

39

40

¤

CHAPTER 1

FUNCTIONS AND LIMITS

29. By Theorem 5, the polynomial 1 − 2 is continuous on (−∞ ∞). By Theorem 7, cos is continuous on its domain, R. By



Theorem 9, cos 1 − 2 is continuous on its domain, which is R.

31.  () =



+1
+1
1 is defined when
≥ 0 ⇒  + 1 ≥ 0 and   0 or  + 1 ≤ 0 and   0 ⇒   0
1+ =




or  ≤ −1, so  has domain (−∞ −1] ∪ (0 ∞).  is the composite of a root function and a rational function, so it is continuous at every number in its domain by Theorems 7 and 9.
33.

=

1 is undefined and hence discontinuous when
1 + sin 

1 + sin  = 0 ⇔ sin  = −1 ⇔  = −  + 2,  an
2
integer. The figure shows discontinuities for  = −1, 0, and 1; that

√
√
 is continuous on [0 ∞),  + 5 is continuous on [−5 ∞), so the

al

35. Because we are dealing with root functions, 5 +

5

3
≈ −785, − ≈ −157, and
≈ 471.
2
2
2

e

is, −

rS

√
5+  is continuous on [0 ∞). Since  is continuous at  = 4, lim  () =  (4) = 7 . quotient  () = √
3
→4
5+
37. Because  and cos  are continuous on R, so is  () =  cos2 . Since  is continuous at  =

→4

39.  () =



2
√



4

=


4

 √ 2
 1
2

= · = .
2
4 2
8

Fo

lim  () = 


,
4

if   1 if  ≥ 1

lim  () = lim

→1+

→1−

→1−

√
√
 = 1. Thus, lim () exists and equals 1. Also,  (1) = 1 = 1. Thus,  is continuous at  = 1.
→1

N

→1+

ot

By Theorem 5, since  () equals the polynomial 2 on (−∞ 1),  is continuous on (−∞ 1). By Theorem 7, since  ()
√
equals the root function  on (1 ∞)  is continuous on (1 ∞). At  = 1, lim () = lim 2 = 1 and

We conclude that  is continuous on (−∞ ∞).

 1 + 2


41.  () = 2 − 


 ( − 2)2

if  ≤ 0

if 0   ≤ 2 if   2

 is continuous on (−∞ 0), (0 2), and (2 ∞) since it is a polynomial on each of these intervals. Now lim  () = lim (1 + 2 ) = 1 and lim () = lim (2 − ) = 2 so  is
→0−

→0−

→0+

→0+

discontinuous at 0. Since  (0) = 1,  is continuous from the left at 0 Also, lim  () = lim (2 − ) = 0
→2−

→2−

lim  () = lim ( − 2)2 = 0, and  (2) = 0, so  is continuous at 2 The only number at which  is discontinuous is 0.

→2+

→2+

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 1.8


 + 2


43.  () = 22



2−

CONTINUITY

¤

41

if   0 if 0 ≤  ≤ 1 if   1

 is continuous on (−∞ 0), (0 1), and (1 ∞) since on each of

these intervals it is a polynomial. Now lim () = lim ( + 2) = 2 and
→0−

→0−

lim  () = lim 2 = 0, so  is discontinuous at 0. Since  (0) = 0,  is continuous from the right at 0. Also
2

→0+

→0+

lim  () = lim 22 = 2 and lim  () = lim (2 − ) = 1, so  is discontinuous at 1. Since  (1) = 2,

→1−

→1−

→1+

→1+

 is continuous from the left at 1.
2 + 2
3

 − 

if   2 if  ≥ 2

 is continuous on (−∞ 2) and (2 ∞). Now lim  () = lim
→2−


 3
 −  = 8 − 2. So  is continuous ⇔ 4 + 4 = 8 − 2 ⇔ 6 = 4 ⇔  = 2 . Thus, for 
3

to be continuous on (−∞ ∞),  = 2 .
3
47. (a)  () =

al

→2+

 2

 + 2 = 4 + 4 and

rS

lim  () = lim

→2+

→2−

e

45.  () =



4 − 1
(2 + 1)(2 − 1)
(2 + 1)( + 1)( − 1)
=
=
= (2 + 1)( + 1) [or 3 + 2 +  + 1]
−1
−1
−1

(b)  () =

Fo

for  6= 1. The discontinuity is removable and () = 3 + 2 +  + 1 agrees with  for  6= 1 and is continuous on R.
(2 −  − 2)
( − 2)( + 1)
3 − 2 − 2
=
=
= ( + 1) [or 2 + ] for  6= 2. The discontinuity
−2
−2
−2

ot

is removable and () = 2 +  agrees with  for  6= 2 and is continuous on R.
(c) lim  () = lim [[sin ]] = lim 0 = 0 and lim  () = lim [[sin ]] = lim (−1) = −1, so lim  () does not
→−

→ −

→ −

→ +

→ +

→ +

→

N

exist. The discontinuity at  =  is a jump discontinuity.
49.  () = 2 + 10 sin  is continuous on the interval [31 32],  (31) ≈ 957, and (32) ≈ 1030. Since 957  1000  1030,

there is a number c in (31 32) such that  () = 1000 by the Intermediate Value Theorem. Note: There is also a number c in
(−32 −31) such that  () = 1000
51.  () = 4 +  − 3 is continuous on the interval [1 2]  (1) = −1, and (2) = 15. Since −1  0  15, there is a number 

in (1 2) such that  () = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation 4 +  − 3 = 0 in the interval (1 2)
53.  () = cos  −  is continuous on the interval [0 1],  (0) = 1, and (1) = cos 1 − 1 ≈ −046. Since −046  0  1, there

is a number  in (0 1) such that  () = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation cos  −  = 0, or cos  = , in the interval (0 1). c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

42

¤

CHAPTER 1

FUNCTIONS AND LIMITS

55. (a)  () = cos  − 3 is continuous on the interval [0 1],  (0) = 1  0, and  (1) = cos 1 − 1 ≈ −046  0. Since

1  0  −046, there is a number  in (0 1) such that  () = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation cos  − 3 = 0, or cos  = 3 , in the interval (0 1).
(b)  (086) ≈ 0016  0 and  (087) ≈ −0014  0, so there is a root between 086 and 087, that is, in the interval
(086 087).
57. (a) Let  () = 5 − 2 − 4. Then (1) = 15 − 12 − 4 = −4  0 and (2) = 25 − 22 − 4 = 24  0. So by the

Intermediate Value Theorem, there is a number  in (1 2) such that  () = 5 − 2 − 4 = 0.

al

e

(b) We can see from the graphs that, correct to three decimal places, the root is  ≈ 1434.



lim  ( + ) =  lim ( + ) =  ().

→0

→0

rS

59. (⇒) If  is continuous at , then by Theorem 8 with () =  + , we have

(⇐) Let   0. Since lim ( + ) =  (), there exists   0 such that 0  ||  

⇒

Fo

→0

| ( + ) −  ()|  . So if 0  | − |  , then | () −  ()| = | ( + ( − )) −  ()|  .
Thus, lim  () =  () and so  is continuous at .
→

61. As in the previous exercise, we must show that lim cos( + ) = cos  to prove that the cosine function is continuous.

ot

→0

lim cos( + ) = lim (cos  cos  − sin  sin ) = lim (cos  cos ) − lim (sin  sin )

→0

→0

63.  () =



→0

0 if  is rational

1 if  is irrational

→0

→0


 

 lim cos  lim cos  − lim sin  lim sin  = (cos )(1) − (sin )(0) = cos 

N

=



→0

→0

→0

is continuous nowhere. For, given any number  and any   0, the interval ( −   + )

contains both infinitely many rational and infinitely many irrational numbers. Since  () = 0 or 1, there are infinitely many numbers  with 0  | − |   and | () −  ()| = 1. Thus, lim () 6=  (). [In fact, lim  () does not even exist.]
→

65. If there is such a number, it satisfies the equation 3 + 1 = 

→

⇔ 3 −  + 1 = 0. Let the left-hand side of this equation be

called  (). Now  (−2) = −5  0, and  (−1) = 1  0. Note also that  () is a polynomial, and thus continuous. So by the
Intermediate Value Theorem, there is a number  between −2 and −1 such that () = 0, so that  = 3 + 1.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 1 REVIEW

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43

67.  () = 4 sin(1) is continuous on (−∞ 0) ∪ (0 ∞) since it is the product of a polynomial and a composite of a

trigonometric function and a rational function. Now since −1 ≤ sin(1) ≤ 1, we have −4 ≤ 4 sin(1) ≤ 4 . Because lim (−4 ) = 0 and lim 4 = 0, the Squeeze Theorem gives us lim (4 sin(1)) = 0, which equals (0). Thus,  is

→0

→0

→0

continuous at 0 and, hence, on (−∞ ∞).
69. Define () to be the monk’s distance from the monastery, as a function of time  (in hours), on the first day, and define ()

to be his distance from the monastery, as a function of time, on the second day. Let  be the distance from the monastery to the top of the mountain. From the given information we know that (0) = 0, (12) = , (0) =  and (12) = 0. Now consider the function  − , which is clearly continuous. We calculate that ( − )(0) = − and ( − )(12) = .
So by the Intermediate Value Theorem, there must be some time 0 between 0 and 12 such that ( − )(0 ) = 0 ⇔

e

(0 ) = (0 ). So at time 0 after 7:00 AM, the monk will be at the same place on both days.

rS

al

1 REVIEW

1. (a) A function  is a rule that assigns to each element  in a set  exactly one element, called (), in a set . The set  is

called the domain of the function. The range of  is the set of all possible values of  () as  varies throughout the domain. Fo

(b) If  is a function with domain , then its graph is the set of ordered pairs {(  ()) |  ∈ }.
(c) Use the Vertical Line Test on page 15.

2. The four ways to represent a function are: verbally, numerically, visually, and algebraically. An example of each is given

ot

below.
Verbally: An assignment of students to chairs in a classroom (a description in words)
Numerically: A tax table that assigns an amount of tax to an income (a table of values)

N

Visually: A graphical history of the Dow Jones average (a graph)
Algebraically: A relationship between distance, rate, and time:  =  (an explicit formula)
3. (a) If a function  satisfies  (−) =  () for every number  in its domain, then  is called an even function. If the graph of

a function is symmetric with respect to the -axis, then  is even. Examples of an even function: () = 2 ,
 () = 4 + 2 ,  () = ||,  () = cos .
(b) If a function  satisfies  (−) = − () for every number  in its domain, then  is called an odd function. If the graph of a function is symmetric with respect to the origin, then  is odd. Examples of an odd function: () = 3 ,
√
 () = 3 + 5 ,  () = 3 ,  () = sin .
4. A function  is called increasing on an interval  if  (1 )   (2 ) whenever 1  2 in .
5. A mathematical model is a mathematical description (often by means of a function or an equation) of a real-world

phenomenon. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

44

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CHAPTER 1

FUNCTIONS AND LIMITS

6. (a) Linear function: () = 2 + 1,  () =  + 

7.

(b) Power function: () =  , () = 
2



(c) Exponential function:  () = 2 ,  () = 
(d) Quadratic function:  () = 2 +  + 1,  () = 2 +  + 
(e) Polynomial of degree 5: () = 5 + 2
(f ) Rational function:  () =


 ()
,  () = where  () and
+2
()

() are polynomials
(b)

(d)

Fo

rS

(c)

al

e

8. (a)

(f )

N

ot

(e )

9. (a) The domain of  +  is the intersection of the domain of  and the domain of ; that is,  ∩ .

(b) The domain of   is also  ∩ .

(c) The domain of   must exclude values of  that make  equal to 0; that is, { ∈  ∩  | () 6= 0}.

10. Given two functions  and , the composite function  ◦  is defined by ( ◦ ) () =  ( ()). The domain of  ◦  is the

set of all  in the domain of  such that () is in the domain of  .
11. (a) If the graph of  is shifted 2 units upward, its equation becomes  =  () + 2.

(b) If the graph of  is shifted 2 units downward, its equation becomes  =  () − 2.

(c) If the graph of  is shifted 2 units to the right, its equation becomes  =  ( − 2).

(d) If the graph of  is shifted 2 units to the left, its equation becomes  =  ( + 2).
(e) If the graph of  is reflected about the -axis, its equation becomes  = − ().

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 1 REVIEW

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45

(f ) If the graph of  is reflected about the -axis, its equation becomes  =  (−).
(g) If the graph of  is stretched vertically by a factor of 2, its equation becomes  = 2 ().
(h) If the graph of  is shrunk vertically by a factor of 2, its equation becomes  = 1  ().
2
 
(i) If the graph of  is stretched horizontally by a factor of 2, its equation becomes  =  1  .
2
(j) If the graph of  is shrunk horizontally by a factor of 2, its equation becomes  =  (2).

12. (a) lim  () = : See Definition 1.5.1 and Figures 1 and 2 in Section 1.5.
→

(b) lim  () = : See the paragraph after Definition 1.5.2 and Figure 9(b) in Section 1.5.
→+

(c) lim () = : See Definition 1.5.2 and Figure 9(a) in Section 1.5.
→−

(d) lim  () = ∞: See Definition 1.5.4 and Figure 12 in Section 1.5.
→

(e) lim  () = −∞: See Definition 1.5.5 and Figure 13 in Section 1.5.
→

e

13. In general, the limit of a function fails to exist when the function does not approach a fixed number. For each of the following

Fo

rS

al

functions, the limit fails to exist at  = 2.

The left- and right-hand

There is an

There are an infinite

limits are not equal.

infinite discontinuity.

number of oscillations.

ot

14. See Definition 1.5.6 and Figures 12–14 in Section 1.5.
15. (a) – (g) See the statements of Limit Laws 1– 6 and 11 in Section 1.6.

N

16. See Theorem 3 in Section 1.6.

17. (a) A function  is continuous at a number  if () approaches  () as  approaches ; that is, lim  () =  ().
→

(b) A function  is continuous on the interval (−∞ ∞) if  is continuous at every real number . The graph of such a function has no breaks and every vertical line crosses it.

18. See Theorem 1.8.10.

1. False.

Let  () = 2 ,  = −1, and  = 1. Then  ( + ) = (−1 + 1)2 = 02 = 0, but

 () + () = (−1)2 + 12 = 2 6= 0 =  ( + ).
3. False.

Let  () = 2 . Then  (3) = (3)2 = 92 and 3 () = 32 . So  (3) 6= 3 (). c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

46

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CHAPTER 1

FUNCTIONS AND LIMITS

5. True.

See the Vertical Line Test.

7. False.

Limit Law 2 applies only if the individual limits exist (these don’t).

9. True.

Limit Law 5 applies.

11. False.

Consider lim

→5

( − 5) sin( − 5) or lim
. The first limit exists and is equal to 5. By Example 3 in Section 1.5,
→5
−5
−5

we know that the latter limit exists (and it is equal to 1).
13. True.

Suppose that lim [ () + ()] exists. Now lim () exists and lim () does not exist, but
→

→

→

lim () = lim {[ () + ()] −  ()} = lim [ () + ()] − lim () [by Limit Law 2], which exists, and

→

→

→

→

we have a contradiction. Thus, lim [ () + ()] does not exist.
→

A polynomial is continuous everywhere, so lim () exists and is equal to ().
→


1( − 1) if  6= 1

e

15. True.

Consider  () =

19. True.

Use Theorem 1.8.8 with  = 2,  = 5, and () = 42 − 11. Note that  (4) = 3 is not needed.

rS

if  = 1

2

21. True, by the definition of a limit with  = 1.
23. True.

al

17. False.

 () = 10 − 102 + 5 is continuous on the interval [0 2],  (0) = 5,  (1) = −4, and  (2) = 989. Since

Fo

−4  0  5, there is a number  in (0 1) such that  () = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation 10 − 102 + 5 = 0 in the interval (0 1). Similarly, there is a root in (1 2).
See Exercise 68(c) in Section 1.8.

ot

25. False

N

1. (a) When  = 2,  ≈ 27. Thus,  (2) ≈ 27.

(c) The domain of  is −6 ≤  ≤ 6, or [−6 6].

(b)  () = 3 ⇒  ≈ 23, 56
(d) The range of  is −4 ≤  ≤ 4, or [−4 4].

(e)  is increasing on [−4 4], that is, on −4 ≤  ≤ 4.
(f)  is odd since its graph is symmetric about the origin.
3.  () = 2 − 2 + 3, so  ( + ) = ( + )2 − 2( + ) + 3 = 2 + 2 + 2 − 2 − 2 + 3, and

(2 + 2 + 2 − 2 − 2 + 3) − (2 − 2 + 3)
(2 +  − 2)
 ( + ) − ()
=
=
= 2 +  − 2.



 


5.  () = 2(3 − 1).
Domain: 3 − 1 6= 0 ⇒ 3 6= 1 ⇒  6= 1 .  = −∞ 1 ∪ 1  ∞
3
3
3
Range:

7.  = 1 + sin .

all reals except 0 ( = 0 is the horizontal asymptote for  .)  = (−∞ 0) ∪ (0 ∞)

Domain: R.
Range:

−1 ≤ sin  ≤ 1 ⇒ 0 ≤ 1 + sin  ≤ 2 ⇒ 0 ≤  ≤ 2.  = [0 2]

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 1 REVIEW

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47

9. (a) To obtain the graph of  =  () + 8, we shift the graph of  =  () up 8 units.

(b) To obtain the graph of  =  ( + 8), we shift the graph of  =  () left 8 units.
(c) To obtain the graph of  = 1 + 2 (), we stretch the graph of  =  () vertically by a factor of 2, and then shift the resulting graph 1 unit upward.
(d) To obtain the graph of  =  ( − 2) − 2, we shift the graph of  =  () right 2 units (for the “−2” inside the parentheses), and then shift the resulting graph 2 units downward.

(e) To obtain the graph of  = − (), we reflect the graph of  =  () about the -axis.
(f) To obtain the graph of  = 3 −  (), we reflect the graph of  =  () about the -axis, and then shift the resulting graph
3 units upward.

13.  = 1 + 1 3 : Start with the
2

graph of  = 3 , compress

shift 1 unit upward.

1
:
+2

ot

15.  () =

Fo

vertically by a factor of 2, and

rS

al

e

11.  = − sin 2: Start with the graph of  = sin , compress horizontally by a factor of 2, and reflect about the -axis.

Start with the graph of  () = 1

N

and shift 2 units to the left.

17. (a) The terms of  are a mixture of odd and even powers of , so  is neither even nor odd.

(b) The terms of  are all odd powers of , so  is odd.


(c)  (−) = cos (−)2 = cos(2 ) =  (), so  is even.

(d)  (−) = 1 + sin(−) = 1 − sin . Now  (−) 6=  () and (−) 6= −(), so  is neither even nor odd.
19.  () =

√
,  = [0 ∞); () = sin ,  = R.

(a) ( ◦ )() =  (()) = (sin ) =

√
√
sin . For sin  to be defined, we must have sin  ≥ 0 ⇔

 ∈ [0 ], [2 3], [−2 −], [4 5], [−4 −3],    , so  = { |  ∈ [2  + 2] , where  is an integer}.
√
√
√
(b) ( ◦ )() = ( ()) = (  ) = sin .  must be greater than or equal to 0 for  to be defined, so  = [0 ∞). c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

48

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CHAPTER 1

FUNCTIONS AND LIMITS

√
√
√
4
(c) ( ◦  )() = ( ()) =  (  ) =
 = .  = [0 ∞).

(d) ( ◦ )() = (()) = (sin ) = sin(sin ).  = R.

Many models appear to be plausible. Your choice depends on whether you

21.

think medical advances will keep increasing life expectancy, or if there is bound to be a natural leveling-off of life expectancy. A linear model,
 = 02493 − 4234818, gives us an estimate of 776 years for the

year 2010.
23. (a) (i) lim  () = 3

(ii)

→2+

lim  () = 0

→−3+

(iii) lim  () does not exist since the left and right limits are not equal. (The left limit is −2.)
→−3

(iv) lim  () = 2

e

→4

(v) lim  () = ∞

(vi) lim  () = −∞

→0

(b) The equations of the vertical asymptotes are  = 0 and  = 2.

al

→2−

25. lim cos( + sin ) = cos
→0

→−3

29. lim

→0

 lim ( + sin ) [by Theorem 1.8.8] = cos 0 = 1

→0

2 − 9
( + 3)( − 3)
−3
−3 − 3
−6
3
= lim
= lim
=
=
=
2 + 2 − 3 →−3 ( + 3)( − 1) →−3  − 1
−3 − 1
−4
2

Fo

27. lim



rS

(c)  is discontinuous at  = −3, 0, 2, and 4. The discontinuities are jump, infinite, infinite, and removable, respectively.

 3



 − 32 + 3 − 1 + 1
( − 1)3 + 1
3 − 32 + 3
= lim
= lim
= lim 2 − 3 + 3 = 3
→0
→0
→0




N

ot

Another solution: Factor the numerator as a sum of two cubes and then simplify.


[( − 1) + 1] ( − 1)2 − 1( − 1) + 12
( − 1)3 + 1
( − 1)3 + 13
= lim
= lim lim →0
→0
→0





= lim ( − 1)2 −  + 2 = 1 − 0 + 2 = 3
→0

√
√


4
+
31. lim
= ∞ since ( − 9) → 0 as  → 9 and
 0 for  6= 9.
→9 ( − 9)4
( − 9)4
33. lim

→1 3

4 − 1
(2 + 1)(2 − 1)
(2 + 1)( + 1)( − 1)
(2 + 1)( + 1)
2(2)
4
= lim
= lim
= lim
=
=
2 − 6
2 + 5 − 6)
→1 (
→1
→1
+ 5
( + 6)( − 1)
( + 6)
1(7)
7

√
√
4− 
4− 
−1
1
−1
√
= lim √
=−
= lim √
= √
→16  − 16
→16 (  + 4)(  − 4)
→16
8
+4
16 + 4

35. lim

37. lim

→0

1−

√
√
1 − (1 − 2 )
1 − 2 1 + 1 − 2
2

√
√
√
√
 = lim 
 = lim
= lim 
=0
·
→0  1 +
→0  1 +
→0 1 +

1 + 1 − 2
1 − 2
1 − 2
1 − 2

39. Since 2 − 1 ≤ () ≤ 2 for 0    3 and lim (2 − 1) = 1 = lim 2 , we have lim  () = 1 by the Squeeze Theorem.
→1

→1

→1

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 1 REVIEW

41. Given   0, we need   0 such that if 0  | − 2|  , then |(14 − 5) − 4|  . But |(14 − 5) − 4|  

|−5 + 10|  

⇔

|−5| | − 2|   ⇔ | − 2|  5. So if we choose  = 5, then 0  | − 2|  

¤

⇔
⇒

|(14 − 5) − 4|  . Thus, lim (14 − 5) = 4 by the definition of a limit.
→2





43. Given   0, we need   0 so that if 0  | − 2|  , then 2 − 3 − (−2)  . First, note that if | − 2|  1, then

−1   − 2  1, so 0   − 1  2 ⇒ | − 1|  2. Now let  = min {2 1}. Then 0  | − 2|  
 2

 − 3 − (−2) = |( − 2)( − 1)| = | − 2| | − 1|  (2)(2) = .

⇒

Thus, lim (2 − 3) = −2 by the definition of a limit.
→2

45. (a)  () =

√
− if   0,  () = 3 −  if 0 ≤   3,  () = ( − 3)2 if   3.
(ii) lim  () = lim

(i) lim  () = lim (3 − ) = 3
→0+

→0−

→0+

(iv) lim  () = lim (3 − ) = 0

→0

→3−

2

(v) lim  () = lim ( − 3) = 0

→3−

(vi) Because of (iv) and (v), lim  () = 0.

→3+

(b)  is discontinuous at 0 since lim  () does not exist.

→3

(c)

rS

→0

al

→3+

√
− = 0

e

(iii) Because of (i) and (ii), lim  () does not exist.

→0−

 is discontinuous at 3 since  (3) does not exist.

The root function domain, [0 ∞).

Fo

47. 3 is continuous on R since it is a polynomial and cos  is also continuous on R, so the product 3 cos  is continuous on R.

√
√
4
4
 is continuous on its domain, [0 ∞), and so the sum, () =  + 3 cos , is continuous on its

ot

49.  () = 5 − 3 + 3 − 5 is continuous on the interval [1 2],  (1) = −2, and (2) = 25. Since −2  0  25, there is a

number  in (1 2) such that  () = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation

N

5 − 3 + 3 − 5 = 0 in the interval (1 2).

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

49

N ot e

al

rS

Fo

PRINCIPLES OF PROBLEM SOLVING
1. Remember that || =  if  ≥ 0 and that || = − if   0. Thus,

 + || =


2 if  ≥ 0

and

if   0

0

We will consider the equation  + || =  + || in four cases.
(1)  ≥ 0  ≥ 0
2 = 2

(2)  ≥ 0,   0
2 = 0

=

=0

(3)   0,  ≥ 0
0 = 2

 + || =


2
0

if  ≥ 0

if   0

(4)   0   0
0=0

0=

Case 1 gives us the line  =  with nonnegative  and .
Case 2 gives us the portion of the -axis with  negative.

e

Case 3 gives us the portion of the -axis with  negative.

3. 0 () = 2 and +1 () = 0 ( ()) for  = 0 1 2   .

al

Case 4 gives us the entire third quadrant.

rS

   2
1 () = 0 (0 ()) = 0 2 = 2 = 4 , 2 () = 0 (1 ()) = 0 (4 ) = (4 )2 = 8 ,

+1

3 () = 0 (2 ()) = 0 (8 ) = (8 )2 = 16 ,   . Thus, a general formula is  () = 2
5. Let  =

.

√
6
, so  = 6 . Then  → 1 as  → 1, so

Fo

√
3
−1
2 − 1
( − 1)( + 1)
+1
1+1
2
= lim 3 lim √
= lim
= lim 2
= 2
= .
→1
→1  − 1
→1 ( − 1) (2 +  + 1)
→1  +  + 1
1 +1+1
3
−1
√

√
√
3
2 + 3  + 1 .
Another method: Multiply both the numerator and the denominator by (  + 1)

Therefore, lim

we have 2 − 1  0 and 2 + 1  0, so |2 − 1| = −(2 − 1) and |2 + 1| = 2 + 1.

|2 − 1| − |2 + 1|
−(2 − 1) − (2 + 1)
−4
= lim
= lim
= lim (−4) = −4.
→0
→0
→0




N

→0

1
,
2

ot

7. For − 1   
2

9. (a) For 0    1, [[]] = 0, so

lim

→0−

[[]]
= lim

→0−



−1




[[]]
[[]]
−1
[[]]
= 0, and lim
= 0. For −1    0, [[]] = −1, so
=
, and



→0+ 

= ∞. Since the one-sided limits are not equal, lim

→0

[[]] does not exist.


(b) For   0, 1 − 1 ≤ [[1]] ≤ 1 ⇒ (1 − 1) ≤ [[1]] ≤ (1) ⇒ 1 −  ≤ [[1]] ≤ 1.
As  → 0+ , 1 −  → 1, so by the Squeeze Theorem, lim [[1]] = 1.
→0+

For   0, 1 − 1 ≤ [[1]] ≤ 1 ⇒ (1 − 1) ≥ [[1]] ≥ (1) ⇒ 1 −  ≥ [[1]] ≥ 1.
As  → 0− , 1 −  → 1, so by the Squeeze Theorem, lim [[1]] = 1.
→0−

Since the one-sided limits are equal, lim [[1]] = 1.
→0

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

51

52

¤

CHAPTER 1 PRINCIPLES OF PROBLEM SOLVING

11.  is continuous on (−∞ ) and ( ∞). To make  continuous on R, we must have continuity at . Thus,

lim  () = lim  () ⇒
→−

→+

lim 2 = lim ( + 1) ⇒ 2 =  + 1 ⇒ 2 −  − 1 = 0 ⇒
→−

→+

√ 

[by the quadratic formula]  = 1 ± 5 2 ≈ 1618 or −0618.

13. lim  () = lim
→

→

=

1
2

1
2

·2+

and lim () = lim
→

[ () + ()] +

→

1
2




[() − ()] =

1
2

· 1 = 3,
2

+ ()] +

1 lim [ ()
2 →


[ () + ()] −  () = lim [ () + ()] − lim () = 2 −
→




So lim [ ()()] = lim () lim () =
→

1 lim [ ()
2 →

→

→

3
2

·

1
2

→

3
2

− ()]

= 1.
2

= 3.
4

Another solution: Since lim [ () + ()] and lim [ () − ()] exist, we must have
→

→

→



lim [ () ()] = lim 1 [ () + ()]2 − [ () − ()]2
4
→

=

1
4



[because all of the  2 and 2 cancel]



 lim [ () + ()]2 − lim [ () − ()]2 = 1 22 − 12 = 3 .
4
4

→

→

rS

→

→

e

lim [ () + ()]2 =

→

→

2

2 lim [ () + ()] and lim [ () − ()]2 = lim [ () − ()] , so

al



15. (a) Consider () =  ( + 180◦ ) −  (). Fix any number . If () = 0, we are done: Temperature at  = Temperature

at  + 180◦ . If ()  0, then ( + 180◦ ) =  ( + 360◦ ) −  ( + 180◦ ) =  () −  ( + 180◦ ) = −()  0.

Fo

Also,  is continuous since temperature varies continuously. So, by the Intermediate Value Theorem,  has a zero on the interval [  + 180◦ ]. If ()  0, then a similar argument applies.
(b) Yes. The same argument applies.

ot

(c) The same argument applies for quantities that vary continuously, such as barometric pressure. But one could argue that

N

altitude above sea level is sometimes discontinuous, so the result might not always hold for that quantity.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

2

DERIVATIVES

2.1 Derivatives and Rates of Change
1. (a) This is just the slope of the line through two points:   =

∆
 () −  (3)
=
.
∆
−3

(b) This is the limit of the slope of the secant line   as  approaches  :  = lim

→3

() −  (3)
.
−3

3. (a) (i) Using Definition 1 with  () = 4 − 2 and  (1 3),
→

 () −  ()
(4 − 2 ) − 3
−(2 − 4 + 3)
−( − 1)( − 3)
= lim
= lim
= lim
→1
→1
→1
−
−1
−1
−1

e

 = lim

= lim (3 − ) = 3 − 1 = 2
(ii) Using Equation 2 with  () = 4 − 2 and  (1 3),

al

→1



4(1 + ) − (1 + )2 − 3
 ( + ) − ()
 (1 + ) −  (1)
= lim
= lim
→0
→0
→0




rS

 = lim

4 + 4 − 1 − 2 − 2 − 3
−2 + 2
(− + 2)
= lim
= lim
= lim (− + 2) = 2
→0
→0
→0
→0




= lim

or  = 2 + 1.
(c)

Fo

(b) An equation of the tangent line is  − () =  0 ()( − ) ⇒  −  (1) =  0 (1)( − 1) ⇒  − 3 = 2( − 1),
The graph of  = 2 + 1 is tangent to the graph of  = 4 − 2 at the

point (1 3). Now zoom in toward the point (1 3) until the parabola and

N

ot

the tangent line are indistiguishable.

5. Using (1) with  () = 4 − 32 and  (2 −4) [we could also use (2)],



4 − 32 − (−4)
 () −  ()
−32 + 4 + 4
 = lim
= lim
= lim
→
→2
→2
−
−2
−2
= lim

→2

(−3 − 2)( − 2)
= lim (−3 − 2) = −3(2) − 2 = −8
→2
−2

Tangent line:  − (−4) = −8( − 2) ⇔  + 4 = −8 + 16 ⇔  = −8 + 12.
√
√
√
√
(  − 1)(  + 1)
− 1
−1
1
1
√
√
7. Using (1),  = lim
= lim
= lim √
= .
= lim
→1
→1 ( − 1)(  + 1)
→1 ( − 1)(  + 1)
→1
−1
2
+1
Tangent line:  − 1 = 1 ( − 1) ⇔
2

 = 1 +
2

1
2

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53

54

¤

CHAPTER 2

DERIVATIVES

9. (a) Using (2) with  =  () = 3 + 42 − 23 ,

 ( + ) −  ()
3 + 4( + )2 − 2( + )3 − (3 + 42 − 23 )
= lim
→0
→0



 = lim

3 + 4(2 + 2 + 2 ) − 2(3 + 32  + 32 + 3 ) − 3 − 42 + 23
→0


= lim

3 + 42 + 8 + 42 − 23 − 62  − 62 − 23 − 3 − 42 + 23
→0


= lim

8 + 42 − 62  − 62 − 23
(8 + 4 − 62 − 6 − 22 )
= lim
→0
→0



= lim

= lim (8 + 4 − 62 − 6 − 22 ) = 8 − 62
→0

(b) At (1 5):  = 8(1) − 6(1)2 = 2, so an equation of the tangent line

(c)

is  − 5 = 2( − 1) ⇔  = 2 + 3.

al

line is  − 3 = −8( − 2) ⇔  = −8 + 19.

e

At (2 3):  = 8(2) − 6(2)2 = −8, so an equation of the tangent

rS

11. (a) The particle is moving to the right when  is increasing; that is, on the intervals (0 1) and (4 6). The particle is moving to

the left when  is decreasing; that is, on the interval (2 3). The particle is standing still when  is constant; that is, on the intervals (1 2) and (3 4).

Fo

(b) The velocity of the particle is equal to the slope of the tangent line of the

graph. Note that there is no slope at the corner points on the graph. On the interval (0 1) the slope is

3−0
= 3. On the interval (2 3), the slope is
1−0

13. Let () = 40 − 162 .

ot

1−3
3−1
= −2. On the interval (4 6), the slope is
= 1.
3−2
6−4

N





40 − 162 − 16
−8 22 − 5 + 2
() − (2)
−162 + 40 − 16
= lim
= lim
= lim
→2
→2
→2
→2
−2
−2
−2
−2

(2) = lim

= lim

→2

−8( − 2)(2 − 1)
= −8 lim (2 − 1) = −8(3) = −24
→2
−2

Thus, the instantaneous velocity when  = 2 is −24 fts.
1
1
2 − ( + )2
− 2
( + ) − ()
( + )2

2 ( + )2
2 − (2 + 2 + 2 )
= lim
= lim
= lim
15. () = lim
→0
→0
→0
→0



2 ( + )2
−(2 + 2 )
−(2 + )
−(2 + )
−2
−2
= lim
= lim 2
= 2 2 = 3 ms
→0 2 ( + )2
→0 2 ( + )2
→0  ( + )2
 ·


= lim

So  (1) =

−2
−2
−2
1
2 ms. = −2 ms, (2) = 3 = − ms, and (3) = 3 = −
13
2
4
3
27

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 2.1

DERIVATIVES AND RATES OF CHANGE

¤

17.  0 (0) is the only negative value. The slope at  = 4 is smaller than the slope at  = 2 and both are smaller than the slope

at  = −2. Thus, 0 (0)  0  0 (4)   0 (2)   0 (−2).
19. For the tangent line  = 4 − 5: when  = 2,  = 4(2) − 5 = 3 and its slope is 4 (the coefficient of ). At the point of

tangency, these values are shared with the curve  =  (); that is,  (2) = 3 and  0 (2) = 4.
21. We begin by drawing a curve through the origin with a

slope of 3 to satisfy  (0) = 0 and  0 (0) = 3. Since
 0 (1) = 0, we will round off our figure so that there is a horizontal tangent directly over  = 1. Last, we make sure that the curve has a slope of −1 as we pass over  = 2. Two of the many possibilities are shown.

= lim

→0

 (1 + ) −  (1)
[3(1 + )2 − (1 + )3 ] − 2
= lim
→0



al

→0

(3 + 6 + 32 ) − (1 + 3 + 32 + 3 ) − 2
3 − 3
(3 − 2 )
= lim
= lim
→0
→0




rS

 0 (1) = lim

e

23. Using (4) with  () = 32 − 3 and  = 1,

= lim (3 − 2 ) = 3 − 0 = 3
→0

Tangent line:  − 2 = 3( − 1) ⇔  − 2 = 3 − 3 ⇔  = 3 − 1

Fo

25. (a) Using (4) with  () = 5(1 + 2 ) and the point (2 2), we have

(b)

5(2 + )
−2
 (2 + ) −  (2)
1 + (2 + )2
 (2) = lim
= lim
→0
→0


0

ot

5 + 10
5 + 10 − 2(2 + 4 + 5)
−2
+ 4 + 5
2 + 4 + 5
= lim
→0



= lim

−22 − 3
(−2 − 3)
−2 − 3
−3
= lim
= lim
=
(2 + 4 + 5) →0 (2 + 4 + 5) →0 2 + 4 + 5
5

N

→0

2

= lim

→0

So an equation of the tangent line at (2 2) is  − 2 = − 3 ( − 2) or  = − 3  +
5
5

16
.
5

27. Use (4) with  () = 32 − 4 + 1.

 ( + ) − ()
[3( + )2 − 4( + ) + 1] − (32 − 4 + 1)]
= lim
→0
→0



 0 () = lim

32 + 6 + 32 − 4 − 4 + 1 − 32 + 4 − 1
6 + 32 − 4
= lim
→0
→0



= lim

= lim

→0

(6 + 3 − 4)
= lim (6 + 3 − 4) = 6 − 4
→0


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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

55

56

¤

CHAPTER 2

DERIVATIVES

29. Use (4) with  () = (2 + 1)( + 3).

2( + ) + 1 2 + 1
−
 ( + ) −  ()
( + ) + 3
+3
(2 + 2 + 1)( + 3) − (2 + 1)( +  + 3)
 0 () = lim
= lim
= lim
→0
→0
→0


( +  + 3)( + 3)
= lim

(22 + 6 + 2 + 6 +  + 3) − (22 + 2 + 6 +  +  + 3)
( +  + 3)( + 3)

= lim

5
5
5
= lim
=
( +  + 3)( + 3) →0 ( +  + 3)( + 3)
( + 3)2

→0

→0

31. Use (4) with  () =

√
1 − 2.

( + ) −  ()
 () = lim
= lim
→0
→0

0


√
1 − 2( + ) − 1 − 2


→0

(1 − 2 − 2) − (1 − 2)
−2

 = lim 

√
√
→0

1 − 2( + ) + 1 − 2

1 − 2( + ) + 1 − 2

rS

= lim

al

e

2 √

2


√
√
1 − 2( + ) −
1 − 2
1 − 2( + ) − 1 − 2
1 − 2( + ) + 1 − 2


= lim
·
= lim
√
√
→0

1 − 2( + ) + 1 − 2 →0 
1 − 2( + ) + 1 − 2

−2
−2
−2
−1
√
= lim 
= √
= √
= √
√
→0
1 − 2 + 1 − 2
2 1 − 2
1 − 2
1 − 2( + ) + 1 − 2

Fo

Note that the answers to Exercises 33 – 38 are not unique.

(1 + )10 − 1
=  0 (1), where  () = 10 and  = 1.
→0


33. By (4), lim

→0

(1 + )10 − 1
=  0 (0), where  () = (1 + )10 and  = 0.


ot

Or: By (4), lim

2 − 32
=  0 (5), where () = 2 and  = 5.
→5  − 5

35. By Equation 5, lim

→0

cos( + ) + 1
=  0 (), where  () = cos  and  = .


Or: By (4), lim

→0

N

37. By (4), lim

cos( + ) + 1
=  0 (0), where  () = cos( + ) and  = 0.


 (5 + ) −  (5)
[100 + 50(5 + ) − 49(5 + )2 ] − [100 + 50(5) − 49(5)2 ]
= lim
→0


(100 + 250 + 50 − 492 − 49 − 1225) − (100 + 250 − 1225)
−492 + 
= lim
= lim
→0
→0


(−49 + 1)
= lim
= lim (−49 + 1) = 1 ms
→0
→0


39. (5) =  0 (5) = lim

→0

The speed when  = 5 is |1| = 1 ms.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 2.1

DERIVATIVES AND RATES OF CHANGE

¤

57

41. The sketch shows the graph for a room temperature of 72◦ and a refrigerator

temperature of 38◦ . The initial rate of change is greater in magnitude than the rate of change after an hour.

43. (a) (i) [2002 2006]:

233 − 141
92
(2006) − (2002)
=
=
= 23 millions of cell phone subscribers per year
2006 − 2002
4
4

(ii) [2002 2004]:

(2004) − (2002)
182 − 141
41
=
=
= 205 millions of cell phone subscribers per year
2004 − 2002
2
2

(iii) [2000 2002]:

141 − 109
32
(2002) − (2000)
=
=
= 16 millions of cell phone subscribers per year
2002 − 2000
2
2
205 + 16
= 1825 millions of cell phone subscribers per year.
2

e

(b) Using the values from (ii) and (iii), we have

175 − 107
68
=
= 17 millions of cell phone subscribers per
2004 − 2000
4

rS

is

al

(c) Estimating A as (2000 107) and B as (2004 175), the slope at 2002

45. (a) (i)

Fo

year.

(105) − (100)
660125 − 6500
∆
=
=
= $2025unit.
∆
105 − 100
5

N

ot

(101) − (100)
652005 − 6500
∆
=
=
= $2005unit.
∆
101 − 100
1


5000 + 10(100 + ) + 005(100 + )2 − 6500
20 + 0052
(100 + ) − (100)
=
=
(b)



= 20 + 005,  6= 0
(ii)

So the instantaneous rate of change is lim

→0

(100 + ) − (100)
= lim (20 + 005) = $20unit.
→0


47. (a)  0 () is the rate of change of the production cost with respect to the number of ounces of gold produced. Its units are

dollars per ounce.
(b) After 800 ounces of gold have been produced, the rate at which the production cost is increasing is $17ounce. So the cost of producing the 800th (or 801st) ounce is about $17.
(c) In the short term, the values of  0 () will decrease because more efficient use is made of start-up costs as  increases. But eventually  0 () might increase due to large-scale operations.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

58

¤

CHAPTER 2

DERIVATIVES

49.  0 (8) is the rate at which the temperature is changing at 8:00 AM . To estimate the value of  0 (8), we will average the

difference quotients obtained using the times  = 6 and  = 10.
Let  =

75 − 84
 (10) −  (8)
90 − 84
 (6) −  (8)
=
= 45 and  =
=
= 3. Then
6−8
−2
10 − 8
2

 0 (8) = lim

→8

 () −  (8)
+
45 + 3
≈
=
= 375◦ Fh.
−8
2
2

51. (a)  0 ( ) is the rate at which the oxygen solubility changes with respect to the water temperature. Its units are (mgL)◦ C.

(b) For  = 16◦ C, it appears that the tangent line to the curve goes through the points (0 14) and (32 6). So
6 − 14
8
=−
= −025 (mgL)◦ C. This means that as the temperature increases past 16◦ C, the oxygen
32 − 0
32

 0 (16) ≈

e

solubility is decreasing at a rate of 025 (mgL)◦ C.

 0 (0) = lim

→0

al

53. Since  () =  sin(1) when  6= 0 and  (0) = 0, we have

 (0 + ) −  (0)
 sin(1) − 0
= lim
= lim sin(1). This limit does not exist since sin(1) takes the
→0
→0



rS

values −1 and 1 on any interval containing 0. (Compare with Example 4 in Section 1.5.)

Fo

2.2 The Derivative as a Function

1. It appears that  is an odd function, so  0 will be an even function—that

is,  0 (−) =  0 ().
(a)  0 (−3) ≈ −02

(c)  0 (−1) ≈ 1

(d)  0 (0) ≈ 2

(e)  0 (1) ≈ 1

(f)  0 (2) ≈ 0

(g)  0 (3) ≈ −02

N

ot

(b)  0 (−2) ≈ 0

3. (a)0 = II, since from left to right, the slopes of the tangents to graph (a) start out negative, become 0, then positive, then 0, then

negative again. The actual function values in graph II follow the same pattern.
(b)0 = IV, since from left to right, the slopes of the tangents to graph (b) start out at a fixed positive quantity, then suddenly become negative, then positive again. The discontinuities in graph IV indicate sudden changes in the slopes of the tangents.
(c)0 = I, since the slopes of the tangents to graph (c) are negative for   0 and positive for   0, as are the function values of graph I.
(d)0 = III, since from left to right, the slopes of the tangents to graph (d) are positive, then 0, then negative, then 0, then positive, then 0, then negative again, and the function values in graph III follow the same pattern.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 2.2

THE DERIVATIVE AS A FUNCTION

¤

Hints for Exercises 4 –11: First plot -intercepts on the graph of  0 for any horizontal tangents on the graph of  . Look for any corners on the graph of  — there will be a discontinuity on the graph of  0 . On any interval where  has a tangent with positive (or negative) slope, the graph of  0 will be positive (or negative). If the graph of the function is linear, the graph of  0 will be a horizontal line.
5.

al

e

7.

11.

ot

Fo

rS

9.

N

13. (a)  0 () is the instantaneous rate of change of percentage

of full capacity with respect to elapsed time in hours.
(b) The graph of  0 () tells us that the rate of change of percentage of full capacity is decreasing and approaching 0.
15. It appears that there are horizontal tangents on the graph of  for  = 1963

and  = 1971. Thus, there are zeros for those values of  on the graph of
 0 . The derivative is negative for the years 1963 to 1971.

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59

60

¤

CHAPTER 2

DERIVATIVES

17. (a) By zooming in, we estimate that  0 (0) = 0,  0

and  0 (2) = 4.

1
2

= 1,  0 (1) = 2,

 
(b) By symmetry,  0 (−) = − 0 (). So  0 − 1 = −1,  0 (−1) = −2,
2
and  0 (−2) = −4.

(c) It appears that  0 () is twice the value of , so we guess that  0 () = 2.
 ( + ) −  ()
( + )2 − 2
= lim
→0
→0


 2

 + 2 + 2 − 2
(2 + )
2 + 2
= lim
= lim
= lim
= lim (2 + ) = 2
→0
→0
→0
→0




(d)  0 () = lim

 ( + ) − ()
= lim
19.  () = lim
→0
→0

0

→0



1
→0 2

= lim

=

2

( + ) −

1
3





−

1
2

−

1
3



= lim

1

2

→0

1
2

+ 1 − 1 − 1 +
2
3
2


1
3

e

= lim

1

2

1

al

Domain of  = domain of  0 = R.



5( + ) − 9( + )2 − (5 − 92 )
 ( + ) − ()
= lim
→0
→0



rS

21.  0 () = lim

5 + 5 − 9(2 + 2 + 2 ) − 5 + 92
5 + 5 − 92 − 18 − 92 − 5 + 92
= lim
→0
→0



= lim

5 − 18 − 92
(5 − 18 − 9)
= lim
= lim (5 − 18 − 9) = 5 − 18
→0
→0
→0



Domain of  = domain of  0 = R.

→0

= lim

→0

( + ) −  ()
[( + )2 − 2( + )3 ] − (2 − 23 )
= lim
→0



ot

23.  0 () = lim

Fo

= lim

2 + 2 + 2 − 23 − 62  − 62 − 23 − 2 + 23


→0

N

2 + 2 − 62  − 62 − 23
(2 +  − 62 − 6 − 22 )
= lim
→0
→0


= lim (2 +  − 62 − 6 − 22 ) = 2 − 62

= lim

Domain of  = domain of  0 = R.
( + ) − ()
= lim
25.  () = lim
→0
→0

0

= lim

→0




√
√
9 − ( + ) − 9 − 
9 − ( + ) + 9 − 

√

9 − ( + ) + 9 − 

[9 − ( + )] − (9 − )
−

 = lim 

√
√
→0

9 − ( + ) + 9 − 

9 − ( + ) + 9 − 

−1
−1
= √
= lim 
√
→0
2 9−
9 − ( + ) + 9 − 

Domain of  = (−∞ 9], domain of 0 = (−∞ 9).

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 2.2

THE DERIVATIVE AS A FUNCTION

¤

61

1 − 2( + )
1 − 2
−
( + ) − ()
3 + ( + )
3+
= lim
27. 0 () = lim
→0
→0


[1 − 2( + )](3 + ) − [3 + ( + )](1 − 2)
[3 + ( + )](3 + )
= lim
→0

3 +  − 6 − 22 − 6 − 2 − (3 − 6 +  − 22 +  − 2)
−6 − 
= lim
→0
→0 (3 +  + )(3 + )
[3 + ( + )](3 + )

= lim
= lim

→0

−7
−7
−7
= lim
=
(3 +  + )(3 + ) →0 (3 +  + )(3 + )
(3 + )2

Domain of  = domain of 0 = (−∞ −3) ∪ (−3 ∞).

 4

 + 43  + 62 2 + 43 + 4 − 4
( + ) −  ()
( + )4 − 4
= lim
= lim
29.  () = lim
→0
→0
→0



3
2 2
3
4
 3

4  + 6  + 4 + 
= lim
= lim 4 + 62  + 42 + 3 = 43
→0
→0


e

0

→0

( + ) −  ()
[( + )4 + 2( + )] − (4 + 2)
= lim
→0



rS

31. (a)  0 () = lim

al

Domain of  = domain of  0 = R.

= lim

4 + 43  + 62 2 + 43 + 4 + 2 + 2 − 4 − 2


= lim

43  + 62 2 + 43 + 4 + 2
(43 + 62  + 42 + 3 + 2)
= lim
→0



→0

→0

Fo

= lim (43 + 62  + 42 + 3 + 2) = 43 + 2
→0

(b) Notice that  0 () = 0 when  has a horizontal tangent,  0 () is positive when the tangents have positive slope, and  0 () is

N

ot

negative when the tangents have negative slope.

33. (a)  0 () is the rate at which the unemployment rate is changing with respect to time. Its units are percent per year.

(b) To find  0 (), we use lim

→0

For 1999:  0 (1999) ≈

 ( + ) − ()
( + ) −  ()
≈
for small values of .



40 − 42
 (2000) − (1999)
=
= −02
2000 − 1999
1

For 2000: We estimate  0 (2000) by using  = −1 and  = 1, and then average the two results to obtain a final estimate.
 = −1 ⇒  0 (2000) ≈
 = 1 ⇒  0 (2000) ≈

42 − 40
 (1999) −  (2000)
=
= −02;
1999 − 2000
−1

47 − 40
 (2001) −  (2000)
=
= 07.
2001 − 2000
1

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

[continued]

62

¤

CHAPTER 2

DERIVATIVES

So we estimate that  0 (2000) ≈ 1 [(−02) + 07] = 025.
2


1999

2000

2001

2002

2003

2004

2005

2006

2007

2008

 0 ()

−02

025

09

065

−015

−045

−045

−025

06

12

35.  is not differentiable at  = −4, because the graph has a corner there, and at  = 0, because there is a discontinuity there.
37.  is not differentiable at  = −1, because the graph has a vertical tangent there, and at  = 4, because the graph has a corner

there.

39. As we zoom in toward (−1 0), the curve appears more and more like a straight

line, so  () =  +


|| is differentiable at  = −1. But no matter how much

we zoom in toward the origin, the curve doesn’t straighten out—we can’t

e

eliminate the sharp point (a cusp). So  is not differentiable at  = 0.

al

41.  =  ,  =  0 ,  =  00 . We can see this because where  has a horizontal tangent,  = 0, and where  has a horizontal tangent,

 = 0. We can immediately see that  can be neither  nor  0 , since at the points where  has a horizontal tangent, neither 

rS

nor  is equal to 0.

43. We can immediately see that  is the graph of the acceleration function, since at the points where  has a horizontal tangent,

neither  nor  is equal to 0. Next, we note that  = 0 at the point where  has a horizontal tangent, so  must be the graph of

45.  0 () = lim

→0

Fo

the velocity function, and hence, 0 = . We conclude that  is the graph of the position function.
( + ) −  ()
[3( + )2 + 2( + ) + 1] − (32 + 2 + 1)
= lim
→0


(32 + 6 + 32 + 2 + 2 + 1) − (32 + 2 + 1)
6 + 32 + 2
= lim
→0



= lim

(6 + 3 + 2)
= lim (6 + 3 + 2) = 6 + 2
→0


→0

→0

ot

= lim

= lim

→0

N

 0 ( + ) −  0 ()
[6( + ) + 2] − (6 + 2)
(6 + 6 + 2) − (6 + 2)
= lim
= lim
→0
→0
→0




 00 () = lim

6
= lim 6 = 6
→0


We see from the graph that our answers are reasonable because the graph of
 0 is that of a linear function and the graph of  00 is that of a constant function. 

2( + )2 − ( + )3 − (22 − 3 )
 ( + ) −  ()
= lim
→0
→0



47.  0 () = lim

(4 + 2 − 32 − 3 − 2 )
= lim (4 + 2 − 32 − 3 − 2 ) = 4 − 32
→0
→0


= lim

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 2.2

THE DERIVATIVE AS A FUNCTION

¤

63



4( + ) − 3( + )2 − (4 − 32 )
 0 ( + ) −  0 ()
(4 − 6 − 3)
= lim
= lim
 00 () = lim
→0
→0
→0



= lim (4 − 6 − 3) = 4 − 6
→0

 00 ( + ) −  00 ()
[4 − 6( + )] − (4 − 6)
−6
= lim
= lim
= lim (−6) = −6
→0
→0
→0 
→0



 000 () = lim

 (4) () = lim

→0

 000 ( + ) −  000 ()
−6 − (−6)
0
= lim
= lim = lim (0) = 0
→0
→0 
→0


The graphs are consistent with the geometric interpretations of the derivatives because  0 has zeros where  has a local minimum and a local maximum,  00 has a zero where  0 has a local maximum, and  000 is a

e

constant function equal to the slope of  00 .

= lim

→

(b)  0(0) = lim

→0

rS

→

 () −  ()
13 − 13
13 − 13
= lim
= lim 13
13 )(23 + 13 13 + 23 )
→
→ (
−
−
−
1
1
= 23 or 1 −23
3
23 + 13 13 + 23
3
(0 + ) − (0)
= lim
→0


√
3
−0
1
= lim 23 . This function increases without bound, so the limit does not
→0 


Fo

 0() = lim

al

49. (a) Note that we have factored  −  as the difference of two cubes in the third step.

exist, and therefore  0(0) does not exist.
(c) lim | 0 ()| = lim

→0

1
= ∞ and  is continuous at  = 0 (root function), so  has a vertical tangent at  = 0.
323


−6

ot

→0

−( − 6) if  − 6  0

N

51.  () = | − 6| =

So the right-hand limit is lim

→6+

is lim

→6−

if  − 6 ≥ 6

=


 − 6 if  ≥ 6
6 −  if   6

 () −  (6)
| − 6| − 0
−6
= lim
= lim
= lim 1 = 1, and the left-hand limit
−6
−6
→6+
→6+  − 6
→6+

() − (6)
| − 6| − 0
6−
= lim
= lim
= lim (−1) = −1. Since these limits are not equal,
−6
−6
→6−
→6−  − 6
→6−

 () −  (6) does not exist and  is not differentiable at 6.
−6

1
if   6
0
0
However, a formula for  is  () =
−1 if   6

 0 (6) = lim

→6

Another way of writing the formula is  0 () =

−6
.
| − 6|

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64

¤

CHAPTER 2 DERIVATIVES

53. (a)  () =  || =

 2


if  ≥ 0

(b) Since  () = 2 for  ≥ 0, we have  0 () = 2 for   0.

if   0

2

−

[See Exercise 17(d).] Similarly, since  () = −2 for   0, we have  0 () = −2 for   0. At  = 0, we have
 0 (0) = lim

→0

 () −  (0)
 ||
= lim
= lim || = 0
→0 
→0
−0

So  is differentiable at 0. Thus,  is differentiable for all .

(c) From part (b), we have  () =
0



2 if  ≥ 0

−2 if   0



= 2 ||.

55. (a) If  is even, then

 (− + ) −  (−)
 [−( − )] −  (−)
= lim
→0



 0 (−) = lim

e

→0

→0

= − lim

∆→0

 ( + ∆) −  ()
= − 0 ()
∆

rS

Therefore,  0 is odd.
(b) If  is odd, then

 (− + ) −  (−)
 [−( − )] −  (−)
= lim
→0



 0 (−) = lim

→0

→0

= lim

∆→0

Fo

−( − ) +  ()
 ( − ) − ()
= lim
→0

−

= lim

[let ∆ = −]

 ( + ∆) −  ()
=  0 ()
∆

ot

Therefore,  0 is even.

[let ∆ = −]

al

 ( − ) − ()
 ( − ) −  ()
= − lim
→0

−

= lim

In the right triangle in the diagram, let ∆ be the side opposite angle  and ∆

57.

N

the side adjacent to angle . Then the slope of the tangent line  is  = ∆∆ = tan . Note that 0   


2.

We know (see Exercise 17)

that the derivative of () = 2 is  0 () = 2. So the slope of the tangent to the curve at the point (1 1) is 2. Thus,  is the angle between 0 and


2

tangent is 2; that is,  = tan−1 2 ≈ 63◦ .

2.3 Differentiation Formulas
1.  () = 240 is a constant function, so its derivative is 0, that is,  0 () = 0.
3.  () = 2 − 2 
3

⇒  0 () = 0 −

5.  () = 3 − 4 + 6

2
3

= −2
3

⇒  0 () = 32 − 4(1) + 0 = 32 − 4

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

whose

SECTION 2.3 DIFFERENTIATION FORMULAS

7. () = 2 (1 − 2) = 2 − 23
9. () = 2−34
11. () = −
13. () =

⇒

65

⇒  0 () = 2 − 2(32 ) = 2 − 62

 0 () = 2(− 3 −74 ) = − 3 −74
4
2

12
= −12−5
5

⇒ 0 () = −12(−5−6 ) = 60−6

⇒ 0 () = 9(2) + 6(1) + 0 = 18 + 6

2 + 4 + 3
√
= 32 + 412 + 3−12


⇒

 
 
 0 = 3 12 + 4 1 −12 + 3 − 1 −32 =
2
2
2


√  note that 32 = 22 · 12 =  

√
2
3
√
+ √ −
 2 

e

3
2

32
4
3
32 + 4 − 3
√ +
√ −
√ =
√
.
2 
2 
2 
2 

al

The last expression can be written as

or 606

1
√ −1
2 

√
 −  = 12 −  ⇒  0 () = 1 −12 − 1 or
2

15. () = (3 + 1)2 = 92 + 6 + 1
17.  =

¤

19. We first expand using the Binomial Theorem (see Reference Page 1).

0

−2

2

 () = 3 + 3 + 3(−1

) + (−3

√
√
5
 + 4 5 = 15 + 452

2

) = 3 + 3 − 3

−2

− 3



⇒ 0 = 1 −45 + 4 5 32 = 1 −45 + 1032
5
2
5

23. Product Rule: () = (1 + 22 )( − 2 )

⇒

−4

Fo

21.  =

−4

rS

() = ( + −1 )3 = 3 + 32 −1 + 3(−1 )2 + (−1 )3 = 3 + 3 + 3−1 + −3

 √ 
√
5 or 1 5 4 + 10 3

⇒

 0 () = (1 + 22 )(1 − 2) + ( − 2 )(4) = 1 − 2 + 22 − 43 + 42 − 43 = 1 − 2 + 62 − 83 .

ot

Multiplying first:  () = (1 + 22 )( − 2 ) =  − 2 + 23 − 24
PR

⇒

N

25.  () = (23 + 3)(4 − 2)

⇒  0 () = 1 − 2 + 62 − 83 (equivalent).

 0 () = (23 + 3)(43 − 2) + (4 − 2)(62 ) = (86 + 83 − 6) + (66 − 123 ) = 146 − 43 − 6
27.  () =




1
3
PR
− 4 ( + 53 ) = ( −2 − 3 −4 )( + 5 3 ) ⇒
2


 0 () = ( −2 − 3 −4 )(1 + 15 2 ) + ( + 53 )(−2 −3 + 12 −5 )
= ( −2 + 15 − 3 −4 − 45 −2 ) + (−2 −2 + 12 −4 − 10 + 60−2 )
= 5 + 14−2 + 9 −4 or 5 + 14 2 + 9 4
29. () =

31.  =

1 + 2
3 − 4

3
1 − 2

QR

QR

⇒ 0 () =

⇒ 0 =

(3 − 4)(2) − (1 + 2)(−4)
6 − 8 + 4 + 8
10
=
=
(3 − 4)2
(3 − 4)2
(3 − 4)2

(1 − 2 ) (32 ) − 3 (−2)
2 (3 − 32 + 22 )
2 (3 − 2 )
=
=
(1 − 2 )2
(1 − 2 )2
(1 − 2 )2

c
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66

¤

CHAPTER 2 DERIVATIVES

33.  =

 3 − 2


√


= 2 − 2

√
 =  2 − 2 12

 
⇒  0 = 2 − 2 1  −12 = 2 −  −12 .
2

√
1
2  − 1
2 32 − 1
√
√
=
We can change the form of the answer as follows: 2 −  −12 = 2 − √ =




=

QR

⇒

(4 − 32 + 1)(2) − (2 + 2)(43 − 6)
2[(4 − 32 + 1) − (2 + 2)(22 − 3)]
=
4 − 32 + 1)2
(
(4 − 32 + 1)2
2(4 − 32 + 1 − 24 − 42 + 32 + 6)
2(−4 − 42 + 7)
=
(4 − 32 + 1)2
(4 − 32 + 1)2
⇒  0 = 2 + 

37.  = 2 +  + 



√
(2 + 12 )(2) − 2 1 −12
2
4 + 212 − 12
4 + 12
4+ 
√
√
√
√
⇒  0 () =
=
= or (2 +  )2
(2 +  )2
(2 +  )2
(2 +  )2

2
√
39.  () =
2+ 
41.  =

e

0 =

2 + 2
4 − 32 + 1

QR

√ 2
3
 ( +  + −1 ) = 13 (2 +  + −1 ) = 73 + 43 + −23

al

35.  =

⇒


 + 

⇒  0 () =

( + )(1) − (1 − 2 )
 +  −  + 
2
2
2
=
=

2
2
 2
2
2 · 2 =


(2 + )2
( + )
 +
+

2


Fo

43.  () =

rS

 0 = 7 43 + 4 13 − 2 −53 = 1 −53 (793 + 463 − 2) = (73 + 42 − 2)(353 )
3
3
3
3

45.  () =   + −1 −1 + · · · + 2 2 + 1  + 0

⇒  0 () = 4514 − 152 .

ot

47.  () = 315 − 53 + 3

⇒  0 () =  −1 + ( − 1)−1 −2 + · · · + 22  + 1

Notice that  0 () = 0 when  has a horizontal tangent,  0 is positive

49. (a)

N

when  is increasing, and  0 is negative when  is decreasing.

(b) From the graph in part (a), it appears that  0 is zero at 1 ≈ −125, 2 ≈ 05, and 3 ≈ 3. The slopes are negative (so  0 is negative) on (−∞ 1 ) and
(2  3 ). The slopes are positive (so  0 is positive) on (1  2 ) and (3  ∞).

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 2.3 DIFFERENTIATION FORMULAS

¤

(c)  () = 4 − 33 − 62 + 7 + 30 ⇒
 0 () = 43 − 92 − 12 + 7

51.  =

2
+1

⇒ 0 =

( + 1)(2) − (2)(1)
2
=
.
( + 1)2
( + 1)2

At (1 1),  0 = 1 , and an equation of the tangent line is  − 1 = 1 ( − 1), or  = 1  + 1 .
2
2
2
2
(b)

⇒

(1 + 2 )(0) − 1(2)
−2
=
. So the slope of the
(1 + 2 )2
(1 + 2 )2



2 tangent line at the point −1 1 is  0 (−1) = 2 =
2
2

equation is  −

1
2

= 1 ( + 1) or  = 1  + 1.
2
2

1
2

and its

al

 0 () =

1
1 + 2

e

53. (a)  =  () =

rS

√
√
 ⇒ 0 = 1 + 1 −12 = 1 + 1(2 ) . At (1 2),  0 = 3 , and an equation of the tangent line is
2
2

55.  =  +

 − 2 = 3 ( − 1), or  = 3  + 1 . The slope of the normal line is − 2 , so an equation of the normal line is
2
2
2
3

57.  =

3 + 1
2 + 1

⇒ 0 =

Fo

 − 2 = − 2 ( − 1), or  = − 2  + 8 .
3
3
3

(2 + 1)(3) − (3 + 1)(2)
6−8
1
. At (1 2),  0 =
= − , and an equation of the tangent line
(2 + 1)2
22
2

is  − 2 = − 1 ( − 1), or  = − 1  + 5 . The slope of the normal line is 2, so an equation of the normal line is
2
2
2

ot

 − 2 = 2( − 1), or  = 2.

⇒  0 () = 43 − 92 + 16 ⇒  00 () = 122 − 18

N

59.  () = 4 − 33 + 16
61.  () =

2
1 + 2

 00 () =
=

⇒  0 () =

(1 + 2)(2) − 2 (2)
2 + 42 − 22
22 + 2
=
=
(1 + 2)2
(1 + 2)2
(1 + 2)2

⇒

(1 + 2)2 (4 + 2) − (22 + 2)(1 + 4 + 42 )0
2(1 + 2)2 (2 + 1) − 2( + 1)(4 + 8)
=
2 ]2
[(1 + 2)
(1 + 2)4

2(1 + 2)[(1 + 2)2 − 4( + 1)]
2(1 + 4 + 42 − 42 − 4)
2
=
=
(1 + 2)4
(1 + 2)3
(1 + 2)3

63. (a)  = 3 − 3

⇒ () = 0 () = 32 − 3 ⇒ () = 0 () = 6

(b) (2) = 6(2) = 12 ms2
(c) () = 32 − 3 = 0 when 2 = 1, that is,  = 1 [ ≥ 0] and (1) = 6 ms2 .
65. (a)  =

53
53
 and  = 50 when  = 0106, so  =   = 50(0106) = 53. Thus,  = and  =
.




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67

68

¤

CHAPTER 2 DERIVATIVES

(b)  = 53 −1

⇒

53

53

= 53(−1 −2 ) = − 2 . When  = 50,
= − 2 = −000212. The derivative is the



50

instantaneous rate of change of the volume with respect to the pressure at 25 ◦ C. Its units are m3kPa.
67. We are given that (5) = 1,  0 (5) = 6, (5) = −3, and 0 (5) = 2.

(a) ( )0 (5) =  (5) 0 (5) + (5) 0 (5) = (1)(2) + (−3)(6) = 2 − 18 = −16
 0
(5) 0 (5) − (5)0 (5)
(−3)(6) − (1)(2)

20
(b)
(5) =
=
=−

[(5)]2
(−3)2
9
(c)

 0
 (5) 0 (5) − (5) 0 (5)
(1)(2) − (−3)(6)

(5) =
=
= 20

[ (5)]2
(1)2

69.  () =

√
√
√
1
1
1
 () ⇒  0 () =   0 () + () · −12 , so  0 (4) = 4 0 (4) + (4) · √ = 2 · 7 + 8 · = 16.
2
4
2 4

e

71. (a) From the graphs of  and , we obtain the following values:  (1) = 2 since the point (1 2) is on the graph of  ;

(2 4) is

al

(1) = 1 since the point (1 1) is on the graph of ;  0 (1) = 2 since the slope of the line segment between (0 0) and
0−4
4−0
= 2;  0 (1) = −1 since the slope of the line segment between (−2 4) and (2 0) is
= −1.
2−0
2 − (−2)

⇒

 0 = 0 () + () · 1 = 0 () + ()

Fo

73. (a)  = ()

rS

Now () =  ()(), so 0 (1) = (1)0 (1) + (1)  0 (1) = 2 · (−1) + 1 · 2 = 0.
 
2 −1 − 3 · 2
−8
2
(5) 0 (5) −  (5)0 (5)
3
3
=
= 3 =−
(b) () = ()(), so  0 (5) =
2
2
[(5)]
2
4
3


()

⇒ 0 =

() · 1 − 0 ()
() −  0 ()
=
[()]2
[()]2

(c)  =

()


⇒ 0 =

 0 () − () · 1
 0 () − ()
=
()2
2

ot

(b)  =

75. The curve  = 23 + 32 − 12 + 1 has a horizontal tangent when  0 = 62 + 6 − 12 = 0

⇔ 6(2 +  − 2) = 0 ⇔

N

6( + 2)( − 1) = 0 ⇔  = −2 or  = 1. The points on the curve are (−2 21) and (1 −6).
77.  = 63 + 5 − 3

⇒  =  0 = 182 + 5, but 2 ≥ 0 for all , so  ≥ 5 for all .

79. The slope of the line 12 −  = 1 (or  = 12 − 1) is 12, so the slope of both lines tangent to the curve is 12.

 = 1 + 3

⇒  0 = 32 . Thus, 32 = 12 ⇒ 2 = 4 ⇒  = ±2, which are the -coordinates at which the tangent

lines have slope 12. The points on the curve are (2 9) and (−2 −7), so the tangent line equations are  − 9 = 12( − 2) or  = 12 − 15 and  + 7 = 12( + 2) or  = 12 + 17.
81. The slope of  = 2 − 5 + 4 is given by  =  0 = 2 − 5. The slope of  − 3 = 5

⇔  = 1 −
3

5
3

is 1 ,
3

so the desired normal line must have slope 1 , and hence, the tangent line to the parabola must have slope −3. This occurs if
3
2 − 5 = −3 ⇒ 2 = 2 ⇒  = 1. When  = 1,  = 12 − 5(1) + 4 = 0, and an equation of the normal line is
 − 0 = 1 ( − 1) or  = 1  − 1 .
3
3
3
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SECTION 2.3 DIFFERENTIATION FORMULAS

¤



Let  2 be a point on the parabola at which the tangent line passes

83.

through the point (0 −4). The tangent line has slope 2 and equation


 − (−4) = 2( − 0) ⇔  = 2 − 4. Since  2 also lies on the

line, 2 = 2() − 4, or 2 = 4. So  = ±2 and the points are (2 4) and (−2 4).
85. (a) ( )0 = [()]0 = ( )0  + ( )0 = ( 0  +  0 ) + ( )0 =  0  +  0  +  0

(b) Putting  =  =  in part (a), we have
(c)  = (4 + 33 + 17 + 82)3


[ ()]3 = (  )0 =  0   +  0  +    0 = 3   0 = 3[()]2  0 ().


⇒  0 = 3(4 + 33 + 17 + 82)2 (43 + 92 + 17)

al

 0 (2) = 3 ⇒ 2(1)(2) +  = 3 ⇒ 4 +  = 3 ⇒  = −1.

⇒ 2 = 2 ⇒  = 1.

e

87. Let  () = 2 +  + . Then  0 () = 2 +  and  00 () = 2.  00 (2) = 2

 (2) = 5 ⇒ 1(2)2 + (−1)(2) +  = 5 ⇒ 2 +  = 5 ⇒  = 3. So  () = 2 −  + 3.
⇒  0 () = 32 + 2 + . The point (−2 6) is on , so  (−2) = 6 ⇒

rS

89.  = () = 3 + 2 +  + 

−8 + 4 − 2 +  = 6 (1). The point (2 0) is on  , so  (2) = 0 ⇒ 8 + 4 + 2 +  = 0 (2). Since there are horizontal tangents at (−2 6) and (2 0),  0 (±2) = 0.  0 (−2) = 0 ⇒ 12 − 4 +  = 0 (3) and  0 (2) = 0 ⇒

Fo

12 + 4 +  = 0 (4). Subtracting equation (3) from (4) gives 8 = 0 ⇒  = 0. Adding (1) and (2) gives 8 + 2 = 6, so  = 3 since  = 0. From (3) we have  = −12, so (2) becomes 8 + 4(0) + 2(−12) + 3 = 0 ⇒ 3 = 16 ⇒
3
3
3
 = 16 . Now  = −12 = −12 16 = − 9 and the desired cubic function is  = 16 3 − 9  + 3.
4
4

ot

91. If  () denotes the population at time  and () the average annual income, then  () =  ()() is the total personal

income. The rate at which  () is rising is given by  0 () =  ()0 () + () 0 () ⇒

N

 0 (1999) =  (1999)0 (1999) + (1999) 0 (1999) = (961,400)($1400yr) + ($30,593)(9200yr)
= $1,345,960,000yr + $281,455,600yr = $1,627,415,600yr

So the total personal income was rising by about $1.627 billion per year in 1999.
The term  ()0 () ≈ $1.346 billion represents the portion of the rate of change of total income due to the existing population’s increasing income. The term () 0 () ≈ $281 million represents the portion of the rate of change of total income due to increasing population.

93.  () =



2 + 1

if   1

+1

if  ≥ 1

Calculate the left- and right-hand derivatives as defined in Exercise 2.2.54:
0
− (1) = lim

→0−

(1 + ) − (1)
[(1 + )2 + 1] − (1 + 1)
2 + 2
= lim
= lim
= lim ( + 2) = 2 and
−
−



→0
→0
→0−

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69

70

¤

CHAPTER 2 DERIVATIVES

0
+ (1) = lim

→0+

 (1 + ) −  (1)
[(1 + ) + 1] − (1 + 1)

= lim
= lim
= lim 1 = 1.
+
+ 


→0
→0
→0+

Since the left and right limits are different, lim →0

(1 + ) − (1) does not exist, that is,  0 (1)


does not exist. Therefore,  is not differentiable at 1.

 2
  −9


 () =
−2 + 9

 2

 −9

⇔ ||  3 ⇔ −3    3. So

if  ≤ −3

if −3    3 if  ≥ 3


 2

0
−2
⇒  () =


2

To show that  0 (3) does not exist we investigate lim

→0

if   −3

if −3    3 if   3

=



2
−2

if ||  3

if ||  3

 (3 + ) −  (3) by computing the left- and right-hand derivatives


al

defined in Exercise 2.2.54.

e

95. (a) Note that 2 − 9  0 for 2  9

 (3 + ) −  (3)
[−(3 + )2 + 9] − 0
= lim
= lim (−6 − ) = −6 and
−


→0
→0−


(3 + )2 − 9 − 0
 (3 + ) −  (3)
6 + 2
0
= lim
= lim
= lim (6 + ) = 6.
+ (3) = lim



→0+
→0+
→0+
→0+
0
− (3) = lim

rS

→0−

Since the left and right limits are different,

Fo

lim

→0

(b)

 (3 + ) −  (3) does not exist, that is,  0 (3)


does not exist. Similarly,  0 (−3) does not exist.

⇒  0 () = 2.

So the slope of the tangent to the parabola at  = 2 is  = 2(2) = 4. The slope

N

97.  = () = 2

ot

Therefore,  is not differentiable at 3 or at −3.

of the given line, 2 +  =  ⇔  = −2 + , is seen to be −2, so we must have 4 = −2 ⇔  = − 1 . So when
2
 = 2, the point in question has -coordinate − 1 · 22 = −2. Now we simply require that the given line, whose equation is
2
2 +  = , pass through the point (2 −2): 2(2) + (−2) =  ⇔  = 2. So we must have  = − 1 and  = 2.
2
√
 is  0 =


√ and the slope of the tangent line  = 3  + 6 is 3 . These must be equal at the
2
2


√ 
√

3
point of tangency    , so √ =
⇒  = 3 . The -coordinates must be equal at  = , so
2
2 
 √ √
√
√
3 
 = 3  + 6 ⇒ 3 = 3  + 6 ⇒ 3  = 6 ⇒  = 4. Since  = 3 , we have
  = 3 + 6 ⇒
2
2
2
2

99. The slope of the curve  = 

=3

2

√
4 = 6.

101.  =  

⇒  = 

⇒  0 =  0  +  0

⇒ 0 =

 0 − ( ) 0
 0  −  0
 0 −  0
=
=


2

c
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SECTION 2.4 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS

103. Solution 1:

Let  () = 1000 . Then, by the definition of a derivative,  0 (1) = lim

→1

¤

71

 () −  (1)
1000 − 1
= lim
.
→1
−1
−1

But this is just the limit we want to find, and we know (from the Power Rule) that  0 () = 1000999 , so
 0 (1) = 1000(1)999 = 1000. So lim

→1

− 1) = ( − 1)(999 + 998 + 997 + · · · + 2 +  + 1). So

1000

Solution 2: Note that ( lim →1

1000 − 1
= 1000.
−1

1000 − 1
( − 1)(999 + 998 + 997 + · · · + 2 +  + 1)
= lim
= lim (999 + 998 + 997 + · · · + 2 +  + 1)
→1
→1
−1
−1
= 1 + 1 + 1 + · · · + 1 + 1 + 1 = 1000, as above.



1000 ones

⇒  0 = 2, so the slope of a tangent line at the point ( 2 ) is  0 = 2 and the slope of a normal line is −1(2),

105.  = 2

2 − 
1
2 − 
, so
=−
−0

2

e

for  6= 0. The slope of the normal line through the points ( 2 ) and (0 ) is

⇒

al

⇒ 2 =  − 1 . The last equation has two solutions if   1 , one solution if  = 1 , and no solution if
2
2
2

2 −  = − 1
2
if  

and one normal line if  ≤ 1 .
2

1
2

rS

  1 . Since the -axis is normal to  = 2 regardless of the value of  (this is the case for  = 0), we have three normal lines
2

1.  () = 32 − 2 cos 
3.  () = sin  +

1
2

⇒  0 () = 6 − 2(− sin ) = 6 + 2 sin 

cot  ⇒  0 () = cos  −

1
2

csc2 

⇒ 0 = sec  (sec2 ) + tan  (sec  tan ) = sec  (sec2  + tan2 ). Using the identity

ot

5.  = sec  tan 

Fo

2.4 Derivatives of Trigonometric Functions

1 + tan2  = sec2 , we can write alternative forms of the answer as sec  (1 + 2 tan2 ) or sec  (2 sec2  − 1).

9.  =


2 − tan 

11.  () =

sec 
1 + sec 

 0 () =

13.  =

0 =

⇒

 0 = (− sin ) + 2 (cos ) + sin  (2) = − sin  + ( cos  + 2 sin )

N

7.  =  cos  + 2 sin 

⇒ 0 =

(2 − tan )(1) − (− sec2 )
2 − tan  +  sec2 
=
(2 − tan )2
(2 − tan )2

⇒

(1 + sec )(sec  tan ) − (sec )(sec  tan )
(sec  tan ) [(1 + sec ) − sec ] sec  tan 
=
=
(1 + sec )2
(1 + sec )2
(1 + sec )2

 sin 
1+

⇒

(1 + )( cos  + sin ) −  sin (1)
 cos  + sin  + 2 cos  +  sin  −  sin 
(2 + ) cos  + sin 
=
=
(1 + )2
(1 + )2
(1 + )2

15. () =  csc  − cot 



⇒ 0 () = (− csc  cot ) + (csc ) · 1 − − csc2  = csc  −  csc  cot  + csc2 

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

72

¤

17.



(csc ) =



19.

 sin2  + cos2 
1
  cos   (sin )(− sin ) − (cos )(cos )
=−
= − 2 = − csc2 
(cot ) =
=
2

 sin  sin  sin2  sin 

CHAPTER 2 DERIVATIVES



1 sin 



=

(sin )(0) − 1(cos )
− cos 
1
cos 
=
=−
·
= − csc  cot  sin  sin  sin2  sin2 

√
⇒ 0 = sec  tan , so  0 (  ) = sec  tan  = 2 3. An equation of the tangent line to the curve  = sec 
3
3
3
√
√ 
√


 at the point   2 is  − 2 = 2 3  −  or  = 2 3  + 2 − 2 3 .
3
3
3

21.  = sec 

23.  = cos  − sin 

⇒  0 = − sin  − cos , so  0 () = − sin  − cos  = 0 − (−1) = 1. An equation of the tangent

line to the curve  = cos  − sin  at the point ( −1) is  − (−1) = 1( − ) or  =  −  − 1.


⇒ 0 = 2( cos  + sin  · 1). At    ,
2


 0 = 2  cos  + sin  = 2(0 + 1) = 2, and an equation of the
2
2
2


tangent line is  −  = 2  −  , or  = 2.
2

(b)

al

27. (a)  () = sec  − 

e

25. (a)  = 2 sin 

⇒  0 () = sec  tan  − 1

rS

(b)

Note that  0 = 0 where  has a minimum. Also note that  0 is negative

Fo

when  is decreasing and  0 is positive when  is increasing.

⇒  0 () =  (cos ) + (sin ) · 1 =  cos  + sin 

29. () =  sin 
00

⇒

 () =  (− sin ) + (cos ) · 1 + cos  = − sin  + 2 cos 

tan  − 1
⇒
sec 
2
1 + tan  sec (sec ) − (tan  − 1)(sec  tan ) sec (sec2  − tan2  + tan )
 0 () =
=
=
2
(sec ) sec 2  sec 

N

ot

31. (a)  () =

sin  − cos  sin 
−1
tan  − 1 cos  cos 
=
= sin  − cos  ⇒  0 () = cos  − (− sin ) = cos  + sin 
(b)  () =
=
1
1
sec  cos  cos 
1
tan 
1 + tan 
=
+
= cos  + sin , which is the expression for  0 () in part (b).
(c) From part (a),  0 () = sec  sec  sec 
33.  () =  + 2 sin  has a horizontal tangent when  0 () = 0

=

2
3

+ 2 or

4
3

+ 2, where  is an integer. Note that

⇔
4
3

and

1 + 2 cos  = 0 ⇔ cos  = − 1
2
2
3

⇔

are ±  units from . This allows us to write the
3

solutions in the more compact equivalent form (2 + 1) ±  ,  an integer.
3
⇒ () = 0 () = 8 cos  ⇒ () = 00 () = −8 sin 
√ 
√
 
 
 
(b) The mass at time  = 2 has position  2 = 8 sin 2 = 8 23 = 4 3, velocity  2 = 8 cos 2 = 8 − 1 = −4,
3
3
3
3
3
2

35. (a) () = 8 sin 

√ 
√
 
  and acceleration  2 = −8 sin 2 = −8 23 = −4 3. Since  2  0, the particle is moving to the left.
3
3
3
c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 2.4 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS

¤

73

From the diagram we can see that sin  = 10 ⇔  = 10 sin . We want to find the rate

37.

of change of  with respect to , that is, . Taking the derivative of  = 10 sin , we get
 = 10(cos ). So when  =

39. lim

→0

sin 3
3 sin 3
= lim
→0

3
sin 3
= 3 lim
3→0
3 sin 
= 3 lim
→0

= 3(1)

 
,
3 

 
= 10 cos  = 10 1 = 5 ftrad.
3
2

[multiply numerator and denominator by 3]
[as  → 0, 3 → 0]
[let  = 3]
[Equation 2]



= 6 lim

→0

sin 6
1

·
·

cos 6 sin 2



sin 3
3
· 2
3
5 − 4

45. Divide numerator and denominator by .

6 sin 6
1
2
· lim
· lim
→0 cos 6 →0 2 sin 2
6



sin 3
3
= lim
· lim
= 1·
→0
→0 52 − 4
3



3
−4



=−

3
4

(sin  also works.)

sin  sin  lim 1
1
→0


=
=
=
sin  sin 
1
1
1+1·1
2
1+
1 + lim
·
lim
→0
 cos 
 →0 cos 

ot

1 − tan 
= lim sin  − cos  →4



sin 
1−
· cos 
√
cos  cos  − sin 
−1
−1
= lim
= lim
= √ =− 2
(sin  − cos ) · cos  →4 (sin  − cos ) cos  →4 cos 
1 2


(sin ) = cos  ⇒


2
(sin ) = − sin  ⇒
2

N

49.

lim

→0

Fo

sin  lim = lim
→0  + tan 
→0

→4

= lim

sin 6
1
2
1
1 1
· lim
· lim
= 6(1) · · (1) = 3
→0 cos 6
6
2 →0 sin 2
1 2

sin 3
= lim
43. lim
→0 53 − 4
→0

47.



al

→0

tan 6
= lim
→0
sin 2

rS

41. lim

e

=3

3
(sin ) = − cos  ⇒
3

The derivatives of sin  occur in a cycle of four. Since 99 = 4(24) + 3, we have
51.  =  sin  +  cos 

4
(sin ) = sin .
4

99
3
(sin ) =
(sin ) = − cos .
99

3

⇒  0 =  cos  −  sin  ⇒  00 = − sin  −  cos . Substituting these

expressions for ,  0 , and  00 into the given differential equation 00 +  0 − 2 = sin  gives us
(− sin  −  cos ) + ( cos  −  sin ) − 2( sin  +  cos ) = sin  ⇔
−3 sin  −  sin  +  cos  − 3 cos  = sin  ⇔ (−3 − ) sin  + ( − 3) cos  = 1 sin , so we must have

−3 −  = 1 and  − 3 = 0 (since 0 is the coefficient of cos  on the right side). Solving for  and , we add the first
1
3 equation to three times the second to get  = − 10 and  = − 10 .

53. (a)

 sin 

tan  =

 cos 

⇒ sec2  =

cos2  + sin2  cos  cos  − sin  (− sin )
=
.
2
cos cos2 

So sec2  =

1
.
cos2 

c
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74

¤

CHAPTER 2 DERIVATIVES

(b)


 1 sec  =

 cos 

(c)

 1 + cot 

(sin  + cos ) =

 csc  cos  − sin  =
=

⇒ sec  tan  =

(cos )(0) − 1(− sin )
.
cos2 

So sec  tan  =

sin 
.
cos2 

⇒

csc  (− csc2 ) − (1 + cot )(− csc  cot ) csc  [− csc2  + (1 + cot ) cot ]
=
csc2  csc2 
− csc2  + cot2  + cot 
−1 + cot 
=
csc  csc 

So cos  − sin  =

cot  − 1
.
csc 

55. By the definition of radian measure,  = , where  is the radius of the circle. By drawing the bisector of the angle , we can

2

=
2




2 · (2)
2

⇒  = 2 sin . So lim
= lim
= lim
= lim
= 1.
→0 sin(2)
2
→0+ 
→0+ 2 sin(2)
→0+ 2 sin(2)

[This is just the reciprocal of the limit lim

sin 


= 1 combined with the fact that as  → 0,

1. Let  = () = 1 + 4 and  =  () =

√

4
 
3
. Then
.
=
= ( 1 −23 )(4) = 
3

 
3 3 (1 + 4)2

5. Let  = () = sin  and  =  () =

√ cos 
 

cos 
.
=
= 1 −12 cos  = √ = √
. Then
2

 
2 sin 
2 

⇒  0 () = 5(4 + 32 − 2)4 ·

ot

7.  () = (4 + 32 − 2)5



or 10(4 + 32 − 2)4 (22 + 3)

N

√
1 − 2 = (1 − 2)12

1
= ( 2 + 1)−1
2 + 1

13.  = cos(3 + 3 )

 

=
= (sec2 )() =  sec2 .

 

Fo

3. Let  = () =  and  =  () = tan . Then

11.  () =

→ 0 also]

⇒

⇒


  4
 + 32 − 2 = 5(4 + 32 − 2)4 (43 + 6)


1
 0 () = 1 (1 − 2)−12 (−2) = − √
2
1 − 2

 0 () = −1( 2 + 1)−2 (2) = −

⇒  0 = − sin(3 + 3 ) · 32

15. Use the Product Rule.

rS

2.5 The Chain Rule

9.  () =


2

al

→0

e

see that sin

2
( 2 + 1)2

[3 is just a constant] = −32 sin(3 + 3 )

 =  sec  ⇒ 0 =  (sec  tan  · ) + sec  · 1 = sec  ( tan  + 1)

17.  () = (2 − 3)4 (2 +  + 1)5

⇒

 0 () = (2 − 3)4 · 5(2 +  + 1)4 (2 + 1) + (2 +  + 1)5 · 4(2 − 3)3 · 2
= (2 − 3)3 (2 +  + 1)4 [(2 − 3) · 5(2 + 1) + (2 +  + 1) · 8]
= (2 − 3)3 (2 +  + 1)4 (202 − 20 − 15 + 82 + 8 + 8) = (2 − 3)3 (2 +  + 1)4 (282 − 12 − 7) c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 2.5 THE CHAIN RULE

19. () = ( + 1)23 (22 − 1)3

¤

⇒

0 () = ( + 1)23 · 3(22 − 1)2 · 4 + (22 − 1)3 · 2 ( + 1)−13 = 2 ( + 1)−13 (22 − 1)2 [18( + 1) + (22 − 1)]
3
3
= 2 ( + 1)−13 (22 − 1)2 (202 + 18 − 1)
3
21.  =



2 + 1
2 − 1

3

⇒

 2
2


 2
2
(2 − 1)(2) − (2 + 1)(2)
 +1
 2 + 1
 +1
·
·
 =3 2
=3 2
2 −1
 −1
 
 −1
(2 − 1)2
0

 2
 2
2
2
 +1
2[2 − 1 − (2 + 1)]
 +1
2(−2)
−12(2 + 1)2
=3 2
·
=3 2
· 2
=
2 − 1)2
2
 −1
(
 −1
( − 1)
(2 − 1)4



=



−1
+1

12

⇒


−12



12
1 +1
1 −1
 −1
( + 1)(1) − ( − 1)(1)
=
·
·
2 +1
  + 1
2 −1
( + 1)2

al

 0 () =

−1
=
+1

1 ( + 1)12  + 1 −  + 1
1 ( + 1)12
2
1
·
=
·
=
2
12
2 ( − 1)
( + 1)
2 ( − 1)12 ( + 1)2
( − 1)12 ( + 1)32


2 + 1

27.  = √

rS

25.  () =

e

⇒  0 = cos( cos ) · [(− sin ) + cos  · 1] = (cos  −  sin ) cos( cos )

23.  = sin( cos )

Fo

⇒

ot

√
2
√
2 + 1 − √
2 + 1 (1) −  · 1 (2 + 1)−12 (2)
2 + 1
2
=
=
0 =
2
2
√
√
2 +1
2 +1



 2
 + 1 − 2
1
= √ or (2 + 1)−32
3 = 2
2 +1
( + 1)32


√
√
2 + 1 2 + 1 − 2
√
2 + 1
2
√
2 + 1

N

Another solution: Write  as a product and make use of the Product Rule.  = (2 + 1)−12

⇒

 0 =  · − 1 (2 + 1)−32 (2) + (2 + 1)−12 · 1 = (2 + 1)−32 [−2 + (2 + 1)1 ] = (2 + 1)−32 (1) = (2 + 1)−32 .
2
The step that students usually have trouble with is factoring out (2 + 1)−32 . But this is no different than factoring out 2 from 2 + 5 ; that is, we are just factoring out a factor with the smallest exponent that appears on it. In this case, − 3 is
2
smaller than − 1 .
2
29.  = sin

√
1 + 2

31.  = sin(tan 2)

⇒  0 = cos

√
√

 √
1 + 2 · 1 (1 + 2 )−12 · 2 =  cos 1 + 2  1 + 2
2

⇒  0 = cos(tan 2) ·



(tan 2) = cos(tan 2) · sec2 (2) ·
(2) = 2 cos(tan 2) sec2 (2)



33.  = sec2  + tan2  = (sec )2 + (tan )2

⇒

 0 = 2(sec )(sec  tan ) + 2(tan )(sec2 ) = 2 sec2  tan  + 2 sec2  tan  = 4 sec2  tan 

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

75

76

¤

CHAPTER 2 DERIVATIVES

35.  =



1 − cos 2
1 + cos 2

4

⇒


3
(1 + cos 2)(2 sin 2) + (1 − cos 2)(−2 sin 2)
1 − cos 2
 =4
·
1 + cos 2
(1 + cos 2)2
3

1 − cos 2
2 sin 2 (1 + cos 2 + 1 − cos 2) 4(1 − cos 2)3 2 sin 2 (2)
16 sin 2 (1 − cos 2)3
=4
·
=
=
2
3 (1 + cos 2)2
1 + cos 2
(1 + cos 2)
(1 + cos 2)
(1 + cos 2)5
0

37.  = cot2 (sin ) = [cot(sin )]2

 0 = 2[cot(sin )] ·

⇒


[cot(sin )] = 2 cot(sin ) · [− csc2 (sin ) · cos ] = −2 cos  cot(sin ) csc2 (sin )


39.  = [2 + (1 − 3)5 ]3

⇒

 0 = 3[2 + (1 − 3)5 ]2 (2 + 5(1 − 3)4 (−3)) = 3[2 + (1 − 3)5 ]2 [2 − 15(1 − 3)4 ]

43. () = (2 sin  + )
45.  = cos

( +


√ −12 
)
1 + 1 −12 =
2

1

√
2 + 


1+



e

1
2

1
√
2 

al


√
 +  ⇒ 0 =

⇒  0 () = (2 sin  + )−1 (2 cos  · ) = (2 sin  + )−1 (22 cos )

 sin(tan ) = cos(sin(tan ))12

⇒

rS

41.  =



(sin(tan ))12 = − sin(sin(tan ))12 · 1 (sin(tan ))−12 ·
(sin(tan ))
2




− sin sin(tan )
− sin sin(tan )



=
· cos(tan ) ·
· cos(tan ) · sec2 () ·  tan  =

2 sin(tan )
2 sin(tan )

− cos(tan ) sec2 () sin sin(tan )

=
2 sin(tan )

47.  = cos(2 )

⇒

ot

Fo

 0 = − sin(sin(tan ))12 ·

 0 = − sin(2 ) · 2 = −2 sin(2 )

⇒

49. () = tan 3

N

 00 = −2 cos(2 ) · 2 + sin(2 ) · (−2) = −42 cos(2 ) − 2 sin(2 )
⇒  0 () = 3 sec2 3 ⇒

 00 () = 2 · 3 sec 3
51.  = (1 + 2)10


(sec 3) = 6 sec 3 (3 sec 3 tan 3) = 18 sec2 3 tan 3


⇒  0 = 10(1 + 2)9 · 2 = 20(1 + 2)9 .

At (0 1), 0 = 20(1 + 0)9 = 20, and an equation of the tangent line is  − 1 = 20( − 0), or  = 20 + 1.
53.  = sin(sin )

⇒  0 = cos(sin ) · cos . At ( 0), 0 = cos(sin ) · cos  = cos(0) · (−1) = 1(−1) = −1, and an

equation of the tangent line is  − 0 = −1( − ), or  = − + .

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 2.5 THE CHAIN RULE



55. (a)  =  () = tan

2
4



⇒  0 () = sec2



2
4

The slope of the tangent at (1 1) is thus
 
 0 (1) = sec2   = 2 ·  = , and its equation
4 2
2

  
2· 4 .

¤

77

(b)

is  − 1 = ( − 1) or  =  −  + 1.

57. (a)  () = 

√
2 − 2 = (2 − 2 )12

⇒



2 − 22
 0 () =  · 1 (2 − 2 )−12 (−2) + (2 − 2 )12 · 1 = (2 − 2 )−12 −2 + (2 − 2 ) = √
2
2 − 2
 0 = 0 when  has a horizontal tangent line,  0 is negative when  is

(b)

e

decreasing, and  0 is positive when  is increasing.

⇒  0 () = 2 cos  + 2 sin  cos  = 0 ⇔

al

59. For the tangent line to be horizontal,  0 () = 0. () = 2 sin  + sin2 

rS

2 cos (1 + sin ) = 0 ⇔ cos  = 0 or sin  = −1, so  =  + 2 or 3 + 2, where  is any integer. Now
2
2
 3 





 2 = 3 and  2 = −1, so the points on the curve with a horizontal tangent are  + 2 3 and 3 + 2 −1 ,
2
2 where  is any integer.

⇒  0 () =  0 (()) ·  0 (), so  0 (5) =  0 ((5)) ·  0 (5) =  0 (−2) · 6 = 4 · 6 = 24

63. (a) () =  (())

Fo

61.  () =  (())

⇒ 0 () =  0 (()) ·  0 (), so 0 (1) =  0 ((1)) ·  0 (1) =  0 (2) · 6 = 5 · 6 = 30.

(b) () = ( ()) ⇒  0 () =  0 ( ()) ·  0 (), so  0 (1) =  0 ( (1)) ·  0 (1) =  0 (3) · 4 = 9 · 4 = 36.
65. (a) () =  (())

⇒ 0 () =  0 (()) 0 (). So 0 (1) =  0 ((1)) 0 (1) =  0 (3) 0 (1). To find  0 (3), note that  is

ot

1
3−4
= − . To find 0 (1), note that  is linear from (0 6) to (2 0), so its slope
6−2
4
 1
0−6
= −3. Thus,  0 (3)0 (1) = − 4 (−3) = 3 . is 4
2−0

N

linear from (2 4) to (6 3), so its slope is

(b) () = ( ()) ⇒  0 () =  0 ( ()) 0 (). So  0 (1) =  0 ( (1)) 0 (1) = 0 (2) 0 (1), which does not exist since
 0 (2) does not exist.

(c) () = (()) ⇒ 0 () =  0 (()) 0 (). So 0 (1) = 0 ((1)) 0 (1) = 0 (3) 0 (1). To find 0 (3), note that  is linear from (2 0) to (5 2), so its slope is

 
2
2−0
= . Thus, 0 (3) 0 (1) = 2 (−3) = −2.
3
5−2
3

67. The point (3 2) is on the graph of  , so  (3) = 2. The tangent line at (3 2) has slope

() =

−4
2
∆
=
=− .
∆
6
3


() ⇒  0 () = 1 [ ()]−12 ·  0 () ⇒
2

√
1
 0 (3) = 1 [ (3)]−12 ·  0 (3) = 1 (2)−12 (− 2 ) = − √ or − 1 2.
2
2
3
6
3 2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

78

¤

CHAPTER 2 DERIVATIVES

69. () =  ((()))

⇒ 0 () =  0 ((())) ·  0 (()) · 0 (), so

0 (1) =  0 (((1))) ·  0 ((1)) · 0 (1) =  0 ((2)) · 0 (2) · 4 =  0 (3) · 5 · 4 = 6 · 5 · 4 = 120
71.  () =  (3 (4 ()))

⇒

 0 () =  0 (3 (4 ())) ·



(3 (4 ())) =  0 (3 (4 ())) · 3 0 (4()) ·
(4())



=  0 (3 (4 ())) · 3 0 (4()) · 4 0 (), so
 0 (0) =  0 (3 (4(0))) · 3 0 (4(0)) · 4 0 (0) =  0 (3 (4 · 0)) · 3 0 (4 · 0) · 4 · 2 =  0 (3 · 0) · 3 · 2 · 4 · 2 = 2 · 3 · 2 · 4 · 2 = 96.
73. Let  () = cos . Then  (2) = 2 0 (2), 2  (2) = 22  00 (2), 3  (2) = 23  000 (2),    ,

()  (2) = 2  () (2). Since the derivatives of cos  occur in a cycle of four, and since 103 = 4(25) + 3, we have

77. (a) () = 40 + 035 sin

(b) At  = 1,

2
54

⇒


=



7
2
=
cos
≈ 016.

54
54

79. By the Chain Rule, () =

1
4

cos(10)(10) =

5
2

cos(10) cms.

al

sin(10) ⇒ the velocity after  seconds is () = 0 () =




2
2
07
2
7
2
035 cos
=
cos
=
cos
54
54
54
54
54
54

rS

1
4


 


=
=
() = () . The derivative  is the rate of change of the velocity

 



Fo

75. () = 10 +

e

 (103) () =  (3) () = sin  and 103 cos 2 = 2103  (103) (2) = 2103 sin 2.

with respect to time (in other words, the acceleration) whereas the derivative  is the rate of change of the velocity with respect to the displacement.

( − 2)8
( − 2)9
− 18
, and the simplification command results in the expression given by Derive.
9
(2 + 1)
(2 + 1)10

N

 0 () = 9

45( − 2)8 without simplifying. With either Maple or Mathematica, we first get
(2 + 1)10

ot

81. (a) Derive gives  0 () =

(b) Derive gives  0 = 2(3 −  + 1)3 (2 + 1)4 (173 + 62 − 9 + 3) without simplifying. With either Maple or
Mathematica, we first get  0 = 10(2 + 1)4 (3 −  + 1)4 + 4(2 + 1)5 (3 −  + 1)3 (32 − 1). If we use
Mathematica’s Factor or Simplify, or Maple’s factor, we get the above expression, but Maple’s simplify gives the polynomial expansion instead. For locating horizontal tangents, the factored form is the most helpful.
83. (a) If  is even, then  () =  (−). Using the Chain Rule to differentiate this equation, we get

 0 () =  0 (−)


(−) = − 0 (−). Thus,  0 (−) = − 0 (), so  0 is odd.


(b) If  is odd, then () = − (−). Differentiating this equation, we get  0 () = − 0 (−)(−1) =  0 (−), so  0 is even. c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 2.6 IMPLICIT DIFFERENTIATION

85. (a)


(sin  cos ) =  sin−1  cos  cos  + sin  (− sin )


[Product Rule]

=  sin−1  (cos  cos  − sin  sin )
=  sin−1  cos( + )

[Addition Formula for cosine]

=  sin−1  cos[( + 1)]
(b)

[factor out  sin−1 ]

[factor out ]


(cos  cos ) =  cos−1  (− sin ) cos  + cos  (− sin )


[Product Rule]

= − cos−1  (cos  sin  + sin  cos )
= − cos−1  sin( + )


180

[factor out ]




 

 rad, we have
(sin ◦ ) = sin 180  =




180

 cos 180  =


180

cos ◦ .

e



[Addition Formula for sine]

= − cos−1  sin[( + 1)]
87. Since  ◦ =

[factor out − cos−1 ]


 
=
, so

 
 

 
 
 
 

2 
  
  
 
=
=
+
=
2
 
  
 

  
 2
 


 2 
2  
 2 
   
+
=
+
=
2
2
   
 


 2

al

89. The Chain Rule says that

1. (a)

rS

Fo

2.6 Implicit Differentiation



9
(92 −  2 ) =
(1) ⇒ 18 − 2  0 = 0 ⇒ 2  0 = 18 ⇒  0 =



⇒

ot

(b) 92 −  2 = 1 ⇒ 2 = 92 − 1

(b)






1
1
+



7.



=

1
1
+ =1 ⇒



(c)  0 = −
5.

√
9
 = ± 92 − 1, so  0 = ± 1 (92 − 1)−12 (18) = ± √
.
2
92 − 1

9
9
= √
, which agrees with part (b).

± 92 − 1

N

(c) From part (a), 0 =

3. (a)

[Product Rule]


1
1
1
1
(1) ⇒ − 2 − 2  0 = 0 ⇒ − 2  0 = 2





1
1
−1
=1− =




⇒ =

⇒ 0 = −

2
2


( − 1)(1) − ()(1)
−1
, so  0 =
=
.
−1
( − 1)2
( − 1)2

2
[( − 1)] 2
2
1
=−
=− 2
=−
2
2
 ( − 1)2
( − 1)2


  3

 + 3 =
(1) ⇒ 32 + 3 2 ·  0 = 0 ⇒ 3 2  0 = −32



⇒ 0 = −

2
2



(2 +  − 2 ) =
(4) ⇒ 2 +  ·  0 +  · 1 − 2  0 = 0 ⇒


 0 − 2  0 = −2 − 

⇒ ( − 2)  0 = −2 − 

⇒ 0 =

2 + 
−2 − 
=
 − 2
2 − 

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

79

9.

¤

CHAPTER 2 DERIVATIVES



  4
  2
 ( + ) =
 (3 − ) ⇒ 4 (1 +  0 ) + ( + ) · 43 =  2 (3 −  0 ) + (3 − ) · 2  0



4 + 4  0 + 44 + 43  = 3 2 −  2  0 + 6 0 − 2 2  0
(4 + 3 2 − 6)  0 = 3 2 − 54 − 43 

⇒ cos  · 0 − 2  0 = 2 +  sin  ⇒

2 +  sin  cos  − 2



(4 cos  sin ) =
(1) ⇒ 4 [cos  · cos  · 0 + sin  · (− sin )] = 0 ⇒


 0 (4 cos  cos ) = 4 sin  sin 

15.

4 sin  sin 
= tan  tan 
4 cos  cos 

0 =

⇒

 · 1 −  · 0


= 1 + 0
[tan()] =
( + ) ⇒ sec2 () ·


2

⇒

 sec2 () −  sec2 () ·  0 =  2 +  2  0

⇒  sec2 () −  2 =  2  0 +  sec2 () ⇒



 sec2 () −  2 =  2 +  sec2 () · 0

⇒ 0 =



19.

0




 − 2
2 



1
()−12 ( 0
2


= 2 − 
2 

rS

⇒

+  · 1) = 0 + 2  0 +  · 2 ⇒

Fo


 = 1 + 2 

 sec2 () −  2
 2 +  sec2 ()

⇒ 

0



2



 0 + 
= 2  0 + 2

2 

 

 − 22 
4  − 


=
2 
2 

⇒ 0 =



( cos ) =
(1 + sin()) ⇒ (− sin ) + cos  ·  0 = cos() · (0 +  · 1) ⇒



⇒


 − 

 − 22 

4

ot

17.

⇒

3 2 − 54 − 43 
4 + 3 2 − 6

⇒ 0 =


 2
( cos ) =
( +  2 ) ⇒ (− sin ) + cos  ·  0 = 2 + 2  0


 0 (cos  − 2) = 2 +  sin  ⇒  0 =

13.

⇒ 4  0 + 3 2  0 − 6  0 = 3 2 − 54 − 43 

al

11.

⇒

e

80

cos  · 0 −  cos() ·  0 =  sin  +  cos() ⇒ [cos  −  cos()] 0 =  sin  +  cos() ⇒

21.

 sin  +  cos() cos  −  cos()

N

0 =


 

 () + 2 [ ()]3 =
(10) ⇒  0 () + 2 · 3[ ()]2 ·  0 () + [ ()]3 · 2 = 0. If  = 1, we have



 0 (1) + 12 · 3[ (1)]2 ·  0 (1) + [ (1)]3 · 2(1) = 0 ⇒  0 (1) + 1 · 3 · 22 ·  0 (1) + 23 · 2 = 0 ⇒
 0 (1) + 12 0 (1) = −16 ⇒ 13 0 (1) = −16 ⇒  0 (1) = − 16 .
13
23.

 4 2

(  − 3  + 2 3 ) =
(0) ⇒ 4 · 2 +  2 · 43 0 − (3 · 1 +  · 32 0 ) + 2( · 3 2 +  3 · 0 ) = 0 ⇒


43  2 0 − 32  0 + 2 3 0 = −24  + 3 − 6 2
0 =

⇒ (43  2 − 32  + 23 ) 0 = −24  + 3 − 6 2

−24  + 3 − 6 2

= 3 2

4  − 32  + 2 3

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

⇒

SECTION 2.6 IMPLICIT DIFFERENTIATION

25.  sin 2 =  cos 2

⇒

 · cos 2 · 2 + sin 2 · 0 = (− sin 2 · 2 0 ) + cos(2) · 1

sin 2 ·  0 + 2 sin 2 ·  0 = −2 cos 2 + cos 2
0 (sin 2 + 2 sin 2) = −2 cos 2 + cos 2
0 =

0 =

81

⇒

⇒
−2 cos 2 + cos 2
. When  = sin 2 + 2 sin 2

0 =

⇒

(−2)(−1) + 0
2
1
=
= , so an equation of the tangent line is  −
0+·1

2

27. 2 +  +  2 = 3

¤


4


2

and  =


,
4

we have

= 1 ( −  ), or  = 1 .
2
2
2

⇒ 2 +   0 +  · 1 + 20 = 0 ⇒  0 + 2  0 = −2 − 

⇒ 0 ( + 2) = −2 − 

⇒

−3
−2 − 
−2 − 1
. When  = 1 and  = 1, we have  0 =
=
= −1, so an equation of the tangent line is
 + 2
1+2·1
3

 − 1 = −1( − 1) or  = − + 2.
⇒ 2 + 2 0 = 2(22 + 2 2 − )(4 + 4  0 − 1). When  = 0 and  = 1 , we have
2

e

29. 2 +  2 = (22 + 2 2 − )2

0 +  0 = 2( 1 )(2 0 − 1) ⇒ 0 = 2 0 − 1 ⇒ 0 = 1, so an equation of the tangent line is  −
2

4( +   0 )(2 +  2 ) = 25( −   0 )

⇒

rS

⇒ 4(2 +  2 )(2 + 2  0 ) = 25(2 − 2 0 ) ⇒

31. 2(2 +  2 )2 = 25(2 −  2 )

4  0 (2 +  2 ) + 250 = 25 − 4(2 +  2 ) ⇒

25 − 4(2 +  2 )
. When  = 3 and  = 1, we have  0 =
25 + 4(2 +  2 )

Fo

0 =

75 − 120
25 + 40

9
9
so an equation of the tangent line is  − 1 = − 13 ( − 3) or  = − 13  +

33. (a)  2 = 54 − 2

= 1( − 0)

al

or  =  + 1 .
2

1
2

⇒ 2  0 = 5(43 ) − 2 ⇒  0 =

40
.
13

(b)

10(1)3 − 1
9
= , and an equation
2
2

ot

So at the point (1 2) we have  0 =

103 − 
.


9
= − 45 = − 13 ,
65

N

of the tangent line is  − 2 = 9 ( − 1) or  = 9  − 5 .
2
2
2
⇒ 18 + 2 0 = 0 ⇒ 2 0 = −18 ⇒  0 = −9 ⇒




 − (−9)
 2 + 92
 · 1 −  · 0
9
 00 = −9
= −9
= −9 ·
= −9 · 3 [since x and y must satisfy the
2
2


3


35. 92 +  2 = 9

original equation, 92 +  2 = 9]. Thus,  00 = −81 3 .
37. 3 +  3 = 1

 00 = −

⇒ 32 + 3 2  0 = 0 ⇒  0 = −

2
2

⇒

 2 (2) − 2 · 2  0
2 2 − 22 (−2 2 )
2 4 + 24 
2( 3 + 3 )
2
=−
=−
=−
=− 5,
2 )2
4
6
(


6


since  and  must satisfy the original equation, 3 +  3 = 1.

c
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82

¤

CHAPTER 2 DERIVATIVES

39. If  = 0 in  +  3 = 1, then we get  3 = 1

⇒  = 1, so the point where  = 0 is (0 1). Differentiating implicitly with

respect to  gives us 0 +  · 1 + 32  0 = 0. Substituting 0 for  and 1 for  gives us 1 + 30 = 0 ⇒ 0 = − 1 .
3
Differentiating  0 +  + 32  0 = 0 implicitly with respect to  gives us  00 +  0 +  0 + 3(2  00 +  0 · 2  0 ) = 0. Now



    substitute 0 for , 1 for , and − 1 for  0 . 0 − 1 − 1 + 3  00 + − 1 · 2 − 1 = 0 ⇒ 3 00 + 2 = 2 ⇒
3
3
3
3
3
9
3
 00 +

2
9

=

2
9

⇒  00 = 0.

41. (a) There are eight points with horizontal tangents: four at  ≈ 157735 and

four at  ≈ 042265.
(b)  0 =

32 − 6 + 2
2(2 3 − 3 2 −  + 1)

⇒ 0 = −1 at (0 1) and  0 =

1
3

at (0 2).

(c)  0 = 0 ⇒ 32 − 6 + 2 = 0 ⇒  = 1 ±

1
3

√
3

al

(d) By multiplying the right side of the equation by  − 3, we obtain the first

e

Equations of the tangent lines are  = − + 1 and  = 1  + 2.
3

graph. By modifying the equation in other ways, we can generate the other

( 2 − 1)( − 2)
= ( − 1)( − 2)( − 3)

N

ot

Fo

rS

graphs.

( 2 − 4)( − 2)

= ( − 1)( − 2)

( + 1)(2 − 1)( − 2)
= ( − 1)( − 2)

( + 1)( 2 − 1)( − 2)
= ( − 1)( − 2)

( 2 + 1)( − 2)

= (2 − 1)( − 2)

( + 1)( 2 − 1)( − 2)
= ( − 1)( − 2)

( + 1)( 2 − 2)

= ( − 1)(2 − 2)

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 2.6 IMPLICIT DIFFERENTIATION

43. From Exercise 31, a tangent to the lemniscate will be horizontal if  0 = 0

[25 − 4(2 +  2 )] = 0

⇒

45.

25
8

2
2
− 2 =1 ⇒
2


 − 0 = we have

25
4

83

⇒ 25 − 4(2 +  2 ) = 0 ⇒

(1). (Note that when  is 0,  is also 0, and there is no horizontal tangent

for 2 +  2 in the equation of the lemniscate, 2(2 +  2 )2 = 25(2 −  2 ), we get
 √

(2). Solving (1) and (2), we have 2 = 75 and  2 = 25 , so the four points are ± 5 4 3  ± 5 .
16
16
4

at the origin.) Substituting
2 −  2 =

2 +  2 =

¤

25
4

2
2 0
2 
− 2 = 0 ⇒ 0 = 2
2


 

⇒ an equation of the tangent line at (0  0 ) is

2
2 0
0
0 
0
0  2
0
( − 0 ). Multiplying both sides by 2 gives 2 − 2 = 2 − 2 . Since (0  0 ) lies on the hyperbola,
2
 0






2
0 
2
0
0 
0
− 2 = 2 − 2 = 1.
2




0
−1
0
. The negative reciprocal of that slope is
=
, which is the slope of  , so the tangent line at
0
−0 0
0

al

at  (0  0 ) is −


⇒ 2 + 2 0 = 0 ⇒ 0 = − , so the slope of the tangent line


e

47. If the circle has radius , its equation is 2 +  2 = 2

 is perpendicular to the radius  .

and  are not both zero]. 2 +  2 = 2

rS

49. 2 +  2 = 2 is a circle with center  and  +  = 0 is a line through  [assume 

⇒ 2 + 2 0 = 0 ⇒ 0 = −, so the

slope of the tangent line at 0 (0  0 ) is −0 0 . The slope of the line 0 is 0 0 ,

Fo

which is the negative reciprocal of −0 0 . Hence, the curves are orthogonal, and the families of curves are orthogonal trajectories of each other.

⇒  0 = 2 and 2 + 2 2 =  [assume   0] ⇒ 2 + 4 0 = 0 ⇒

20 = −


1

=−
=−
, so the curves are orthogonal if
2()
2(2 )
2

ot

51.  = 2

⇒ 0 = −

N

 6= 0. If  = 0, then the horizontal line  = 2 = 0 intersects 2 + 2 2 =  orthogonally

 √ at ±  0 , since the ellipse 2 + 2 2 =  has vertical tangents at those two points.
53. Since 2  2 , we are assured that there are four points of intersection.

(1)

2
2
+ 2 =1 ⇒
2


2
2 0
+ 2 =0 ⇒
2


0

=− 2
2


⇒  0 = 1 = −

2
.
2
[continued]

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

84

¤

(2)

CHAPTER 2 DERIVATIVES

2
2
− 2 =1 ⇒
2



Now 1 2 = −

2
20
− 2 =0 ⇒
2



 0

= 2
2


⇒  0 = 2 =

 2
.
2

2  2
2  2 2
2
2
2
2
·
= − 2 2 · 2 (3). Subtracting equations, (1) − (2), gives us 2 + 2 − 2 + 2 = 0 ⇒
2
2
 
  





2
2
2
2
+ 2 = 2 − 2
2




 2  2 +  2 2
2 2 − 2 2
=
2  2
2 2

⇒

⇒

 2 (2 +  2 )
2 (2 − 2 )
=
(4). Since
2  2
2 2

2 − 2 = 2 +  2 , we have 2 − 2 = 2 +  2 . Thus, equation (4) becomes

2
2
= 2 2
2  2
 

2
2 2
= 2 2 , and
2
 

⇒

2
2  2 2 2 in equation (3) gives us 1 2 = − 2 2 · 2 2 = −1. Hence, the ellipse and hyperbola are orthogonal
2
   

substituting for trajectories. ⇒   −   +

2 
3 
= 
−

2
⇒

al



(  −   + 2  −1 − 3  −2 ) =
( )



⇒

e



2 
55. (a)  + 2 ( − ) = 


  0 +  · 1 −  − 2  −2 ·  0 + 23  −3 ·  0 = 0 ⇒  0 ( − 2  −2 + 23  −3 ) =  − 
−

 3 ( −  )
 − 

= or 3  −3
3 − 2  + 23 
+ 2



rS

0 =

⇒

2  −2

(b) Using the last expression for   from part (a), we get

−995733 L4
≈ −404 L atm
2464386541 L3 - atm

ot

=

Fo

(10 L)3 [(1 mole)(004267 Lmole) − 10 L]

= 


(25 atm)(10 L)3 − (1 mole)2 (3592 L2 - atm mole2 )(10 L)


+ 2(1 mole)3 (3592 L2 - atm mole2 )(004267 L mole)

57. To find the points at which the ellipse 2 −  +  2 = 3 crosses the -axis, let  = 0 and solve for .

N

√
 √ 
 = 0 ⇒ 2 − (0) + 02 = 3 ⇔  = ± 3. So the graph of the ellipse crosses the -axis at the points ± 3 0 .

Using implicit differentiation to find  0 , we get 2 −  0 −  + 20 = 0 ⇒  0 (2 − ) =  − 2 ⇔ 0 =
So 0 at

√
√
√ 
 √ 
0−2 3
0+2 3
√ = 2 and  0 at − 3 0 is
√ = 2. Thus, the tangent lines at these points are parallel.
3 0 is
2(0) − 3
2(0) + 3

59. 2  2 +  = 2

⇒ 2 · 2 0 +  2 · 2 +  ·  0 +  · 1 = 0 ⇔ 0 (22  + ) = −22 − 

2

0 = −

 − 2
.
2 − 

⇔

2

2 + 
2 + 
. So − 2
= −1 ⇔ 22 +  = 22  +  ⇔ (2 + 1) = (2 + 1) ⇔
22  + 
2  + 

(2 + 1) − (2 + 1) = 0 ⇔ (2 + 1)( − ) = 0 ⇔  = − 1 or  = . But  = − 1
2
2
2  2 +  =

1
4

−

1
2

⇒

6= 2, so we must have  = . Then 2  2 +  = 2 ⇒ 4 + 2 = 2 ⇔ 4 + 2 − 2 = 0 ⇔

(2 + 2)(2 − 1) = 0. So 2 = −2, which is impossible, or 2 = 1 ⇔  = ±1. Since  = , the points on the curve where the tangent line has a slope of −1 are (−1 −1) and (1 1). c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 2.7

61. (a)  = () and  00 +  0 +  = 0

RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES

¤

85

⇒  00 () +  0 () + () = 0. If  = 0, we have 0 +  0 (0) + 0 = 0,

so  0 (0) = 0.
(b) Differentiating  00 +  0 +  = 0 implicitly, we get  000 +  00 · 1 +  00 +  0 +  · 1 = 0 ⇒
000 + 2 00 +  0 +  = 0, so  000 () + 2 00 () +  0 () + () = 0. If  = 0, we have
0 + 2 00 (0) + 0 + 1 [(0) = 1 is given] = 0 ⇒ 2 00 (0) = −1 ⇒  00 (0) = − 1 .
2

2.7 Rates of Change in the Natural and Social Sciences
⇒ () =  0 () = 32 − 24 + 36 (in fts)

1. (a)  = () = 3 − 122 + 36 (in feet)

e

(b) (3) = 27 − 72 + 36 = −9 fts
(c) The particle is at rest when () = 0. 32 − 24 + 36 = 0 ⇔ 3( − 2)( − 6) = 0 ⇔  = 2 s or 6 s.

al

(d) The particle is moving in the positive direction when ()  0. 3( − 2)( − 6)  0 ⇔ 0 ≤   2 or   6.
(e) Since the particle is moving in the positive direction and in the

(f)

rS

negative direction, we need to calculate the distance traveled in the intervals [0 2], [2 6], and [6 8] separately.
| (2) −  (0)| = |32 − 0| = 32.

Fo

| (6) −  (2)| = |0 − 32| = 32.
| (8) −  (6)| = |32 − 0| = 32.

The total distance is 32 + 32 + 32 = 96 ft.

ot

(g) () = 32 − 24 + 36 ⇒

(h)

() =  0 () = 6 − 24.

N

(3) = 6(3) − 24 = −6 (fts)s or fts2 .

(i) The particle is speeding up when  and  have the same sign. This occurs when 2    4 [ and  are both negative] and when   6 [ and  are both positive]. It is slowing down when  and  have opposite signs; that is, when 0 ≤   2 and when 4    6.
3. (a)  = () = cos(4)

⇒ () =  0 () = − sin(4) · (4)

(b) (3) = −  sin 3 = −  ·
4
4
4

√

2
2

√

= −  8 2 fts [≈ −056]

(c) The particle is at rest when () = 0. −  sin  = 0 ⇒ sin  = 0 ⇒
4
4
4


4

=  ⇒  = 0, 4, 8 s.

(d) The particle is moving in the positive direction when ()  0. −  sin   0 ⇒ sin   0 ⇒ 4    8.
4
4
4
c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

86

¤

CHAPTER 2

DERIVATIVES

(e) From part (c), () = 0 for  = 0 4 8. As in Exercise 1, we’ll

(f)

find the distance traveled in the intervals [0 4] and [4 8].
| (4) −  (0)| = |−1 − 1| = 2
| (8) −  (4)| = |1 − (−1)| = 2.
The total distance is 2 + 2 = 4 ft.
(g) () = −



sin
4
4

(h)

⇒


 
2
 cos · = − cos .
4
4 4
16
4
√
√ 
2
2 
2
2

3

 2
(3) = − cos
=−
−
=
(fts)s or fts2 .
16
4
16
2
32

() =  0 () = −

e

(i) The particle is speeding up when  and  have the same sign. This occurs when 0    2 or 8    10 [ and  are

al

both negative] and when 4    6 [ and  are both positive]. It is slowing down when  and  have opposite signs; that is, when 2    4 and when 6    8.

rS

5. (a) From the figure, the velocity  is positive on the interval (0 2) and negative on the interval (2 3). The acceleration  is

positive (negative) when the slope of the tangent line is positive (negative), so the acceleration is positive on the interval
(0 1), and negative on the interval (1 3). The particle is speeding up when  and  have the same sign, that is, on the

Fo

interval (0 1) when   0 and   0, and on the interval (2 3) when   0 and   0. The particle is slowing down when  and  have opposite signs, that is, on the interval (1 2) when   0 and   0.
(b)   0 on (0 3) and   0 on (3 4).   0 on (1 2) and   0 on (0 1) and (2 4). The particle is speeding up on (1 2)

ot

[  0,   0] and on (3 4) [  0,   0]. The particle is slowing down on (0 1) and (2 3) [  0,   0].
7. (a) () = 2 + 245 − 492

⇒ () = 0 () = 245 − 98. The velocity after 2 s is (2) = 245 − 98(2) = 49 ms

N

and after 4 s is (4) = 245 − 98(4) = −147 ms.

(b) The projectile reaches its maximum height when the velocity is zero. () = 0 ⇔ 245 − 98 = 0 ⇔
=

245
= 25 s.
98



(c) The maximum height occurs when  = 25. (25) = 2 + 245(25) − 49(25)2 = 32625 m or 32 5 m .
8

(d) The projectile hits the ground when  = 0 ⇔ 2 + 245 − 492 = 0 ⇔

−245 ± 2452 − 4(−49)(2)
=
⇒  =  ≈ 508 s [since  ≥ 0]
2(−49)

(e) The projectile hits the ground when  =  . Its velocity is ( ) = 245 − 98 ≈ −253 ms [downward].
9. (a) () = 15 − 1862

⇒

() = 0 () = 15 − 372. The velocity after 2 s is (2) = 15 − 372(2) = 756 ms.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 2.7

¤

RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES


15 ± 152 − 4(186)(25)
(b) 25 =  ⇔ 1862 − 15 + 25 = 0 ⇔  =
2(186)

⇔  = 1 ≈ 235 or  = 2 ≈ 571.

The velocities are (1 ) = 15 − 3721 ≈ 624 ms [upward] and (2 ) = 15 − 3722 ≈ −624 ms [downward].
⇒ 0 () = 2. 0 (15) = 30 mm2 mm is the rate at which

11. (a) () = 2

the area is increasing with respect to the side length as  reaches 15 mm.
(b) The perimeter is  () = 4, so 0 () = 2 = 1 (4) = 1  (). The
2
2

figure suggests that if ∆ is small, then the change in the area of the square is approximately half of its perimeter (2 of the 4 sides) times ∆. From the figure, ∆ = 2 (∆) + (∆)2 . If ∆ is small, then ∆ ≈ 2 (∆) and so ∆∆ ≈ 2.

(ii)

441 − 4
(21) − (2)
=
= 41
21 − 2
01

(b) () = 2

(25) − (2)
625 − 4
=
= 45
25 − 2
05

al

(iii)

(3) − (2)
9 − 4
=
= 5
3−2
1

rS

(i)

e

13. (a) Using () = 2 , we find that the average rate of change is:

⇒ 0 () = 2, so 0 (2) = 4.

(c) The circumference is () = 2 = 0 (). The figure suggests that if ∆ is small, then the change in the area of the circle (a ring around the outside) is approximately

Fo

equal to its circumference times ∆. Straightening out this ring gives us a shape that is approximately rectangular with length 2 and width ∆, so ∆ ≈ 2(∆).
Algebraically, ∆ = ( + ∆) − () = ( + ∆)2 − 2 = 2(∆) + (∆)2 .

15. () = 42
0

ot

So we see that if ∆ is small, then ∆ ≈ 2(∆) and therefore, ∆∆ ≈ 2.
⇒  0 () = 8

(a)  (1) = 8 ft ft

⇒

(b)  0 (2) = 16 ft2 ft

(c)  0 (3) = 24 ft2 ft

N

2

As the radius increases, the surface area grows at an increasing rate. In fact, the rate of change is linear with respect to the radius. 17. The mass is  () = 32 , so the linear density at  is () =  0 () = 6.

(a) (1) = 6 kgm

(b) (2) = 12 kgm

(c) (3) = 18 kgm

Since  is an increasing function, the density will be the highest at the right end of the rod and lowest at the left end.
19. The quantity of charge is () = 3 − 22 + 6 + 2, so the current is 0 () = 32 − 4 + 6.

(a) 0 (05) = 3(05)2 − 4(05) + 6 = 475 A

(b) 0 (1) = 3(1)2 − 4(1) + 6 = 5 A

The current is lowest when 0 has a minimum. 00 () = 6 − 4  0 when   2 . So the current decreases when  
3
increases when   2 . Thus, the current is lowest at  =
3

2
3

s.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

2
3

and

87

88

¤

CHAPTER 2

DERIVATIVES


−12
2
21. With  = 0 1 − 2
,




−12
−32  


2 
2


2
1

− 2
() =  () + 
() = 0 1 − 2
1− 2
()
·  +  · 0 −
=




2




−32 


2
2
0 
2
· 1− 2 + 2 =
= 0 1 − 2



(1 −  2 2 )32

Note that we factored out (1 −  2 2 )−32 since −32 was the lesser exponent. Also note that


() = .


23. (a) To find the rate of change of volume with respect to pressure, we first solve for  in terms of  .

 = 

⇒  =




⇒



= − 2.



110
210
1860 − 1750
2070 − 1860
=
= 11, 2 =
=
= 21,
1920 − 1910
10
1930 − 1920
10

rS

25. (a) 1920: 1 =

al

Thus, the volume is decreasing more rapidly at the beginning.


1


1

1 
=−
− 2 =
=
=
(c)  = −
 


(  )



e

(b) From the formula for   in part (a), we see that as  increases, the absolute value of   decreases.

(1 + 2 )/ 2 = (11 + 21)2 = 16 millionyear

740
830
4450 − 3710
5280 − 4450
=
= 74, 2 =
=
= 83,
1980 − 1970
10
1990 − 1980
10

Fo

1980: 1 =

(1 + 2 )/ 2 = (74 + 83)2 = 785 millionyear

(b)  () = 3 + 2 +  +  (in millions of people), where  ≈ 00012937063,  ≈ −7061421911,  ≈ 12,82297902,

ot

and  ≈ −7,743,770396.

(c)  () = 3 + 2 +  +  ⇒  0 () = 32 + 2 +  (in millions of people per year)

N

(d)  0 (1920) = 3(00012937063)(1920)2 + 2(−7061421911)(1920) + 12,82297902
≈ 14.48 millionyear [smaller than the answer in part (a), but close to it]
 (1980) ≈ 75.29 millionyear (smaller, but close)
0

(e)  0 (1985) ≈ 81.62 millionyear, so the rate of growth in 1985 was about 8162 millionyear.
27. (a) Using  =

() =
(b) () =


(2 − 2 ) with  = 001,  = 3,  = 3000, and  = 0027, we have  as a function of :
4

3000
(0012 − 2 ). (0) = 0925 cms, (0005) = 0694 cms, (001) = 0.
4(0027)3



(2 − 2 ) ⇒  0 () =
(−2) = −
. When  = 3,  = 3000, and  = 0027, we have
4
4
2

0 () = −

3000
. 0 (0) = 0,  0 (0005) = −92592 (cms)cm, and  0 (001) = −185185 (cms)cm.
2(0027)3

(c) The velocity is greatest where  = 0 (at the center) and the velocity is changing most where  =  = 001 cm
(at the edge). c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 2.8

29. (a) () = 1200 + 12 − 012 + 000053

RELATED RATES

¤

89

⇒  0 () = 12 − 02 + 000152 $yard, which is the marginal cost

function.
(b)  0 (200) = 12 − 02(200) + 00015(200)2 = $32yard, and this is the rate at which costs are increasing with respect to the production level when  = 200.  0 (200) predicts the cost of producing the 201st yard
(c) The cost of manufacturing the 201st yard of fabric is (201) − (200) = 36322005 − 3600 ≈ $3220, which is approximately  0 (200).
31. (a) () =

()


⇒ 0 () =

0 () − () · 1
0 () − ()
=
.
2
2

0 ()  0 ⇒ () is increasing; that is, the average productivity increases as the size of the workforce increases.

0 () − ()
 0 ⇒ 0 ()  0.
2

⇒  =


1

=
=
(  ). Using the Product Rule, we have

(10)(00821)
0821

rS

33.   = 

⇒ 0 ()  () ⇒

al

0 () − ()  0 ⇒

()


e

(b) 0 () is greater than the average productivity ⇒ 0 ()  () ⇒ 0 () 

1

1
=
[ () 0 () +  () 0 ()] =
[(8)(−015) + (10)(010)] ≈ −02436 Kmin.

0821
0821


= 0 and
= 0.



Fo

35. (a) If the populations are stable, then the growth rates are neither positive nor negative; that is,

(b) “The caribou go extinct” means that the population is zero, or mathematically,  = 0.
(c) We have the equations



=  −  and
= − +  . Let  =  = 0,  = 005,  = 0001,



ot

 = 005, and  = 00001 to obtain 005 − 0001 = 0 (1) and −005 + 00001 = 0 (2). Adding 10 times
(2) to (1) eliminates the  -terms and gives us 005 − 05 = 0 ⇒  = 10 . Substituting  = 10 into (1)

N

results in 005(10 ) − 0001(10 ) = 0 ⇔ 05 − 001 2 = 0 ⇔ 50 −  2 = 0 ⇔
 (50 −  ) = 0 ⇔  = 0 or 50. Since  = 10 ,  = 0 or 500. Thus, the population pairs (  ) that lead to stable populations are (0 0) and (500 50). So it is possible for the two species to live in harmony.

2.8 Related Rates
1.  = 3

⇒



 
=
= 32

 


3. Let  denote the side of a square. The square’s area  is given by  = 2 . Differentiating with respect to  gives us





= 2 . When  = 16,  = 4. Substitution 4 for  and 6 for gives us
= 2(4)(6) = 48 cm2s.




5.  = 2  = (5)2  = 25

⇒



= 25



⇒ 3 = 25




⇒


3
= mmin. 
25

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90

¤

CHAPTER 2

7. (a)  =

(b)  =

DERIVATIVES

√

=3 ⇒
2 + 1 and

√
2 + 1

⇒


 
1
3

3
=
= (2 + 1)−12 · 2 · 3 = √
= √ = 1.
. When  = 4,

 
2

2 + 1
9

 2 = 2 + 1 ⇒ 2 = 2 − 1 ⇒  = 1  2 −
2

1
2

and


=5 ⇒


√

 

=
=  · 5 = 5. When  = 12,  = 25 = 5, so
= 5(5) = 25.

 

9.

 2









( +  2 +  2 ) =
(9) ⇒ 2
+ 2
+ 2
=0 ⇒ 
+
+
= 0. If
= 5,
= 4 and










(  ) = (2 2 1), then 2(5) + 2(4) + 1


=0 ⇒



= −18.


11. (a) Given: a plane flying horizontally at an altitude of 1 mi and a speed of 500 mih passes directly over a radar station.

If we let  be time (in hours) and  be the horizontal distance traveled by the plane (in mi), then we are given

e

that  = 500 mih.
(c)

al

(b) Unknown: the rate at which the distance from the plane to the station is increasing

when it is 2 mi from the station. If we let  be the distance from the plane to the station,

rS

then we want to find  when  = 2 mi.

(d) By the Pythagorean Theorem,  2 = 2 + 1 ⇒ 2 () = 2 ().

√
√
√

3

 

=
= (500). Since  2 = 2 + 1, when  = 2,  = 3, so
=
(500) = 250 3 ≈ 433 mih.

 


2

Fo

(e)

13. (a) Given: a man 6 ft tall walks away from a street light mounted on a 15-ft-tall pole at a rate of 5 fts. If we let  be time (in s)

and  be the distance from the pole to the man (in ft), then we are given that  = 5 fts.
(c)

ot

(b) Unknown: the rate at which the tip of his shadow is moving when he is 40 ft from the pole. If we let  be the distance from the man to the tip of his

( + ) when  = 40 ft.


N

shadow (in ft), then we want to find
(d) By similar triangles,

+
15
=
6


⇒ 15 = 6 + 6

(e) The tip of the shadow moves at a rate of

15.

⇒ 9 = 6 ⇒  = 2 .
3





2
5 
( + ) =
+  =
= 5 (5) =
3


3
3 

We are given that

25
3

fts.



= 60 mih and
= 25 mih.  2 = 2 +  2



⇒




1


=

+
.




√
After 2 hours,  = 2 (60) = 120 and  = 2 (25) = 50 ⇒  = 1202 + 502 = 130,



1


120(60) + 50(25) so =

+
=
= 65 mih.




130

2




= 2
+ 2




⇒ 




=
+




⇒

c
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SECTION 2.8

RELATED RATES

¤

91



= 4 fts and
= 5 fts.  2 = ( + )2 + 5002 ⇒





 
2
= 2( + )
+
. 15 minutes after the woman starts, we have



We are given that

17.

 = (4 fts)(20 min)(60 smin) = 4800 ft and  = 5 · 15 · 60 = 4500 ⇒

√
 = (4800 + 4500)2 + 5002 = 86,740,000, so


837

 +  

4800 + 4500
(4 + 5) = √
≈ 899 fts
=
+
= √




86,740,000
8674
19.  =

1
,
2

where  is the base and  is the altitude. We are given that



= 1 cmmin and
= 2 cm2 min. Using the






1


=

+
. When  = 10 and  = 100, we have 100 = 1 (10) ⇒
2

2




1



4 − 20
 = 20, so 2 =
20 · 1 + 10
⇒ 4 = 20 + 10
⇒
=
= −16 cmmin.
2



10

1
2

= 10 ⇒

al

e

Product Rule, we have



= 35 kmh and
= 25 kmh.  2 = ( + )2 + 1002 ⇒





 
2
= 2( + )
+
. At 4:00 PM,  = 4(35) = 140 and  = 4(25) = 100 ⇒




√
 = (140 + 100)2 + 1002 = 67,600 = 260, so


 +  

140 + 100
720

=
+
=
(35 + 25) =
≈ 554 kmh.




260
13
We are given that

Fo

rS

21.

23. If  = the rate at which water is pumped in, then


=  − 10,000, where


 1 2
 =
3

 3

27

⇒



=
2
6

⇒ =

1
 ⇒
3




= 2
. When  = 200 cm,

9


N

 = 1
3

ot

 = 1 2  is the volume at time . By similar triangles,
3

800,000


= 20 cmmin, so  − 10,000 = (200)2 (20) ⇒  = 10,000 +
 ≈ 289,253 cm3min.

9
9
The figure is labeled in meters. The area  of a trapezoid is

25.

1
(base1
2

+ base2 )(height), and the volume  of the 10-meter-long trough is 10.

Thus, the volume of the trapezoid with height  is  = (10) 1 [03 + (03 + 2)].
2
By similar triangles,
Now


 
=

 

025
1

=
= , so 2 =  ⇒  = 5(06 + ) = 3 + 52 .

05
2
⇒

02 = (3 + 10)




⇒


02
=
. When  = 03,

3 + 10

02
02
1
10

=
= mmin = mmin or cmmin. 
3 + 10(03)
6
30
3

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92

¤

CHAPTER 2

DERIVATIVES

27. We are given that

 2

1
1
3

=
= 30 ft3 min.  = 2  = 

3
3
2
12
2 
4 

 

=

 

⇒ 30 =

When  = 10 ft,

⇒

120
6

= 2 =
≈ 038 ftmin.

10 
5

29.  =

1
,
2

but  = 5 m and sin  =

⇒


4


120
.
=

2

⇒  = 4 sin , so  = 1 (5)(4 sin ) = 10 sin .
2



 
= 006 rads, so
=
= (10 cos )(006) = 06 cos .


 

 

 
= 06 cos
= (06) 1 = 03 m2s.
When  = ,
2
3 
3
We are given


= −015 m s, and


e

31. From the figure and given information, we have 2 +  2 = 2 ,

⇒ 2





+ 2
=0 ⇒ 
= − . Substituting the given





rS

2 +  2 = 2

al


= 02 m s when  = 3 m. Differentiating implicitly with respect to , we get


information gives us (−015) = −3(02) ⇒  = 4 m. Thus, 32 + 42 = 2
2

Fo

 = 25 ⇒  = 5 m.

⇒

33. Differentiating both sides of   =  with respect to  and using the Product Rule gives us 



+
=0 ⇒




 


600
=−
. When  = 600,  = 150 and
= 20, so we have
=−
(20) = −80. Thus, the volume is

 


150

ot

decreasing at a rate of 80 cm3min.

1
1
180
9
400
1
1
1
1
1
1
=
+
=
=
, so  =
. Differentiating
=
+
=
+

1
2
80 100
8000
400
9

1
2


1 
1 1
1 1
1 2

1 2
2
with respect to , we have − 2
=− 2
− 2
⇒
=
+ 2
. When 1 = 80 and
2
 
1 
2 

1 
2 


4002 1

1
107
= 2
≈ 0132 Ωs.
(03) +
(02) =
2 = 100,

9
802
1002
810

N

35. With 1 = 80 and 2 = 100,

37. We are given  = 2◦min =
2

2

2


90

radmin. By the Law of Cosines,

 = 12 + 15 − 2(12)(15) cos  = 369 − 360 cos 
2



= 360 sin 



=

⇒

⇒


180 sin  
=
. When  = 60◦ ,




√
√
√
√
√
7
180 sin 60◦ 
 3

√
=
= √ =
≈ 0396 mmin.
369 − 360 cos 60◦ = 189 = 3 21, so

21
3 21 90
3 21

c
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SECTION 2.8

RELATED RATES

¤

93

39. (a) By the Pythagorean Theorem, 40002 +  2 = 2 . Differentiating with respect to ,




= 2 . We know that
= 600 fts, so when  = 3000 ft,



√
√
 = 40002 + 30002 = 25,000,000 = 5000 ft we obtain 2

and

 
3000
1800

=
=
(600) =
= 360 fts.

 
5000
5

(b) Here tan  =
 = 3000 ft,


4000


  
(tan ) =

 4000

⇒

⇒

sec2 

1 

=

4000 

⇒

 cos2  
=
. When

4000 

4000
4000
4

(45)2

= 600 fts,  = 5000 and cos  =
=
= , so
=
(600) = 0096 rads.


5000
5

4000




1 
 2   1 
−
⇒ − csc2 
=
⇒ − csc
=
5

5 
3
6
5 


2

5 2
10
√
=
 kmmin [≈ 130 mih]
=

6
9
3

41. cot  =

al

e

⇒


= 300 kmh. By the Law of Cosines,

 
 2 = 2 + 12 − 2(1)() cos 120◦ = 2 + 1 − 2 − 1 = 2 +  + 1, so
2

 
= 2
+




=

⇒


2 + 1 
=
. After 1 minute,  =

2 

Fo

2

rS

43. We are given that

√
√
52 + 5 + 1 = 31 km ⇒

300
60

= 5 km ⇒

2(5) + 1

1650
√
=
(300) = √
≈ 296 kmh.

2 31
31

45. Let the distance between the runner and the friend be . Then by the Law of Cosines,

ot

 2 = 2002 + 1002 − 2 · 200 · 100 · cos  = 50,000 − 40,000 cos  (). Differentiating

N

implicitly with respect to , we obtain 2



= −40,000(− sin ) . Now if  is the



distance run when the angle is  radians, then by the formula for the length of an arc on a circle,  = , we have  = 100, so  =

1

100

⇒


1 
7
=
=
. To substitute into the expression for

100 
100


, we must know sin  at the time when  = 200, which we find from (): 2002 = 50,000 − 40,000 cos 


√
√ 

 

2
15
15
7
cos  = 1 ⇒ sin  = 1 − 1 = 4 . Substituting, we get 2(200)
⇒
= 40,000 4 100
4
4


 =

7

√
15
4

⇔

≈ 678 ms. Whether the distance between them is increasing or decreasing depends on the direction in which

the runner is running.

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94

¤

CHAPTER 2 DERIVATIVES

2.9 Linear Approximations and Differentials
1.  () = 4 + 32

⇒  0 () = 43 + 6, so  (−1) = 4 and  0 (−1) = −10.

Thus, () =  (−1) +  0 (−1)( − (−1)) = 4 + (−10)( + 1) = −10 − 6.
√
√
 ⇒  0 () = 1 −12 = 1(2  ), so  (4) = 2 and  0 (4) = 1 . Thus,
2
4

3.  () =

() =  (4) +  0 (4)( − 4) = 2 + 1 ( − 4) = 2 + 1  − 1 = 1  + 1.
4
4
4
√
1 −  ⇒  0 () =

2

−1
√
, so  (0) = 1 and  0 (0) = − 1 .
2
1−

al

Therefore,
 
√
1 −  =  () ≈ (0) +  0 (0)( − 0) = 1 + − 1 ( − 0) = 1 − 1 .
2
2
√
√
1
So 09 = 1 − 01 ≈ 1 − 2 (01) = 095
√
√ and 099 = 1 − 001 ≈ 1 − 1 (001) = 0995.
2

e

5.  () =

√
4
1 + 2 ⇒  0 () = 1 (1 + 2)−34 (2) = 1 (1 + 2)−34 , so
4
2

7.  () =

rS

 (0) = 1 and  0 (0) = 1 . Thus,  () ≈  (0) +  0 (0)( − 0) = 1 + 1 .
2
2
√
√
We need 4 1 + 2 − 01  1 + 1   4 1 + 2 + 01, which is true when
2

1
= (1 + 2)−4
(1 + 2)4

9.  () =

 0 () = −4(1 + 2)−5 (2) =

⇒

Fo

−0368    0677.

−8
, so  (0) = 1 and  0 (0) = −8.
(1 + 2)5

We need

ot

Thus, () ≈  (0) +  0 (0)( − 0) = 1 + (−8)( − 0) = 1 − 8.

1
1
− 01  1 − 8 
+ 01, which is true
(1 + 2)4
(1 + 2)4

N

when − 0045    0055.

11. (a) The differential  is defined in terms of  by the equation  =  0 () . For  =  () = 2 sin 2,

 0 () = 2 cos 2 · 2 + sin 2 · 2 = 2( cos 2 + sin 2), so  = 2( cos 2 + sin 2) .


−12

1 + 2
(2)  = √

1 + 2
√
√
√ 1 −12
√ 0 sec2  sec2 
2
√ , so  =
√ .
13. (a) For  =  () = tan ,  () = sec
· 
=
2
2 
2 
(b)  =

√
1 + 2

⇒  =

(b) For  =  () =

1
2

1 − 2
,
1 + 2

 0 () =

(1 +  2 )(−2) − (1 − 2 )(2)
−2[(1 + 2 ) + (1 −  2 )]
−2(2)
−4
=
=
=
,
(1 +  2 )2
(1 +  2 )2
(1 +  2 )2
(1 +  2 )2

so  =

−4
.
(1 +  2 )2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 2.9 LINEAR APPROXIMATIONS AND DIFFERENTIALS

15. (a)  = tan 

√
3 + 2

95

 = sec2  

⇒

(b) When  = 4 and  = −01,  = [sec(4)]2 (−01) =
17. (a)  =

¤

⇒  =

√ 2
2 (−01) = −02.

1


(3 + 2 )−12 (2)  = √
2
3 + 2

1
1
(b)  = 1 and  = −01 ⇒  = √
(−01) = (−01) = −005.
2
3 + 12
19.  =  () = 2 − 2 ,  = 2, ∆ = −04

⇒

∆ =  (16) −  (2) = 064 − 0 = 064

 = −

2
5

−

2
4

= −01

2
2
 = − 2 (1) = −0125
2
4

rS

∆ =  (5) − (4) =

⇒

al

21.  =  () = 2,  = 4, ∆ = 1

e

 = (2 − 2)  = (2 − 4)(−04) = 08

Fo

23. To estimate (1999)4 , we’ll find the linearization of  () = 4 at  = 2. Since  0 () = 43 ,  (2) = 16, and

 0 (2) = 32, we have () = 16 + 32( − 2). Thus, 4 ≈ 16 + 32( − 2) when  is near 2, so
(1999)4 ≈ 16 + 32(1999 − 2) = 16 − 0032 = 15968.
25.  =  () =

√
3
 ⇒  = 1 −23 . When  = 1000 and  = 1,  = 1 (1000)−23 (1) =
3
3

ot

√
3
1001 =  (1001) ≈  (1000) +  = 10 +

1
300

1
300 ,

so

= 10003 ≈ 10003.

⇒  = sec2  . When  = 45◦ and  = −1◦ ,
√ 2
 = sec2 45◦ (−180) =
2 (−180) = −90, so tan 44◦ =  (44◦ ) ≈  (45◦ ) +  = 1 − 90 ≈ 0965.

N

27.  =  () = tan 

29.  =  () = sec 

⇒  0 () = sec  tan , so  (0) = 1 and  0 (0) = 1 · 0 = 0. The linear approximation of  at 0 is

 (0) +  0 (0)( − 0) = 1 + 0() = 1. Since 008 is close to 0, approximating sec 008 with 1 is reasonable.
31. (a) If  is the edge length, then  = 3

⇒  = 32 . When  = 30 and  = 01,  = 3(30)2 (01) = 270, so the

maximum possible error in computing the volume of the cube is about 270 cm3 . The relative error is calculated by dividing the change in  , ∆ , by  . We approximate ∆ with  .


∆

32 
01

Relative error =
≈
=
=3
= 001.
=3


3

30
Percentage error = relative error × 100% = 001 × 100% = 1%.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

96

¤

CHAPTER 2 DERIVATIVES

(b)  = 62

⇒  = 12 . When  = 30 and  = 01,  = 12(30)(01) = 36, so the maximum possible error in

computing the surface area of the cube is about 36 cm2 .



12 
01

∆
≈
=
=2
= 0006.
=2
Relative error =


62

30
Percentage error = relative error × 100% = 0006 × 100% = 06%.
33. (a) For a sphere of radius , the circumference is  = 2 and the surface area is  = 42 , so


2

 
2

2
⇒  = 4
=
2


so the maximum error is about

⇒  =

1 2
 . When  = 84 and  = 05,
22

e

 =

2
2
84
 . When  = 84 and  = 05,  = (84)(05) =
,




84
84
1

≈ 27 cm2 . Relative error ≈
= 2
=
≈ 0012 = 12%


84 
84

 
3

4 3
4
3
 = 
=
3
3
2
62

(b)  =

⇒  =

1
1764
1764
(84)2 (05) =
, so the maximum error is about 2 ≈ 179 cm3 .
22
2


35. (a)  = 2 


17642
1
=
=
≈ 0018 = 18%.

(84)3(62 )
56

rS

The relative error is approximately

al

=

⇒ ∆ ≈  = 2  = 2 ∆

(b) The error is

37.  = 

⇒ =




⇒  = −

Fo

∆ −  = [( + ∆)2  − 2 ] − 2 ∆ = 2  + 2 ∆ + (∆)2  − 2  − 2 ∆ = (∆)2 .
∆


−( 2 ) 

. The relative error in calculating  is
≈
=
=−
.
2


 


Hence, the relative error in calculating  is approximately the same (in magnitude) as the relative error in .

 = 0  = 0


(b) () =

ot

39. (a)  =








+
 =
 +
 =  + 





N


( + )  =
(c) ( + ) =




()  = 
 =  




(d) () =
()  =









+
 = 
 + 
 =   +  












−
 − 

 
  −  

  = 

=
 =
=
(e) 

 
2
2
2
(f )  ( ) =


( )  = −1 


41. (a) The graph shows that  0 (1) = 2, so () =  (1) +  0 (1)( − 1) = 5 + 2( − 1) = 2 + 3.

 (09) ≈ (09) = 48 and  (11) ≈ (11) = 52.
(b) From the graph, we see that  0 () is positive and decreasing. This means that the slopes of the tangent lines are positive, but the tangents are becoming less steep. So the tangent lines lie above the curve. Thus, the estimates in part (a) are too large. c
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CHAPTER 2 REVIEW

¤

97

2 Review

1. See Definition 2.1.1.
2. See the paragraph containing Formula 3 in Section 2.1.
3. (a) The average rate of change of  with respect to  over the interval [1  2 ] is

(b) The instantaneous rate of change of  with respect to  at  = 1 is lim

2 →1

 (2 ) − (1 )
.
2 − 1

 (2 ) −  (1 )
.
2 − 1

4. See Definition 2.1.4. The pages following the definition discuss interpretations of  0 () as the slope of a tangent line to the

graph of  at  =  and as an instantaneous rate of change of  () with respect to  when  = .
(b)

e

5. (a) A function  is differentiable at a number  if its derivative  0 exists

at  = ; that is, if  () exists.

al

0

(c) See Theorem 2.2.4. This theorem also tells us that if  is not

rS

continuous at , then  is not differentiable at .
6. See the discussion and Figure 7 on page 120.

7. The second derivative of a function  is the rate of change of the first derivative  0 . The third derivative is the derivative (rate

Fo

of change) of the second derivative. If  is the position function of an object,  0 is its velocity function,  00 is its acceleration function, and  000 is its jerk function.

8. (a) The Power Rule: If  is any real number, then


( ) = −1 . The derivative of a variable base raised to a constant


ot

power is the power times the base raised to the power minus one.
(b) The Constant Multiple Rule: If  is a constant and  is a differentiable function, then



[ ()] = 
 ().



N

The derivative of a constant times a function is the constant times the derivative of the function.
(c) The Sum Rule: If  and  are both differentiable, then




[ () + ()] =
 () +
(). The derivative of a sum




of functions is the sum of the derivatives.
(d) The Difference Rule: If  and  are both differentiable, then




[ () − ()] =
 () −
(). The derivative of a




difference of functions is the difference of the derivatives.
(e) The Product Rule: If  and  are both differentiable, then




[ () ()] =  ()
() + ()
 (). The




derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function.

c
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98

¤

CHAPTER 2 DERIVATIVES


 ()   () −  ()  ()
  ()


=
(f ) The Quotient Rule: If  and  are both differentiable, then
.
 ()
[ ()]2
The derivative of a quotient of functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.
(g) The Chain Rule: If  and  are both differentiable and  =  ◦  is the composite function defined by  () =  (()), then  is differentiable and  0 is given by the product  0 () =  0 (())  0 (). The derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
9. (a)  = 

⇒  0 = −1

(b)  = sin  ⇒  0 = cos 

(c)  = cos  ⇒ 0 = − sin 

(d)  = tan  ⇒  0 = sec2 

(e)  = csc  ⇒  0 = − csc  cot 

(f)  = sec  ⇒ 0 = sec  tan 

e

(g)  = cot  ⇒  0 = − csc2 

10. Implicit differentiation consists of differentiating both sides of an equation involving  and  with respect to , and then

al

solving the resulting equation for  0 .

rS

11. See the examples in Section 2.7 as well as the text following Example 8.
12. (a) The linearization  of  at  =  is () =  () +  0 ()( − ).

(b) If  =  (), then the differential  is given by  =  0 () .

Fo

(c) See Figure 5 in Section 2.9.

See the note after Theorem 4 in Section 2.2.

3. False.

See the warning before the Product Rule.

7. False.

 0 ()
 

1
 () =
[ ()]12 = [ ()]−12  0 () = 


2
2 ()

N

5. True.

ot

1. False.





 () = 2 +  = 2 +  for  ≥ 0 or  ≤ −1 and 2 +  = −(2 + ) for −1    0.

So  0 () = 2 + 1 for   0 or   −1 and  0 () = −(2 + 1) for −1    0. But |2 + 1| = 2 + 1 for  ≥ − 1 and |2 + 1| = −2 − 1 for   − 1 .
2
2
9. True.

() = 5 lim →2

11. False.

⇒ 0 () = 54

⇒  0 (2) = 5(2)4 = 80, and by the definition of the derivative,

() − (2)
=  0 (2) = 80.
−2

A tangent line to the parabola  = 2 has slope  = 2, so at (−2 4) the slope of the tangent is 2(−2) = −4 and an equation of the tangent line is  − 4 = −4( + 2). [The given equation,  − 4 = 2( + 2), is not even

linear!]

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 2 REVIEW

¤

1. (a)  = () = 1 + 2 + 2 4. The average velocity over the time interval [1 1 + ] is

ave =


1 + 2(1 + ) + (1 + )2 4 − 134
10 + 2
10 + 
(1 + ) − (1)
=
=
=
(1 + ) − 1

4
4

So for the following intervals the average velocities are:
(i) [1 3]:  = 2, ave = (10 + 2)4 = 3 ms

(ii) [1 2]:  = 1, ave = (10 + 1)4 = 275 ms

(iii) [1 15]:  = 05, ave = (10 + 05)4 = 2625 ms
(b) When  = 1, the instantaneous velocity is lim

→0

(iv) [1 11]:  = 01, ave = (10 + 01)4 = 2525 ms

(1 + ) − (1)
10 + 
10
= lim
=
= 25 ms.
→0

4
4

5. The graph of  has tangent lines with positive slope for   0 and negative

e

3.

al

slope for   0, and the values of  fit this pattern, so  must be the graph of the derivative of the function for . The graph of  has horizontal tangent

rS

lines to the left and right of the -axis and  has zeros at these points. Hence,
 is the graph of the derivative of the function for . Therefore,  is the graph

ot

Fo

of  ,  is the graph of  0 , and  is the graph of  00 .

7. (a)  0 () is the rate at which the total cost changes with respect to the interest rate. Its units are dollars(percent per year).

N

(b) The total cost of paying off the loan is increasing by $1200(percent per year) as the interest rate reaches 10%. So if the interest rate goes up from 10% to 11%, the cost goes up approximately $1200.
(c) As  increases,  increases. So  0 () will always be positive.
9.  0 (1990) is the rate at which the total value of US currency in circulation is changing in billions of dollars per year. To

estimate the value of  0 (1990), we will average the difference quotients obtained using the times  = 1985 and  = 1995.
Let  =
=

1873 − 2719
−846
(1985) − (1990)
=
=
= 1692 and
1985 − 1990
−5
−5

(1995) − (1990)
4093 − 2719
1374
=
=
= 2748. Then
1995 − 1990
5
5

 0 (1990) = lim

→1990

() − (1990)
+
1692 + 2748
444
≈
=
=
= 222 billion dollarsyear.
 − 1990
2
2
2

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99

¤

CHAPTER 2 DERIVATIVES

11.  () = 3 + 5 + 4

 0 () = lim

→0

= lim

→0

 ( + ) −  ()
( + )3 + 5( + ) + 4 − (3 + 5 + 4)
= lim
→0


32  + 32 + 3 + 5
= lim (32 + 3 + 2 + 5) = 32 + 5
→0

⇒  0 = 4(2 + 3 )3 (2 + 32 ) = 4(2 )3 (1 + )3 (2 + 3) = 47 ( + 1)3 (3 + 2)

13.  = (2 + 3 )4

2 −  + 2
√
= 32 − 12 + 2−12


4 − 1
4 + 1

⇒ 0 =

(4 + 1)43 − (4 − 1)43
43 [(4 + 1) − (4 − 1)]
83
=
= 4
4 + 1)2
4 + 1)2
(
(
( + 1)2

√




√
sec2 1 − 
1
√
(−1) = − √
⇒  0 = sec2 1 − 
2 1−
2 1−

√

al

21.  = tan 1 − 
23.

3 12 1 −12
3√
1
1

− 
− −32 =
− √ − √
2
2
2
2 
3

⇒  0 = 2 (cos ) + (sin )(2) = ( cos  + 2 sin )

17.  = 2 sin 
19.  =

⇒ 0 =

e

15.  =

⇒



( 4 + 2 ) =
( + 3) ⇒  · 4 3  0 +  4 · 1 + 2 ·  0 +  · 2 = 1 + 3 0


 0 (4 3 + 2 − 3) = 1 −  4 − 2 sec 2
1 + tan 2

⇒

⇒

1 −  4 − 2
4 3 + 2 − 3

Fo

25.  =

⇒ 0 =

rS

100

(1 + tan 2)(sec 2 tan 2 · 2) − (sec 2)(sec2 2 · 2)
2 sec 2 [(1 + tan 2) tan 2 − sec2 2]
=
(1 + tan 2)2
(1 + tan 2)2
2
2


2 sec 2 (tan 2 + tan 2 − sec 2)
2 sec 2 (tan 2 − 1)
1 + tan2  = sec2 
=
=
(1 + tan 2)2
(1 + tan 2)2

27.  = (1 − −1 )−1

⇒

ot

0 =

N

 0 = −1(1 − −1 )−2 [−(−1−2 )] = −(1 − 1)−2 −2 = −(( − 1))−2 −2 = −( − 1)−2
29. sin() = 2 − 

⇒ cos()(0 +  · 1) = 2 −  0

 0 [ cos() + 1] = 2 −  cos() ⇒ 0 =
31.  = cot(32 + 5)

⇒  cos() 0 +  0 = 2 −  cos() ⇒

2 −  cos()
 cos() + 1

⇒  0 = − csc2 (32 + 5)(6) = −6 csc2 (32 + 5)

√
√
 cos  ⇒


√ 
√ 
√ 0
√ √ 0 √ 
√ 
 0 =  cos  + cos 
 =  − sin  1 −12 + cos  1 −12
2
2

33.  =

=

1 −12
2

√
√
√
 √
√
√  cos  −  sin 
√
−  sin  + cos  =
2 
2

35.  = tan2 (sin ) = [tan(sin )]

⇒  0 = 2[tan(sin )] · sec2 (sin ) · cos 

c
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CHAPTER 2 REVIEW

37.  = ( tan )15



39.  = sin tan
41.  () =

¤

⇒  0 = 1 ( tan )−45 (tan  +  sec2 )
5

√
√
√



 √


⇒  0 = cos tan 1 + 3 sec2 1 + 3 32 2 1 + 3
1 + 3

√
4 + 1 ⇒  0 () =

1
2 (4

+ 1)−12 · 4 = 2(4 + 1)−12

⇒

4
 00 () = 2(− 1 )(4 + 1)−32 · 4 = −4(4 + 1)32 , so  00 (2) = −4932 = − 27 .
2

⇒ 65 + 6 5  0 = 0 ⇒  0 = −5 5

43. 6 +  6 = 1

⇒





54  4  − (−5 5 )
54 ( 6 + 6 ) 5
 5 (54 ) − 5 (5 4  0 )
54
00
 =−
=−
=−
= − 11
( 5 )2
 10
6

→0

sec  sec 0
1
=
=
=1
1 − sin 
1 − sin 0
1−0



⇒  0 = 4 · 2 sin  cos . At   1 ,  0 = 8 ·
6
√ 
√
√

is  − 1 = 2 3  −  , or  = 2 3  + 1 −  33.
6

·

√
3
2

=2

√
3, so an equation of the tangent line

√
2 cos 
1 + 4 sin  ⇒ 0 = 1 (1 + 4 sin )−12 · 4 cos  = √
.
2
1 + 4 sin 

rS

49.  =

1
2

al

47.  = 4 sin2 

e

45. lim

2
At (0 1), 0 = √ = 2, so an equation of the tangent line is  − 1 = 2( − 0), or  = 2 + 1.
1
The slope of the normal line is − 1 , so an equation of the normal line is  − 1 = − 1 ( − 0), or  = − 1  + 1.
2
2
2

Fo

√
5− ⇒
√


√
√
−
2(5 − )
1
−
2 5−
 0 () =  (5 − )−12 (−1) + 5 −  = √
+ 5−· √
= √
+ √
2
2 5−
2 5−
2 5−
2 5−

51. (a)  () = 

10 − 3
− + 10 − 2
√
= √
2 5−
2 5−

ot

=

(b) At (1 2):  0 (1) = 7 .
4

(c)

N

So an equation of the tangent line is  − 2 = 7 ( − 1) or  = 7  + 1 .
4
4
4
At (4 4):  0 (4) = − 2 = −1.
2
So an equation of the tangent line is  − 4 = −1( − 4) or  = − + 8.
The graphs look reasonable, since  0 is positive where  has tangents with

(d)

positive slope, and  0 is negative where  has tangents with negative slope.

⇒  0 = cos  − sin  = 0 ⇔ cos  = sin  and 0 ≤  ≤ 2
√ 

 √  are   2 and 5  − 2 .
4
4

53.  = sin  + cos 

⇔ =


4

or

5
,
4

so the points

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101

102

¤

CHAPTER 2 DERIVATIVES

55.  = () = 2 +  + 

⇒  0 () = 2 + . We know that  0 (−1) = 6 and  0 (5) = −2, so −2 +  = 6 and

10 +  = −2. Subtracting the first equation from the second gives 12 = −8 ⇒  = − 2 . Substituting − 2 for  in the
3
3 first equation gives  =

14
.
3

Now  (1) = 4 ⇒ 4 =  +  + , so  = 4 +

57.  () = ( − )( − )( − )

So

2
3

−

14
3

= 0 and hence,  () = − 2 2 +
3

14
.
3

⇒  0 () = ( − )( − ) + ( − )( − ) + ( − )( − ).

 0 ()
( − )( − ) + ( − )( − ) + ( − )( − )
1
1
1
=
=
+
+
.
()
( − )( − )( − )
−
−
−
⇒ 0 () =  ()  0 () + ()  0 () ⇒

59. (a) () =  () ()

0 (2) = (2)  0 (2) + (2)  0 (2) = (3)(4) + (5)(−2) = 12 − 10 = 2
(b)  () =  (()) ⇒  0 () =  0 (())  0 () ⇒  0 (2) =  0 ((2)) 0 (2) =  0 (5)(4) = 11 · 4 = 44
⇒  0 () = 2  0 () + ()(2) = [0 () + 2()]

63.  () = [ ()]2

⇒  0 () = 2[ ()] ·  0 () = 2() 0 ()

65.  () = (())

⇒  0 () = 0 (())  0 ()

67.  () = (sin )

⇒  0 () = 0 (sin ) · cos 
⇒

al rS 0 () =

 () ()
 () + ()

[ () + ()] [ ()  0 () + ()  0 ()] −  () () [ 0 () +  0 ()]
[ () + ()]2

Fo

69. () =

e

61.  () = 2 ()

[ ()]2  0 () +  () ()  0 () +  () () 0 () + [ ()]2  0 () −  () ()  0 () −  () ()  0 ()
[() + ()]2

=

 0 () [ ()]2 +  0 () [ ()]2
[ () + ()]2

ot

=

71. Using the Chain Rule repeatedly, () =  ((sin 4))

⇒



((sin 4)) =  0 ((sin 4)) ·  0 (sin 4) ·
(sin 4) =  0 ((sin 4)) 0 (sin 4)(cos 4)(4).



N

0 () =  0 ((sin 4)) ·
73. (a)  = 3 − 12 + 3

⇒ () =  0 = 32 − 12 ⇒ () =  0 () = 6

(b) () = 3(2 − 4)  0 when   2, so it moves upward when   2 and downward when 0 ≤   2.
(c) Distance upward = (3) − (2) = −6 − (−13) = 7,
Distance downward = (0) − (2) = 3 − (−13) = 16. Total distance = 7 + 16 = 23.
(d)

(e) The particle is speeding up when  and  have the same sign, that is, when   2. The particle is slowing down when  and  have opposite signs; that is, when 0    2.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 2 REVIEW

¤

75. The linear density  is the rate of change of mass  with respect to length .


√ 
 =  1 +  =  + 32
77. If  = edge length, then  = 3

⇒  =  = 1 +

3
2

√
√
, so the linear density when  = 4 is 1 + 3 4 = 4 kgm.
2

⇒   = 32  = 10 ⇒  = 10(32 ) and  = 62

 = (12)  = 12[10(32 )] = 40. When  = 30,  =
79. Given  = 5 and  = 15, find .  2 = 2 + 2




2
= 2
+ 2




40
30

=

4
3

⇒

cm2 min.

⇒


1
⇒
= (15 + 5). When  = 3,


√
 = 45 + 3(5) = 60 and  = 15(3) = 45 ⇒  = 452 + 602 = 75,
1

=
[15(45) + 5(60)] = 13 fts.

75

 = 400 cot 

⇒

⇒



= −400 csc2  . When  =




,
6

rS


= −400(2)2 (−025) = 400 fth.

√
3
1 + 3 = (1 + 3)13

e

81. We are given  = −025 radh. tan  = 400

al

so

⇒  0 () = (1 + 3)−23 , so the linearization of  at  = 0 is
√
() =  (0) +  0 (0)( − 0) = 113 + 1−23  = 1 + . Thus, 3 1 + 3 ≈ 1 +  ⇒

√
3
103 = 3 1 + 3(001) ≈ 1 + (001) = 101.

Fo

83. (a)  () =

√
(b) The linear approximation is 3 1 + 3 ≈ 1 + , so for the required accuracy
√
√ we want 3 1 + 3 − 01  1 +   3 1 + 3 + 01. From the graph,

ot

it appears that this is true when −0235    0401.

N



 1 2 

= 1 +  2 ⇒  = 2 +   . When  = 60
2
8
4


and  = 01,  = 2 +  60(01) = 12 + 3 , so the maximum error is
4
2

85.  = 2 + 1 
2

approximately 12 +

87. lim

→0

3
2

≈ 167 cm2 .

√



4

16 +  − 2
 √
1
1
1
4
=

= −34 
=  √ 3 =



4
32
4 4 16
 = 16
 = 16

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103

104

¤

CHAPTER 2 DERIVATIVES

√
√

√
√
√
√
1 + tan  − 1 + sin 
1 + tan  + 1 + sin 
1 + tan  − 1 + sin 

√
√
89. lim
= lim
→0
→0
3
3
1 + tan  + 1 + sin 
= lim

→0

= lim

→0

3
3

cos 
(1 + tan ) − (1 + sin ) sin  (1 cos  − 1)
 = lim √
·
√
√
√
→0 3 cos 
1 + tan  + 1 + sin 
1 + tan  + 1 + sin 
1 + cos  sin  (1 − cos )

√
√
·
1 + tan  + 1 + sin  cos  1 + cos 

sin  · sin2 
√

√
→0 3
1 + tan  + 1 + sin  cos  (1 + cos )
3
 sin 
1

√
lim √
= lim
→0
→0

1 + tan  + 1 + sin  cos  (1 + cos )

= lim

 1 2
= 1 2 , so  0 () = 1 2 .
2
8
8

e

⇒  0 (2) · 2 = 2

⇒  0 (2) = 1 2 . Let  = 2. Then  0 () =
2

1
2

ot

Fo

rS

al


[ (2)] = 2


N

91.

1
1
√ 
= 13 · √
=
4
1 + 1 · 1 · (1 + 1)

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PROBLEMS PLUS
1. Let  be the -coordinate of . Since the derivative of  = 1 − 2 is  0 = −2, the slope at  is −2. But since the triangle
√
√
√
√
31, so the slope at  is − 3. Therefore, we must have that −2 = − 3 ⇒  = 23 .
√

 √

 √ 2   √
3
= 23  1 and by symmetry,  has coordinates − 23  1 .
Thus, the point  has coordinates 2  1 − 23
4
4

is equilateral,  =

We must show that  (in the figure) is halfway between  and , that is,

3.

 = ( + )2. For the parabola  = 2 +  + , the slope of the tangent line is given by  0 = 2 + . An equation of the tangent line at  =  is
 − (2 +  + ) = (2 + )( − ). Solving for  gives us

e

 = (2 + ) − 22 −  + (2 +  + ) or (1)

al

 = (2 + ) +  − 2

Similarly, an equation of the tangent line at  =  is

rS

 = (2 + ) +  −  2

(2)

We can eliminate  and solve for  by subtracting equation (1) from equation (2).
[(2 + ) − (2 + )] −  2 + 2 = 0

Fo

(2 − 2) =  2 − 2
2( − ) = ( 2 − 2 )
=

+
( + )( − )
=
2( − )
2

ot

Thus, the -coordinate of the point of intersection of the two tangent lines, namely , is ( + )2.
5. Using  0 () = lim

→

 () −  () sec  − sec 
, we recognize the given expression,  () = lim
, as
→
−
−

N

 0 () with () = sec . Now  0 (  ) = 00 (  ), so we will find 00 ().
4
4

 0 () = sec  tan  ⇒  00 () = sec  sec2  + tan  sec  tan  = sec (sec2  + tan2 ), so
√ √ 2
√
√
 00 (  ) = 2( 2 + 12 ) = 2(2 + 1) = 3 2.
4
7. We use mathematical induction. Let  be the statement that


(sin4  + cos4 ) = 4−1 cos(4 + 2).


1 is true because



(sin4  + cos4 ) = 4 sin3  cos  − 4 cos3  sin  = 4 sin  cos  sin2  − cos2  


= −4 sin  cos  cos 2 = −2 sin 2 cos 2 = − sin 4 = sin(−4)






= cos  − (−4) = cos  + 4 = 4−1 cos 4 +   when  = 1
2
2
2

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[continued]

105

106

¤

CHAPTER 2 PROBLEMS PLUS




  4 sin  + cos4  = 4−1 cos 4 +   . Then
2



 



  −1

(sin4  + cos4 ) =
(sin4  + cos4 ) = cos 4 +  
4
2

 


Now assume  is true, that is,
+1
+1

which shows that +1 is true.
Therefore,

  




4 +   = −4 sin 4 +  
= −4−1 sin 4 +   ·
2
2
2











= 4 sin −4 −  2 = 4 cos  − −4 −   = 4 cos 4 + ( + 1) 
2
2
2




(sin4  + cos4 ) = 4−1 cos 4 +   for every positive integer , by mathematical induction.
2


Another proof: First write

sin4  + cos4  = (sin2  + cos2 )2 − 2 sin2  cos2  = 1 −


(sin4  + cos4 ) =




sin2 2 = 1 − 1 (1 − cos 4) =
4

3
4

+

1
4

cos 4




3
1
1


+ cos 4 = · 4 cos 4 + 
= 4−1 cos 4 +  .
4
4
4
2
2

e

Then we have



1
2



1 and pass through ±0  2 respectively, so the
0
±20

rS



points ±0  2 . The normals to  = 2 at  = ±0 have slopes −
0

al

9. We must find a value 0 such that the normal lines to the parabola  = 2 at  = ±0 intersect at a point one unit from the

1
1
1
( − 0 ) and  − 2 =
( + 0 ). The common -intercept is 2 + .
0
0
20
20
2

 

We want to find the value of 0 for which the distance from 0 2 + 1 to 0  2 equals 1. The square of the distance is
0
0
2

normals have the equations  − 2 = −
0

√

Fo



2
= 2 +
(0 − 0)2 + 2 − 2 + 1
0
0
0
2
 5 the center of the circle is at 0 4 .

1
4

= 1 ⇔ 0 = ±

3
.
2

For these values of 0 , the -intercept is 2 +
0

1
2

= 5 , so
4

ot

Another solution: Let the center of the circle be (0 ). Then the equation of the circle is 2 + ( − )2 = 1.

Solving with the equation of the parabola,  = 2 , we get 2 + (2 − )2 = 1 ⇔ 2 + 4 − 22 + 2 = 1 ⇔

N

4 + (1 − 2)2 + 2 − 1 = 0. The parabola and the circle will be tangent to each other when this quadratic equation in 2 has equal roots; that is, when the discriminant is 0. Thus, (1 − 2)2 − 4(2 − 1) = 0 ⇔


1 − 4 + 42 − 42 + 4 = 0 ⇔ 4 = 5, so  = 5 . The center of the circle is 0 5 .
4
4

11. We can assume without loss of generality that  = 0 at time  = 0, so that  = 12 rad. [The angular velocity of the wheel

is 360 rpm = 360 · (2 rad)(60 s) = 12 rads.] Then the position of  as a function of time is
40 sin  sin 
1

=
=
= sin 12.
12 m
120
3
3
1


= · 12 · cos 12 = 4 cos . When  = , we have
(a) Differentiating the expression for sin , we get cos  ·

3
3

√
√
 √ 2 
4 cos 
3
3
11
1
4 3

2
3
= sin  = sin  =
≈ 656 rads.
= √
, so cos  = 1 − and =
= 
3
6
6
12

cos 
11
1112

 = (40 cos  40 sin ) = (40 cos 12 40 sin 12), so sin  =

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 2 PROBLEMS PLUS

(b) By the Law of Cosines, | |2 = ||2 + | |2 − 2 || | | cos 

¤

107

⇒

1202 = 402 + | |2 − 2 · 40 | | cos  ⇒ | |2 − (80 cos ) | | − 12,800 = 0 ⇒
√
√




√
| | = 1 80 cos  ± 6400 cos2  + 51,200 = 40 cos  ± 40 cos2  + 8 = 40 cos  + 8 + cos2  cm
2
√
[since | |  0]. As a check, note that | | = 160 cm when  = 0 and | | = 80 2 cm when  =  .
2

√


(c) By part (b), the -coordinate of  is given by  = 40 cos  + 8 + cos2  , so





 
2 cos  sin  cos 
=
= 40 − sin  − √
· 12 = −480 sin  1 + √ cms. 
 
2 8 + cos2 
8 + cos2 
In particular,  = 0 cms when  = 0 and  = −480 cms when  =


.
2

13. It seems from the figure that as  approaches the point (0 2) from the right,  → ∞ and  → 2+ . As  approaches the

point (3 0) from the left, it appears that  → 3+ and  → ∞. So we guess that  ∈ (3 ∞) and  ∈ (2 ∞). It is

e

more difficult to estimate the range of values for  and  . We might perhaps guess that  ∈ (0 3),

al

and  ∈ (−∞ 0) or (−2 0).

In order to actually solve the problem, we implicitly differentiate the equation of the ellipse to find the equation of the
2
2
+
=1 ⇒
9
4

2
2 0
4
+
 = 0, so 0 = −
. So at the point (0  0 ) on the ellipse, an equation of the
9
4
9

rS

tangent line:

2
2
4 0
0  0 
2
+
= 0 + 0 = 1,
( − 0 ) or 40  + 90  = 42 + 90 . This can be written as
0
9 0
9
4
9
4
0  0 
+
= 1. because (0  0 ) lies on the ellipse. So an equation of the tangent line is
9
4

Fo

tangent line is  − 0 = −

Therefore, the -intercept  for the tangent line is given by by 0 
9
= 1 ⇔  =
, and the -intercept  is given
9
0

0 
4
= 1 ⇔  = .
4
0

ot

So as 0 takes on all values in (0 3),  takes on all values in (3 ∞), and as 0 takes on all values in (0 2),  takes on
At the point (0  0 ) on the ellipse, the slope of the normal line is −

9 0
1
=
, and its
 0 (0  0 )
4 0

N

all values in (2 ∞).

equation is  − 0 =

9 0
9 0
( − 0 ). So the -intercept  for the normal line is given by 0 − 0 =
( − 0 ) ⇒
4 0
4 0

40
50
9 0
90
50
(0 − 0 ) ⇒  = −
+ 0 =
, and the -intercept  is given by  − 0 =
+ 0 = −
.
9
9
4 0
4
4


So as 0 takes on all values in (0 3),  takes on all values in 0 5 , and as 0 takes on all values in (0 2),  takes on
3


all values in − 5  0 .
2
 = −

15. (a) If the two lines 1 and 2 have slopes 1 and 2 and angles of

inclination 1 and 2 , then 1 = tan 1 and 2 = tan 2 . The triangle in the figure shows that 1 +  + (180◦ − 2 ) = 180◦ and so
 = 2 − 1 . Therefore, using the identity for tan( − ), we have tan  = tan(2 − 1 ) =

tan 2 − tan 1
2 − 1 and so tan  =
.
1 + tan 2 tan 1
1 + 1 2

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108

¤

CHAPTER 2 PROBLEMS PLUS

(b) (i) The parabolas intersect when 2 = ( − 2)2

⇒  = 1. If  = 2 , then  0 = 2, so the slope of the tangent

to  = 2 at (1 1) is 1 = 2(1) = 2. If  = ( − 2)2 , then  0 = 2( − 2), so the slope of the tangent to
 = ( − 2)2 at (1 1) is 2 = 2(1 − 2) = −2. Therefore, tan  = so  = tan−1

4
3

4
2 − 1
−2 − 2
= and
=
1 + 1 2
1 + 2(−2)
3

≈ 53◦ [or 127◦ ].

(ii) 2 −  2 = 3 and 2 − 4 +  2 + 3 = 0 intersect when 2 − 4 + (2 − 3) + 3 = 0 ⇔ 2( − 2) = 0 ⇒
 = 0 or 2, but 0 is extraneous. If  = 2, then  = ±1. If 2 −  2 = 3 then 2 − 2 0 = 0 ⇒  0 =  and
2 − 4 +  2 + 3 = 0 ⇒ 2 − 4 + 2 0 = 0 ⇒ 0 =
2 = 0, so tan  =

= −2 ⇒  ≈ 117◦ . At (2 −1) the slopes are 1 = −2 and 2 = 0

17. Since ∠ = ∠ = , the triangle  is isosceles, so

e

0 − (−2)
= 2 ⇒  ≈ 63◦ [or 117◦ ].
1 + (−2)(0)

al

so tan  =

0−2
1+2·0

2−
. At (2 1) the slopes are 1 = 2 and


|| = || = . By the Law of Cosines, 2 = 2 + 2 − 2 cos . Hence,


2
=
. Note that as  → 0+ ,  → 0+ (since
2 cos 
2 cos 

sin  = ), and hence  →

rS

2 cos  = 2 , so  =



= . Thus, as  is taken closer and closer
2 cos 0
2

→0

sin( + 2) − 2 sin( + ) + sin 
2
sin  cos 2 + cos  sin 2 − 2 sin  cos  − 2 cos  sin  + sin 
= lim
→0
2 sin  (cos 2 − 2 cos  + 1) + cos  (sin 2 − 2 sin )
= lim
→0
2

ot

19. lim

Fo

to the -axis, the point  approaches the midpoint of the radius .

sin  (2 cos2  − 1 − 2 cos  + 1) + cos  (2 sin  cos  − 2 sin )
→0
2 sin  (2 cos )(cos  − 1) + cos  (2 sin )(cos  − 1)
= lim
→0
2
2(cos  − 1)[sin  cos  + cos  sin ](cos  + 1)
= lim
→0
2 (cos  + 1)

2
−2 sin2  [sin( + )] sin( + 0) sin  sin( + )
= −2 lim
= −2(1)2
= − sin 
·
= lim
→0
→0
2 (cos  + 1)

cos  + 1 cos 0 + 1

N

= lim

21.  = 4 − 22 − 

⇒  0 = 43 − 4 − 1. The equation of the tangent line at  =  is

 − (4 − 22 − ) = (43 − 4 − 1)( − ) or  = (43 − 4 − 1) + (−34 + 22 ) and similarly for  = . So if at
 =  and  =  we have the same tangent line, then 43 − 4 − 1 = 43 − 4 − 1 and −34 + 22 = −34 + 22 . The first equation gives 3 − 3 =  −  ⇒ ( − )(2 +  + 2 ) = ( − ). Assuming  6= , we have 1 = 2 +  + 2 .
The second equation gives 3(4 − 4 ) = 2(2 − 2 ) ⇒ 3(2 − 2 )(2 + 2 ) = 2(2 − 2 ) which is true if  = −.

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¤

CHAPTER 2 PROBLEMS PLUS

Substituting into 1 = 2 +  + 2 gives 1 = 2 − 2 + 2

109

⇒  = ±1 so that  = 1 and  = −1 or vice versa. Thus,

the points (1 −2) and (−1 0) have a common tangent line.
As long as there are only two such points, we are done. So we show that these are in fact the only two such points.
Suppose that 2 − 2 6= 0. Then 3(2 − 2 )(2 + 2 ) = 2(2 − 2 ) gives 3(2 + 2 ) = 2 or 2 + 2 = 2 .
3
Thus,  = (2 +  + 2 ) − (2 + 2 ) = 1 −

2
1
2
1
1
= , so  =
. Hence, 2 + 2 = , so 94 + 1 = 62
3
3
3
9
3
1
3

0 = 94 − 62 + 1 = (32 − 1)2 . So 32 − 1 = 0 ⇒ 2 =

⇒ 2 =

⇒

1
1
= = 2 , contradicting our assumption
92
3

that 2 6= 2 .
23. Because of the periodic nature of the lattice points, it suffices to consider the points in the 5 × 2 grid shown. We can see that

the minimum value of  occurs when there is a line with slope

2
5

which touches the circle centered at (3 1) and the circles

rS

al

e

centered at (0 0) and (5 2).

 = − 5  ⇒ 2 +
2

25
4

Fo

To find  , the point at which the line is tangent to the circle at (0 0), we simultaneously solve 2 +  2 = 2 and
2 = 2

⇒ 2 =

4
29

2

⇒ =

√2
29

,  = − √5 . To find , we either use symmetry or
29

solve ( − 3)2 + ( − 1)2 = 2 and  − 1 = − 5 ( − 3). As above, we get  = 3 −
2

√
√
29 + 50 = 6 29 − 8

N

5

1+

√5
29

√2
29

ot

the line   is 2 , so   =
5



 − − √5 
29

3−

−

⇔ 58 =

√2
29



=

1+
3−

√
29 ⇔  =

√

10
√
29
√4
29

29
.
58




=

√2
29

,  = 1 +

√
29 + 10
2
√
=
5
3 29 − 4

√5
29

. Now the slope of

⇒

So the minimum value of  for which any line with slope

intersects circles with radius  centered at the lattice points on the plane is  =

√

29
58

≈ 0093.


5

=
⇒ =
. The volume of the cone is
5
16
16
 2
5
25 2 
25 3

2
1
1
 , so
=

. Now the rate of
=
 = 3   = 3 
16
768

256

By similar triangles,

25.

change of the volume is also equal to the difference of what is being added
(2 cm3 min) and what is oozing out (, where  is the area of the cone and  is a proportionality constant). Thus,
Equating the two expressions for


= 2 − .



10

5(10)
25

= and substituting  = 10,
= −03,  =
=
, and √


16
8
16
281

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

⇔

2
5

¤

CHAPTER 2 PROBLEMS PLUS

√
125 281
750
=2+
. Solving for  gives us
64
256

=

5√
25
25 5 √
281, we get
281 ⇔
(10)2 (−03) = 2 − 
·
8
256
8 8

=

256 + 375
√
. To maintain a certain height, the rate of oozing, , must equal the rate of the liquid being poured in;
250 281

Fo

rS

al

e


= 0. Thus, the rate at which we should pour the liquid into the container is

√
256 + 375
25 5 281
256 + 375
√
 =
··
·
=
≈ 11204 cm3min
8
8
128
250 281

ot

that is,

N

110

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

3

APPLICATIONS OF DIFFERENTIATION

3.1 Maximum and Minimum Values
1. A function  has an absolute minimum at  =  if () is the smallest function value on the entire domain of , whereas

 has a local minimum at  if () is the smallest function value when  is near .
3. Absolute maximum at , absolute minimum at , local maximum at , local minima at  and , neither a maximum nor a

minimum at  and .

 (6) = 4; local minimum values are  (2) = 2 and (1) =  (5) = 3.

local maximum at 3, local minima at 2 and 4

Fo

rS

local minimum at 4

9. Absolute maximum at 5, absolute minimum at 2,

al

7. Absolute minimum at 2, absolute maximum at 3,

e

5. Absolute maximum value is (4) = 5; there is no absolute minimum value; local maximum values are  (4) = 5 and

(b)

(c)

N

ot

11. (a)

13. (a) Note: By the Extreme Value Theorem,

(b)

 must not be continuous; because if it were, it would attain an absolute minimum. c
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111

112

¤

CHAPTER 3

15.  () =

1
(3
2

APPLICATIONS OF DIFFERENTIATION

− 1),  ≤ 3. Absolute maximum

17.  () = 1,  ≥ 1. Absolute maximum  (1) = 1;

no local maximum. No absolute or local minimum.

 (3) = 4; no local maximum. No absolute or local minimum. 19.  () = sin , 0 ≤   2. No absolute or local

21.  () = sin , −2 ≤  ≤ 2. Absolute maximum

maximum. Absolute minimum (0) = 0; no local



rS

al

minimum.

e

= 1; no local maximum. Absolute minimum
2
 
 − 2 = −1; no local minimum.


25.  () = 1 −

Fo

23.  () = 1 + ( + 1)2 , −2 ≤   5. No absolute or local

no local maximum. No absolute or local minimum.

27.  () =



1−

N

ot

maximum. Absolute and local minimum  (−1) = 1.

√
. Absolute maximum  (0) = 1;

2 − 4

if 0 ≤   2

if 2 ≤  ≤ 3

Absolute maximum (3) = 2; no local maximum.

29.  () = 4 + 1  − 1 2
3
2

⇒  0 () =

1
3

− .

 0 () = 0 ⇒  = 1 . This is the only critical number.
3

No absolute or local minimum.

c
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SECTION 3.1

MAXIMUM AND MINIMUM VALUES

¤

113

⇒  0 () = 62 − 6 − 36 = 6(2 −  − 6) = 6( + 2)( − 3).

31.  () = 23 − 32 − 36

 0 () = 0 ⇔  = −2, 3. These are the only critical numbers.
⇒  0 () = 43 + 32 + 2 = (42 + 3 + 2). Using the quadratic formula, we see that

33. () = 4 + 3 + 2 + 1

42 + 3 + 2 = 0 has no real solution (its discriminant is negative), so  0 () = 0 only if  = 0. Hence, the only critical number is 0.
35. () =

 0 () =

−1
2 −  + 1

⇒

( 2 −  + 1)(1) − ( − 1)(2 − 1)
 2 −  + 1 − (22 − 3 + 1)
− 2 + 2
(2 − )
=
= 2
= 2
.
( 2 −  + 1)2
( 2 −  + 1)2
( −  + 1)2
( −  + 1)2

 0 () = 0 ⇒  = 0, 2. The expression  2 −  + 1 is never equal to 0, so  0 () exists for all real numbers.

√
=2

39.  () = 45 ( − 4)2

⇒

√
=

3

√
−2
√
.
4
4 3

al

0 () = 0 ⇒ 3

⇒ 0 () = 3 −14 − 2 −34 = 1 −34 (312 − 2) =
4
4
4

⇒  = 4 . 0 () does not exist at  = 0, so the critical numbers are 0 and 4 .
9
9

2
3

rS

37. () = 34 − 214

e

The critical numbers are 0 and 2.

⇒

 0 () = 45 · 2( − 4) + ( − 4)2 · 4 −15 = 1 −15 ( − 4)[5 ·  · 2 + ( − 4) · 4]
5
5
( − 4)(14 − 16)
2( − 4)(7 − 8)
=
515
515

Fo

=

 0 () = 0 ⇒  = 4, 8 .  0 (0) does not exist. Thus, the three critical numbers are 0, 8 , and 4.
7
7
41.  () = 2 cos  + sin2 

⇒  0 () = −2 sin  + 2 sin  cos .  0 () = 0 ⇒ 2 sin  (cos  − 1) = 0 ⇒ sin  = 0

ot

or cos  = 1 ⇒  =  [ an integer] or  = 2. The solutions  =  include the solutions  = 2, so the critical

N

numbers are  = .

43. A graph of  0 () = 1 +

210 sin  is shown. There are 10 zeros
2 − 6 + 10

between −25 and 25 (one is approximately −005).  0 exists everywhere, so  has 10 critical numbers.

45.  () = 12 + 4 − 2 , [0 5].

 0 () = 4 − 2 = 0 ⇔  = 2.  (0) = 12,  (2) = 16, and  (5) = 7.

So  (2) = 16 is the absolute maximum value and  (5) = 7 is the absolute minimum value.
47.  () = 23 − 32 − 12 + 1, [−2 3].

 0 () = 62 − 6 − 12 = 6(2 −  − 2) = 6( − 2)( + 1) = 0 ⇔

 = 2 −1.  (−2) = −3,  (−1) = 8,  (2) = −19, and  (3) = −8. So (−1) = 8 is the absolute maximum value and
 (2) = −19 is the absolute minimum value. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

114

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

49.  () = 34 − 43 − 122 + 1, [−2 3].

 0 () = 123 − 122 − 24 = 12(2 −  − 2) = 12( + 1)( − 2) = 0 ⇔

 = −1, 0, 2.  (−2) = 33,  (−1) = −4,  (0) = 1,  (2) = −31, and  (3) = 28. So (−2) = 33 is the absolute maximum value and (2) = −31 is the absolute minimum value.
1
1
2 − 1
( + 1)( − 1)
=
= 0 ⇔  = ±1, but  = −1 is not in the given
, [02 4].  0 () = 1 − 2 =


2
2

51.  () =  +

interval, [02 4].  0 () does not exist when  = 0, but 0 is not in the given interval, so 1 is the only critical nuumber.
 (02) = 52,  (1) = 2, and  (4) = 425. So  (02) = 52 is the absolute maximum value and  (1) = 2 is the absolute minimum value.
53.  () = 

√
4 − 2 , [−1 2].

√
√ 
2 = 2 is the absolute maximum value and  (−1) = − 3 is the absolute minimum value.

rS

 (2) = 0. So 

al

e

√
−2
−2 + (4 − 2 )
4 − 22
√
 0 () =  · 1 (4 − 2 )−12 (−2) + (4 − 2 )12 · 1 = √
+ 4 − 2 =
= √
.
2
2
2
4−
4−
4 − 2
√
√
 0 () = 0 ⇒ 4 − 22 = 0 ⇒ 2 = 2 ⇒  = ± 2, but  = − 2 is not in the given interval, [−1 2].
√
√ 
 0 () does not exist if 4 − 2 = 0 ⇒  = ±2, but −2 is not in the given interval.  (−1) = − 3,  2 = 2, and

55.  () = 2 cos  + sin 2, [0, 2].

So  (  ) =
6

3
2

Fo

 0 () = −2 sin  + cos 2 · 2 = −2 sin  + 2(1 − 2 sin2 ) = −2(2 sin2  + sin  − 1) = −2(2 sin  − 1)(sin  + 1).
√
√
√
 0 () = 0 ⇒ sin  = 1 or sin  = −1 ⇒  =  .  (0) = 2,  (  ) = 3 + 1 3 = 3 3 ≈ 260, and  (  ) = 0.
2
6
6
2
2
2
√
3 is the absolute maximum value and  (  ) = 0 is the absolute minimum value.
2

57.  () =  (1 − ) , 0 ≤  ≤ 1,   0,   0.

ot

 0 () =  · (1 − )−1 (−1) + (1 − ) · −1 = −1 (1 − )−1 [ · (−1) + (1 − ) · ]

N

= −1 (1 − )−1 ( −  − )

At the endpoints, we have (0) =  (1) = 0 [the minimum value of  ]. In the interval (0 1),  0 () = 0 ⇔  =





+

So 
59. (a)





=


+






+

=

 
1−


+



=


( + )



+−
+



=



+



 
·
=
.
( + ) ( + )
( + )+

  is the absolute maximum value.
( + )+
From the graph, it appears that the absolute maximum value is about
 (−077) = 219, and the absolute minimum value is about  (077) = 181.

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SECTION 3.1

MAXIMUM AND MINIMUM VALUES

¤

115


(b)  () = 5 − 3 + 2 ⇒  0 () = 54 − 32 = 2 (52 − 3). So  0 () = 0 ⇒  = 0, ± 3 .
5

        
5
3
 3 2  3
3
3
3
3
3
 − 5 = − 5 − − 5 +2=− 5
5 + 5
5 +2


 3

9
6
= 3 − 25
+ 2 = 25 3 + 2 (maximum)
5
5
5

 
3
6
= − 25 3 + 2 (minimum). and similarly, 
5
5

61. (a)

From the graph, it appears that the absolute maximum value is about
 (075) = 032, and the absolute minimum value is  (0) = (1) = 0; that is, at both endpoints.

√
 − 2

⇒  0 () =  ·

√
1 − 2
( − 22 ) + (2 − 22 )
3 − 42
√
√
+  − 2 =
= √
.
2  − 2
2  − 2
2  − 2

e

(b)  () = 

63. The density is defined as  =

mass
1000
=
(in gcm3 ). But a critical point of  will also be a critical point of  volume  ( )



= −1000 −2 and  is never 0], and  is easier to differentiate than .



Fo

[since

rS

al

So  0 () = 0 ⇒ 3 − 42 = 0 ⇒ (3 − 4) = 0 ⇒  = 0 or 3 .
4


√
 
 2
3
 (0) =  (1) = 0 (minimum), and  3 = 3 3 − 3 = 3 16 = 3163 (maximum).
4
4
4
4
4

 ( ) = 99987 − 006426 + 00085043 2 − 00000679 3

⇒  0 ( ) = −006426 + 00170086 − 00002037 2 .

ot

Setting this equal to 0 and using the quadratic formula to find  , we get
√
−00170086 ± 001700862 − 4 · 00002037 · 006426
 =
≈ 39665◦ C or 795318◦ C. Since we are only interested
2(−00002037)
in the region 0◦ C ≤  ≤ 30◦ C, we check the density  at the endpoints and at 39665◦ C: (0) ≈

1000
1000
≈ 099625; (39665) ≈
≈ 1000255. So water has its maximum density at
10037628
9997447

N

(30) ≈

1000
≈ 100013;
99987

about 39665◦ C.

65. Let  = −0000 032 37,  = 0000 903 7,  = −0008 956,  = 003629,  = −004458, and  = 04074.

Then () = 5 + 4 + 3 + 2 +  +  and  0 () = 54 + 43 + 32 + 2 + .
We now apply the Closed Interval Method to the continuous function  on the interval 0 ≤  ≤ 10. Since  0 exists for all , the only critical numbers of  occur when  0 () = 0. We use a rootfinder on a CAS (or a graphing device) to find that
 0 () = 0 when 1 ≈ 0855, 2 ≈ 4618, 3 ≈ 7292, and 4 ≈ 9570. The values of  at these critical numbers are
(1 ) ≈ 039, (2 ) ≈ 043645, (3 ) ≈ 0427, and (4 ) ≈ 043641. The values of  at the endpoints of the interval are
(0) ≈ 041 and (10) ≈ 0435. Comparing the six numbers, we see that sugar was most expensive at 2 ≈ 4618
(corresponding roughly to March 1998) and cheapest at 1 ≈ 0855 (June 1994).

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

116

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

⇒  0 () = 20  − 32 .  0 () = 0 ⇒ (20 − 3) = 0 ⇒




3
 = 0 or 2 0 (but 0 is not in the interval). Evaluating  at 1 0 , 2 0 , and 0 , we get  1 0 = 1 0 ,  2 0 =
3
2
3
2
8
3

67. (a) () = (0 − )2 = 0 2 − 3

and (0 ) = 0. Since

4
27

3
4
0 ,
27

 1 ,  attains its maximum value at  = 2 0 . This supports the statement in the text.
8
3

(b) From part (a), the maximum value of  is

69.  () = 101 + 51 +  + 1

3
4
0 .
27

(c)

⇒  0 () = 101100 + 5150 + 1 ≥ 1 for all , so  0 () = 0 has no solution. Thus,  ()

e

has no critical number, so  () can have no local maximum or minimum.

al

71. If  has a local minimum at , then () = − () has a local maximum at , so  0 () = 0 by the case of Fermat’s Theorem

proved in the text. Thus,  0 () = − 0 () = 0.

1.  () = 5 − 12 + 32 , [1, 3].

rS

3.2 The Mean Value Theorem

Since  is a polynomial, it is continuous and differentiable on R, so it is continuous on [1 3]

Fo

and differentiable on (1 3). Also  (1) = −4 =  (3).  0 () = 0 ⇔ −12 + 6 = 0 ⇔  = 2, which is in the open interval (1 3), so  = 2 satisfies the conclusion of Rolle’s Theorem.
3.  () =

√
 − 1 , [0 9].  , being the difference of a root function and a polynomial, is continuous and differentiable
3

on [0 ∞), so it is continuous on [0 9] and differentiable on (0 9). Also, (0) = 0 =  (9).  0 () = 0 ⇔
√
√
3
1
1
√ − =0 ⇔ 2 =3 ⇔
=
3
2


ot

2

⇒ =

9
9
, which is in the open interval (0 9), so  = satisfies the
4
4

N

conclusion of Rolle’s Theorem.
5.  () = 1 − 23 .

 (−1) = 1 − (−1)23 = 1 − 1 = 0 =  (1).  0 () = − 2 −13 , so  0 () = 0 has no solution. This
3

does not contradict Rolle’s Theorem, since  0 (0) does not exist, and so  is not differentiable on (−1 1).
7.  0 () =

 0 () =

3−0
3
 (8) −  (0)
=
= . It appears that
8−0
8
8

3
8

when  ≈ 03, 3, and 63.

9.  () = 22 − 3 + 1, [0 2].

differentiable on R.  0 () = is in (0 2)

 is continuous on [0 2] and differentiable on (0 2) since polynomials are continuous and
 () −  ()
−

⇔ 4 − 3 =

 (2) −  (0)
3−1
=
= 1 ⇔ 4 = 4 ⇔  = 1, which
2−0
2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.2

11.  () =

√
3
, [0 1].

THE MEAN VALUE THEOREM

323 = 1 ⇔ 23 =

1
1
 () −  ()
 (1) −  (0)
1−0
⇔
⇔
=
=
−
1−0
1
323
323

√
√
 1 3
1
1
⇔ 2 = 3 = 27 ⇔  = ± 27 = ± 93 , but only 93 is in (0 1).

1
3

√
 (4) −  (0)
, [0 4].  0 () =
4−0

1
1
√ =
2
2 

⇔

117

 is continuous on R and differentiable on (−∞ 0) ∪ (0 ∞), so  is continuous on [0 1]

and differentiable on (0 1).  0 () =

13.  () =

¤

⇔

1
2−0
√ =
4
2 

⇔

⇔

√
 = 1 ⇔  = 1. The secant line and the tangent line

are parallel.

1
1
−2
−
=
·3 ⇒
12
(−2)2
( − 3)3

⇒ ( − 3)3 = −8 ⇒  − 3 = −2 ⇒  = 1, which is not in the open interval (1 4). This does not

al

−6
3
=
4
( − 3)3

⇒  0 () = −2 ( − 3)−3 .  (4) −  (1) =  0 ()(4 − 1) ⇒

e

15.  () = ( − 3)−2

contradict the Mean Value Theorem since  is not continuous at  = 3.

rS

17. Let  () = 2 + cos . Then (−) = −2 − 1  0 and (0) = 1  0. Since  is the sum of the polynomial 2 and the

trignometric function cos ,  is continuous and differentiable for all . By the Intermediate Value Theorem, there is a number
 in (− 0) such that  () = 0. Thus, the given equation has at least one real root. If the equation has distinct real roots  and

Fo

 with   , then  () =  () = 0. Since  is continuous on [ ] and differentiable on ( ), Rolle’s Theorem implies that there is a number  in ( ) such that  0 () = 0. But  0 () = 2 − sin   0 since sin  ≤ 1. This contradiction shows that the given equation can’t have two distinct real roots, so it has exactly one root.

ot

19. Let  () = 3 − 15 +  for  in [−2 2]. If  has two real roots  and  in [−2 2], with   , then  () =  () = 0. Since

the polynomial  is continuous on [ ] and differentiable on ( ), Rolle’s Theorem implies that there is a number  in ( )

N

such that  0 () = 0. Now  0 () = 32 − 15. Since  is in ( ), which is contained in [−2 2], we have ||  2, so 2  4.
It follows that 32 − 15  3 · 4 − 15 = −3  0. This contradicts  0 () = 0, so the given equation can’t have two real roots in [−2 2]. Hence, it has at most one real root in [−2 2].
21. (a) Suppose that a cubic polynomial  () has roots 1  2  3  4 , so  (1 ) =  (2 ) =  (3 ) =  (4 ).

By Rolle’s Theorem there are numbers 1 , 2 , 3 with 1  1  2 , 2  2  3 and 3  3  4 and
 0 (1 ) =  0 (2 ) =  0 (3 ) = 0. Thus, the second-degree polynomial  0 () has three distinct real roots, which is impossible. (b) We prove by induction that a polynomial of degree  has at most  real roots. This is certainly true for  = 1. Suppose that the result is true for all polynomials of degree  and let  () be a polynomial of degree  + 1. Suppose that  () has more than  + 1 real roots, say 1  2  3  · · ·  +1  +2 . Then  (1 ) =  (2 ) = · · · =  (+2 ) = 0.
By Rolle’s Theorem there are real numbers 1      +1 with 1  1  2      +1  +1  +2 and

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

118

¤

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

 0 (1 ) = · · · =  0 (+1 ) = 0. Thus, the th degree polynomial  0 () has at least  + 1 roots. This contradiction shows that  () has at most  + 1 real roots.
23. By the Mean Value Theorem,  (4) −  (1) =  0 ()(4 − 1) for some  ∈ (1 4). But for every  ∈ (1 4) we have

 0 () ≥ 2. Putting  0 () ≥ 2 into the above equation and substituting  (1) = 10, we get
 (4) =  (1) +  0 ()(4 − 1) = 10 + 3 0 () ≥ 10 + 3 · 2 = 16. So the smallest possible value of  (4) is 16.
25. Suppose that such a function  exists. By the Mean Value Theorem there is a number 0    2 with

 0 () =

5
 (2) −  (0)
= . But this is impossible since  0 () ≤ 2 
2−0
2

27. We use Exercise 26 with  () =

5
2

for all , so no such function can exist.

√
1 + , () = 1 + 1 , and  = 0. Notice that  (0) = 1 = (0) and
2

√
1
1
√
 =  0 () for   0. So by Exercise 26,  ()  () ⇒
1 +   1 + 1  for   0.
2
2
1+
√
√
Another method: Apply the Mean Value Theorem directly to either  () = 1 + 1  − 1 +  or () = 1 +  on [0 ].
2
2

al

e

 0 () =

29. Let  () = sin  and let   . Then  () is continuous on [ ] and differentiable on ( ). By the Mean Value Theorem,

rS

there is a number  ∈ ( ) with sin  − sin  =  () −  () =  0 ()( − ) = (cos )( − ). Thus,
|sin  − sin | ≤ |cos | | − | ≤ | − |. If   , then |sin  − sin | = |sin  − sin | ≤ | − | = | − |. If  = , both sides of the inequality are 0.

Fo

31. For   0,  () = (), so  0 () = 0 (). For   0,  0 () = (1)0 = −12 and 0 () = (1 + 1)0 = −12 , so

again  0 () =  0 (). However, the domain of () is not an interval [it is (−∞ 0) ∪ (0 ∞)] so we cannot conclude that
 −  is constant (in fact it is not).

33. Let () and () be the position functions of the two runners and let  () = () − (). By hypothesis,

ot

 (0) = (0) − (0) = 0 and  () = () − () = 0, where  is the finishing time. Then by the Mean Value Theorem,

N

there is a time , with 0    , such that  0 () =

 () −  (0)
. But () = (0) = 0, so  0 () = 0. Since
−0

 0 () =  0 () − 0 () = 0, we have 0 () = 0 (). So at time , both runners have the same speed  0 () = 0 ().

3.3 How Derivatives Affect the Shape of a Graph
1. (a)  is increasing on (1 3) and (4 6).

(c)  is concave upward on (0 2).

(b)  is decreasing on (0 1) and (3 4).
(d)  is concave downward on (2 4) and (4 6).

(e) The point of inflection is (2 3).
3. (a) Use the Increasing/Decreasing (I/D) Test.

(b) Use the Concavity Test.

(c) At any value of  where the concavity changes, we have an inflection point at ( ()).
5. (a) Since  0 ()  0 on (1 5),  is increasing on this interval. Since  0 ()  0 on (0 1) and (5 6),  is decreasing on these

intervals. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.3

HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH

¤

119

(b) Since  0 () = 0 at  = 1 and  0 changes from negative to positive there,  changes from decreasing to increasing and has a local minimum at  = 1. Since  0 () = 0 at  = 5 and  0 changes from positive to negative there,  changes from increasing to decreasing and has a local maximum at  = 5.
7. (a) There is an IP at  = 3 because the graph of  changes from CD to CU there. There is an IP at  = 5 because the graph

of  changes from CU to CD there.
(b) There is an IP at  = 2 and at  = 6 because  0 () has a maximum value there, and so  00 () changes from positive to negative there. There is an IP at  = 4 because  0 () has a minimum value there and so  00 () changes from negative to positive there.
(c) There is an inflection point at  = 1 because  00 () changes from negative to positive there, and so the graph of  changes from concave downward to concave upward. There is an inflection point at  = 7 because  00 () changes from positive to

⇒  0 () = 62 + 6 − 36 = 6(2 +  − 6) = 6( + 3)( − 2).

al

9. (a)  () = 23 + 32 − 36

e

negative there, and so the graph of  changes from concave upward to concave downward.

We don’t need to include the “6” in the chart to determine the sign of  0 ().
−2

 0 ()

−

−

+

+

Interval

+

+

rS

+3

  −3

−3    2

+

2

−

−



increasing on (−∞ −3) decreasing on (−3 2) increasing on (2 ∞)

Fo

(b)  changes from increasing to decreasing at  = −3 and from decreasing to increasing at  = 2. Thus,  (−3) = 81 is a local maximum value and  (2) = −44 is a local minimum value.

ot

(c)  0 () = 62 + 6 − 36 ⇒  00 () = 12 + 6.  00 () = 0 at  = − 1 ,  00 ()  0 ⇔   − 1 , and
2
2

 1


00
1
 ()  0 ⇔   − 2 . Thus,  is concave upward on − 2  ∞ and concave downward on −∞ − 1 . There is an
2

  

inflection point at − 1   − 1 = − 1  37 .
2
2
2 2


⇒  0 () = 43 − 4 = 4 2 − 1 = 4( + 1)( − 1).

N

11. (a)  () = 4 − 22 + 3

Interval

+1



−1

 0 ()



  −1

−

−

−

−

decreasing on (−∞ −1)

01

+

+

+

−

decreasing on (0 1)

1

−

−1    0

+

−
+

−
+

+
+

increasing on (−1 0) increasing on (1 ∞)

(b)  changes from increasing to decreasing at  = 0 and from decreasing to increasing at  = −1 and  = 1. Thus,
 (0) = 3 is a local maximum value and  (±1) = 2 are local minimum values.
√ 
√
√
√ 



(c)  00 () = 122 − 4 = 12 2 − 1 = 12  + 1 3  − 1 3 .  00 ()  0 ⇔   −1 3 or   1 3 and
3
√
√
√


√

 00 ()  0 ⇔ −1 3    1 3. Thus,  is concave upward on −∞ − 33 and
33 ∞ and concave
√
 √

 √

downward on − 33 33 . There are inflection points at ± 33 22 .
9
c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

120

¤

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

13. (a)  () = sin  + cos , 0 ≤  ≤ 2.

 0 () = cos  − sin  = 0 ⇒ cos  = sin  ⇒ 1 =

sin  cos 

⇒

Thus,  0 ()  0 ⇔ cos  − sin   0 ⇔ cos   sin  ⇔ 0     or
4
 

 5
5
   2 and  0 ()  0 ⇔ cos   sin  ⇔     5 . So  is increasing on 0 4 and 4  2 and 
4
4
4


is decreasing on   5 .
4
4 tan  = 1 ⇒  =


4

or

5
4 .

(b)  changes from increasing to decreasing at  = local maximum value and 

 5 
4


4

and from decreasing to increasing at  =

3
4

or

√
2 is a

Thus, 

7
.
4

Divide the interval

√
= − 2 is a local minimum value.

(c)  00 () = − sin  − cos  = 0 ⇒ − sin  = cos  ⇒ tan  = −1 ⇒  =



5
.
4

4

=

(0 2) into subintervals with these numbers as endpoints and complete a second derivative chart.

15.  () = 1 + 32 − 23

4




 0 and 7  0 .
4

1
2

√
30

e

 00 () = 1  0


 00 11 = 1 −
6
2

Concavity downward upward

al

 3

 00 () = − sin  − cos 
 
 00  = −1  0
2

downward

rS

There are inflection points at

Interval
 3 
0 4
 3 7 
 4
4
 7

 2
4

⇒  0 () = 6 − 62 = 6(1 − ).

Fo

First Derivative Test:  0 ()  0 ⇒ 0    1 and  0 ()  0 ⇒   0 or   1. Since  0 changes from negative to positive at  = 0,  (0) = 1 is a local minimum value; and since  0 changes from positive to negative at  = 1, (1) = 2 is a local maximum value.

Second Derivative Test:  00 () = 6 − 12.  0 () = 0 ⇔  = 0 1.  00 (0) = 6  0 ⇒  (0) = 1 is a local

ot

minimum value.  00 (1) = −6  0 ⇒  (1) = 2 is a local maximum value.
Preference: For this function, the two tests are equally easy.

N

√
√
1
 − 4  ⇒  0 () = −12 −
2
√
First Derivative Test: 2 4  − 1  0 ⇒

17.  () =

√
1 −34
1
24 −1
√

= −34 (214 − 1) =
4
4
4
4 3



Since  0 changes from negative to positive at  =

1
,
16

1
,
16

so  0 ()  0 ⇒  

1
 ( 16 ) =

1
4

−

1
2

1
16

and  0 ()  0 ⇒ 0   

1
.
16

= − 1 is a local minimum value.
4

1
3
1
3
Second Derivative Test:  00 () = − −32 + −74 = − √ + √ .
4
4
16
4 3
16 7
1
1
1
 0 () = 0 ⇔  = 16 .  00 16 = −16 + 24 = 8  0 ⇒  16 = − 1 is a local minimum value.
4
Preference: The First Derivative Test may be slightly easier to apply in this case.

19. (a) By the Second Derivative Test, if  0 (2) = 0 and  00 (2) = −5  0,  has a local maximum at  = 2.

(b) If  0 (6) = 0, we know that  has a horizontal tangent at  = 6. Knowing that  00 (6) = 0 does not provide any additional information since the Second Derivative Test fails. For example, the first and second derivatives of  = ( − 6)4 , c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.3

HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH

¤

121

 = −( − 6)4 , and  = ( − 6)3 all equal zero for  = 6, but the first has a local minimum at  = 6, the second has a local maximum at  = 6, and the third has an inflection point at  = 6.
⇒ horizontal tangents at  = 0, 2, 4.

21.  0 (0) =  0 (2) =  0 (4) = 0

 0 ()  0 if   0 or 2    4 ⇒  is increasing on (−∞ 0) and (2 4).
 0 ()  0 if 0    2 or   4 ⇒  is decreasing on (0 2) and (4 ∞).
 00 ()  0 if 1    3 ⇒  is concave upward on (1 3).
 00 ()  0 if   1 or   3 ⇒  is concave downward on (−∞ 1) and (3 ∞). There are inflection points when  = 1 and 3.
23.  0 ()  0 if ||  2

⇒  is increasing on (−2 2).

 0 (−2) = 0 ⇒ horizontal tangent at

 = −2. lim | 0 ()| = ∞ ⇒ there is a vertical
→2

al

and (2 ∞).

e

 0 ()  0 if ||  2 ⇒  is decreasing on (−∞ −2)

rS

asymptote or vertical tangent (cusp) at  = 2.  00 ()  0 if  6= 2 ⇒  is concave upward on (−∞ 2) and (2 ∞).
25. The function must be always decreasing (since the first derivative is always

negative) and concave downward (since the second derivative is always

Fo

negative).

ot

27. (a)  is increasing where  0 is positive, that is, on (0 2), (4 6), and (8 ∞); and decreasing where  0 is negative, that is, on

(2 4) and (6 8).

N

(b)  has local maxima where  0 changes from positive to negative, at  = 2 and at  = 6, and local minima where  0 changes from negative to positive, at  = 4 and at  = 8.
(c)  is concave upward (CU) where  0 is increasing, that is, on (3 6) and (6 ∞), and concave downward (CD) where  0 is decreasing, that is, on (0 3).
(d) There is a point of inflection where  changes from

(e)

being CD to being CU, that is, at  = 3.

29. (a)  () = 3 − 12 + 2

⇒  0 () = 32 − 12 = 3(2 − 4) = 3( + 2)( − 2).  0 ()  0 ⇔   −2 or   2

and  0 ()  0 ⇔ −2    2. So  is increasing on (−∞ −2) and (2 ∞) and  is decreasing on (−2 2).

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

122

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(b)  changes from increasing to decreasing at  = −2, so  (−2) = 18 is a local maximum value.  changes from decreasing to increasing at  = 2, so  (2) = −14 is a local minimum value.
(c)  00 () = 6.  00 () = 0 ⇔  = 0.  00 ()  0

(d)

on (0 ∞) and  00 ()  0 on (−∞ 0). So  is concave upward on (0 ∞) and  is concave downward on (−∞ 0). There is an inflection point at (0 2).

31. (a)  () = 2 + 22 − 4

⇒  0 () = 4 − 43 = 4(1 − 2 ) = 4(1 + )(1 − ).  0 ()  0 ⇔   −1 or

0    1 and  0 ()  0 ⇔ −1    0 or   1. So  is increasing on (−∞ −1) and (0 1) and  is decreasing

e

on (−1 0) and (1 ∞).

al

(b)  changes from increasing to decreasing at  = −1 and  = 1, so  (−1) = 3 and  (1) = 3 are local maximum values.
 changes from decreasing to increasing at  = 0, so  (0) = 2 is a local minimum value.
(d)

Fo

rS

(c)  00 () = 4 − 122 = 4(1 − 32 ).  00 () = 0 ⇔ 1 − 32 = 0 ⇔
√
√
√ 

2 = 1 ⇔  = ±1 3.  00 ()  0 on −1 3 1 3 and  00 ()  0
3
√ 
 √

 on −∞ −1 3 and 1 3 ∞ . So  is concave upward on
√
√ 
√ 


−1 3 1 3 and  is concave downward on −∞ −1 3 and
√ 
 
 √
1 3 ∞ .  ±1 3 = 2 + 2 − 1 = 23 . There are points of inflection
3
9
9
√ 23 

at ±1 3 9 .

⇒ 0 () = 5( + 1)4 − 5. 0 () = 0 ⇔ 5( + 1)4 = 5 ⇔ ( + 1)4 = 1 ⇒

ot

33. (a) () = ( + 1)5 − 5 − 2

( + 1)2 = 1 ⇒  + 1 = 1 or  + 1 = −1 ⇒  = 0 or  = −2. 0 ()  0 ⇔   −2 or   0 and

N

0 ()  0 ⇔ −2    0. So  is increasing on (−∞ −2) and (0 ∞) and  is decreasing on (−2 0).
(b) (−2) = 7 is a local maximum value and (0) = −1 is a local minimum value.

(d)

(c) 00 () = 20( + 1)3 = 0 ⇔  = −1. 00 ()  0 ⇔   −1 and
00 ()  0 ⇔   −1, so  is CU on (−1 ∞) and  is CD on (−∞ −1).
There is a point of inflection at (−1 (−1)) = (−1 3).

√

35. (a)  () =  6 − 

⇒

 0 () =  · 1 (6 − )−12 (−1) + (6 − )12 (1) = 1 (6 − )−12 [− + 2(6 − )] =
2
2

−3 + 12
√
.
2 6−

 0 ()  0 ⇔ −3 + 12  0 ⇔   4 and  0 ()  0 ⇔ 4    6. So  is increasing on (−∞ 4) and  is decreasing on (4 6). c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.3

HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH

¤

123

√
(b)  changes from increasing to decreasing at  = 4, so  (4) = 4 2 is a local maximum value. There is no local minimum value. (c)  0 () = − 3 ( − 4)(6 − )−12 ⇒
2




00
 () = − 3 ( − 4) − 1 (6 − )−32 (−1) + (6 − )−12 (1)
2
2

(d)

3 1
3( − 8)
= − · (6 − )−32 [( − 4) + 2(6 − )] =
2 2
4(6 − )32

 00 ()  0 on (−∞ 6), so  is CD on (−∞ 6). There is no inflection point.
⇒  0 () = 4 13 + 4 −23 = 4 −23 ( + 1) =
3
3
3

37. (a) () = 13 ( + 4) = 43 + 413

4( + 1)
√
.  0 ()  0 if
3
3 2

−1    0 or   0 and  0 ()  0 for   −1, so  is increasing on (−1 ∞) and  is decreasing on (−∞ −1).
(d)
4( − 2)
√
.
3
9 5

al

(c)  00 () = 4 −23 − 8 −53 = 4 −53 ( − 2) =
9
9
9

e

(b) (−1) = −3 is a local minimum value.

 00 ()  0 for 0    2 and  00 ()  0 for   0 and   2, so  is

39. (a)  () = 2 cos  + cos2 , 0 ≤  ≤ 2

rS

concave downward on (0 2) and concave upward on (−∞ 0) and (2 ∞).
√ 

There are inflection points at (0 0) and 2 6 3 2 ≈ (2 756).

⇒  0 () = −2 sin  + 2 cos  (− sin ) = −2 sin  (1 + cos ).

Fo

 0 () = 0 ⇔  = 0  and 2.  0 ()  0 ⇔     2 and  0 ()  0 ⇔ 0    . So  is increasing on ( 2) and  is decreasing on (0 ).
(b)  () = −1 is a local minimum value.

ot

(c)  0 () = −2 sin  (1 + cos ) ⇒

 00 () = −2 sin  (− sin ) + (1 + cos )(−2 cos ) = 2 sin2  − 2 cos  − 2 cos2 

N

= 2(1 − cos2 ) − 2 cos  − 2 cos2  = −4 cos2  − 2 cos  + 2
= −2(2 cos2  + cos  − 1) = −2(2 cos  − 1)(cos  + 1)

(d)

Since −2(cos  + 1)  0 [for  6= ],  00 ()  0 ⇒ 2 cos  − 1  0 ⇒ cos   1 ⇒     5 and
2
3
3
  5 
 
00
1

5
 ()  0 ⇒ cos   2 ⇒ 0    3 or 3    2. So  is CU on 3  3 and  is CD on 0 3 and



       5 
  
 5
= 3  4 and 5   5 = 5  5 .
3  2 . There are points of inflection at 3   3
3
3
3 4

41. The nonnegative factors ( + 1)2 and ( − 6)4 do not affect the sign of  0 () = ( + 1)2 ( − 3)5 ( − 6)4 .

So  0 ()  0 ⇒ ( − 3)5  0 ⇒  − 3  0 ⇒   3. Thus,  is increasing on the interval (3 ∞). c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

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APPLICATIONS OF DIFFERENTIATION

43. (a)

From the graph, we get an estimate of (1) ≈ 141 as a local maximum value, and no local minimum value.
1−
.
(2 + 1)32
√
2
 0 () = 0 ⇔  = 1.  (1) = √2 = 2 is the exact value.

+1
 () = √
2 + 1

⇒  0 () =

(b) From the graph in part (a),  increases most rapidly somewhere between  = − 1 and  = − 1 . To find the exact value,
2
4 we need to find the maximum value of  0 , which we can do by finding the critical numbers of  0 .
√
√
3 + 17
22 − 3 − 1
3 ± 17
 00 () =
. = corresponds to the minimum value of  0 .
=0 ⇔ =
4
4
(2 + 1)52
The maximum value of  0 occurs at  =
1
2

√
4

17

≈ −028.

cos 2 ⇒  0 () = − sin  − sin 2 ⇒  00 () = − cos  − 2 cos 2

e

45.  () = cos  +

3−

(a)

al

From the graph of  , it seems that  is CD on (0 1), CU on (1 25), CD on
(25 37), CU on (37 53), and CD on (53 2). The points of inflection

rS

appear to be at (1 04), (25 −06), (37 −06), and (53 04).

(b)

From the graph of  00 (and zooming in near the zeros), it seems that  is CD

Fo

on (0 094), CU on (094 257), CD on (257 371), CU on (371 535),

and CD on (535 2). Refined estimates of the inflection points are

ot

(094 044), (257 −063), (371 −063), and (535 044).

4 + 3 + 1
. In Maple, we define  and then use the command
2 +  + 1

47.  () = √

N

plot(diff(diff(f,x),x),x=-2..2);. In Mathematica, we define  and then use Plot[Dt[Dt[f,x],x],{x,-2,2}]. We see that  00  0 for
  −06 and   00 [≈ 003] and  00  0 for −06    00. So  is CU on (−∞ −06) and (00 ∞) and CD on (−06 00).
49. (a) The rate of increase of the population is initially very small, then gets larger until it reaches a maximum at about

 = 8 hours, and decreases toward 0 as the population begins to level off.
(b) The rate of increase has its maximum value at  = 8 hours.
(c) The population function is concave upward on (0 8) and concave downward on (8 18).
(d) At  = 8, the population is about 350, so the inflection point is about (8 350).

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.3

HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH

¤

125

51. Most students learn more in the third hour of studying than in the eighth hour, so (3) − (2) is larger than (8) − (7).

In other words, as you begin studying for a test, the rate of knowledge gain is large and then starts to taper off, so  0 () decreases and the graph of  is concave downward.
⇒  0 () = 32 + 2 + .

53.  () = 3 + 2 +  + 

We are given that (1) = 0 and  (−2) = 3, so  (1) =  +  +  +  = 0 and
 (−2) = −8 + 4 − 2 +  = 3. Also  0 (1) = 3 + 2 +  = 0 and
 0 (−2) = 12 − 4 +  = 0 by Fermat’s Theorem. Solving these four equations, we get


 = 2 ,  = 1 ,  = − 4 ,  = 7 , so the function is  () = 1 23 + 32 − 12 + 7 .
9
3
3
9
9
1
 0 ( √3 ) = 0 ⇒ 1 +

√
√
⇒  0 () = 32 + 2 + .  has the local minimum value − 2 3 at  = 1 3, so
9

2
√ 
3

√
1
 ( √3 ) = − 2 3
9

+  = 0 (1) and

⇒

1
9

√
√
√
3 + 1  + 1 3 = − 2 3 (2).
3
3
9

e

55. (a)  () = 3 + 2 + 

2
3

al

Rewrite the system of equations as
√
3 +
+

(3)

(4)

rS

1

3

 = −1
√
√
1
3 = − 1 3
3
3

√ and then multiplying (4) by −2 3 gives us the system

√
3 +  = −1
√
− 2 3 − 2 = 2
3

Fo

2
3

Adding the equations gives us − = 1 ⇒  = −1. Substituting −1 for  into (3) gives us
√
√
2
3 − 1 = −1 ⇒ 2 3 = 0 ⇒  = 0. Thus,  () = 3 − .
3
3

ot

(b) To find the smallest slope, we want to find the minimum of the slope function,  0 , so we’ll find the critical numbers of  0 .  () = 3 −  ⇒  0 () = 32 − 1 ⇒  00 () = 6.  00 () = 0 ⇔  = 0.

N

At  = 0,  = 0,  0 () = −1, and  00 changes from negative to positive. Thus, we have a minimum for  0 and
 − 0 = −1( − 0), or  = −, is the tangent line that has the smallest slope.
57.  =  sin 

⇒  0 =  cos  + sin  ⇒ 00 = − sin  + 2 cos .  00 = 0 ⇒ 2 cos  =  sin  [which is ] ⇒

(2 cos )2 = ( sin )2 cos2 (4 + 2 ) = 2

⇒ 4 cos2  = 2 sin2  ⇒ 4 cos2  = 2 (1 − cos2 ) ⇒ 4 cos2  + 2 cos2  = 2
⇒ 4 cos2 (2 + 4) = 42

⇒

⇒  2 (2 + 4) = 42 since  = 2 cos  when 00 = 0.

59. (a) Since  and  are positive, increasing, and CU on  with  00 and  00 never equal to 0, we have   0,  0 ≥ 0,  00  0,

  0, 0 ≥ 0,  00  0 on . Then ( )0 =  0  +   0

⇒ ()00 =  00  + 2 0  0 +   00 ≥  00  +  00  0 on 

⇒

  is CU on .
(b) In part (a), if  and  are both decreasing instead of increasing, then  0 ≤ 0 and  0 ≤ 0 on , so we still have 2 0  0 ≥ 0 on . Thus, ( )00 =  00  + 2 0 0 +  00 ≥  00  +   00  0 on 

⇒   is CU on  as in part (a).

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

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APPLICATIONS OF DIFFERENTIATION

(c) Suppose  is increasing and  is decreasing [with  and  positive and CU]. Then  0 ≥ 0 and 0 ≤ 0 on , so 2 0 0 ≤ 0 on  and the argument in parts (a) and (b) fails.
Example 1.

 = (0 ∞),  () = 3 , () = 1. Then ( )() = 2 , so ( )0 () = 2 and
()00 () = 2  0 on . Thus,   is CU on .

Example 2.

Example 3.

√
√
√
 = (0 ∞),  () = 4 , () = 1. Then ( )() = 4 , so ()0 () = 2  and
√
()00 () = −1 3  0 on . Thus,   is CD on .
 = (0 ∞),  () = 2 , () = 1. Thus, ( )() = , so   is linear on .

61.  () = tan  − 

⇒  0 () = sec2  − 1  0 for 0   



on 0  . Thus,  ()   (0) = 0 for 0   
2


2


2

since sec2   1 for 0   


2.

So  is increasing

⇒ tan  −   0 ⇒ tan    for 0   


2.

⇒  0 () = 32 + 2 +  ⇒  00 () = 6 + 2.

e

63. Let the cubic function be () = 3 + 2 +  + 

al

So  is CU when 6 + 2  0 ⇔   −(3), CD when   −(3), and so the only point of inflection occurs when  = −(3). If the graph has three -intercepts 1 , 2 and 3 , then the expression for  () must factor as

rS

 () = ( − 1 )( − 2 )( − 3 ). Multiplying these factors together gives us

 () = [3 − (1 + 2 + 3 )2 + (1 2 + 1 3 + 2 3 ) − 1 2 3 ]
Equating the coefficients of the 2 -terms for the two forms of  gives us  = −(1 + 2 + 3 ). Hence, the -coordinate of
−(1 + 2 + 3 )
1 + 2 + 3

=−
=
.
3
3
3

Fo

the point of inflection is −

65. By hypothesis  =  0 is differentiable on an open interval containing . Since (  ()) is a point of inflection, the concavity

changes at  = , so  00 () changes signs at  = . Hence, by the First Derivative Test,  0 has a local extremum at  = .

√
√
2 , we have that () =  || =  2

N

67. Using the fact that || =

ot

Thus, by Fermat’s Theorem  00 () = 0.

⇒ 0 () =

√
√
√
2 + 2 = 2 2 = 2 || ⇒

 −12
2
 0 for   0 and 00 ()  0 for   0, so (0 0) is an inflection point. But  00 (0) does not
=
 00 () = 2 2
||

exist.

69. Suppose that  is differentiable on an interval  and  0 ()  0 for all  in  except  = . To show that  is increasing on ,

let 1 , 2 be two numbers in  with 1  2 .
Case 1 1  2  . Let  be the interval { ∈  |   }. By applying the Increasing/Decreasing Test to  on , we see that  is increasing on , so  (1 )   (2 ).
Case 2   1  2 . Apply the Increasing/Decreasing Test to  on  = { ∈  |   }.
Case 3 1  2 = . Apply the proof of the Increasing/Decreasing Test, using the Mean Value Theorem (MVT) on the interval [1  2 ] and noting that the MVT does not require  to be differentiable at the endpoints of [1  2 ]. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.3

HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH

¤

127

Case 4  = 1  2 . Same proof as in Case 3.
Case 5 1    2 . By Cases 3 and 4,  is increasing on [1  ] and on [ 2 ], so  (1 )   ()  (2 ).
In all cases, we have shown that  (1 )  (2 ). Since 1 , 2 were any numbers in  with 1  2 , we have shown that  is increasing on .
71. (a)  () = 4 sin

1


⇒  0 () = 4 cos

1




1
1
1
1
− 2 + sin (43 ) = 43 sin − 2 cos .







1
= 24 + () ⇒  0 () = 83 +  0 ().
() = 4 2 + sin



1
= −24 +  () ⇒ 0 () = −83 +  0 ().
() = 4 −2 + sin


1
4 sin − 0
 () −  (0)
1

= lim
= lim 3 sin . Since
It is given that  (0) = 0, so  (0) = lim
→0
→0
→0
−0



e

0

al

 
 
 
1
− 3  ≤ 3 sin ≤ 3  and lim 3  = 0, we see that  0 (0) = 0 by the Squeeze Theorem. Also,
→0

For 2 =

rS

 0 (0) = 8(0)3 +  0 (0) = 0 and 0 (0) = −8(0)3 +  0 (0) = 0, so 0 is a critical number of  , , and .
1
1
1
[ a nonzero integer], sin
= sin 2 = 0 and cos
= cos 2 = 1, so  0 (2 ) = −2  0.
2
2
2
2
1
1
1
, sin
= sin(2 + 1) = 0 and cos
= cos(2 + 1) = −1, so
(2 + 1)
2+1
2+1

Fo

For 2+1 =

0
 0 (2+1 ) = 2
2+1  0. Thus,  changes sign infinitely often on both sides of 0.

Next,  0 (2 ) = 83 +  0 (2 ) = 83 − 2 = 2 (82 − 1)  0 for 2  1 , but
2
2
2
2
8

sides of 0.

ot

2
2
0
1
 0 (2+1 ) = 83
2+1 + 2+1 = 2+1 (82+1 + 1)  0 for 2+1  − 8 , so  changes sign infinitely often on both

Last, 0 (2 ) = −83 +  0 (2 ) = −83 − 2 = −2 (82 + 1)  0 for 2  − 1 and
2
2
2
2
8

N

2
2
0
1
0 (2+1 ) = −83
2+1 + 2+1 = 2+1 (−82+1 + 1)  0 for 2+1  8 , so  changes sign infinitely often on both

sides of 0.

(b)  (0) = 0 and since sin

1
1
and hence 4 sin is both positive and negative inifinitely often on both sides of 0, and



arbitrarily close to 0,  has neither a local maximum nor a local minimum at 0.


1
1
Since 2 + sin ≥ 1, () = 4 2 + sin
 0 for  6= 0, so (0) = 0 is a local minimum.




1
1
 0 for  6= 0, so (0) = 0 is a local maximum.
Since −2 + sin ≤ −1, () = 4 −2 + sin



c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

128

¤

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

3.4 Limits at Infinity; Horizontal Asymptotes
1. (a) As  becomes large, the values of  () approach 5.

(b) As  becomes large negative, the values of  () approach 3.
3. (a) lim  () = −2

(b) lim  () = 2

→∞

(c) lim  () = ∞

→−∞

→1

(e) Vertical:  = 1,  = 3; horizontal:  = −2,  = 2

(d) lim  () = −∞
→3

5. If  () = 22 , then a calculator gives (0) = 0,  (1) = 05, (2) = 1,  (3) = 1125,  (4) = 1,  (5) = 078125,

 (6) = 05625,  (7) = 03828125,  (8) = 025,  (9) = 0158203125, (10) = 009765625,  (20) ≈ 000038147,
 (50) ≈ 22204 × 10−12 ,  (100) ≈ 78886 × 10−27 .


It appears that lim 22 = 0.
→∞

=

[Limit Law 5]

lim (2 + 5 − 82 )

lim 3 − lim (1) + lim (42 )

→∞

→∞

→∞

lim 2 + lim (5) − lim (82 )

→∞

→∞

→∞

3 − lim (1) + 4 lim (12 )
→∞

3
2

[Theorem 5 of Section 1.8]

lim 3 − 2 lim 1
3 − 2
(3 − 2)
3 − 2
3 − 2(0)
3
→∞
→∞
= lim
= lim
=
=
=
2 + 1 →∞ (2 + 1) →∞ 2 + 1 lim 2 + lim 1
2+0
2

N

→∞

3 − 0 + 4(0)
2 + 5(0) − 8(0)

=

[Limit Laws 7 and 3]

→∞

ot

=

9. lim

→∞

2 + 5 lim (1) − 8 lim (12 )
→∞

[Limit Laws 1 and 2]

Fo

=

e

lim (3 − 1 + 42 )

→∞
→∞

=

al

→∞

[divide both the numerator and denominator by 2
(the highest power of  that appears in the denominator)]

32 −  + 4
(32 −  + 4)2
= lim
2 + 5 − 8
→∞ (22 + 5 − 8)2
2

rS

7. lim

→∞

−2
( − 2)2
1 − 22
= lim
11. lim
= lim
=
→−∞ 2 + 1
→−∞ (2 + 1)2
→−∞ 1 + 12

→∞

lim 1 − 2 lim 12

→−∞

→−∞

lim 1 + lim 12

→−∞

→−∞

=

0 − 2(0)
=0
1+0

√
√
 + 2
(  + 2 )2
132 + 1
0+1
13. lim
= lim
= lim
=
= −1
2
2 )2
→∞ 2 − 
→∞ (2 − 
→∞ 2 − 1
0−1
(22 + 1)2
(22 + 1)2 4
[(22 + 1)2 ]2
= lim
= lim
→∞ ( − 1)2 (2 + )
→∞ [( − 1)2 (2 + )]4
→∞ [(2 − 2 + 1)2 ][(2 + )2 ]

15. lim

(2 + 12 )2
(2 + 0)2
=
=4
2 )(1 + 1)
→∞ (1 − 2 + 1
(1 − 0 + 0)(1 + 0)

= lim

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.4

√
√
96 − 
96 −  3
=
= lim
3 +1
→∞
→∞ (3 + 1)3


17. lim

lim

=

→∞

lim 1 + lim (13 )

→∞

19. lim

→∞


9 − 15


(96 − )6 lim →∞

lim (1 +

→∞

=

→∞



[since 3 =

13 )

lim 9 − lim (15 )

→∞

→∞

1+0

LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES

=

¤

129

√
6 for   0]

√
9−0 =3

√

2
√
√
92 +  − 3
92 +  + 3
92 +  − (3)2
√
√
= lim
→∞
→∞
92 +  + 3
92 +  + 3
 2

9 +  − 92

1
= lim √
·
= lim √
→∞
→∞
92 +  + 3
92 +  + 3 1


√
92 +  − 3 = lim


1
1
1
1
= lim 
= lim 
= √
=
=
→∞
3+3
6
9+3
92 2 + 2 + 3 →∞ 9 + 1 + 3
→∞

e

→∞

√
√
√
 √

2 +  − 2 + 
2 +  + 2 + 
√
√
2 +  + 2 + 

al

√
√

21. lim
2 +  − 2 +  = lim

rS

(2 + ) − (2 + )
[( − )]
√
= lim √
= lim √
√
 √
2 +  +
2 + 
2 +  +
→∞
→∞



2 +   2
−
−
−

√
= √
=
= lim 
→∞
2
1+0+ 1+0
1 +  + 1 + 

→∞



divide by the highest power of  in the denominator

Fo

4 − 32 + 
(4 − 32 + )3
= lim
→∞ (3 −  + 2)3
3 −  + 2

23. lim



 − 3 + 12
=∞
→∞ 1 − 12 + 23

= lim

since the numerator increases without bound and the denominator approaches 1 as  → ∞.
1
lim (4 + 5 ) = lim 5 (  + 1) [factor out the largest power of ] = −∞ because 5 → −∞ and 1 + 1 → 1

→−∞

→−∞

as  → −∞.
Or:

lim

27. lim

→∞

 4

 + 5 = lim 4 (1 + ) = −∞.
→−∞

N

→−∞

ot

25.



√ √
√
√
√ 
 − 1 = ∞ since  → ∞ and  − 1 → ∞ as  → ∞.
 −  = lim 

29. If  =

→∞

1 sin 
1
1
, then lim  sin = lim sin  = lim
= 1.
→∞

 →0+ 

→0+

31. (a)

(b)


()

−10,000

−04999625

−1,000,000

−04999996

−100,000

From the graph of  () =

√
2 +  + 1 + , we

−04999962

From the table, we estimate the limit to be −05.

estimate the value of lim  () to be −05.
→−∞

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

130

¤

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

 2


√ 2
√
√


 +  + 1 − 2
 ++1−
2 +  + 1 +  = lim
2 +  + 1 +  √
= lim √
→−∞
→−∞
→−∞
2 +  + 1 − 
2 +  + 1 − 

(c) lim

( + 1)(1)
1 + (1)


= lim √
= lim
→−∞
2 +  + 1 −  (1) →−∞ − 1 + (1) + (12 ) − 1

1
1+0
√
=−
2
− 1+0+0−1
√
Note that for   0, we have 2 = || = −, so when we divide the radical by , with   0, we get

1 √ 2
1√ 2
 +  + 1 = −√
 +  + 1 = − 1 + (1) + (12 ).
2




1
1
1
2 + 1 lim 2 + lim 2 + lim
2+

2 + 1
→∞
→∞ 

 = →∞ 
 =
= lim
33. lim
= lim
2
2
→∞  − 2
→∞  − 2
→∞
2
1−
lim 1 − lim lim 1 −
→∞
→∞ 


→∞

=

2+0
= 2, so  = 2 is a horizontal asymptote.
1−0

investigate  =  () =

2 + 1 as  approaches 2.
−2

al

The denominator  − 2 is zero when  = 2 and the numerator is not zero, so we

e

=

lim  () = −∞ because as

→2−

rS

 → 2− the numerator is positive and the denominator approaches 0 through

negative values. Similarly, lim  () = ∞. Thus,  = 2 is a vertical asymptote.
→2+

The graph confirms our work.

22 +  − 1
(2 − 1)( + 1)
=
, so lim () = ∞,
2 +  − 2
( + 2)( − 1)
→−2−

N

 =  () =

ot

Fo



1
1
22 +  − 1
1
1 lim 2 + − 2
2+ − 2


22 +  − 1
2

 = →∞ 

= lim
35. lim
= lim
2
1
2
→∞ 2 +  − 2
→∞  +  − 2
→∞
1
2
1+ − 2 lim 1 + − 2


→∞


2
1
1
− lim 2 lim 2 + lim
2+0−0
→∞
→∞ 
→∞ 
= 2, so  = 2 is a horizontal asymptote.
=
=
1
1
1 + 0 − 2(0) lim 1 + lim
− 2 lim 2
→∞
→∞ 
→∞ 

lim  () = −∞, lim  () = −∞, and lim  () = ∞. Thus,  = −2
→1−

→−2+

→1+

and  = 1 are vertical asymptotes. The graph confirms our work.
37.  =  () =

2

(2 − 1)
( + 1)( − 1)
( + 1)
3 − 
=
=
=
= () for  6= 1.
− 6 + 5
( − 1)( − 5)
( − 1)( − 5)
−5

The graph of  is the same as the graph of  with the exception of a hole in the graph of  at  = 1. By long division, () =

2 + 
30
=+6+
.
−5
−5

As  → ±∞, () → ±∞, so there is no horizontal asymptote. The denominator of  is zero when  = 5. lim () = −∞ and lim () = ∞, so  = 5 is a
→5−

→5+

vertical asymptote. The graph confirms our work. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.4

LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES

¤

131

39. From the graph, it appears  = 1 is a horizontal asymptote.

33 + 5002
3 + 500
3 + (500)
3
= lim 3 lim = lim
→∞ 3 + 5002 + 100 + 2000
→∞  + 5002 + 100 + 2000
→∞ 1 + (500) + (1002 ) + (20003 )
3
3+0
= 3, so  = 3 is a horizontal asymptote.
=
1+0+0+0
3

2

The discrepancy can be explained by the choice of the viewing window. Try
[−100,000 100,000] by [−1 4] to get a graph that lends credibility to our calculation that  = 3 is a horizontal asymptote.

lim () = 0 ⇒ degree of numerator  degree of denominator

→±∞

al

(1)

e

41. Let’s look for a rational function.

(2) lim  () = −∞ ⇒ there is a factor of 2 in the denominator (not just , since that would produce a sign
→0

rS

change at  = 0), and the function is negative near  = 0.

(3) lim  () = ∞ and lim () = −∞ ⇒ vertical asymptote at  = 3; there is a factor of ( − 3) in the
→3−

→3+

denominator.

Fo

(4)  (2) = 0 ⇒ 2 is an -intercept; there is at least one factor of ( − 2) in the numerator.
Combining all of this information and putting in a negative sign to give us the desired left- and right-hand limits gives us
2−
as one possibility.
2 ( − 3)

ot

 () =

43. (a) We must first find the function  . Since  has a vertical asymptote  = 4 and -intercept  = 1,  − 4 is a factor of the

N

denominator and  − 1 is a factor of the numerator. There is a removable discontinuity at  = −1, so  − (−1) =  + 1 is a factor of both the numerator and denominator. Thus,  now looks like this:  () =

be determined. Then lim  () = lim
→−1

 = 5. Thus  () =

 (0) =

→−1

( − 1)( + 1)
, where  is still to
( − 4)( + 1)

( − 1)( + 1)
( − 1)
(−1 − 1)
2
2
= lim
=
= , so  = 2, and
→−1  − 4
( − 4)( + 1)
(−1 − 4)
5
5

5( − 1)( + 1) is a ratio of quadratic functions satisfying all the given conditions and
( − 4)( + 1)

5
5(−1)(1)
= .
(−4)(1)
4
2 − 1
(2 2 ) − (12 )
1−0
= 5 lim
=5
= 5(1) = 5
→∞ 2 − 3 − 4
→∞ (2 2 ) − (32 ) − (42 )
1−0−0

(b) lim  () = 5 lim
→∞

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

132

¤

45.  =

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

1− has domain (−∞ −1) ∪ (−1 ∞).
1+

lim

→±∞

1 − 1
1−
0−1
= lim
=
= −1, so  = −1 is a HA.
1 +  →±∞ 1 + 1
0+1

The line  = −1 is a VA.
0 =

(1 + )(−1) − (1 − )(1)
−2
=
 0 for  6= 1. Thus,
(1 + )2
(1 + )2

(−∞ −1) and (−1 ∞) are intervals of decrease.
 00 = −2 ·

−2(1 + )
4
=
 0 for   −1 and  00  0 for   −1, so the curve is CD on (−∞ −1) and CU on
(1 + )3
[(1 + )2 ]2

(−1 ∞). Since  = −1 is not in the domain, there is no IP. lim →±∞


1
0
= lim
= 0, so  = 0 is a
=
2 + 1 →±∞ 1 + 12
1+0

2 + 1 − (2)
1 − 2
= 2
= 0 when  = ±1 and 0  0 ⇔
(2 + 1)2
( + 1)2

rS

0 =

al

horizontal asymptote.

e

47.

2  1 ⇔ −1    1, so  is increasing on (−1 1) and decreasing on (−∞ −1) and (1 ∞).

√
√
(1 + 2 )2 (−2) − (1 − 2 )2(2 + 1)2
2(2 − 3)
=
 0 ⇔   3 or − 3    0, so  is CU on
(1 + 2 )4
(1 + 2 )3
√ 
√

 √ 

 √ 
3 ∞ and − 3 0 and CD on −∞ − 3 and 0 3 .

Fo

 00 =

The -intercept is

ot

49.  =  () = 4 − 6 = 4 (1 − 2 ) = 4 (1 + )(1 − ).

 (0) = 0. The -intercepts are 0, −1, and 1 [found by solving  () = 0 for ].

N

Since 4  0 for  6= 0,  doesn’t change sign at  = 0. The function does change sign at  = −1 and  = 1. As  → ±∞,  () = 4 (1 − 2 ) approaches −∞ because 4 → ∞ and (1 − 2 ) → −∞.
51.  =  () = (3 − )(1 + )2 (1 − )4 .

The -intercept is  (0) = 3(1)2 (1)4 = 3.

The -intercepts are 3, −1, and 1. There is a sign change at 3, but not at −1 and 1.
When  is large positive, 3 −  is negative and the other factors are positive, so lim  () = −∞. When  is large negative, 3 −  is positive, so

→∞

lim () = ∞.

→−∞

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.4

LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES

¤

133

53. First we plot the points which are known to be on the graph: (2 −1) and

(0 0). We can also draw a short line segment of slope 0 at  = 2, since we are given that  0 (2) = 0. Now we know that  0 ()  0 (that is, the function is decreasing) on (0 2), and that  00 ()  0 on (0 1) and
 00 ()  0 on (1 2). So we must join the points (0 0) and (2 −1) in such a way that the curve is concave down on (0 1) and concave up on (1 2). The curve must be concave up and increasing on (2 4) and concave down and increasing toward  = 1 on (4 ∞). Now we just need to reflect the curve in the -axis, since we are given that  is an even function [the condition that  (−) =  () for all ].
55. We are given that  (1) =  0 (1) = 0. So we can draw a short horizontal line at the point (1 0) to represent this situation. We

are given that  = 0 and  = 2 are vertical asymptotes, with lim  () = −∞, lim () = ∞ and lim  () = −∞, so
→0

→2−

e

we can draw the parts of the curve which approach these asymptotes.

→2+

al

On the interval (−∞ 0), the graph is concave down, and  () → ∞ as
 → −∞. Between the asymptotes the graph is concave down. On the

rS

interval (2 ∞) the graph is concave up, and  () → 0 as  → ∞, so

 = 0 is a horizontal asymptote. The diagram shows one possible function satisfying all of the given conditions.

sin 
1
1
≤
≤ for   0. As  → ∞, −1 → 0 and 1 → 0, so by the Squeeze




Fo

57. (a) Since −1 ≤ sin  ≤ 1 for all  −

Theorem, (sin ) → 0. Thus, lim

→∞

sin 
= 0.


ot

(b) From part (a), the horizontal asymptote is  = 0. The function
 = (sin ) crosses the horizontal asymptote whenever sin  = 0;

N

that is, at  =  for every integer . Thus, the graph crosses the asymptote an infinite number of times.

59. (a) Divide the numerator and the denominator by the highest power of  in ().

(a) If deg   deg , then the numerator → 0 but the denominator doesn’t. So lim [ ()()] = 0.
→∞

(b) If deg   deg , then the numerator → ±∞ but the denominator doesn’t, so lim [ ()()] = ±∞
→∞

(depending on the ratio of the leading coefficients of  and ).
61. lim

→∞





4 − 1
42 + 3
1
3
= lim 4 +
= lim 4 −
= 4 and lim
= 4. Therefore, by the Squeeze Theorem,
→∞
→∞
→∞


2


lim  () = 4.

→∞

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

134

¤

CHAPTER 3

63. Let () =

32 + 1 and  () = |() − 15|. Note that
22 +  + 1

lim () =

→∞

APPLICATIONS OF DIFFERENTIATION

3
2

and lim () = 0. We are interested in finding the
→∞

-value at which  ()  005. From the graph, we find that  ≈ 14804, so we choose  = 15 (or any larger number).
√
 42 + 1

65. For  = 05, we need to find  such that 




− (−2)  05 ⇔


+1
√
42 + 1
−25 
 −15 whenever  ≤ . We graph the three parts of this
+1

choose  = −6 (or any smaller number).
√
42 + 1
 −19 whenever  ≤  From the
For  = 01, we need −21 
+1

al

graph, it seems that this inequality holds for  ≤ −22. So we choose  = −22

rS

(or any smaller number).

67. (a) 12  00001

e

inequality on the same screen, and see that the inequality holds for  ≤ −6 So we

⇔ 2  100001 = 10 000 ⇔   100 (  0)

Fo

√
√
(b) If   0 is given, then 12   ⇔ 2  1 ⇔   1 . Let  = 1 .



1
1
1
1
⇒  2 − 0 = 2  , so lim 2 = 0.
Then    ⇒   √
 

→∞ 


69. For   0, |1 − 0| = −1. If   0 is given, then −1  

⇒   −1 ⇒ |(1) − 0| = −1  , so lim (1) = 0.
→−∞

ot

Take  = −1. Then   

⇔   −1.

71. Suppose that lim  () = . Then for every   0 there is a corresponding positive number  such that | () − |  

N

→∞

whenever   . If  = 1, then   

⇔ 0  1  1

⇔

0    1. Thus, for every   0 there is a

corresponding   0 (namely 1) such that |(1) − |   whenever 0    . This proves that lim  (1) =  = lim ().

→0+

→∞

Now suppose that lim  () = . Then for every   0 there is a corresponding negative number  such that
→−∞

| () − |   whenever   . If  = 1, then   

⇔ 1  1  0 ⇔ 1    0. Thus, for every

  0 there is a corresponding   0 (namely −1) such that | (1) − |   whenever −    0. This proves that lim  (1) =  = lim  ().

→0−

→−∞

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.5 SUMMARY OF CURVE SKETCHING

3.5 Summary of Curve Sketching
1.  =  () = 3 − 122 + 36 = (2 − 12 + 36) = ( − 6)2

A.  is a polynomial, so  = R.

B. -intercepts are 0 and 6, -intercept =  (0) = 0 C. No symmetry D. No asymptote
E.  0 () = 32 − 24 + 36 = 3(2 − 8 + 12) = 3( − 2)( − 6)  0 ⇔

H.

2    6, so  is decreasing on (2 6) and increasing on (−∞ 2) and (6 ∞).
F. Local maximum value  (2) = 32, local minimum value  (6) = 0
G.  00 () = 6 − 24 = 6( − 4)  0 ⇔   4, so  is CU on (4 ∞) and
CD on (−∞ 4). IP at (4 16)

3.  =  () = 4 − 4 = (3 − 4)

A.  = R B. -intercepts are 0 and

√
3
4,

H.

e

-intercept =  (0) = 0 C. No symmetry D. No asymptote
E.  0 () = 43 − 4 = 4(3 − 1) = 4( − 1)(2 +  + 1)  0 ⇔   1, so

al

 is increasing on (1 ∞) and decreasing on (−∞ 1). F. Local minimum value
 (1) = −3, no local maximum G.  00 () = 122  0 for all , so  is CU on

5.  =  () = ( − 4)3

rS

(−∞ ∞). No IP

A.  = R B. -intercepts are 0 and 4, -intercept  (0) = 0 C. No symmetry
H.

Fo

D. No asymptote

E.  () =  · 3( − 4) + ( − 4) · 1 = ( − 4) [3 + ( − 4)]
0

2

3

2

= ( − 4)2 (4 − 4) = 4( − 1)( − 4)2  0

⇔

  1, so  is increasing on (1 ∞) and decreasing on (−∞ 1).

ot

F. Local minimum value (1) = −27, no local maximum value
G.  00 () = 4[( − 1) · 2( − 4) + ( − 4)2 · 1] = 4( − 4)[2( − 1) + ( − 4)]
= 4( − 4)(3 − 6) = 12( − 4)( − 2)  0

⇔

N

2    4, so  is CD on (2 4) and CU on (−∞ 2) and (4 ∞). IPs at (2 −16) and (4 0)
7.  =  () =

1 5
5



− 8 3 + 16 =  1 4 − 8 2 + 16 A.  = R B. -intercept 0, -intercept = (0) = 0
3
5
3

C.  (−) = − (), so  is odd; the curve is symmetric about the origin. D. No asymptote
E.  0 () = 4 − 82 + 16 = (2 − 4)2 = ( + 2)2 ( − 2)2  0 for all 

H.

except ±2, so  is increasing on R. F. There is no local maximum or minimum value.

G.  00 () = 43 − 16 = 4(2 − 4) = 4( + 2)( − 2)  0 ⇔
−2    0 or   2, so  is CU on (−2 0) and (2 ∞), and  is CD on




(−∞ −2) and (0 2). IP at −2 − 256 , (0 0), and 2 256
15
15

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

135

136

¤

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

A.  = { |  6= 1} = (−∞ 1) ∪ (1 ∞) B. -intercept = 0, -intercept =  (0) = 0



= 1, so  = 1 is a HA. lim
= −∞, lim
= ∞, so  = 1 is a VA.
C. No symmetry D. lim
→±∞  − 1
→1−  − 1
→1+  − 1

9.  =  () = ( − 1)

E.  0 () =

( − 1) − 
−1
=
 0 for  6= 1, so  is
( − 1)2
( − 1)2

H.

decreasing on (−∞ 1) and (1 ∞)  F. No extreme values
G.  00 () =

2
 0 ⇔   1, so  is CU on (1 ∞) and
( − 1)3

CD on (−∞ 1). No IP

11.  =  () =

 − 2
(1 − )

=
=
for  6= 1. There is a hole in the graph at (1 1).
2 − 3 + 2
(1 − )(2 − )
2−

(2 − )(1) − (−1)
2
=
 0 [ 6= 1 2], so  is
(2 − )2
(2 − )2

G.  0 () = 2(2 − )−2

⇒

 00 () = −4(2 − )−3 (−1) =

H.

rS

increasing on (−∞ 1), (1 2), and (2 ∞). F. No extrema

al

E.  0 () =

e

A.  = { |  6= 1 2} = (−∞ 1) ∪ (1 2) ∪ (2 ∞) B. -intercept = 0, -intercept =  (0) = 0 C. No symmetry



= −1, so  = −1 is a HA. lim
= ∞, lim
= −∞, so  = 2 is a VA.
D. lim
→±∞ 2 − 
→2− 2 − 
→2+ 2 − 

4
 0 ⇔   2, so  is CU on
(2 − )3

Fo

(−∞ 1) and (1 2), and  is CD on (2 ∞). No IP

A.  = { |  6= ±3} = (−∞ −3) ∪ (−3 3) ∪ (3 ∞) B. -intercept =  (0) = − 1 , no
9

13.  = () = 1(2 − 9)

-intercept C.  (−) =  () ⇒  is even; the curve is symmetric about the -axis. D.
→3−

1
1
1
1
= −∞, lim 2
= ∞, lim
= ∞, lim
= −∞, so  = 3 and  = −3
2
2
2 − 9
→3+  − 9
→−3−  − 9
→−3+  − 9
2
 0 ⇔   0 ( 6= −3) so  is increasing on (−∞ −3) and (−3 0) and
(2 − 9)2

N

are VA. E.  0 () = −

decreasing on (0 3) and (3 ∞). F. Local maximum value  (0) = − 1 .
9
G.  00 =

1
= 0, so  = 0
2 − 9

ot

is a HA. lim

lim

→±∞

H.

−2(2 − 9)2 + (2)2(2 − 9)(2)
6(2 + 3)
= 2
0 ⇔
2 − 9)4
(
( − 9)3

2  9 ⇔   3 or   −3, so  is CU on (−∞ −3) and (3 ∞) and
CD on (−3 3). No IP
15.  =  () = (2 + 9)

A.  = R B. -intercept:  (0) = 0; -intercept: () = 0 ⇔  = 0

C.  (−) = − (), so  is odd and the curve is symmetric about the origin. D.
HA; no VA E.  0 () =

lim [(2 + 9)] = 0, so  = 0 is a

→±∞

(2 + 9)(1) − (2)
9 − 2
(3 + )(3 − )
= 2
=
 0 ⇔ −3    3, so  is increasing
2 + 9)2
(
( + 9)2
(2 + 9)2

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SECTION 3.5 SUMMARY OF CURVE SKETCHING

¤

137

on (−3 3) and decreasing on (−∞ −3) and (3 ∞). F. Local minimum value  (−3) = − 1 , local maximum
6
value  (3) =

1
6



(2)(2 + 9) −(2 + 9) − 2(9 − 2 )
(2 + 9)2 (−2) − (9 − 2 ) · 2(2 + 9)(2)
=
[(2 + 9)2 ]2
(2 + 9)4
√
√
2(2 − 27)
=
= 0 ⇔  = 0, ± 27 = ±3 3
H.
(2 + 9)3
√
√
 √ 
 00 ()  0 ⇔ −3 3    0 or   3 3, so  is CU on −3 3 0
√ 
√ 



 √ and 3 3 ∞ , and CD on −∞ −3 3 and 0 3 3 . There are three
√ 
 √
1
inflection points: (0 0) and ±3 3 ± 12 3 .
G.  00 () =

−1
2

A.  = { |  6= 0} = (−∞ 0) ∪ (0 ∞) B. No -intercept; -intercept:  () = 0 ⇔  = 1

C. No symmetry D.

−1
−1
= 0, so  = 0 is a HA. lim
= −∞, so  = 0 is a VA.
→0 2
2

2 · 1 − ( − 1) · 2
−( − 2)
−2 + 2
=
=
, so  0 ()  0 ⇔ 0    2 and  0 ()  0 ⇔
(2 )2
4
3

al

E.  0 () =

lim

→±∞

e

17.  =  () =

  0 or   2. Thus,  is increasing on (0 2) and decreasing on (−∞ 0)

H.

rS

and (2 ∞). F. No local minimum, local maximum value (2) = 1 .
4

3 · (−1) − [−( − 2)] · 32
23 − 62
2( − 3)
=
=
.
3 )2
(
6
4

G.  00 () =

19.  =  () =

Fo

 00 () is negative on (−∞ 0) and (0 3) and positive on (3 ∞), so  is CD

 on (−∞ 0) and (0 3) and CU on (3 ∞). IP at 3 2
9
(2 + 3) − 3
3
2
=
=1− 2
+3
2 + 3
 +3

2

A.  = R B. -intercept: (0) = 0;

D.

ot

-intercepts:  () = 0 ⇔  = 0 C.  (−) =  (), so  is even; the graph is symmetric about the -axis.
−2
6
2
= 2
.
= 1, so  = 1 is a HA. No VA. E. Using the Reciprocal Rule,  0 () = −3 · 2
+3
( + 3)2
( + 3)2

lim

→±∞ 2

N

 0 ()  0 ⇔   0 and  0 ()  0 ⇔   0, so  is decreasing on (−∞ 0) and increasing on (0 ∞).
F. Local minimum value (0) = 0, no local maximum.
G.  00 () =
=

(2 + 3)2 · 6 − 6 · 2(2 + 3) · 2
[(2 + 3)2 ]2

H.

6(2 + 3)[(2 + 3) − 42 ]
6(3 − 32 )
−18( + 1)( − 1)
=
=
2 + 3)4
(
(2 + 3)3
(2 + 3)3

 00 () is negative on (−∞ −1) and (1 ∞) and positive on (−1 1),

 so  is CD on (−∞ −1) and (1 ∞) and CU on (−1 1). IP at ±1 1
4
√

21.  =  () = ( − 3)  = 32 − 312

A.  = [0 ∞) B. -intercepts: 0 3; -intercept =  (0) = 0 C. No

symmetry D. No asymptote E.  0 () = 3 12 − 3 −12 = 3 −12 ( − 1) =
2
2
2

3( − 1)
√
0
2 

⇔

  1,

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138

¤

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

H.

so  is increasing on (1 ∞) and decreasing on (0 1).
F. Local minimum value (1) = −2, no local maximum value
G.  00 () = 3 −12 + 3 −32 = 3 −32 ( + 1) =
4
4
4

3( + 1)
 0 for   0,
432

so  is CU on (0 ∞). No IP
23.  =  () =


√
2 +  − 2 = ( + 2)( − 1) A.  = { | ( + 2)( − 1) ≥ 0} = (−∞ −2] ∪ [1 ∞)

B. -intercept: none; -intercepts: −2 and 1 C. No symmetry D. No asymptote
E.  0 () = 1 (2 +  − 2)−12 (2 + 1) =
2

2

2 + 1
√
,  0 () = 0 if  = − 1 , but − 1 is not in the domain.
2
2
2 +  − 2

 0 ()  0 ⇒   − 1 and  0 ()  0 ⇒   − 1 , so (considering the domain)  is increasing on (1 ∞) and
2
2

 is decreasing on (−∞ −2). F. No local extrema

2(2 +  − 2)12 (2) − (2 + 1) · 2 · 1 (2 +  − 2)−12 (2 + 1)
2
 √
2
2 2 +  − 2


(2 +  − 2)−12 4(2 +  − 2) − (42 + 4 + 1)
=
4(2 +  − 2)

al

−9
0
4(2 +  − 2)32

so  is CD on (−∞ −2) and (1 ∞). No IP

A.  = R B. -intercept:  (0) = 0; -intercepts:  () = 0 ⇒  = 0

Fo

√

25.  =  () =  2 + 1

rS

=

H.

e

G.  00 () =

C.  (−) = − (), so  is odd; the graph is symmetric about the origin.




1
1
√ = lim 
=1
= √
D. lim  () = lim √
= lim √
= lim √
→∞
→∞
→∞
1+0
2 + 1 →∞ 2 + 1 →∞ 2 + 1 2
1 + 12

ot

and




1
 √  = lim

= lim √
= lim √ lim () = lim √
→−∞
→−∞ −
2 + 1 →−∞ 2 + 1 →−∞ 2 + 1 − 2
1 + 12

→−∞

No VA.
0

E.  () =

1
√
= −1 so  = ±1 are HA.
− 1+0

N

=

√
2 + 1 −  ·

2
√
2 + 1 − 2
1
2 + 1
=
=
 0 for all , so  is increasing on R.
2 + 1)12 ]2
[(
(2 + 1)32
(2 + 1)32
2

H.

F. No extreme values
G.  00 () = − 3 (2 + 1)−52 · 2 =
2

−3
, so  00 ()  0 for   0
(2 + 1)52

and  00 ()  0 for   0. Thus,  is CU on (−∞ 0) and CD on (0 ∞).
IP at (0 0)

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SECTION 3.5 SUMMARY OF CURVE SKETCHING

¤

√
1 − 2  A.  = { | || ≤ 1,  6= 0} = [−1 0) ∪ (0 1] B. -intercepts ±1, no -intercept
√
√
1 − 2
1 − 2
C.  (−) = − (), so the curve is symmetric about (0 0)  D. lim
= ∞, lim
= −∞,
+
−


→0
→0
 √
 2√
−  1 − 2 − 1 − 2
1
0 so  = 0 is a VA. E.  () =
=− √
 0, so  is decreasing
2
2
 1 − 2

27.  = () =

H.

on (−1 0) and (0 1). F. No extreme values
G.  00 () =

2 − 32

2 )32



 0 ⇔ −1    − 2 or 0    2 , so
3
3

3 (1 −
 
  
 




2
1 .
 is CU on −1 − 2 and 0 2 and CD on − 2  0 and
3
3
3
3

 

1
IP at ± 2  ± √2
3

A.  = R B. -intercept:  (0) = 0; -intercepts:  () = 0 ⇒  = 313 ⇒
√
3 = 27 ⇒ 3 − 27 = 0 ⇒ (2 − 27) = 0 ⇒  = 0, ±3 3 C.  (−) = − (), so  is odd;

al

e

29.  =  () =  − 313

rS

the graph is symmetric about the origin. D. No asymptote E.  0 () = 1 − −23 = 1 −

1
23

=

23 − 1
.
23

 0 ()  0 when ||  1 and  0 ()  0 when 0  ||  1, so  is increasing on (−∞ −1) and (1 ∞), and decreasing on (−1 0) and (0 1) [hence decreasing on (−1 1) since  is

H.

continuous on (−1 1)]. F. Local maximum value  (−1) = 2, local minimum

Fo

value  (1) = −2 G.  00 () = 2 −53  0 when   0 and  00 ()  0
3
when   0, so  is CD on (−∞ 0) and CU on (0 ∞). IP at (0 0)
31.  =  () =

√
3
2 − 1 A.  = R B. -intercept:  (0) = −1; -intercepts:  () = 0 ⇔ 2 − 1 = 0 ⇔

ot

 = ±1 C.  (−) =  (), so the curve is symmetric about the -axis D. No asymptote

N

E.  0 () = 1 (2 − 1)−23 (2) =
3

2

.  0 ()  0 ⇔   0 and  0 ()  0 ⇔   0, so  is
3
3 (2 − 1)2

increasing on (0 ∞) and decreasing on (−∞ 0). F. Local minimum value  (0) = −1
G.  00 () =

=

2
23
2
−13
(2)
2 ( − 1) (1) −  · 2 ( − 1)
3
·
3
[(2 − 1)23 ]2

H.

2 (2 − 1)−13 [3(2 − 1) − 42 ]
2(2 + 3)
·
=−
2 − 1)43
9
(
9(2 − 1)53

 00 ()  0 ⇔ −1    1 and  00 ()  0 ⇔   −1 or   1, so
 is CU on (−1 1) and  is CD on (−∞ −1) and (1 ∞). IP at (±1 0)
33.  =  () = sin3 

A.  = R B. -intercepts:  () = 0 ⇔  = ,  an integer; -intercept =  (0) = 0

C.  (−) = − (), so  is odd and the curve is symmetric about the origin. Also, ( + 2) = (), so  is periodic with period 2, and we determine E–G for 0 ≤  ≤ . Since  is odd, we can reflect the graph of  on [0 ] about the

c
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139

¤

140

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

origin to obtain the graph of  on [− ], and then since  has period 2, we can extend the graph of  for all real numbers.
D. No asymptote E.  0 () = 3 sin2  cos   0 ⇔ cos   0 and sin  6= 0 ⇔ 0     , so  is increasing on
2

 




 
0 2 and  is decreasing on 2   . F. Local maximum value  2 = 1 local minimum value  −  = −1
2

G.  00 () = 3 sin2  (− sin ) + 3 cos  (2 sin  cos ) = 3 sin  (2 cos2  − sin2 )

= 3 sin  [2(1 − sin2 ) − sin2 ] = 3 sin (2 − 3 sin2 )  0 ⇔

 
  sin   0 and sin2   2 ⇔ 0     and 0  sin   2 ⇔ 0    sin−1 2 let  = sin−1 2 or
3
3
3
3

 −     , so  is CU on (0 ) and ( −  ), and  is CD on (  − ). There are inflection points at  = 0, , ,

and  =  − .

35.  =  () =  tan , −    
2

symmetric about the -axis. D.

rS

al

e

H.



B. Intercepts are 0 C.  (−) =  (), so the curve is
A.  = −   
2 2


2

lim

→(2)−

 tan  = ∞ and

lim

→−(2)+

 tan  = ∞, so  =

Fo



E.  0 () = tan  +  sec2   0 ⇔ 0     , so  increases on 0 
2
2

 and decreases on −   0 . F. Absolute and local minimum value  (0) = 0.
2

1

2

H.

so  is

− sin , 0    3

A.  = (0 3) B. No -intercept. The -intercept, approximately 19, can be

N

37.  =  () =


,
2

and  = −  are VA.
2

ot

G.  00 = 2 sec2  + 2 tan  sec2   0 for −    
2
  
CU on − 2  2 . No IP


2

found using Newton’s Method. C. No symmetry D. No asymptote E.  0 () = 1 − cos   0 ⇔ cos  
2
 7



  5 



5
7
and 3  3 and decreasing on 0  and 5  7 .
3    3 or 3    3, so  is increasing on 3  3
3
3
3
F. Local minimum value 



 5 
3

=

5
6

+

√
3
,
2


3

=


6

−

√

3
,
2

local minimum value 

local maximum value

 7 
3

=

7
6

−

39.  =  () =

sin 
1 + cos 

when

cos  6= 1
 =

⇔

H.

√

3
2

G.  00 () = sin   0 ⇔ 0     or 2    3, so  is CU on


(0 ) and (2 3) and CD on ( 2). IPs at   and (2 )
2


1
2



1 − cos  sin  (1 − cos ) sin 
1 − cos 

·
=
= csc  − cot 
=
1 + cos  1 − cos  sin  sin2 

A. The domain of  is the set of all real numbers except odd integer multiples of ; that is, all reals except (2 + 1), where c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.5 SUMMARY OF CURVE SKETCHING

¤

141

 is an integer. B. -intercept:  (0) = 0; -intercepts:  = 2,  an integer. C.  (−) = − (), so  is an odd function; the graph is symmetric about the origin and has period 2. D. When  is an odd integer, lim →()−

 () = ∞ and

E.  0 () =

lim

→()+

 () = −∞, so  =  is a VA for each odd integer . No HA.

(1 + cos ) · cos  − sin (− sin )
1 + cos 
1
=
=
.  0 ()  0 for all  except odd multiples of
(1 + cos )2
(1 + cos )2
1 + cos 

, so  is increasing on ((2 − 1) (2 + 1)) for each integer . F. No extreme values
G.  00 () =

sin 
 0 ⇒ sin   0 ⇒
(1 + cos )2

H.

 ∈ (2 (2 + 1)) and  00 ()  0 on ((2 − 1) 2) for each integer .  is CU on (2 (2 + 1)) and CD on ((2 − 1) 2)

e

for each integer .  has IPs at (2 0) for each integer .

al

0
. The m-intercept is  (0) = 0 . There are no -intercepts. lim  () = ∞, so  =  is a VA.
→−
1 −  2 2

41.  = () = 

0 
0 
0 
= 2 2
= 2
 0, so  is
2 32
2 (1 −  2 2 )32
( −  2 )32
 ( −  )
3
increasing on (0 ). There are no local extreme values.

=

(2 −  2 )32 (0 ) − 0  · 3 (2 −  2 )12 (−2)
2
[(2 −  2 )32 ]2

Fo

 00 () =

rS

 0 () = − 1 0 (1 −  2 2 )−32 (−22 ) =
2

0 (2 −  2 )12 [(2 −  2 ) + 3 2 ]
0 (2 + 2 2 )
=
 0,
2 −  2 )3
(
(2 −  2 )52

43.  = −


 4
  3  2 2
 2 2
 +
 −
 =−
  − 2 + 2
24
12
24
24

− 2
 ( − )2 = 2 ( − )2
24

N

=

ot

so  is CU on (0 ). There are no inflection points.

where  = −

 is a negative constant and 0 ≤  ≤ . We sketch
24

 () = 2 ( − )2 for  = −1.  (0) =  () = 0.
 0 () = 2 [2( − )] + ( − )2 (2) = 2( − )[ + ( − )] = 2( − )(2 − ). So for 0    ,
 0 ()  0 ⇔ ( − )(2 − )  0 [since   0] ⇔ 2     and  0 ()  0 ⇔ 0    2.
Thus,  is increasing on (2 ) and decreasing on (0 2), and there is a local and absolute

 minimum at the point (2  (2)) = 2 4 16 .  0 () = 2[( − )(2 − )] ⇒

 00 () = 2[1( − )(2 − ) + (1)(2 − ) + ( − )(2)] = 2(62 − 6 + 2 ) = 0 ⇔
√
√
6 ± 122
= 1  ± 63 , and these are the -coordinates of the two inflection points.
=
2
12

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

142

¤

45.  =

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

2 + 1
. Long division gives us:
+1

−1

2

+1 

+1

2 + 
−+1
−−1
2
2
2
2 + 1
=−1+
and  () − ( − 1) =
=
Thus,  =  () =
+1
+1
+1

2

1+

[for  6= 0] → 0 as  → ±∞.

1


So the line  =  − 1 is a slant asymptote (SA).
47.  =

43 − 22 + 5
. Long division gives us:
22 +  − 3

22 +  − 3 43 − 22

2 − 2
+5

e

43 + 22 − 6

al

− 42 + 6 + 5
− 42 − 2 + 6

rS

8 − 1

8 − 1
8 − 1
43 − 22 + 5
= 2 − 2 + 2 and  () − (2 − 2) = 2
=
Thus,  =  () =
22 +  − 3
2 +  − 3
2 +  − 3

8
1
− 2


1
3
2+ − 2



49.  =  () =

1
2
=+1+
−1
−1

Fo

[for  6= 0] → 0 as  → ±∞. So the line  = 2 − 2 is a SA.

A.  = (−∞ 1) ∪ (1 ∞) B. -intercept:  () = 0 ⇔  = 0;

-intercept: (0) = 0 C. No symmetry D. lim  () = −∞ and lim () = ∞, so  = 1 is a VA.
→1−

→±∞

→±∞

1
( − 1)2 − 1
2 − 2
( − 2)
=
=
=
 0 for
2
2
( − 1)
( − 1)
( − 1)2
( − 1)2

H.

N

E.  0 () = 1 −

1
= 0, so the line  =  + 1 is a SA.
−1

ot

lim [ () − ( + 1)] = lim

→1+

  0 or   2, so  is increasing on (−∞ 0) and (2 ∞), and  is decreasing on (0 1) and (1 2). F. Local maximum value  (0) = 0, local minimum value
 (2) = 4 G.  00 () =

2
 0 for   1, so  is CU on (1 ∞) and 
( − 1)3

is CD on (−∞ 1). No IP

51.  =  () =

3 + 4
4
=+ 2
2


√
A.  = (−∞ 0) ∪ (0 ∞) B. -intercept:  () = 0 ⇔  = − 3 4; no -intercept

C. No symmetry D. lim  () = ∞, so  = 0 is a VA.
→0

lim [ () − ] = lim

→±∞

→±∞

4
= 0, so  =  is a SA.
2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.5 SUMMARY OF CURVE SKETCHING

E.  0 () = 1 −

8
3 − 8
=
 0 for   0 or   2, so  is increasing on
3

3

¤

143

H.

(−∞ 0) and (2 ∞), and  is decreasing on (0 2). F. Local minimum value
 (2) = 3, no local maximum value G.  00 () =

24
 0 for  6= 0, so  is CU
4

on (−∞ 0) and (0 ∞). No IP

53.  = () =

−2
23 + 2 + 1
= 2 + 1 + 2
2 + 1
 +1

A.  = R B. -intercept:  (0) = 1; -intercept: () = 0 ⇒

0 = 23 + 2 + 1 = ( + 1)(22 −  + 1) ⇒  = −1 C. No symmetry D. No VA lim [ () − (2 + 1)] = lim

→±∞

→±∞

(2 + 1)(−2) − (−2)(2)
2(4 + 22 + 1) − 22 − 2 + 42
24 + 62
22 (2 + 3)
=
=
=
2 + 1)2
2 + 1)2
2 + 1)2
(
(
(
(2 + 1)2

e

E.  0 () = 2 +

−2
−2
= lim
= 0, so the line  = 2 + 1 is a slant asymptote.
2 + 1 →±∞ 1 + 12

so  0 ()  0 if  6= 0. Thus,  is increasing on (−∞ 0) and (0 ∞). Since  is continuous at 0,  is increasing on R.
(2 + 1)2 · (83 + 12) − (24 + 62 ) · 2(2 + 1)(2)
[(2 + 1)2 ]2

rS

G.  00 () =

al

F. No extreme values

4(2 + 1)[(2 + 1)(22 + 3) − 24 − 62 ]
4(−2 + 3)
=
(2 + 1)4
(2 + 1)3
√
√ so  00 ()  0 for   − 3 and 0    3, and  00 ()  0 for
H.
√
√
√ 

 √ 
− 3    0 and   3.  is CU on −∞ − 3 and 0 3 ,
 √ 
√
 and CD on − 3 0 and
3 ∞ . There are three IPs: (0 1),
√

 √
− 3 − 3 3 + 1 ≈ (−173 −160), and
2

√ 3 √
3 2 3 + 1 ≈ (173 360).

ot

Fo

=

√
4
42 + 9 ⇒  0 () = √
⇒
42 + 9
√
√
42 + 9 · 4 − 4 · 4 42 + 9
4(42 + 9) − 162
36
=
.  is defined on (−∞ ∞).
 00 () =
=
2 +9
4
(42 + 9)32
(42 + 9)32

N

55.  =  () =

 (−) = (), so  is even, which means its graph is symmetric about the -axis. The -intercept is (0) = 3. There are no
-intercepts since  ()  0 for all 
√
√

√

42 + 9 − 2
42 + 9 + 2
2 + 9 − 2 = lim
√
4 lim →∞
→∞
42 + 9 + 2

(42 + 9) − 42
9
= lim √
=0
= lim √
→∞
→∞
42 + 9 + 2
42 + 9 + 2

and, similarly, lim

→−∞

√

9
42 + 9 + 2 = lim √
= 0,
→−∞
42 + 9 − 2

so  = ±2 are slant asymptotes.  is decreasing on (−∞ 0) and increasing on (0 ∞) with local minimum  (0) = 3.

 00 ()  0 for all , so  is CU on R.

c
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144

57.

¤

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

2
2
√ 2
− 2 =1 ⇒  =±
 − 2 . Now
2



√


√
 2 − 2 + 
−2
√ 2



2 −
2 − 2 −  √
 = · lim
 −

= · lim √
= 0, lim →∞ 

 →∞
 →∞ 2 − 2 + 
2 − 2 + 

 is a slant asymptote. Similarly,




−2
√ 2



2 − −  lim −
= − · lim √
 −
= 0, so  = −  is a slant asymptote.
→∞


 →∞ 2 − 2 + 

which shows that  =



4 + 1
1
4
() − 3 = lim
−
= lim
= 0, so the graph of  is asymptotic to that of  = 3 
→±∞
→±∞
→±∞ 




1
= −∞ and
A.  = { |  6= 0} B. No intercept C.  is symmetric about the origin. D. lim 3 +

→0−


1
lim 3 +
= ∞, so  = 0 is a vertical asymptote, and as shown above, the graph of  is asymptotic to that of  = 3 .

→0+




1
1
1
E.  0 () = 32 − 12  0 ⇔ 4  1 ⇔ ||  √3 , so  is increasing on −∞ − √ and √  ∞ and
4
3
4
4
3
3




1
1
H.
decreasing on − √  0 and 0 √ . F. Local maximum value
4
4
3
3




1
1
−54
 −√
, local minimum value  √
= −4 · 3
= 4 · 3−54
4
4
3
3 lim rS

al

e

59.

on (−∞ 0). No IP

Fo

G.  00 () = 6 + 23  0 ⇔   0, so  is CU on (0 ∞) and CD

ot

3.6 Graphing with Calculus and Calculators
1.  () = 44 − 323 + 892 − 95 + 29

⇒  0 () = 163 − 962 + 178 − 95 ⇒  00 () = 482 − 192 + 178.

N

 () = 0 ⇔  ≈ 05, 160;  0 () = 0 ⇔  ≈ 092, 25, 258 and  00 () = 0

⇔

 ≈ 146, 254.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.6

GRAPHING WITH CALCULUS AND CALCULATORS

¤

145

From the graphs of  0 , we estimate that  0  0 and that  is decreasing on (−∞ 092) and (25 258), and that  0  0 and  is increasing on (092 25) and (258 ∞) with local minimum values  (092) ≈ −512 and  (258) ≈ 3998 and local maximum value  (25) = 4. The graphs of  0 make it clear that  has a maximum and a minimum near  = 25, shown more clearly in the fourth graph.
From the graph of  00 , we estimate that  00  0 and that  is CU on
(−∞ 146) and (254 ∞), and that  00  0 and  is CD on (146 254).
There are inflection points at about (146 −140) and (254 3999).

3.  () = 6 − 105 − 4004 + 25003
00

4

3

⇒  0 () = 65 − 504 − 16003 + 75002

⇒

2

ot

Fo

rS

al

e

 () = 30 − 200 − 4800 + 1500

From the graph of  0 , we estimate that  is decreasing on (−∞ −15), increasing on (−15 440), decreasing

N

on (440 1893), and increasing on (1893 ∞), with local minimum values of  (−15) ≈ −9,700,000 and
 (18.93) ≈ −12,700,000 and local maximum value  (440) ≈ 53,800. From the graph of  00 , we estimate that  is CU on
(−∞ −1134), CD on (−1134 0), CU on (0 292), CD on (292 1508), and CU on (1508 ∞). There is an inflection point at (0 0) and at about (−1134 −6,250,000), (292 31,800), and (1508 −8,150,000).
5.  () =


3 + 2 + 1

⇒  0 () = −

23 + 2 − 1
(3 + 2 + 1)2

⇒  00 () =

2(34 + 33 + 2 − 6 − 3)
(3 + 2 + 1)3

[continued]

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

146

¤

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

From the graph of  , we see that there is a VA at  ≈ −147. From the graph of  0 , we estimate that  is increasing on

(−∞ −147), increasing on (−147 066), and decreasing on (066 ∞), with local maximum value (066) ≈ 038.

From the graph of  00 , we estimate that  is CU on (−∞ −147), CD on (−147 −049), CU on (−049 0), CD on
(0 110), and CU on (110 ∞). There is an inflection point at (0 0) and at about (−049 −044) and (110 031).
7.  () = 6 sin  + cot , − ≤  ≤ 

⇒  0 () = 6 cos  − csc2  ⇒  00 () = −6 sin  + 2 csc2  cot 

e

From the graph of  , we see that there are VAs at  = 0 and  = ±.  is an odd function, so its graph is symmetric about the origin. From the graph of  0 , we estimate that  is decreasing on (− −140), increasing on (−140 −044), decreasing

al

on (−044 0), decreasing on (0 044), increasing on (044 140), and decreasing on (140 ), with local minimum values
 (−140) ≈ −609 and (044) ≈ 468, and local maximum values  (−044) ≈ −468 and  (140) ≈ 609.

rS

From the graph of  00 , we estimate that  is CU on (− −077), CD on (−077 0), CU on (0 077), and CD on
(077 ). There are IPs at about (−077 −522) and (077 522).
⇒  0 () = −

1
16
3
1
− 3 − 4 = − 4 (2 + 16 + 3) ⇒
2




2
48
12
2
+ 4 + 5 = 5 (2 + 24 + 6).
3




N

ot

 00 () =

8
1
1
+ 2 + 3




Fo

9.  () = 1 +

From the graphs, it appears that  increases on (−158 −02) and decreases on (−∞ −158), (−02 0), and (0 ∞); that  has a local minimum value of (−158) ≈ 097 and a local maximum value of (−02) ≈ 72; that  is CD on (−∞ −24) and (−025 0) and is CU on (−24 −025) and (0 ∞); and that  has IPs at (−24 097) and (−025 60).
√
√
−16 ± 256 − 12
To find the exact values, note that  0 = 0 ⇒  =
= −8 ± 61 [≈ −019 and −1581].
2
√
√ 
√ 


 0 is positive ( is increasing) on −8 − 61 −8 + 61 and  0 is negative ( is decreasing) on −∞ −8 − 61 ,
√
√
√


−24 ± 576 − 24
−8 + 61 0 , and (0 ∞).  00 = 0 ⇒  =
= −12 ± 138 [≈ −025 and −2375].  00 is
2
√
√
√



 positive ( is CU) on −12 − 138 −12 + 138 and (0 ∞) and  00 is negative ( is CD) on −∞ −12 − 138
√


and −12 + 138 0 . c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.6

11.

 () =

GRAPHING WITH CALCULUS AND CALCULATORS

¤

147

( + 4)( − 3)2 has VA at  = 0 and at  = 1 since lim  () = −∞,
→0
4 ( − 1)

lim  () = −∞ and lim  () = ∞.

→1−

→1+

 + 4 ( − 3)2 

·
(1 + 4)(1 − 3)2 dividing numerator

2
→0
 () =
=
3
4
and denominator by 
( − 1)

· ( − 1)
3
as  → ±∞, so  is asymptotic to the -axis.
Since  is undefined at  = 0, it has no -intercept. () = 0 ⇒ ( + 4)( − 3)2 = 0 ⇒  = −4 or  = 3, so  has

-intercepts −4 and 3. Note, however, that the graph of  is only tangent to the -axis and does not cross it at  = 3, since  is

rS

al

e

positive as  → 3− and as  → 3+ .

From these graphs, it appears that  has three maximum values and one minimum value. The maximum values are approximately  (−56) = 00182,  (082) = −2815 and (52) = 00145 and we know (since the graph is tangent to the

2 ( + 1)3
( − 2)2 ( − 4)4

⇒  0 () = −

( + 1)2 (3 + 182 − 44 − 16)
( − 2)3 ( − 4)5

[from CAS].

N

ot

13.  () =

Fo

-axis at  = 3) that the minimum value is (3) = 0.

From the graphs of  0 , it seems that the critical points which indicate extrema occur at  ≈ −20, −03, and 25, as estimated in Example 3. (There is another critical point at  = −1, but the sign of  0 does not change there.) We differentiate again, obtaining  00 () = 2

( + 1)(6 + 365 + 64 − 6283 + 6842 + 672 + 64)
.
( − 2)4 ( − 4)6

[continued] c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

148

¤

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APPLICATIONS OF DIFFERENTIATION

From the graphs of  00 , it appears that  is CU on (−353 −50), (−1 −05), (−01 2), (2 4) and (4 ∞) and CD on (−∞ −353), (−50 −1) and (−05 −01). We check back on the graphs of  to find the -coordinates of the inflection points, and find that these points are approximately (−353 −0015), (−50 −0005), (−1 0), (−05 000001), and (−01 00000066).
15.  () =

3 + 52 + 1
−(5 + 104 + 63 + 42 − 3 − 22)
. From a CAS,  0 () = and 4 + 3 − 2 + 2
(4 + 3 − 2 + 2)2
2(9 + 158 + 187 + 216 − 95 − 1354 − 763 + 212 + 6 + 22)
(4 + 3 − 2 + 2)3

al

e

 00 () =

rS

The first graph of  shows that  = 0 is a HA. As  → ∞,  () → 0 through positive values. As  → −∞, it is not clear if
 () → 0 through positive or negative values. The second graph of  shows that  has an -intercept near −5, and will have a

ot

Fo

local minimum and inflection point to the left of −5.

From the two graphs of  0 , we see that  0 has four zeros. We conclude that  is decreasing on (−∞ −941), increasing on

N

(−941 −129), decreasing on (−129 0), increasing on (0 105), and decreasing on (105 ∞). We have local minimum values (−941) ≈ −0056 and  (0) = 05, and local maximum values (−129) ≈ 749 and  (105) ≈ 235.

From the two graphs of  00 , we see that  00 has five zeros. We conclude that  is CD on (−∞ −1381), CU on
(−1381 −155), CD on (−155 −103), CU on (−103 060), CD on (060 148), and CU on (148 ∞). There are five inflection points: (−1381 −005), (−155 564), (−103 539), (060 152), and (148 193). c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.6

17.  =  () =

GRAPHING WITH CALCULUS AND CALCULATORS

¤

149

√
 + 5 sin ,  ≤ 20.

5 cos  + 1
10 cos  + 25 sin2  + 10 sin  + 26
√
. and  00 = −
4( + 5 sin )32
2  + 5 sin 

We’ll start with a graph of () =  + 5 sin . Note that () = () is only defined if () ≥ 0. () = 0 ⇔  = 0
From a CAS, 0 =

or  ≈ −491, −410, 410, and 491. Thus, the domain of  is [−491 −410] ∪ [0 410] ∪ [491 20].

e

 
From the expression for  0 , we see that  0 = 0 ⇔ 5 cos  + 1 = 0 ⇒ 1 = cos−1 − 1 ≈ 177 and
5

al

2 = 2 − 1 ≈ −451 (not in the domain of  ). The leftmost zero of  0 is 1 − 2 ≈ −451. Moving to the right, the zeros of  0 are 1 , 1 + 2, 2 + 2, 1 + 4, and 2 + 4. Thus,  is increasing on (−491 −451), decreasing on

rS

(−451 −410), increasing on (0 177), decreasing on (177 410), increasing on (491 806), decreasing on (806 1079), increasing on (1079 1434), decreasing on (1434 1708), and increasing on (1708 20). The local maximum values are

and  (1708) ≈ 349.

Fo

 (−451) ≈ 062,  (177) ≈ 258,  (806) ≈ 360, and (1434) ≈ 439. The local minimum values are  (1079) ≈ 243

 is CD on (−491 −410), (0 410), (491 960), CU on (960 1225),
CD on (1225 1581), CU on (1581 1865), and CD on (1865 20). There are

N

19.

ot

inflection points at (960 295), (1225 327), (1581 391), and (1865 420).

From the graph of  () = sin( + sin 3) in the viewing rectangle [0 ] by [−12 12], it looks like  has two maxima and two minima. If we calculate and graph  0 () = [cos( + sin 3)] (1 + 3 cos 3) on [0 2], we see that the graph of  0 appears to be almost tangent to the -axis at about  = 07. The graph of
 00 = − [sin( + sin 3)] (1 + 3 cos 3)2 + cos( + sin 3)(−9 sin 3) is even more interesting near this -value: it seems to just touch the -axis.
[continued]
c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

150

¤

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

e

If we zoom in on this place on the graph of  00 , we see that  00 actually does cross the axis twice near  = 065, indicating a change in concavity for a very short interval. If we look at the graph of  0 on the same interval, we see that it

al

changes sign three times near  = 065, indicating that what we had thought was a broad extremum at about  = 07 actually

rS

consists of three extrema (two maxima and a minimum). These maximum values are roughly  (059) = 1 and  (068) = 1, and the minimum value is roughly  (064) = 099996. There are also a maximum value of about  (196) = 1 and minimum values of about  (146) = 049 and  (273) = −051. The points of inflection on (0 ) are about (061 099998),

Fo

(066 099998), (117 072), (175 077), and (228 034). On ( 2), they are about (401 −034), (454 −077),
(511 −072), (562 −099998), and (567 −099998). There are also IP at (0 0) and ( 0). Note that the function is odd and periodic with period 2, and it is also rotationally symmetric about all points of the form ((2 + 1) 0),  an integer.
√
√
(22 + )
4 + 2 = || 2 +  ⇒  0 () = √
4 + 2

⇒  00 () =

4 (22 + 3)
(4 + 2 )32

N

ot

21.  () =

√
-intercepts: When  ≥ 0, 0 is the only -intercept. When   0, the -intercepts are ± −.
-intercept =  (0) = 0 when  ≥ 0. When   0, there is no -intercept.

 is an even function, so its graph is symmetric with respect to the -axis.
 0 () =

22 + 
22 + 
(22 + )
√
= −√ for   0 and √ for   0, so  has a corner or “point” at  = 0 that gets
2 +
2 +
|| 

2 + 

sharper as  increases. There is an absolute minimum at 0 for  ≥ 0. There are no other maximums nor minimums.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.6

GRAPHING WITH CALCULUS AND CALCULATORS

¤

151



4 (22 + 3)
, so  00 () = 0 ⇒  = ± −32 [only for   0].  00 changes sign at ± −32, so  is
3
2 + )32
|| (



 
√




√ 
−32 ∞ , and  is CD on − −32 − − and
− −32 . There are
CU on −∞ − −32 and

 00 () =

 


IPs at ± −32 32 4 . The more negative  becomes, the farther the IPs move from the origin. The only transitional value is  = 0.

23. Note that  = 0 is a transitional value at which the graph consists of the -axis. Also, we can see that if we substitute − for ,

 will be reflected in the -axis, so we investigate only positive values of  (except  = −1, as a
1 + 2 2

demonstration of this reflective property). Also,  is an odd function. for all . We calculate  0 () =

lim  () = 0, so  = 0 is a horizontal asymptote

→±∞

(1 + 2 2 ) − (22 )
(2 2 − 1) 0
=−
.  () = 0 ⇔ 2 2 − 1 = 0 ⇔
(1 + 2 2 )2
(1 + 2 2 )2

 = ±1. So there is an absolute maximum value of  (1) =

1
2

and an absolute minimum value of  (−1) = − 1 .
2

e

the function  () =

=

(−23 )(1 + 2 2 )2 − (−3 2 + )[2(1 + 2 2 )(22 )]
(1 + 2 2 )4

rS

 00 () =

al

These extrema have the same value regardless of , but the maximum points move closer to the -axis as  increases.

(−23 )(1 + 2 2 ) + (3 2 − )(42 )
23 (2 2 − 3)
=
2 2 )3
(1 + 
(1 + 2 2 )3

√
 00 () = 0 ⇔  = 0 or ± 3, so there are inflection points at (0 0) and

points approach the -axis.
25.  () =  + sin 

Fo

√
 √

at ± 3 ± 34 . Again, the -coordinate of the inflection points does not depend on , but as  increases, both inflection
⇒  0 () =  + cos  ⇒  00 () = − sin 

ot

 (−) = − (), so  is an odd function and its graph is symmetric with respect to the origin.
 () = 0 ⇔ sin  = −, so 0 is always an -intercept.

N

 0 () = 0 ⇔ cos  = −, so there is no critical number when ||  1. If || ≤ 1, then there are infinitely many critical numbers. If 1 is the unique solution of cos  = − in the interval [0 ], then the critical numbers are 2 ± 1 , where  ranges over the integers. (Special cases: When  = −1, 1 = 0; when  = 0,  =


;
2

and when  = 1, 1 = .)

 00 ()  0 ⇔ sin   0, so  is CD on intervals of the form (2 (2 + 1)).  is CU on intervals of the form
((2 − 1) 2). The inflection points of  are the points ( ), where  is an integer.
If  ≥ 1, then  0 () ≥ 0 for all , so  is increasing and has no extremum. If  ≤ −1, then  0 () ≤ 0 for all , so  is decreasing and has no extremum. If ||  1, then  0 ()  0 ⇔ cos   − ⇔  is in an interval of the form
(2 − 1  2 + 1 ) for some integer . These are the intervals on which  is increasing. Similarly, we find that  is decreasing on the intervals of the form (2 + 1  2( + 1) − 1 ). Thus,  has local maxima at the points
√
2 + 1 , where  has the values (2 + 1 ) + sin 1 = (2 + 1 ) + 1 − 2 , and  has local minima at the points
√
2 − 1 , where we have  (2 − 1 ) = (2 − 1 ) − sin 1 = (2 − 1 ) − 1 − 2 .

[continued]

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152

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APPLICATIONS OF DIFFERENTIATION

The transitional values of  are −1 and 1. The inflection points move vertically, but not horizontally, when  changes.
When || ≥ 1, there is no extremum. For ||  1, the maxima are spaced
2 apart horizontally, as are the minima. The horizontal spacing between maxima and adjacent minima is regular (and equals ) when  = 0, but the horizontal space between a local maximum and the nearest local minimum shrinks as || approaches 1.
27. (a)  () = 4 − 22 + 1. For  = 0, () = −22 + 1, a parabola whose vertex, (0 1), is the absolute maximum. For

  0,  () = 4 − 22 + 1 opens upward with two minimum points. As  → 0, the minimum points spread apart and move downward; they are below the -axis for 0    1 and above for   1. For   0, the graph opens downward, and

e

has an absolute maximum at  = 0 and no local minimum.

al

(b)  0 () = 43 − 4 = 4(2 − 1) [ 6= 0]. If  ≤ 0, 0 is the only critical number.
 00 () = 122 − 4, so  00 (0) = −4 and there is a local maximum at

rS

(0  (0)) = (0 1), which lies on  = 1 − 2 . If   0, the critical
√
numbers are 0 and ±1 . As before, there is a local maximum at

Fo

(0  (0)) = (0 1), which lies on  = 1 − 2 .

√ 
 00 ±1  = 12 − 4 = 8  0, so there is a local minimum at

√ 
√
 = ±1 . Here  ±1  = (12 ) − 2 + 1 = −1 + 1.

ot




√
√ 2
But ±1  −1 + 1 lies on  = 1 − 2 since 1 − ±1 
= 1 − 1.

3.7 Optimization Problems
First Number
1

Second Number

N

1. (a)

Product

22

22

21

42

3

20

60

4

19
18

90

6

17

102

7

16

112

8

15

than the second, since we can just interchange the numbers

76

5

We needn’t consider pairs where the first number is larger

120

2

9

14
13

130

11

12

have considered only integers in the table.

126

10

in such cases. The answer appears to be 11 and 12, but we

132

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SECTION 3.7

OPTIMIZATION PROBLEMS

¤

153

(b) Call the two numbers  and . Then  +  = 23, so  = 23 − . Call the product  . Then
 =  = (23 − ) = 23 − 2 , so we wish to maximize the function  () = 23 − 2 . Since  0 () = 23 − 2, we see that  0 () = 0 ⇔  =

23
2

= 115. Thus, the maximum value of  is  (115) = (115)2 = 13225 and it

occurs when  =  = 115.
Or: Note that  00 () = −2  0 for all , so  is everywhere concave downward and the local maximum at  = 115 must be an absolute maximum.
3. The two numbers are  and

100
100
2 − 100
100
. The critical
, where   0. Minimize  () =  +
.  0 () = 1 − 2 =



2

number is  = 10. Since  0 ()  0 for 0    10 and  0 ()  0 for   10, there is an absolute minimum at  = 10.
The numbers are 10 and 10.

 0 () = 1 − 2 = 0 ⇔  = 1 . (−1) = 0, 
2

1
2

e

5. Let the vertical distance be given by () = ( + 2) − 2 , −1 ≤  ≤ 2.

= 9 , and (2) = 0, so
4

rS

al

there is an absolute maximum at  = 1 . The maximum distance is
2
1 1
1
9
 2 = 2 + 2 − 4 = 4.

7. If the rectangle has dimensions  and , then its perimeter is 2 + 2 = 100 m, so  = 50 − . Thus, the area is

 =  = (50 − ). We wish to maximize the function () = (50 − ) = 50 − 2 , where 0    50. Since

Fo

0 () = 50 − 2 = −2( − 25), 0 ()  0 for 0    25 and 0 ()  0 for 25    50. Thus,  has an absolute maximum at  = 25, and (25) = 252 = 625 m2 . The dimensions of the rectangle that maximize its area are  =  = 25 m.
(The rectangle is a square.)

 0 () =

 () =


1 + 2

⇒

ot

9. We need to maximize  for  ≥ 0.

(1 +  2 ) − (2)
(1 −  2 )
(1 + )(1 − )
=
=
.  0 ()  0 for 0    1 and  0 ()  0
2 )2
(1 + 
(1 +  2 )2
(1 +  2 )2

11. (a)

N

for   1. Thus,  has an absolute maximum of  (1) = 1  at  = 1.
2

The areas of the three figures are 12,500, 12,500, and 9000 ft2 . There appears to be a maximum area of at least 12,500 ft2 .
(b) Let  denote the length of each of two sides and three dividers.
Let  denote the length of the other two sides.
(c) Area  = length × width =  · 
(d) Length of fencing = 750 ⇒ 5 + 2 = 750



(e) 5 + 2 = 750 ⇒  = 375 − 5  ⇒ () = 375 − 5   = 375 − 5 2
2
2
2
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154

¤

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

(f ) 0 () = 375 − 5 = 0 ⇒  = 75. Since 00 () = −5  0 there is an absolute maximum when  = 75. Then
 
 = 375 = 1875. The largest area is 75 375 = 14,0625 ft2 . These values of  and  are between the values in the first
2
2 and second figures in part (a). Our original estimate was low.

 = 15 × 106 , so  = 15 × 106. Minimize the amount of fencing, which is

13.

3 + 2 = 3 + 2(15 × 106) = 3 + 3 × 106 =  ().
 0 () = 3 − 3 × 1062 = 3(2 − 106 )2 . The critical number is  = 103 and
 0 ()  0 for 0    103 and  0 ()  0 if   103 , so the absolute minimum occurs when  = 103 and  = 15 × 103 .
The field should be 1000 feet by 1500 feet with the middle fence parallel to the short side of the field.
15. Let  be the length of the base of the box and  the height. The surface area is 1200 = 2 + 4

⇒  = (1200 − 2 )(4).

al

e

The volume is  = 2  = 2 (1200 − 2 )4 = 300 − 34 ⇒  0 () = 300 − 3 2 .
4
√
0
2
0
3 2
⇒  = 400 ⇒  = 400 = 20. Since  ()  0 for 0    20 and  0 ()  0 for
 () = 0 ⇒ 300 = 4 
  20, there is an absolute maximum when  = 20 by the First Derivative Test for Absolute Extreme Values (see page 253).

rS

If  = 20, then  = (1200 − 202 )(4 · 20) = 10, so the largest possible volume is 2  = (20)2 (10) = 4000 cm3 .
10 = (2)() = 22 , so  = 52 . The cost is

17.

() = 10(22 ) + 6[2(2) + 2] + 6(22 )

Fo

= 322 + 36 = 322 + 180

 0 () = 64 − 1802 = 4(163 − 45)2

3

45
.
16

The minimum cost is 


3

45
16

ot

 0 ()  0 for  

⇒ =




3

45
16

is the critical number.  0 ()  0 for 0   

√
= 32(28125)23 + 180 3 28125 ≈ $19128.

19. The distance  from the origin (0 0) to a point ( 2 + 3) on the line is given by  =

3

45
16

and


( − 0)2 + (2 + 3 − 0)2 and the

square of the distance is  = 2 = 2 + (2 + 3)2 .  0 = 2 + 2(2 + 3)2 = 10 + 12 and  0 = 0 ⇔  = − 6 . Now
5
 6
 6 3
 00 = 10  0, so we know that  has a minimum at  = − 6 . Thus, the -value is 2 − 5 + 3 = 3 and the point is − 5  5 .
5
5

N

21.



From the figure, we see that there are two points that are farthest away from

(1 0). The distance  from  to an arbitrary point  ( ) on the ellipse is

 = ( − 1)2 + ( − 0)2 and the square of the distance is

 =  2 = 2 − 2 + 1 +  2 = 2 − 2 + 1 + (4 − 42 ) = −32 − 2 + 5.
 0 = −6 − 2 and  0 = 0 ⇒  = − 1 . Now  00 = −6  0, so we know
3
that  has a maximum at  = − 1 . Since −1 ≤  ≤ 1, (−1) = 4,
3

 1
 − 3 = 16 , and (1) = 0, we see that the maximum distance is 16 . The corresponding -values are
3
3



√
√ 
 2

 = ± 4 − 4 − 1 = ± 32 = ± 4 2 ≈ ±189. The points are − 1  ± 4 2 .
3
9
3
3
3

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SECTION 3.7

¤

OPTIMIZATION PROBLEMS

The area of the rectangle is (2)(2) = 4. Also 2 = 2 +  2 so
√
√
 = 2 − 2 , so the area is () = 4 2 − 2 . Now


√
2
2 − 22
2 − 2 − √
. The critical number is
0 () = 4
=4√
2 − 2
2 − 2

23.

=
=



2 −

and 2 =



1
√ 
2

√
2 .

2

=



1 2
2

=

1
√ 
2

1
√ .
2

Clearly this gives a maximum.

= , which tells us that the rectangle is a square. The dimensions are 2 =

The height  of the equilateral triangle with sides of length  is

25.

since 2 + (2)2 = 2

√
3 =

Using similar triangles,

√
3
2 

3

2

−

√
3

2

e

√
3
.
2

√

=

−

=


2
√
⇒  = 23  − 3  ⇒  =
√

al

=

⇒ 2 = 2 − 1 2 = 3 2
4
4

√
3
2

√
2

,

⇒
√
3 ⇒

√
3
2 (

− 2).

√
3

2

−

√

3

4

=

√
3
,
4

so the dimensions are 2 and

Fo

 = 4, and  =

rS

√
√
√
The area of the inscribed rectangle is () = (2) = 3 ( − 2) = 3  − 2 3 2 , where 0 ≤  ≤ 2. Now
√
√
√  √ 
0 = 0 () = 3  − 4 3  ⇒  = 3  4 3 = 4. Since (0) = (2) = 0, the maximum occurs when
√

3
.
4

The area of the triangle is

27.

ot

() = 1 (2)( + ) = ( + ) =
2

N

√
2 + 
√
= 2 − 2
2 − 2


√
2 − 2 ( + ). Then

√
−2
−2
0 = 0 () =  √
+ 2 − 2 +  √
2 − 2
2 
2 2 − 2
√
2 + 
= −√
+ 2 − 2 ⇒
2 − 2


⇒ 2 +  = 2 − 2

⇒ 0 = 22 +  − 2 = (2 − )( + ) ⇒

 = 1  or  = −. Now () = 0 = (−) ⇒ the maximum occurs where  = 1 , so the triangle has
2
2


√
 1 2 height  + 1  = 3  and base 2 2 − 2  = 2 3 2 = 3 .
2
2
4
29.

The cylinder has volume  =  2 (2). Also 2 +  2 = 2

⇒  2 = 2 − 2 , so

 () = (2 − 2 )(2) = 2(2  − 3 ), where 0 ≤  ≤ .
√


 0 () = 2 2 − 32 = 0 ⇒  =  3. Now  (0) =  () = 0, so there is a

√
 √ 
 √ 
 √  maximum when  =  3 and   3 = (2 − 2 3) 2 3 = 43  3 3 .

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155

156

¤

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

The cylinder has surface area

31.

2(area of the base) + (lateral surface area) = 2(radius)2 + 2(radius)(height)
= 22 + 2(2)
√
 = 2 − 2 , so the surface area is
√
() = 2(2 − 2 ) + 4 2 − 2  0 ≤  ≤ 

 √
= 22 − 22 + 4  2 − 2


 0 () = 0 − 4 + 4  · 1 (2 − 2 )−12 (−2) + (2 − 2 )12 · 1
Thus,
2
√


√
2
− 2 − 2 − 2 + 2 − 2
2 − 2
√
= 4 − − √
= 4 ·
+ 
2 − 2
2 − 2
√
 √
2
 0 () = 0 ⇒  2 − 2 = 2 − 22 () ⇒  2 − 2 = (2 − 22 )2 ⇒

2 (2 − 2 ) = 4 − 42 2 + 44

⇒  2 = 2 − 2

⇒

⇒ 2 2 − 4 = 4 − 42 2 + 44

√
5± 5 2
 ,
10

but we reject the root with the + sign since it
 √ doesn’t satisfy (). [The right side is negative and the left side is positive.] So  = 5 − 5 . Since (0) = () = 0, the
10

the surface area is

 √  √
 √ 
2 5 + 5 2 + 4 5 − 5 5 + 5 2 = 2 2 ·
10
10
10


√
5− 5 2

10

⇒  2 = 2 −

rS

maximum surface area occurs at the critical number and 2 =

al

This is a quadratic equation in 2 . By the quadratic formula, 2 =

⇒ 54 − 52 2 + 4 = 0.

e

Now 2 +  2 = 2

√
5+ 5
10

+4

√

(5−



√
5)(5+ 5)

10

Fo

√
√
2 5+ 5 + 2·2 5
5



=

= 2





√
5+5 5
5

= 2



√
5− 5 2

10



√
5+ 5
5

+

=

√
5+ 5 2

10

⇒

√ 
2 20
5

√ 

= 2 1 + 5 .

 = 384 ⇒  = 384. Total area is

33.

ot

() = (8 + )(12 + 384) = 12(40 +  + 256), so

0 () = 12(1 − 2562 ) = 0 ⇒  = 16. There is an absolute minimum

N

when  = 16 since 0 ()  0 for 0    16 and 0 ()  0 for   16.

When  = 16,  = 38416 = 24, so the dimensions are 24 cm and 36 cm.

Let  be the length of the wire used for the square. The total area is
  2 1  10 −   √3  10 −  
+
() =
4
2
3
2
3

35.

=
√

√

√

1 2
16 

+

√

3
36 (10

− )2 , 0 ≤  ≤ 10
√

3
9
3
0 () = 1  − 18 (10 − ) = 0 ⇔ 72  + 4723  − 40 3 = 0 ⇔  = 9 40 4√3 .
8
72
+
√ 
 √ 
3
3
Now (0) = 36 100 ≈ 481, (10) = 100 = 625 and  9 40 4√3 ≈ 272, so
16
+

(a) The maximum area occurs when  = 10 m, and all the wire is used for the square.

(b) The minimum area occurs when  =

√
40 √
3
9+4 3

≈ 435 m.

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SECTION 3.7

37.

OPTIMIZATION PROBLEMS

¤

The volume is  = 2  and the surface area is



2
= 2 +
.
() = 2 + 2 = 2 + 2
2



2
0
3
 () = 2 − 2 = 0 ⇒ 2 = 2 ⇒  = 3 cm. 





This gives an absolute minimum since  0 ()  0 for 0    3 and  0 ()  0 for   3
.







3 
When  =
,=
cm.
=
= 3

2

()23
2 + 2 = 2

39.

 0 () =

⇒  =


(2
3

 2
3 

=

2

3 (

− 32 ) = 0 when  =

− 2 ) =

1
√ .
3

2

3 ( 

− 3 ).

This gives an absolute maximum, since
1
√ .
3

The maximum volume is

By similar triangles,

 −

=



(1). The volume of the inner cone is  = 1 2 ,
3

rS

41.

al

e

1
 0 ()  0 for 0    √3  and  0 ()  0 for  




1
1
1
2
 √3  =  √3 3 − 3√3 3 = 9√3 3 .
3

so we’ll solve (1) for .

 − 


=
= ( − ) (2).




 2 
 · ( − ) =
(2 − 3 ) ⇒
3

3

Fo

=−


= − ⇒


Thus, () =



(2 − 32 ) =
(2 − 3).
3
3
  1  1

 0 () = 0 ⇒  = 0 or 2 = 3 ⇒  = 2  and from (2),  =
 = 3 .
 − 2 =
3
3

 3
 0 () changes from positive to negative at  = 2 , so the inner cone has a maximum volume of
3
 2 2  1 
2
2
1
1
4
1
 = 27 · 3  , which is approximately 15% of the volume of the larger cone.
 = 3   = 3  3 
3

N

ot

 0 () =

43.  () =

2
( + )2

⇒

 0 () =
=
 0 () = 0 ⇒  = 

( + )2 ·  2 −  2  · 2( + )
(2 + 2 + 2 ) 2 − 2 2 2 − 2 2 
=
2 ]2
[( + )
( + )4
 2 2 −  2 2
 2 (2 − 2 )
 2 ( + )( − )
 2 ( − )
=
=
=
( + )4
( + )4
( + )4
( + )3
⇒  () =

2 
2
2
=
=
.
2
2
( + )
4
4

The expression for  0 () shows that  0 ()  0 for    and  0 ()  0 for   . Thus, the maximum value of the power is  2 (4), and this occurs when  = .
45.  = 6 − 3 2 cot  + 32
2

(a)

√
3
2

csc 

√
√



= 3 2 csc2  − 32 23 csc  cot  or 3 2 csc  csc  − 3 cot  .
2
2

c
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157

158

¤

(b)

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

√

= 0 when csc  − 3 cot  = 0 ⇒


√ cos 
1
− 3
= 0 ⇒ cos  = sin  sin 
 
1
that the minimum surface area occurs when  = cos−1 √3 ≈ 55◦ .

(c)

If cos  =

1
√ ,
3

then cot  =

1
 = 6 − 3 2 √2 +
2

1
√
2

and csc  =

√
√3 ,
2

√ √
32 23 √3
2

= 6 −


6
1
= 6 + 2√2 2 = 6  + 2√2 

The First Derivative Test shows

so the surface area is
3
√ 2
2 2

+

9
√ 2
2 2

√
√
2 + 25
5−

1
+
, 0 ≤  ≤ 5 ⇒  0 () = √
− = 0 ⇔ 8 = 6 2 + 25 ⇔
2 + 25
6
8
8
6 

162 = 9(2 + 25) ⇔  =

15
√ .
7

But

15
√
7

 5, so  has no critical number. Since  (0) ≈ 146 and  (5) ≈ 118, he

e

47. Here  () =

1
√ .
3

49. There are (6 − ) km over land and

√
2 + 4 km under the river.

al

should row directly to .

rS

We need to minimize the cost  (measured in $100,000) of the pipeline.

√
2 + 4 (8) ⇒
() = (6 − )(4) +
8
.
 0 () = −4 + 8 · 1 (2 + 4)−12 (2) = −4 + √
2
2 + 4

ot

Fo

√
8
 0 () = 0 ⇒ 4 = √
2 + 4 = 2 ⇒ 2 + 4 = 42 ⇒ 4 = 32 ⇒ 2 = 4 ⇒
⇒
3
2 +4

√
√
 = 2 3 [0 ≤  ≤ 6]. Compare the costs for  = 0, 2 3, and 6. (0) = 24 + 16 = 40,
√
√
√
√
 √ 
 2 3 = 24 − 8 3 + 32 3 = 24 + 24 3 ≈ 379, and (6) = 0 + 8 40 ≈ 506. So the minimum cost is about
√
$379 million when  is 6 − 2 3 ≈ 485 km east of the refinery.
The total illumination is () =

N

51.

 0 () =

3

+
, 0    10. Then
2
(10 − )2

−6
2
+
= 0 ⇒ 6(10 − )3 = 23
3
(10 − )3

⇒

√
√
√
√
√
3
3 (10 − ) =  ⇒ 10 3 3 − 3 3  =  ⇒ 10 3 3 =  + 3 3  ⇒
√
√
√ 

10 3 3
√ ≈ 59 ft. This gives a minimum since  00 ()  0 for 0    10.
10 3 3 = 1 + 3 3  ⇒  =
1+ 33
3(10 − )3 = 3

53.

⇒

Every line segment in the first quadrant passing through ( ) with endpoints on the and -axes satisfies an equation of the form  −  = ( − ), where   0. By setting



 = 0 and then  = 0, we find its endpoints, (0  − ) and   −   0 . The



distance  from  to  is given by  = [  −  − 0]2 + [0 − ( − )]2 .

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.7

OPTIMIZATION PROBLEMS

¤

159

It follows that the square of the length of the line segment, as a function of , is given by

2
2

2
+ ( − )2 = 2 −
() =  −
+ 2 + 2 2 − 2 + 2 . Thus,




2
22
2
− 3 + 22  − 2 = 3 ( − 2 + 2 4 − 3 )
2



2
2
= 3 [( − ) + 3 ( − )] = 3 ( − )( + 3 )




2


Thus,  0 () = 0 ⇔  =  or  = − 3  . Since   0 and   0,  must equal − 3  . Since 3  0, we see







that  0 ()  0 for   − 3  and  0 ()  0 for   − 3  . Thus,  has its absolute minimum value when  = − 3  .
 0 () =

That value is
2 
   
 2 

2  √
2
√
3
3 2


+ − 3  −  =  + 2 +
 +
 −3  = + 3 


e

= 2 + 243 23 + 23 43 + 43 23 + 223 43 + 2 = 2 + 343 23 + 323 43 + 2

[= ( + )3 ] with  = 23 and  = 23 ,
√
so we can write it as (23 + 23 )3 and the shortest such line segment has length  = (23 + 23 )32 .

−

3


⇒ 0 = −

3
3
, so an equation of the tangent line at the point (  ) is
2

rS

55.  =

al

The last expression is of the form 3 + 32  + 3 2 +  3

3
6
3
3
= − 2 ( − ), or  = − 2  + . The -intercept [ = 0] is 6. The





-intercept [ = 0] is 2. The distance  of the line segment that has endpoints at the

36
(2 − 0)2 + (0 − 6)2 . Let  = 2 , so  = 42 + 2


Fo

intercepts is  =

⇒

√
72
= 8 ⇔ 4 = 9 ⇔ 2 = 3 ⇒  = 3.
3
√
216
 00 = 8 + 4  0, so there is an absolute minimum at  = 3 Thus,  = 4(3) + 36 = 12 + 12 = 24 and
3

√
√ hence,  = 24 = 2 6.
72
. 0 = 0 ⇔
3

ot

 0 = 8 −

()
 0 () − ()
, then, by the Quotient Rule, we have 0 () =
. Now 0 () = 0 when

2

N

57. (a) If () =

()
= (). Therefore, the marginal cost equals the average cost.

√
(b) (i) () = 16,000 + 200 + 432 , (1000) = 16,000 + 200,000 + 40,000 10 ≈ 216,000 + 126,491, so
 0 () − () = 0 and this gives  0 () =

(1000) ≈ $342,491. () = () =
 0 (1000) = 200 + 60

16,000
+ 200 + 412 , (1000) ≈ $34249unit.  0 () = 200 + 612 ,


√
10 ≈ $38974unit.

(ii) We must have  0 () = () ⇔ 200 + 612 =

16,000
+ 200 + 412


⇔ 232 = 16,000 ⇔

 = (8,000)23 = 400 units. To check that this is a minimum, we calculate
0 () =

2
−16,000
2
+ √ = 2 (32 − 8000). This is negative for   (8000)23 = 400, zero at  = 400,
2



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160

¤

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

and positive for   400, so  is decreasing on (0 400) and increasing on (400 ∞). Thus,  has an absolute minimum at  = 400. [Note: 00 () is not positive for all   0.]
(iii) The minimum average cost is (400) = 40 + 200 + 80 = $320unit.
59. (a) We are given that the demand function  is linear and (27,000) = 10, (33,000) = 8, so the slope is
10 − 8
27,000 − 33,000



1
1
= − 3000 and an equation of the line is  − 10 = − 3000 ( − 27,000) ⇒

1
 = () = − 3000  + 19 = 19 − (3000).

(b) The revenue is () = () = 19 − (23000) ⇒ 0 () = 19 − (1500) = 0 when  = 28,500. Since
00 () = −11500  0, the maximum revenue occurs when  = 28,500 ⇒ the price is (28,500) = $950.
61. (a) As in Example 6, we see that the demand function  is linear. We are given that (1000) = 450 and deduce that

1
1
so an equation is  − 450 = − 10 ( − 1000) or () = − 10  + 550.

440 − 450
1100 − 1000

1
= − 10 ,

e

(1100) = 440, since a $10 reduction in price increases sales by 100 per week. The slope for  is

(2750) = 275, so the rebate should be 450 − 275 = $175.

al

1
(b) () = () = − 10 2 + 550. 0 () = − 1  + 550 = 0 when  = 5(550) = 2750.
5

rS

1
1
(c) () = 68,000 + 150 ⇒  () = () − () = − 10 2 + 550 − 68,000 − 150 = − 10 2 + 400 − 68,000,

 0 () = − 1  + 400 = 0 when  = 2000. (2000) = 350. Therefore, the rebate to maximize profits should be
5

Fo

450 − 350 = $100.

Here 2 = 2 + 24, so 2 = 2 − 24. The area is  = 1 
2

63.

ot

Let the perimeter be , so 2 +  =  or  = ( − )2 ⇒


() = 1  ( − )24 − 24 =  2 − 24. Now
2

2 − 2
4
−3 + 2
0
−
 () =
= 
.
4
2 − 2
4 2 − 2


2 − 24.

N

Therefore, 0 () = 0 ⇒ −3 + 2 = 0 ⇒  = 3. Since 0 ()  0 for   3 and 0 ()  0 for   3, there is an absolute maximum when  = 3. But then 2 + 3 = , so  = 3 ⇒  =  ⇒ the triangle is equilateral.
65. Note that || = | | + | |

⇒ 5 =  + | | ⇒ | | = 5 − .

Using the Pythagorean Theorem for ∆  and ∆  gives us


() = | | + | | + | | =  + (5 − )2 + 22 + (5 − )2 + 32
√
√
=  + 2 − 10 + 29 + 2 − 10 + 34 ⇒
−5
−5
0 () = 1 + √
+√
. From the graphs of 
2 − 10 + 29
2 − 10 + 34 and 0 , it seems that the minimum value of  is about (359) = 935 m.

c
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SECTION 3.7

OPTIMIZATION PROBLEMS

¤

161

The total time is

67.

 () = (time from  to ) + (time from  to )

√
2 + ( − )2
2 + 2
+
, 0
=
1
2
 0 () =


−
sin 1 sin 2
√
− 
=
−
1
2
1 2 + 2
2 2 + ( − )2

The minimum occurs when  0 () = 0 ⇒

sin 1 sin 2
=
.
1
2

[Note:  00 ()  0]
 2 = 2 +  2 , but triangles  and  are similar, so
 √

√
8 =  4  − 4
⇒  = 2  − 4. Thus, we minimize

69.

( − 4)(32 ) − 3
2 [3( − 4) − ]
22 ( − 6)
=
=
=0
2
2
( − 4)
( − 4)
( − 4)2

al

 0 () =

e

 () =  2 = 2 + 42 ( − 4) = 3 ( − 4), 4   ≤ 8.

rS

when  = 6.  0 ()  0 when   6,  0 ()  0 when   6, so the minimum occurs when  = 6 in.

It suffices to maximize tan . Now

71.

Fo

3 tan  + tan 
 + tan 
= tan( + ) =
=
. So
1
1 − tan  tan 
1 −  tan 

3(1 −  tan ) =  + tan 

=

1
√
3

2
.
1 + 32





2 1 + 32 − 2(6)
2 1 − 32
⇒  () =
=
= 0 ⇔ 1 − 32 = 0 ⇔
(1 + 32 )2
(1 + 32 )2

since  ≥ 0. Now  0 ()  0 for 0 ≤  

 √ 
2 1 3
1
and tan  =
 √ 2 = √
3
1 + 3 1 3
√


⇒  = .
3 = tan  + 
6
6

N

73.

⇒ tan  =

0

ot

2
Let () = tan  =
1 + 32

⇒ 2 = (1 + 32 ) tan 

⇒ =

1
√
3


6.

and  0 ()  0 for  

1
√ ,
3

so  has an absolute maximum when  =

1
√
3

Substituting for  and  in 3 = tan( + ) gives us

In the small triangle with sides  and  and hypotenuse  , sin  =

 and 

cos  =



. In the triangle with sides  and  and hypotenuse , sin  = and



cos  =


. Thus,  =  sin ,  =  cos ,  =  sin , and  =  cos , so the


area of the circumscribed rectangle is
[continued]

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162

¤

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

() = ( + )( + ) = ( sin  +  cos )( cos  +  sin )
=  2 sin  cos  +   sin2  +  cos2  + 2 sin  cos 
=  sin2  +  cos2  + (2 +  2 ) sin  cos 
=  (sin2  + cos2 ) + (2 +  2 ) ·

1
2

· 2 sin  cos  =  + 1 (2 +  2 ) sin 2, 0 ≤  ≤
2

This expression shows, without calculus, that the maximum value of () occurs when sin 2 = 1 ⇔ 2 =
 
 =  . So the maximum area is   =  + 1 (2 +  2 ) = 1 (2 + 2 +  2 ) = 1 ( +  )2 .
4
4
2
2
2

75. (a)


2


2

⇒

If  = energykm over land, then energykm over water = 14.
√
So the total energy is  = 14 25 + 2 + (13 − ), 0 ≤  ≤ 13,

and so

= 0: 14 = (25 + 2 )12


⇒ 1962 = 2 + 25 ⇒ 0962 = 25 ⇒  =

√5
096

e

Set

14

=
− .

(25 + 2 )12
≈ 51.

al

Testing against the value of  at the endpoints: (0) = 14(5) + 13 = 20, (51) ≈ 179, (13) ≈ 195.
Thus, to minimize energy, the bird should fly to a point about 51 km from .

rS

(b) If  is large, the bird would fly to a point  that is closer to  than to  to minimize the energy used flying over water.

Fo

If  is small, the bird would fly to a point  that is closer to  than to  to minimize the distance of the flight.
√
√
25 + 2



2 + (13 − ) ⇒
= √
=
. By the same sort of
−  = 0 when
 =  25 + 



25 + 2 argument as in part (a), this ratio will give the minimal expenditure of energy if the bird heads for the point  km from .
(c) For flight direct to ,  = 13, so from part (b),  =

√

25 + 132
13

≈ 107. There is no value of  for which the bird

should fly directly to . But note that lim () = ∞, so if the point at which  is a minimum is close to , then
 is large.

ot

→0+

N

(d) Assuming that the birds instinctively choose the path that minimizes the energy expenditure, we can use the equation for
√
 = 0 from part (a) with 14 = ,  = 4, and  = 1: (4) = 1 · (25 + 42 )12 ⇒  = 414 ≈ 16.

3.8 Newton's Method
1. (a)

The tangent line at  = 1 intersects the -axis at  ≈ 23, so
2 ≈ 23. The tangent line at  = 23 intersects the -axis at
 ≈ 3, so 3 ≈ 30.

(b) 1 = 5 would not be a better first approximation than 1 = 1 since the tangent line is nearly horizontal. In fact, the second approximation for 1 = 5 appears to be to the left of  = 1.

c
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SECTION 3.8 NEWTON’S METHOD

¤

163

3. Since the tangent line  = 9 − 2 is tangent to the curve  =  () at the point (2 5), we have 1 = 2, (1 ) = 5, and

 0 (1 ) = −2 [the slope of the tangent line]. Thus, by Equation 2,
2 = 1 −
Note that geometrically

9
2

 (1 )
5
9
=2−
=
 0 (1 )
−2
2

represents the -intercept of the tangent line  = 9 − 2.

5. The initial approximations 1 =  , and  will work, resulting in a second approximation closer to the origin, and lead to the

root of the equation  () = 0, namely,  = 0. The initial approximation 1 =  will not work because it will result in

5 −  − 1

. Now 1 = 1 ⇒
54 − 1


 
(125)5 − 125 − 1
1−1−1
= 1 − − 1 = 125 ⇒ 3 = 125 −
≈ 11785.
4
5−1
5(125)4 − 1

9.  () = 3 +  + 3

Now 1 = −1 ⇒

 0 () = 32 + 1, so +1 =  −

3 +  + 3


32 + 1


(−1)3 + (−1) + 3
−1 − 1 + 3
1
= −1 −
= −1 − = −125.
3(−1)2 + 1
3+1
4

ot

2 = −1 −

⇒

Fo

2 = 1 −

⇒  0 () = 54 − 1, so +1 =  −

rS

7.  () = 5 −  − 1

al

e

successive approximations farther and farther from the origin.

Newton’s method follows the tangent line at (−1 1) down to its intersection with

N

the -axis at (−125 0), giving the second approximation 2 = −125.
11. To approximate  =

+1 =  −

√
5
20 (so that 5 = 20), we can take  () = 5 − 20. So  0 () = 54 , and thus,

√
5 − 20

. Since 5 32 = 2 and 32 is reasonably close to 20 we’ll use 1 = 2. We need to find approximations
54


until they agree to eight decimal places. 1 = 2 ⇒ 2 = 185, 3 ≈ 182148614, 4 ≈ 182056514,
√
5 ≈ 182056420 ≈ 6 . So 5 20 ≈ 182056420, to eight decimal places.
Here is a quick and easy method for finding the iterations for Newton’s method on a programmable calculator.
(The screens shown are from the TI-84 Plus, but the method is similar on other calculators.) Assign  () = 5 − 20 to Y1 , and  0 () = 54 to Y2 . Now store 1 = 2 in X and then enter X − Y1 Y2 → X to get 2 = 185. By successively pressing the ENTER key, you get the approximations 3 , 4 ,    .
[continued]

c
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164

¤

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

In Derive, load the utility file SOLVE. Enter NEWTON(xˆ5-20,x,2) and then APPROXIMATE to get
[2, 1.85, 1.82148614, 1.82056514, 1.82056420]. You can request a specific iteration by adding a fourth argument. For example, NEWTON(xˆ5-20,x,2,2) gives [2 185 182148614].
In Maple, make the assignments  :=  → ˆ5 − 20;,  :=  →  −  ()()();, and  := 2;. Repeatedly execute the command  := (); to generate successive approximations.
In Mathematica, make the assignments  [_ ] := ˆ5 − 20, [_ ] :=  − [] 0 [], and  = 2 Repeatedly execute the

4 − 23 + 52 − 6



. We need to find
43 − 62 + 10



⇒  0 () = 43 − 62 + 10 ⇒ +1 =  −

al

13.  () = 4 − 23 + 52 − 6

e

command  = [] to generate successive approximations.

rS

approximations until they agree to six decimal places. We’ll let 1 equal the midpoint of the given interval, [1 2].
1 = 15 ⇒ 2 = 12625, 3 ≈ 1218808, 4 ≈ 1217563, 5 ≈ 1217562 ≈ 6 . So the root is 1217562 to six decimal places. +1 =  −

⇒  0 () = cos  − 2 ⇒

Fo

15. sin  = 2 , so  () = sin  − 2

sin  − 2

. From the figure, the positive root of sin  = 2 is cos  − 2

near 1. 1 = 1 ⇒ 2 ≈ 0891396, 3 ≈ 0876985, 4 ≈ 0876726 ≈ 5 . So

From the graph, we see that there appear to be points of intersection near
 = −4,  = −2, and  = 1. Solving 3 cos  =  + 1 is the same as solving

N

17.

ot

the positive root is 0876726, to six decimal places.

() = 3 cos  −  − 1 = 0.  0 () = −3 sin  − 1, so

+1 =  −

3 cos  −  − 1
.
−3 sin  − 1

1 = −4

1 = −2

1 = 1

2 ≈ −3682281

2 ≈ −1856218

2 ≈ 0892438

3 ≈ −3638960

3 ≈ −1862356

3 ≈ 0889473

4 ≈ −3637959

4 ≈ −1862365 ≈ 5

4 ≈ 0889470 ≈ 5

5 ≈ −3637958 ≈ 6
To six decimal places, the roots of the equation are −3637958, −1862365, and 0889470.

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.8 NEWTON’S METHOD

¤

From the graph, we see that there appear to be points of intersection near
√
 = −05 and  = 15. Solving 3  = 2 − 1 is the same as solving
√
√
3
3
 () =  − 2 + 1 = 0. () =  − 2 + 1 ⇒
√
3
 − 2 + 1
0
1 −23
− 2, so +1 =  − 1 −23 
.
 () = 3 
− 2
3 

19.

1 = −05

1 = 15

2 ≈ −0471421

2 ≈ 461653

3 ≈ −0471074 ≈ 4

3 ≈ 1461070 ≈ 4

To six decimal places, the roots are −0471074 and 1461070.
21. From the graph, there appears to be a point of intersection near  = 06.

√
√
 is the same as solving  () = cos  −  = 0.
 √ 
√
 () = cos  −  ⇒  0 () = − sin  − 1 2  , so

al

√

 √  . Now 1 = 06 ⇒ 2 ≈ 0641928,
− sin  − 1 2  cos  −

rS

+1 =  −

e

Solving cos  =

3 ≈ 0641714 ≈ 4 . To six decimal places, the root of the equation is 0641714.
() = 6 − 5 − 64 − 2 +  + 10 ⇒

23.

Fo

 0 () = 65 − 54 − 243 − 2 + 1 ⇒
+1 =  −

6 − 5 − 64 − 2 +  + 10




.
65 − 54 − 243 − 2 + 1




From the graph of  , there appear to be roots near −19, −12, 11, and 3.

1 = −19

1 = 11

1 = 3

2 ≈ −122006245

2 ≈ 114111662

2 ≈ 299

3 ≈ −193828380

3 ≈ −121997997 ≈ 4

3 ≈ 113929741

3 ≈ 298984106

4 ≈ 113929375 ≈ 5

4 ≈ 298984102 ≈ 5

ot

1 = −12

2 ≈ −194278290

N

4 ≈ −193822884

5 ≈ −193822883 ≈ 6

To eight decimal places, the roots of the equation are −193822883, −121997997, 113929375, and 298984102.
25.

165

Solving

√
1 − 2
1

− 1 −  = 0.  0 () = 2
+ √
+1
( + 1)2
2 1−
√

− 1 − 
2 +1

=  −
.
1
1 − 2

+ √
2 1 − 
(2 + 1)2


 () =

+1

√

= 1 −  is the same as solving
2 + 1
2

⇒

From the graph, we see that the curves intersect at about 08. 1 = 08 ⇒ 2 ≈ 076757581, 3 ≈ 076682610,

4 ≈ 076682579 ≈ 5 . To eight decimal places, the root of the equation is 076682579.

c
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166

¤

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

27. (a)  () = 2 − 

+1 =  −

⇒  0 () = 2, so Newton’s method gives



2 − 
1

1

1


=  −  +
=  +
=
.
 +
2
2
2
2
2
2


(b) Using (a) with  = 1000 and 1 =
So

√
1000 ≈ 31622777.

29.  () = 3 − 3 + 6

√
900 = 30, we get 2 ≈ 31666667, 3 ≈ 31622807, and 4 ≈ 31622777 ≈ 5 .

⇒  0 () = 32 − 3. If 1 = 1, then  0 (1 ) = 0 and the tangent line used for approximating 2 is

horizontal. Attempting to find 2 results in trying to divide by zero.
31. For  () = 13 ,  0 () =

1 −23

3

and
13

 ( )

=  − 1 −23 =  − 3 = −2 .
 0 ( )

3 

Therefore, each successive approximation becomes twice as large as the

2 = −2(05) = −1, and 3 = −2(−1) = 2.
33. (a)  () = 6 − 4 + 33 − 2

rS

converge to the root, which is 0. In the figure, we have 1 = 05,

al

previous one in absolute value, so the sequence of approximations fails to

e

+1 =  −

⇒  0 () = 65 − 43 + 92 − 2 ⇒

 00 () = 304 − 122 + 18. To find the critical numbers of  , we’ll find the

Fo

zeros of  0 . From the graph of  0 , it appears there are zeros at approximately
 = −13, −04, and 05. Try 1 = −13 ⇒
2 = 1 −

 0 (1 )
≈ −1293344 ⇒ 3 ≈ −1293227 ≈ 4 .
 00 (1 )

ot

Now try 1 = −04 ⇒ 2 ≈ −0443755 ⇒ 3 ≈ −0441735 ⇒ 4 ≈ −0441731 ≈ 5 . Finally try
1 = 05 ⇒ 2 ≈ 0507937 ⇒ 3 ≈ 0507854 ≈ 4 . Therefore,  = −1293227, −0441731, and 0507854 are

N

all the critical numbers correct to six decimal places.

(b) There are two critical numbers where  0 changes from negative to positive, so  changes from decreasing to increasing.
 (−1293227) ≈ −20212 and  (0507854) ≈ −06721, so −20212 is the absolute minimum value of  correct to four decimal places.
35.

 = 2 sin  ⇒  0 = 2 cos  + (sin )(2) ⇒
 00 = 2 (− sin ) + (cos )(2) + (sin )(2) + 2 cos 
= −2 sin  + 4 cos  + 2 sin 

⇒

 000 = −2 cos  + (sin )(−2) + 4(− sin ) + (cos )(4) + 2 cos 
= −2 cos  − 6 sin  + 6 cos .

From the graph of  = 2 sin , we see that  = 15 is a reasonable guess for the -coordinate of the inflection point. Using c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.9

ANTIDERIVATIVES

¤

167

Newton’s method with () =  00 and 0 () =  000 , we get 1 = 15 ⇒ 2 ≈ 1520092, 3 ≈ 1519855 ≈ 4 .
The inflection point is about (1519855 2306964).
37. We need to minimize the distance from (0 0) to an arbitrary point ( ) on the

 curve  = ( − 1)2 .  = 2 +  2 ⇒


() = 2 + [( − 1)2 ]2 = 2 + ( − 1)4 . When 0 = 0,  will be

minimized and equivalently,  = 2 will be minimized, so we will use Newton’s

method with  = 0 and  0 = 00 .
 () = 2 + 4( − 1)3

⇒  0 () = 2 + 12( − 1)2 , so +1 =  −

2 + 4( − 1)3
. Try 1 = 05 ⇒
2 + 12( − 1)2

2 = 04, 3 ≈ 0410127, 4 ≈ 0410245 ≈ 5 . Now (0410245) ≈ 0537841 is the minimum distance and the point on

e

the parabola is (0410245 0347810), correct to six decimal places.

375
[1 − (1 + )−60 ] ⇔ 48 = 1 − (1 + )−60


[multiply each term by (1 + )60 ] ⇔

rS

18,000 =


[1 − (1 + )− ] becomes


al

39. In this case,  = 18,000,  = 375, and  = 5(12) = 60. So the formula  =

48(1 + )60 − (1 + )60 + 1 = 0. Let the LHS be called  (), so that

 0 () = 48(60)(1 + )59 + 48(1 + )60 − 60(1 + )59

+1 =  −

Fo

= 12(1 + )59 [4(60) + 4(1 + ) − 5] = 12(1 + )59 (244 − 1)

48 (1 +  )60 − (1 +  )60 + 1
. An interest rate of 1% per month seems like a reasonable estimate for
12(1 +  )59 (244 − 1)

ot

 = . So let 1 = 1% = 001, and we get 2 ≈ 00082202, 3 ≈ 00076802, 4 ≈ 00076291, 5 ≈ 00076286 ≈ 6 .

N

Thus, the dealer is charging a monthly interest rate of 076286% (or 955% per year, compounded monthly).

3.9 Antiderivatives

1.  () =  − 3 = 1 − 3

⇒  () =

1+1
− 3 +  = 1 2 − 3 + 
2
1+1

Check:  0 () = 1 (2) − 3 + 0 =  − 3 = ()
2
3.  () =

1
2

+ 3 2 − 4 3
4
5

Check:  0 () =

1
2

⇒  () = 1  +
2

+ 1 (32 ) − 1 (43 ) + 0 =
4
5

5.  () = ( + 1)(2 − 1) = 22 +  − 1
7.  () = 725 + 8−45
9.  () =

3 2+1
4 3+1
−
+  = 1  + 1 3 − 1 4 + 
2
4
5
4 2+1
5 3+1
1
2

+ 3 2 − 4 3 =  ()
4
5

⇒  () = 2

1
3


3 + 1 2 −  +  = 2 3 + 1 2 −  + 
2
3
2



⇒  () = 7 5 75 + 8(515 ) +  = 575 + 4015 + 
7

√
√
2 is a constant function, so  () = 2  + .

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

168

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION


5
10−8


+ 1 = − 8 + 1
10
−8
4
−9
11.  () = 9 = 10 has domain (−∞ 0) ∪ (0 ∞), so  () =
 5


−
+ 2
48

if   0 if   0

See Example 1(c) for a similar problem.

13. () =

1 +  + 2
√
= −12 + 12 + 32


15. () = 2 sin  − sec2 

⇒ () = 212 + 2 32 + 2 52 + 
3
5



⇒ () = −2 cos  − tan  +  on the interval  −    +  .
2
2



and  −    +  for integers  ≥ 1. The antiderivative is
2
2




 () = 2 sec  + 12 + 0 on the interval 0  or  () = 2 sec  + 12 +  on the interval  −    +  for
2
2
2


17.  () = 2 sec  tan  + 1 −12 has domain 0
2


2



⇒  () = 5 ·

 (0) = 4 ⇒ 05 −

1
3

5
6
−2·
+  = 5 − 1 6 + .
3
5
6

· 06 +  = 4

⇒

 = 4, so  () = 5 − 1 6 + 4.
3

al

19.  () = 54 − 25

e

integers  ≥ 1.

The graph confirms our answer since  () = 0 when  has a local maximum,  is

rS

positive when  is increasing, and  is negative when  is decreasing.

 3
 2
 4



− 12
+6
+  = 54 − 43 + 32 + 
21.  () = 20 − 12 + 6 ⇒  () = 20
4
3
2
 4
 3
 5



−4
+3
+  +  = 5 − 4 + 3 +  + 
 () = 5
5
4
3
23.  00 () =

0

2

⇒  0 () =

2 23
3

25.  000 () = cos 

2
3



Fo

3

53
53

⇒  00 () = sin  + 1

+  = 2 53 + 
5

⇒  () =

⇒  0 () = − cos  + 1  + 

2
5



83
83



+  +  =

3 83
20 

⇒

+  + 

⇒  () = − sin  + 2 +  + ,

N

where  = 1 1 .
2
√



ot

00

27.  0 () = 1 + 3 



⇒  () =  + 3 2 32 +  =  + 232 + .  (4) = 4 + 2(8) +  and (4) = 25 ⇒
3

20 +  = 25 ⇒  = 5, so  () =  + 232 + 5.
29.  0 () =

√
(6 + 5) = 612 + 532

⇒  () = 432 + 252 + .

 (1) = 6 +  and  (1) = 10 ⇒  = 4, so  () = 432 + 252 + 4.
⇒  () = 2 sin  + tan  +  because −2    2.
√
√
√
√
 √
 
= 2 32 + 3 +  = 2 3 +  and   = 4 ⇒  = 4 − 2 3, so  () = 2 sin  + tan  + 4 − 2 3.
3

31.  0 () = 2 cos  + sec2 




3

33.  00 () = −2 + 12 − 122

⇒  0 () = −2 + 62 − 43 + .  0 (0) =  and  0 (0) = 12 ⇒  = 12, so

 0 () = −2 + 62 − 43 + 12 and hence,  () = −2 + 23 − 4 + 12 + .  (0) =  and (0) = 4 ⇒  = 4, so  () = −2 + 23 − 4 + 12 + 4. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.9

35.  00 () = sin  + cos 

ANTIDERIVATIVES

¤

169

⇒  0 () = − cos  + sin  + .  0 (0) = −1 +  and  0 (0) = 4 ⇒  = 5, so

 0 () = − cos  + sin  + 5 and hence, () = − sin  − cos  + 5 + .  (0) = −1 +  and  (0) = 3 ⇒  = 4, so  () = − sin  − cos  + 5 + 4.
37.  00 () = 4 + 6 + 242

⇒  0 () = 4 + 32 + 83 + 

⇒  () = 22 + 3 + 24 +  + .  (0) =  and

 (0) = 3 ⇒  = 3, so  () = 22 + 3 + 24 +  + 3. (1) = 8 +  and  (1) = 10 ⇒  = 2, so  () = 22 + 3 + 24 + 2 + 3.
39.  00 () = 2 + cos 

⇒  0 () = 2 + sin  +  ⇒  () = 2 − cos  +  + .
 

 
 
 = − 2 4 ⇒
 (0) = −1 +  and  (0) = −1 ⇒  = 0.   =  2 4 +   and   = 0 ⇒
2
2
2
2
 
 = −  , so  () = 2 − cos  −  .
2
2
⇒

12 + 1 +  = 6 ⇒

al

 = 4. Therefore,  () = 2 +  + 4 and (2) = 22 + 2 + 4 = 10.

e

41. Given  0 () = 2 + 1, we have  () = 2 +  + . Since  passes through (1 6),  (1) = 6

rS

43.  is the antiderivative of  . For small ,  is negative, so the graph of its antiderivative must be decreasing. But both  and 

are increasing for small , so only  can be  ’s antiderivative. Also,  is positive where  is increasing, which supports our conclusion. The graph of  must start at (0 1). Where the given graph,  =  (), has a

Fo

45.

local minimum or maximum, the graph of  will have an inflection point.
Where  is negative (positive),  is decreasing (increasing).

Where  changes from positive to negative,  will have a maximum.
Where  is decreasing (increasing),  is concave downward (upward).

N

ot

Where  changes from negative to positive,  will have a minimum.

47.


2

0
 () = 1


−1

if 0 ≤   1 if 1    2 if 2   ≤ 3


 2 + 

⇒  () =  + 


− + 

if 0 ≤   1

if 1    2 if 2   ≤ 3

 (0) = −1 ⇒ 2(0) +  = −1 ⇒  = −1. Starting at the point
(0 −1) and moving to the right on a line with slope 2 gets us to the point (1 1).
The slope for 1    2 is 1, so we get to the point (2 2). Here we have used the fact that  is continuous. We can include the point  = 1 on either the first or the second part of  . The line connecting (1 1) to (2 2) is  = , so  = 0. The slope for

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

170

¤

CHAPTER 3

APPLICATIONS OF DIFFERENTIATION

2   ≤ 3 is −1, so we get to (3 1).

(3) = 1 ⇒ −3 +  = 1 ⇒  = 4. Thus

2 − 1 if 0 ≤  ≤ 1

if 1    2
 () = 


− + 4 if 2 ≤  ≤ 3

Note that  0 () does not exist at  = 1 or at  = 2.
49.  () =

sin 
, −2 ≤  ≤ 2
1 + 2

Note that the graph of  is one of an odd function, so the graph of  will

51. () = 0 () = sin  − cos 

rS

al

e

be one of an even function.

⇒ () = − cos  − sin  + . (0) = −1 +  and (0) = 0 ⇒  = 1, so

() = − cos  − sin  + 1.

⇒ () = 2 +  + . (0) =  and (0) = −2 ⇒  = −2, so () = 2 +  − 2 and

Fo

53. () =  0 () = 2 + 1

() = 1 3 + 1 2 − 2 + . (0) =  and (0) = 3 ⇒  = 3, so () = 1 3 + 1 2 − 2 + 3.
3
2
3
2
55. () =  0 () = 10 sin  + 3 cos 

⇒ () = −10 cos  + 3 sin  + 

⇒ () = −10 sin  − 3 cos  +  + .
6
.


Thus,

ot

(0) = −3 +  = 0 and (2) = −3 + 2 +  = 12 ⇒  = 3 and  =
() = −10 sin  − 3 cos  +

6



+ 3.

N

57. (a) We first observe that since the stone is dropped 450 m above the ground, (0) = 0 and (0) = 450.

0 () = () = −98 ⇒ () = −98 + . Now (0) = 0 ⇒  = 0, so () = −98 ⇒
() = −492 + . Last, (0) = 450 ⇒  = 450 ⇒ () = 450 − 492 .
(b) The stone reaches the ground when () = 0. 450 − 492 = 0 ⇒ 2 = 45049 ⇒ 1 =

(c) The velocity with which the stone strikes the ground is (1 ) = −98 45049 ≈ −939 ms.


45049 ≈ 958 s.

(d) This is just reworking parts (a) and (b) with (0) = −5. Using () = −98 + , (0) = −5 ⇒ 0 +  = −5 ⇒
() = −98 − 5. So () = −492 − 5 +  and (0) = 450 ⇒  = 450 ⇒ () = −492 − 5 + 450.
√


Solving () = 0 by using the quadratic formula gives us  = 5 ± 8845 (−98) ⇒ 1 ≈ 909 s.

59. By Exercise 58 with  = −98, () = −492 + 0  + 0 and () = 0 () = −98 + 0 . So



2
2
2
[()]2 = (−98 + 0 )2 = (98)2 2 − 1960  + 0 = 0 + 96042 − 1960  = 0 − 196 −492 + 0  .

2
But −492 + 0  is just () without the 0 term; that is, () − 0 . Thus, [()]2 = 0 − 196 [() − 0 ].

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 3.9

ANTIDERIVATIVES

¤

171

61. Using Exercise 58 with  = −32, 0 = 0, and 0 =  (the height of the cliff ), we know that the height at time  is

() = −162 + . () = 0 () = −32 and () = −120 ⇒ −32 = −120 ⇒  = 375, so
0 = (375) = −16(375)2 +  ⇒  = 16(375)2 = 225 ft.
63. Marginal cost = 192 − 0002 =  0 ()

⇒ () = 192 − 00012 + . But (1) = 192 − 0001 +  = 562 ⇒

 = 560081. Therefore, () = 192 − 00012 + 560081 ⇒ (100) = 742081, so the cost of producing
100 items is $74208.
65. Taking the upward direction to be positive we have that for 0 ≤  ≤ 10 (using the subscript 1 to refer to 0 ≤  ≤ 10),
0
1 () = − (9 − 09) = 1 () ⇒ 1 () = −9 + 0452 + 0 , but 1 (0) = 0 = −10 ⇒

1 () = −9 + 0452 − 10 = 0 () ⇒ 1 () = − 9 2 + 0153 − 10 + 0 . But 1 (0) = 500 = 0
1
2

⇒

e

1 () = − 9 2 + 0153 − 10 + 500. 1 (10) = −450 + 150 − 100 + 500 = 100, so it takes
2
more than 10 seconds for the raindrop to fall. Now for   10, () = 0 =  0 () ⇒

al

() = constant = 1 (10) = −9(10) + 045(10)2 − 10 = −55 ⇒ () = −55.

rS

At 55 ms, it will take 10055 ≈ 18 s to fall the last 100 m. Hence, the total time is 10 +
67. () = , the initial velocity is 30 mih = 30 ·

50 mih = 50 ·

5280
3600

=

220
3

5280
3600

100
55

=

130
11

≈ 118 s.

= 44 fts, and the final velocity (after 5 seconds) is

fts. So () =  +  and (0) = 44 ⇒  = 44. Thus, () =  + 44 ⇒
220
,
3

so 5 + 44 =

220
3

⇒

5 =

88
3

⇒

=

88
15

≈ 587 fts2 .

Fo

(5) = 5 + 44. But (5) =

69. Let the acceleration be () =  kmh2 . We have (0) = 100 kmh and we can take the initial position (0) to be 0.

We want the time  for which () = 0 to satisfy ()  008 km. In general,  0 () = () = , so () =  + , where  = (0) = 100. Now 0 () = () =  + 100, so () = 1 2 + 100 + , where  = (0) = 0.
2

−


2




1
100
1
5,000
1
100
 −
= 10,000
−
=−
. The condition ( ) must satisfy is
+ 100 −
2


2



N

( ) =

ot

Thus, () = 1 2 + 100. Since ( ) = 0, we have  + 100 = 0 or  = −100, so
2

5,000
5,000
 008 ⇒ −


008

[ is negative] ⇒   −62,500 kmh2 , or equivalently,

  − 3125 ≈ −482 ms2 .
648
71. (a) First note that 90 mih = 90 ×

5280
3600

 = 0. Now 4 = 132 when  =

fts = 132 fts. Then () = 4 fts2

132
4

⇒ () = 4 + , but (0) = 0 ⇒

= 33 s, so it takes 33 s to reach 132 fts. Therefore, taking (0) = 0, we have

() = 22 , 0 ≤  ≤ 33. So (33) = 2178 ft. 15 minutes = 15(60) = 900 s, so for 33   ≤ 933 we have
() = 132 fts ⇒ (933) = 132(900) + 2178 = 120,978 ft = 229125 mi.
(b) As in part (a), the train accelerates for 33 s and travels 2178 ft while doing so. Similarly, it decelerates for 33 s and travels
2178 ft at the end of its trip. During the remaining 900 − 66 = 834 s it travels at 132 fts, so the distance traveled is
132 · 834 = 110,088 ft. Thus, the total distance is 2178 + 110,088 + 2178 = 114,444 ft = 21675 mi. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

172

¤

CHAPTER 3 APPLICATIONS OF DIFFERENTIATION

(c) 45 mi = 45(5280) = 237,600 ft. Subtract 2(2178) to take care of the speeding up and slowing down, and we have
233,244 ft at 132 fts for a trip of 233,244132 = 1767 s at 90 mih. The total time is
1767 + 2(33) = 1833 s = 30 min 33 s = 3055 min.
(d) 375(60) = 2250 s. 2250 − 2(33) = 2184 s at maximum speed. 2184(132) + 2(2178) = 292,644 total feet or
292,6445280 = 55425 mi.

3 Review

1. A function  has an absolute maximum at  =  if  () is the largest function value on the entire domain of  , whereas  has

a local maximum at  if () is the largest function value when  is near . See Figure 6 in Section 3.1.

e

2. (a) See the Extreme Value Theorem on page 199.

al

(b) See the Closed Interval Method on page 202.

(b) See the definition of a critical number on page 201.
4. (a) See Rolle’s Theorem on page 208.

rS

3. (a) See Fermat’s Theorem on page 200.

(b) See the Mean Value Theorem on page 209. Geometric interpretation—there is some point  on the graph of a function 

Fo

[on the interval ( )] where the tangent line is parallel to the secant line that connects (  ()) and (  ()).
5. (a) See the Increasing/Decreasing (I/D) Test on page 214.

(b) If the graph of  lies above all of its tangents on an interval , then it is called concave upward on .

ot

(c) See the Concavity Test on page 217.
(d) An inflection point is a point where a curve changes its direction of concavity. They can be found by determining the points

N

at which the second derivative changes sign.
6. (a) See the First Derivative Test on page 215.

(b) See the Second Derivative Test on page 218.
(c) See the note before Example 7 in Section 3.3.
7. (a) See Definitions 3.4.1 and 3.4.5.

(b) See Definitions 3.4.2 and 3.4.6.
(c) See Definition 3.4.7.
(d) See Definition 3.4.3.
8. Without calculus you could get misleading graphs that fail to show the most interesting features of a function.

See Example 1 in Section 3.6.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 3 REVIEW

¤

173

9. (a) See Figure 2 in Section 3.8.

(b) 2 = 1 −

(1 )
 0 (1 )

(c) +1 =  −

 ( )
 0 ( )

(d) Newton’s method is likely to fail or to work very slowly when  0 (1 ) is close to 0. It also fails when  0 ( ) is undefined, such as with  () = 1 − 2 and 1 = 1.
10. (a) See the definition at the beginning of Section 3.9.

1. False.

e

(b) If 1 and 2 are both antiderivatives of  on an interval , then they differ by a constant.

For example, take  () = 3 , then  0 () = 32 and  0 (0) = 0, but  (0) = 0 is not a maximum or minimum;

For example, () =  is continuous on (0 1) but attains neither a maximum nor a minimum value on (0 1).

rS

3. False.

al

(0 0) is an inflection point.

Don’t confuse this with  being continuous on the closed interval [ ], which would make the statement true.
This is an example of part (b) of the I/D Test.

7. False.

 0 () =  0 () ⇒  () = () + . For example, if () =  + 2 and () =  + 1, then  0 () =  0 () = 1, but  () 6= ().

The graph of one such function is sketched.

11. True.

N

ot

9. True.

Fo

5. True.

Let 1  2 where 1  2 ∈ . Then  (1 )  (2 ) and (1 )  (2 ) [since  and  are increasing on  ],

so ( + )(1 ) =  (1 ) + (1 )   (2 ) + (2 ) = ( + )(2 ).
13. False.

Take  () =  and () =  − 1. Then both  and  are increasing on (0 1). But  () () = ( − 1) is not increasing on (0 1).

15. True.

Let 1  2 ∈  and 1  2 . Then  (1 )   (2 ) [ is increasing] ⇒

1
1

[ is positive] ⇒
 (1 )
 (2 )

(1 )  (2 ) ⇒ () = 1 () is decreasing on .
17. True.

If  is periodic, then there is a number  such that  ( + ) =  () for all . Differentiating gives
 0 () =  0 ( + ) · ( + )0 =  0 ( + ) · 1 =  0 ( + ), so  0 is periodic. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

174

CHAPTER 3 APPLICATIONS OF DIFFERENTIATION

By the Mean Value Theorem, there exists a number  in (0 1) such that  (1) −  (0) =  0 ()(1 − 0) =  0 ().

19. True.

Since  0 () is nonzero, (1) −  (0) 6= 0, so  (1) 6=  (0).

1.  () = 3 − 62 + 9 + 1, [2 4].

 0 () = 32 − 12 + 9 = 3(2 − 4 + 3) = 3( − 1)( − 3).  0 () = 0 ⇒

 = 1 or  = 3, but 1 is not in the interval.  0 ()  0 for 3    4 and  0 ()  0 for 2    3, so  (3) = 1 is a local minimum value. Checking the endpoints, we find  (2) = 3 and  (4) = 5. Thus, (3) = 1 is the absolute minimum value and
 (4) = 5 is the absolute maximum value.
3.  () =

3 − 4
(2 + 1)(3) − (3 − 4)(2)
−(32 − 8 − 3)
−(3 + 1)( − 3)
=
=
.
, [−2 2].  0 () =
2 +1
2 + 1)2

(
(2 + 1)2
(2 + 1)2

al

e

 0 () = 0 ⇒  = − 1 or  = 3, but 3 is not in the interval.  0 ()  0 for − 1    2 and  0 ()  0 for
3
3
 1
1
−5
9
−2    − 3 , so  − 3 = 109 = − 2 is a local minimum value. Checking the endpoints, we find  (−2) = −2 and
5.  () =  + 2 cos , [− ].

6

=

5
6

−


6

+

1
2

⇒ =

 5
, 6.
6

 0 ()  0 for

√
3 ≈ 226 is a local maximum value and

√
3 ≈ 089 is a local minimum value. Checking the endpoints, we find  (−) = − − 2 ≈ −514 and

Fo

 5 

is the absolute maximum value.

 0 () = 1 − 2 sin .  0 () = 0 ⇒ sin  =





 


−  and 5   , and  0 ()  0 for   5 , so   =
6
6
6
6
6



2
5

rS

 
 (2) = 2 . Thus,  − 1 = − 9 is the absolute minimum value and (2) =
5
3
2

 () =  − 2 ≈ 114. Thus,  (−) = − − 2 is the absolute minimum value and  maximum value.

6

=


6

+

√
3 is the absolute

5
4 = 3 + 0 + 0 = 1
1
6−0+0
2
4

ot

1
3+ 3 −
34 +  − 5

= lim
7. lim
2
→∞ 64 − 22 + 1
→∞
6− 2 +




lim

→−∞


√
√
√
4 + 12
42 + 1 2
42 + 1
√
= lim
= lim
→−∞ (3 − 1) 2
→−∞ −3 + 1
3 − 1

N

9.

=

11. lim

→∞

[since − = || =

√
2 for   0]

2
2
=−
−3 + 0
3

√
√
42 + 3 − 2
42 + 3 + 2
(42 + 3) − 42
·√
= lim √
2 + 3 + 2
→∞
→∞
1
4
42 + 3 + 2
√
3
3 2
= lim √
= lim √
 √
→∞
42 + 3 + 2 →∞
42 + 3 + 2  2

√

42 + 3 − 2 = lim

3
= lim 
→∞
4 + 3 + 2
=

[since  = || =

√
2 for   0]

3
3
=
2+2
4

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 3 REVIEW

¤

13.  (0) = 0,  0 (−2) =  0 (1) =  0 (9) = 0, lim  () = 0, lim () = −∞,
→∞

→6

 0 ()  0 on (−∞ −2), (1 6), and (9 ∞),  0 ()  0 on (−2 1) and (6 9),
 00 ()  0 on (−∞ 0) and (12 ∞),  00 ()  0 on (0 6) and (6 12)

15.  is odd,  0 ()  0 for 0    2,

17.  =  () = 2 − 2 − 3

 00 ()  0 for   3, lim→∞  () = −2

A.  = R B. -intercept:  (0) = 2.

H.

e

 00 ()  0 for 0    3,

 0 ()  0 for   2,

al

The -intercept (approximately 0770917) can be found using Newton’s
Method. C. No symmetry D. No asymptote

rS

E.  0 () = −2 − 32 = −(32 + 2)  0, so  is decreasing on R.

F. No extreme value G.  00 () = −6  0 on (0 ∞) and  00 ()  0 on

(−∞ 0), so  is CD on (0 ∞) and CU on (−∞ 0). There is an IP at (0 2).
A.  = R B. -intercept:  (0) = 0; -intercepts: () = 0 ⇔

Fo

19.  =  () = 4 − 33 + 32 −  = ( − 1)3

 = 0 or  = 1 C. No symmetry D.  is a polynomial function and hence, it has no asymptote.

N

ot

E.  0 () = 43 − 92 + 6 − 1. Since the sum of the coefficients is 0, 1 is a root of  0 , so




 0 () = ( − 1) 42 − 5 + 1 = ( − 1)2 (4 − 1).  0 ()  0 ⇒   1 , so  is decreasing on −∞ 1
4
4


H.
and  is increasing on 1  ∞ . F.  0 () does not change sign at  = 1, so
4
1
27
there is not a local extremum there.  4 = − 256 is a local minimum value.
G.  00 () = 122 − 18 + 6 = 6(2 − 1)( − 1).  00 () = 0 ⇔  =

 or 1.  00 ()  0 ⇔ 1    1 ⇒  is CD on 1  1 and CU on
2
2




1
−∞ 1 and (1 ∞). There are inflection points at 1  − 16 and (1 0).
2
2

21.  = () =

D.

lim

→±∞

1
( − 3)2

1
2

A.  = { |  6= 0 3} = (−∞ 0) ∪ (0 3) ∪ (3 ∞) B. No intercepts. C. No symmetry.

1
= 0, so  = 0 is a HA.
( − 3)2

so  = 0 and  = 3 are VA. E.  0 () = −

lim

→0+

1
1
1
= ∞, lim
= −∞, lim
= ∞,
2
→3 ( − 3)2
( − 3)2
→0− ( − 3)

( − 3)2 + 2( − 3)
3(1 − )
= 2
2 ( − 3)4
 ( − 3)3

⇒  0 ()  0 ⇔ 1    3,

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

175

176

¤

CHAPTER 3 APPLICATIONS OF DIFFERENTIATION

so  is increasing on (1 3) and decreasing on (−∞ 0), (0 1), and (3 ∞).
F. Local minimum value (1) =

1
4

G.  00 () =

H.

6(22 − 4 + 3)
.
3 ( − 3)4

Note that 22 − 4 + 3  0 for all  since it has negative discriminant.
So  00 ()  0 ⇔   0 ⇒  is CU on (0 3) and (3 ∞) and
CD on (−∞ 0). No IP
23.  =  () =

→∞

lim

→−8+

A.  = { |  6= −8} B. Intercepts are 0 C. No symmetry

2
64
= ∞, but  () − ( − 8) =
→ 0 as  → ∞, so  =  − 8 is a slant asymptote.
+8
+8

64
( + 16)
2
2
=
0 ⇔
= ∞ and lim
= −∞, so  = −8 is a VA. E.  0 () = 1 −
+8
( + 8)2
( + 8)2
→−8−  + 8

e

D. lim

2
64
= −8+
+8
+8

H.

  0 or   −16, so  is increasing on (−∞ −16) and (0 ∞) and

al

decreasing on (−16 −8) and (−8 0) 

F. Local maximum value  (−16) = −32, local minimum value  (0) = 0

rS

G.  00 () = 128( + 8)3  0 ⇔   −8, so  is CU on (−8 ∞) and
CD on (−∞ −8). No IP

√
2 +  A.  = [−2 ∞) B. -intercept: (0) = 0; -intercepts: −2 and 0 C. No symmetry

Fo

25.  =  () = 

√
3 + 4

1
√
+ 2+= √
[ + 2(2 + )] = √
= 0 when  = − 4 , so  is
3
2 2+
2 2+
2 2+

√



 

decreasing on −2 − 4 and increasing on − 4  ∞ . F. Local minimum value  − 4 = − 4 2 = − 4 9 6 ≈ −109,
3
3
3
3
3
no local maximum

ot

D. No asymptote E.  0 () =

H.

=

N

√
1
2 2 +  · 3 − (3 + 4) √
6(2 + ) − (3 + 4)
2+
=
G.  00 () =
4(2 + )
4(2 + )32
3 + 8
4(2 + )32

 00 ()  0 for   −2, so  is CU on (−2 ∞). No IP
27.  =  () = sin2  − 2 cos 

A.  = R B. -intercept:  (0) = −2 C.  (−) =  (), so  is symmetric with respect

to the -axis.  has period 2. D. No asymptote E. 0 = 2 sin  cos  + 2 sin  = 2 sin  (cos  + 1). 0 = 0 ⇔ sin  = 0 or cos  = −1 ⇔  =  or  = (2 + 1). 0  0 when sin   0, since cos  + 1 ≥ 0 for all .
Therefore, 0  0 [and so  is increasing] on (2 (2 + 1)); 0  0 [and so  is decreasing] on ((2 − 1) 2).
F. Local maximum values are  ((2 + 1)) = 2; local minimum values are (2) = −2.
G.  0 = sin 2 + 2 sin  ⇒

 00 = 2 cos 2 + 2 cos  = 2(2 cos2  − 1) + 2 cos  = 4 cos2  + 2 cos  − 2
= 2(2 cos2  + cos  − 1) = 2(2 cos  − 1)(cos  + 1)

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 3 REVIEW

 00 = 0 ⇔ cos  =

¤

177

H. or −1 ⇔  = 2 ±  or  = (2 + 1).
3
 00

 00  0 [and so  is CU] on 2 −   2 +  ;  ≤ 0 [and so  is CD]
3
3





5 on 2 + 3  2 + 3 . There are inflection points at 2 ±   − 1 .
3
4
1
2



3 (2) − 2 − 1 32
3 − 2
⇒  () =
=
6
4
 3

4
2
2
 (−2) − 3 −  4
2 − 12
=
 00 () =
8
5

2 − 1
29.  () =
3

0

⇒

Estimates: From the graphs of  0 and  00 , it appears that  is increasing on
(−173 0) and (0 173) and decreasing on (−∞ −173) and (173 ∞);
 (−17) = −038;  is CU on (−245 0) and (245 ∞), and CD on
(−∞ −245) and (0 245); and  has inflection points at about

al

(−245 −034) and (245 034).

e

 has a local maximum of about  (173) = 038 and a local minimum of about

3 − 2 is positive for 0  2  3, that is,  is increasing
4
 √ 
 √  on − 3 0 and 0 3 ; and  0 () is negative (and so  is decreasing) on
√ 
√

√

−∞ − 3 and
3 ∞ .  0 () = 0 when  = ± 3.
√
 0 goes from positive to negative at  = 3, so  has a local maximum of


√  (√3 )2 − 1
3 = √ 3 =
( 3)

√
2 3
;
9

Fo

rS

Exact: Now  0 () =

and since  is odd, we know that maxima on the

interval (0 ∞) correspond to minima on (−∞ 0), so  has a local minimum of

N

ot

√
 √ 
22 − 12 is positive (so  is CU) on
 − 3 = − 2 9 3 . Also,  00 () =
5
√ 
√


 √ 
6 ∞ , and negative (so  is CD) on −∞ − 6 and
− 6 0 and
 √
√ √ 
√ 
 √ 
0 6 . There are IP at
6 5366 and − 6 − 5366 .

31.  () = 36 − 55 + 4 − 53 − 22 + 2

⇒  0 () = 185 − 254 + 43 − 152 − 4 ⇒

 00 () = 904 − 1003 + 122 − 30 − 4

From the graphs of  0 and  00 , it appears that  is increasing on (−023 0) and (162 ∞) and decreasing on (−∞ −023) and (0 162);  has a local maximum of (0) = 2 and local minima of about  (−023) = 196 and  (162) = −192; c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

178

¤

CHAPTER 3 APPLICATIONS OF DIFFERENTIATION

 is CU on (−∞ −012) and (124 ∞) and CD on (−012 124); and  has inflection points at about (−012 198) and
(124 −121).

33. Let  () = 3 + 2 cos  + 5. Then (0) = 7  0 and  (−) = −3 − 2 + 5 = −3 + 3 = −3( − 1)  0 and since  is

continuous on R (hence on [− 0]), the Intermediate Value Theorem assures us that there is at least one zero of  in [− 0].
Now  0 () = 3 − 2 sin   0 implies that  is increasing on R, so there is exactly one zero of , and hence, exactly one real

e

root of the equation 3 + 2 cos  + 5 = 0.

√
√
5
√
33 − 5 32
= 5 33 − 2, but 1 −45  0 ⇒
5
33 − 32

rS

(32 33) such that  0 () = 1 −45 =
5

al

35. Since  is continuous on [32 33] and differentiable on (32 33), then by the Mean Value Theorem there exists a number  in

√
5
33 − 2  0 ⇒

 0 is decreasing, so that  0 ()   0 (32) = 1 (32)−45 = 00125 ⇒ 00125   0 () =
5
√
5
33  20125.

37. (a) () =  (2 )

√
5
33  20125.

⇒ 0 () = 2 0 (2 ) by the Chain Rule. Since  0 ()  0 for all  6= 0, we must have  0 (2 )  0 for

Fo

Therefore, 2 

√
5
33 − 2 ⇒

√
5
33  2. Also

 6= 0, so  0 () = 0 ⇔  = 0. Now  0 () changes sign (from negative to positive) at  = 0, since one of its factors,
 0 (2 ), is positive for all , and its other factor, 2, changes from negative to positive at this point, so by the First

ot

Derivative Test,  has a local and absolute minimum at  = 0.
(b)  0 () = 2 0 (2 ) ⇒  00 () = 2[ 00 (2 )(2) +  0 (2 )] = 42  00 (2 ) + 2 0 (2 ) by the Product Rule and the Chain

N

Rule. But 2  0 for all  6= 0,  00 (2 )  0 [since  is CU for   0], and  0 (2 )  0 for all  6= 0, so since all of its factors are positive,  00 ()  0 for  6= 0. Whether  00 (0) is positive or 0 doesn’t matter [since the sign of 00 does not change there];  is concave upward on R.





1 +   = |1 + 1 + | , so assume
√
39. If  = 0, the line is vertical and the distance from  = − to (1  1 ) is 


2 +  2

 6= 0. The square of the distance from (1  1 ) to the line is  () = ( − 1 )2 + ( − 1 )2 where  +  +  = 0, so


2


we minimize  () = ( − 1 )2 + −  −
− 1









− 1
.
⇒  0 () = 2 ( − 1 ) + 2 −  −
−






 2 1 − 1 − 
2
00 and this gives a minimum since  () = 2 1 + 2  0. Substituting
 () = 0 ⇒  =
2 +  2

0

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 3 REVIEW

¤

179

(1 + 1 + )2
, so the minimum distance is
2 +  2

this value of  into  () and simplifying gives  () =

|1 + 1 + |
√
() =
.
2 +  2

By similar triangles,

41.



, so the area of the triangle is
= √
2 − 2



2
⇒
() = 1 (2) =  = √
2
2 − 2
√
√
2 ( − 3)
2 2 − 2 − 2 ( − ) 2 − 2
0 () =
=
=0
2 − 2

(2 − 2)32 when  = 3

e

√
(92 )
0 ()  0 when 2    3, 0 ()  0 when   3. So  = 3 gives a minimum and (3) = √
= 3 3 2 .
3
√
We minimize () = | | + | | + | | = 2 2 + 16 + (5 − ),
√
√
0 ≤  ≤ 5. 0 () = 2 2 + 16 − 1 = 0 ⇔ 2 = 2 + 16 ⇔
 
4
4
42 = 2 + 16 ⇔  = √3 . (0) = 13,  √3 ≈ 119, (5) ≈ 128, so the

rS

al

43.

minimum occurs when  =



+



⇒



= 

2 () + ()



1

− 2



Fo

45.  = 





4
√
3

≈ 23.

=0 ⇔

1

= 2



⇔ 2 =  2

⇔  = .

This gives the minimum velocity since  0  0 for 0     and  0  0 for   .
47. Let  denote the number of $1 decreases in ticket price. Then the ticket price is $12 − $1(), and the average attendance is

ot

11,000 + 1000(). Now the revenue per game is

N

() = (price per person) × (number of people per game)
= (12 − )(11,000 + 1000) = −10002 + 1000 + 132,000

for 0 ≤  ≤ 4 [since the seating capacity is 15,000] ⇒ 0 () = −2000 + 1000 = 0 ⇔  = 05. This is a maximum since 00 () = −2000  0 for all . Now we must check the value of () = (12 − )(11,000 + 1000) at
 = 05 and at the endpoints of the domain to see which value of  gives the maximum value of .
(0) = (12)(11,000) = 132,000, (05) = (115)(11,500) = 132,250, and (4) = (8)(15,000) = 120,000. Thus, the maximum revenue of $132,250 per game occurs when the average attendance is 11,500 and the ticket price is $1150.
49.  () = 5 − 4 + 32 − 3 − 2

⇒  0 () = 54 − 43 + 6 − 3, so +1 =  −

5 − 4 + 32 − 3 − 2



.
54 − 43 + 6 − 3



Now 1 = 1 ⇒ 2 = 15 ⇒ 3 ≈ 1343860 ⇒ 4 ≈ 1300320 ⇒ 5 ≈ 1297396 ⇒

6 ≈ 1297383 ≈ 7 , so the root in [1 2] is 1297383, to six decimal places.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

180

¤

CHAPTER 3 APPLICATIONS OF DIFFERENTIATION

51.  () = cos  +  − 2

⇒  0 () = − sin  + 1 − 2.  0 () exists for all

, so to find the maximum of  , we can examine the zeros of  0 .
From the graph of  0 , we see that a good choice for 1 is 1 = 03.
Use () = − sin  + 1 − 2 and 0 () = − cos  − 2 to obtain
2 ≈ 033535293, 3 ≈ 033541803 ≈ 4 . Since  00 () = − cos  − 2  0 for all , (033541803) ≈ 116718557 is the absolute maximum.
53.  0 () =

√
√
3
3 + 2 = 32 + 23

55.  0 () = 2 − 3 sin 

⇒  () =

52
53
+
+  = 2 52 + 3 53 + 
5
5
52
53

⇒  () = 2 + 3 cos  + .

 (0) = 3 +  and  (0) = 5 ⇒  = 2, so  () = 2 + 3 cos  + 2.
⇒  0 () =  − 32 + 163 + .  0 (0) =  and  0 (0) = 2 ⇒  = 2, so

al

 0 () =  − 32 + 163 + 2 and hence, () = 1 2 − 3 + 44 + 2 + .
2

e

57.  00 () = 1 − 6 + 482

 (0) =  and  (0) = 1 ⇒  = 1, so  () = 1 2 − 3 + 44 + 2 + 1.
2
⇒ () = 2 + cos  + .

rS

59. () = 0 () = 2 − sin 

(0) = 0 + 1 +  =  + 1 and (0) = 3 ⇒  + 1 = 3 ⇒  = 2, so () = 2 + cos  + 2.

ot

Fo

61.  () = 2 sin(2 ), 0 ≤  ≤ 

63. Choosing the positive direction to be upward, we have () = −98

⇒ () = −98 + 0 , but (0) = 0 = 0

⇒

N

() = −98 = 0 () ⇒ () = −492 + 0 , but (0) = 0 = 500 ⇒ () = −492 + 500. When  = 0,


−492 + 500 = 0 ⇒ 1 = 500 ≈ 101 ⇒ (1 ) = −98 500 ≈ −98995 ms. Since the canister has been
49
49

designed to withstand an impact velocity of 100 ms, the canister will not burst.
65. (a)

The cross-sectional area of the rectangular beam is
√
 = 2 · 2 = 4 = 4 100 − 2 , 0 ≤  ≤ 10, so

 

= 4 1 (100 − 2 )−12 (−2) + (100 − 2 )12 · 4
2



4[−2 + 100 − 2 ]
−42
2 12
=
+ 4(100 −  )
=
.
(100 − 2 )12
(100 − 2 )12




= 0 when −2 + 100 − 2 = 0 ⇒


2 = 50 ⇒  =


√
√ 2 √
50 ≈ 707 ⇒  = 100 −
50 = 50.

Since (0) = (10) = 0, the rectangle of maximum area is a square.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 3 REVIEW

(b)

¤

The cross-sectional area of each rectangular plank (shaded in the figure) is
√ 
√
√ 
√

 = 2  − 50 = 2 100 − 2 − 50 , 0 ≤  ≤ 50, so
√ 
√
 

= 2 100 − 2 − 50 + 2 1 (100 − 2 )−12 (−2)
2

= 2(100 − 2 )12 − 2

Set

√
50 −

22
(100 − 2 )12

√
√

= 0: (100 − 2 ) − 50 (100 − 2 )12 − 2 = 0 ⇒ 100 − 22 = 50 (100 − 2 )12


⇒

e

10,000 − 4002 + 44 = 50(100 − 2 ) ⇒ 44 − 3502 + 5000 = 0 ⇒ 24 − 1752 + 2500 = 0 ⇒
√
√
175 ± 10,625
2
 =
≈ 6952 or 1798 ⇒  ≈ 834 or 424. But 834  50, so 1 ≈ 424 ⇒
4

√
√
 − 50 = 100 − 2 − 50 ≈ 199. Each plank should have dimensions about 8 1 inches by 2 inches.
1
2

al

(c) From the figure in part (a), the width is 2 and the depth is 2, so the strength is

10
√ .
3

The dimensions should be

20
√
3

≈ 1155 inches by

√
20 2
√
3

≈ 1633 inches.

N

ot

Fo

maximum strength occurs when  =

rS

 = (2)(2)2 = 8 2 = 8(100 − 2 ) = 800 − 83 , 0 ≤  ≤ 10.  = 800 − 242 = 0 when

√
√
10
√
242 = 800 ⇒ 2 = 100 ⇒  = √3 ⇒  = 200 = 10 3 2 = 2 . Since (0) = (10) = 0, the
3
3

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

181

N ot e

al

rS

Fo

PROBLEMS PLUS
1. Let () = sin  − cos  on [0 2] since  has period 2.  0 () = cos  + sin  = 0

⇔ cos  = − sin  ⇔
  √ tan  = −1 ⇔  = 3 or 7 . Evaluating  at its critical numbers and endpoints, we get  (0) = −1,  3 = 2,
4
4
4
√
√
√
 7 
 4 = − 2, and  (2) = −1. So  has absolute maximum value 2 and absolute minimum value − 2. Thus,
√
√
√
− 2 ≤ sin  − cos  ≤ 2 ⇒ |sin  − cos | ≤ 2.

3.  =

sin 


⇒ 0 =

 cos  − sin 
2

⇒  00 =

−2 sin  − 2 cos  + 2 sin 
. If ( ) is an inflection point,
3

then 00 = 0 ⇒ (2 − 2 ) sin  = 2 cos  ⇒ (2 − 2 )2 sin2  = 42 cos2  ⇒

⇒ (4 + 4)

sin2  sin 
.
= 4 ⇒ 2 (4 + 4) = 4 since  =
2


al

(4 + 4 ) sin2  = 42

e

(2 − 2 )2 sin2  = 42 (1 − sin2 ) ⇒ (4 − 42 + 4 ) sin2  = 42 − 42 sin2  ⇒

At a highest or lowest point,

rS

5. Differentiating 2 +  +  2 = 12 implicitly with respect to  gives 2 +  + 




2 + 
+ 2
= 0, so
=−
.



 + 2


= 0 ⇔  = −2. Substituting −2 for  in the original equation gives


Fo

2 + (−2) + (−2)2 = 12, so 32 = 12 and  = ±2. If  = 2, then  = −2 = −4, and if  = −2 then  = 4. Thus, the highest and lowest points are (−2 4) and (2 −4).
1
1
+
1 + ||
1 + | − 2|

1
 1 +

1 −  1 − ( − 2)




 1
1
+
=
1 +  1 − ( − 2)



 1
1


+

1 +  1 + ( − 2)

if   0

N

ot

7.  () =

if 0 ≤   2 if  ≥ 2

⇒


1
 1

(1 − )2 + (3 − )2




 −1
1
0
+
 () =
2
(3 − )2

(1 + )


 −1
1


−

(1 + )2
( − 1)2

if   0 if 0    2 if   2

We see that  0 ()  0 for   0 and  0 ()  0 for   2. For 0    2, we have
 0 () =

1
1
(2 + 2 + 1) − (2 − 6 + 9)
8 ( − 1)
−
=
=
, so  0 ()  0 for 0    1,
2
2
(3 − )
( + 1)
(3 − )2 ( + 1)2
(3 − )2 ( + 1)2

 0 (1) = 0 and  0 ()  0 for 1    2. We have shown that  0 ()  0 for   0;  0 ()  0 for 0    1;  0 ()  0 for
1    2; and  0 ()  0 for   2. Therefore, by the First Derivative Test, the local maxima of  are at  = 0 and  = 2, where  takes the value 4 . Therefore,
3






4
3

is the absolute maximum value of  .





9.  = 1  2 and  = 2  2 , where 1 and 2 are the solutions of the quadratic equation 2 =  + . Let  =  2
1
2



and set 1 = (1  0), 1 = (2  0), and 1 = ( 0). Let () denote the area of triangle  . Then  () can be expressed

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

183

184

¤

CHAPTER 3 PROBLEMS PLUS

in terms of the areas of three trapezoids as follows:
() = area (1 1 ) − area (1  1 ) − area (1  1 )






= 1 2 + 2 (2 − 1 ) − 1 2 + 2 ( − 1 ) − 1 2 + 2 (2 − )
1
2
1
2
2
2
2

After expanding and canceling terms, we get




 () = 1 2 2 − 1 2 − 2 + 1 2 − 2 2 + 2 = 1 2 (2 − ) + 2 ( − 1 ) + 2 (1 − 2 )
1
2
1
2
1
2
2
2


 0 () = 1 −2 + 2 + 2(1 − 2 ) .  00 () = 1 [2(1 − 2 )] = 1 − 2  0 since 2  1 .
1
2
2
2
 0 () = 0 ⇒ 2(1 − 2 ) = 2 − 2
1
2

⇒  = 1 (1 + 2 ).
2

 2 1




1 2 (2 − 1 ) + 2 1 (2 − 1 ) + 1 (1 + 2 )2 (1 − 2 )
2 2
4



 


 
= 1 1 (2 − 1 ) 2 + 2 − 1 (2 − 1 )(1 + 2 )2 = 1 (2 − 1 ) 2 2 + 2 − 2 + 21 2 + 2
1
2
1
2
1
2
2 2
4
8


= 1 (2 − 1 ) 2 − 21 2 + 2 = 1 (2 − 1 )(1 − 2 )2 = 1 (2 − 1 )(2 − 1 )2 = 1 (2 − 1 )3
1
2
8
8
8
8
1
2

e

 ( ) =

To put this in terms of  and , we solve the system  = 2 and  = 1 + , giving us 2 − 1 −  = 0 ⇒
1
1

al

√
√


√


3
 − 2 + 4 . Similarly, 2 = 1  + 2 + 4 . The area is then 1 (2 − 1 )3 = 1 2 + 4 ,
2
8
8




and is attained at the point    2 =  1  1 2 .

2
4
1
2

rS

1 =

Note: Another way to get an expression for  () is to use the formula for an area of a triangle in terms of the coordinates of

 
 

the vertices:  () = 1 2 2 − 1 2 + 1 2 − 2 + 2 − 2 2 .
1
2
1
2
2






⇒  0 () = − 2 +  − 6 sin 2 (2) + ( − 2). The derivative exists

Fo

11.  () = 2 +  − 6 cos 2 + ( − 2) + cos 1

for all , so the only possible critical points will occur where  0 () = 0 ⇔ 2( − 2)( + 3) sin 2 =  − 2 ⇔ either  = 2 or 2( + 3) sin 2 = 1, with the latter implying that sin 2 =

ot

1
1
 −1 or
 1. Solving these inequalities, we get
2( + 3)
2( + 3)

N

this equation has no solution whenever either
−7    −5.
2
2

1
. Since the range of sin 2 is [−1 1],
2( + 3)

13. (a) Let  = ||,  = ||, and 1 = ||, so that || · || = 1. We compute

the area A of 4 in two ways.
First, A =

1
2

|| || sin 2 =
3

1
2

·1·

√

3
2

=

√

3
.
4

Second,
A = (area of 4) + (area of 4) =
= 1 
2

√

3
2

+ 1 (1)
2

√
3
2

=

√
3
(
4

1
2

|| || sin  +
3

1
2

|| || sin 
3

+ 1)

Equating the two expressions for the area, we get

√

3

4



1
=
+


√

3
4

⇔ =


1
= 2
,   0.
 + 1
 +1

Another method: Use the Law of Sines on the triangles  and . In 4, we have

∠ + ∠ + ∠ = 180◦

⇔ 60◦ +  + ∠ = 180◦

⇔ ∠ = 120◦ − . Thus,

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CHAPTER 3 PROBLEMS PLUS

¤

185

√

√ cos  + 1 sin 

2
⇒
= 23 cot  + 1 , and by a
2
sin 

√

  similar argument with 4, 23 cot  = 2 + 1 . Eliminating cot  gives = 2 + 1 + 1 ⇒
2
2
2


,   0.
= 2
 +1

 sin(120◦ − ) sin 120◦ cos  − cos 120◦ sin 
=
=
=
 sin  sin 

3
2

(b) We differentiate our expression for  with respect to  to find the maximum:

(2 + 1) − (2)
1 − 2
=
= 2
= 0 when  = 1. This indicates a maximum by the First Derivative Test, since

(2 + 1)2
( + 1)2
 0 ()  0 for 0    1 and  0 ()  0 for   1, so the maximum value of  is (1) = 1 .
2
15. (a)  =

1

2

with sin  = , so  = 1  sin . But  is a constant,
2

al

e

so differentiating this equation with respect to , we get



1



=0=
 cos 
+
sin  +
 sin 
⇒

2











1 
1 
 cos 
= − sin  
+
⇒
= − tan 
+
.




 
 

rS

(b) We use the Law of Cosines to get the length of side  in terms of those of  and , and then we differentiate implicitly with

Fo

respect to : 2 = 2 + 2 − 2 cos  ⇒








= 2
+ 2
− 2 (− sin )
+
cos  +
 cos 
⇒
2









1





=

+
+  sin 
−
cos  −  cos  . Now we substitute our value of  from the Law of







Cosines and the value of  from part (a), and simplify (primes signify differentiation by ):
0 + 0 +  sin  [− tan (0 + 0)] − (0 + 0 )(cos )

√
=

2 + 2 − 2 cos 

ot

0 + 0 − [sin2  (0 + 0 ) + cos2  (0 + 0 )] cos 
0 + 0 − (0 + 0 ) sec 
√
√
=
2 + 2 − 2 cos 

2 + 2 − 2 cos 

2 | |
||
2 sec 
 − 2 tan 
, 2 =
+
=
+
,
1
1
2
1
2

N

=

17. (a) Distance = rate × time, so time = distancerate. 1 =

3 =

(b)

2


√
2 + 2 /4
42 + 2
=
.
1
1

2
2
2
=
· sec  tan  − sec2  = 0 when 2 sec 

1
2
1 sin 
1 1
−
=0 ⇒
1 cos 
2 cos 

sin 
1
=
1 cos 
2 cos 



1
1
tan  − sec 
1
2

⇒ sin  =



=0 ⇒

1
. The First Derivative Test shows that this gives
2

a minimum.
(c) Using part (a) with  = 1 and 1 = 026, we have 1 =
2
42 + 2 = 3 2
1

⇒ =

1
2


1

⇒ 1 =

1
026

≈ 385 kms. 3 =

√
42 + 2
1

⇒



1
 2 2 − 2 = 1 (034)2 (1026)2 − 12 ≈ 042 km. To find 2 , we use sin  =
3 1
2
2

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186

¤

CHAPTER 3 PROBLEMS PLUS

from part (b) and 2 =

2 sec 
 − 2 tan 
+
from part (a). From the figure,
1
2

1
2

2
1
⇒ sec  =  2 and tan  =  2
, so
2 − 2
2 − 2
1
1

 2 − 2 − 21
22
2
 1
2 =  2
+
. Using the values for 2 [given as 0.32],
1 2 − 2
2 2 − 2
1
2
1
sin  =


 2 − 2 − 21
22
2
 1
, 1 , and  we can graph Y1 = 2 and Y2 =  2
+
and find their intersection points.
1 2 − 2
2 2 − 2
1
2
1

Doing so gives us 2 ≈ 410 and 766, but if 2 = 410, then  = arcsin(1 2 ) ≈ 696◦ , which implies that point  is to the left of point  in the diagram. So 2 = 766 kms.
Let  = | | and  = | | as shown in the figure.

19.

e

Since  = | | + | |, | | =  − . Now

Let () =
1
2

−12
 2
−12

−

.
(−2) − 1 2 + 2 − 2
(−2) = √
+√
 − 2
2
2 − 2
2 +  2 − 2



 0 () = 0 ⇒

Fo

 0 () =

√
√
2 − 2 +  − 2 + 2 − 2.

rS

al

|| = | | + | | =  +  − 

√
2 − 2 +  − ( − )2 + 2

√
2
√
= 2 − 2 +  − ( − )2 +
2 − 2
√
√
= 2 − 2 +  − 2 − 2 + 2 + 2 − 2



√
= √
2 − 2
2 +  2 − 2



⇒

2
2
= 2
2 − 2
 + 2 − 2

⇒

ot

2 2 + 2 2 − 23 = 2 2 − 2 2 ⇒ 0 = 23 − 22 2 − 2 2 + 2 2 ⇒




⇒ 0 = 22 ( − ) − 2 ( + )( − ) ⇒ 0 = ( − ) 22 − 2 ( + )
0 = 22 ( − ) − 2 2 − 2

N

But     , so  6= . Thus, we solve 22 − 2  − 2 = 0 for :
 

√
− −2 ± (−2 )2 − 4(2)(−2 )
√
2 ± 4 + 8 2 2
=
. Because 4 + 82 2  2 , the “negative” can be
=
2(2)
4
√ √
√
√

2 +  2 + 82
 
2 + 2 2 + 82
=
[  0] =
 + 2 + 82 . The maximum discarded. Thus,  =
4
4
4
value of || occurs at this value of .

21.  =

4
3
3

⇒

Therefore, 42





= 42
. But is proportional to the surface area, so
=  · 42 for some constant .





=  · 42


⇔


=  = constant. An antiderivative of  with respect to  is , so  =  + .


When  = 0, the radius  must equal the original radius 0 , so  = 0 , and  =  + 0 . To find  we use the fact that

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 3 PROBLEMS PLUS
3
3 when  = 3,  = 3 + 0 and  = 1 0 ⇒ 4 (3 + 0 )3 = 1 · 4 0 ⇒ (3 + 0 )3 = 1 0 ⇒
2
3
2
3
2




1
1
1
3 + 0 = √ 0 ⇒  = 1 0 √ − 1 . Since  =  + 0 ,  = 1 0 √ − 1  + 0 . When the snowball
3
3
3
3
3
2

2

has melted completely we have  = 0 ⇒

2

1

3 0

1
√
3
2

√

332
− 1  + 0 = 0 which gives  = √
. Hence, it takes
3
2−1

N

ot

Fo

rS

al

e

√
332
3
√
−3= √
≈ 11 h 33 min longer.
3
3
2−1
2−1



c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

187

N ot e

al

rS

Fo

4

INTEGRALS

4.1 Areas and Distances
1. (a) Since  is increasing, we can obtain a lower estimate by using left

endpoints. We are instructed to use four rectangles, so  = 4.


4

8−0
−
=
=2
4 =
 (−1 ) ∆
∆ =

4
=1
=  (0 ) · 2 +  (1 ) · 2 +  (2 ) · 2 +  (3 ) · 2
= 2[ (0) +  (2) +  (4) + (6)]

Since  is increasing, we can obtain an upper estimate by using right

al

endpoints.
4

4 =
( ) ∆

e

= 2(2 + 375 + 5 + 575) = 2(165) = 33

rS

=1

= (1 ) · 2 + (2 ) · 2 +  (3 ) · 2 +  (4 ) · 2
= 2[ (2) +  (4) + (6) +  (8)]

= 2(375 + 5 + 575 + 6) = 2(205) = 41

Fo

Comparing 4 to 4 , we see that we have added the area of the rightmost upper rectangle,  (8) · 2, to the sum and

subtracted the area of the leftmost lower rectangle,  (0) · 2, from the sum.


8

8−0
=1
 (−1 )∆
∆ =
(b) 8 =
8
=1

ot

= 1[ (0 ) + (1 ) + · · · + (7 )]
=  (0) + (1) + · · · +  (7)

N

≈ 2 + 30 + 375 + 44 + 5 + 54 + 575 + 59
= 352
8 =

8


( )∆ =  (1) +  (2) + · · · +  (8)

 add rightmost upper rectangle,
= 8 + 1 · (8) − 1 ·  (0)
=1

subtract leftmost lower rectangle

= 352 + 6 − 2 = 392
3. (a) 4 =

4


=1

( ) ∆



∆ =

2 − 0

=
4
8

= [ (1 ) +  (2 ) + (3 ) +  (4 )] ∆


= cos  + cos 2 + cos 3 + cos 4 
8
8
8
8 8



=



4


=1


 ( ) ∆

≈ (09239 + 07071 + 03827 + 0)  ≈ 07908
8

Since  is decreasing on [0 2], an underestimate is obtained by using the right endpoint approximation, 4 . c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

189

190

¤

CHAPTER 4

(b) 4 =

4


INTEGRALS

 (−1 ) ∆ =

=1



4


=1


 (−1 ) ∆

= [(0 ) +  (1 ) +  (2 ) +  (3 )] ∆


= cos 0 + cos  + cos 2 + cos 3 
8
8
8 8

≈ (1 + 09239 + 07071 + 03827)  ≈ 11835
8

4 is an overestimate. Alternatively, we could just add the area of the leftmost upper rectangle and subtract the area of the
  rightmost lower rectangle; that is, 4 = 4 +  (0) ·  −   ·  .
8
2
8

2 − (−1)
=1 ⇒
3
3 = 1 ·  (0) + 1 · (1) + 1 ·  (2) = 1 · 1 + 1 · 2 + 1 · 5 = 8.

5. (a)  () = 1 + 2 and ∆ =

2 − (−1)
= 05 ⇒
6
6 = 05[ (−05) +  (0) +  (05) +  (1) +  (15) +  (2)]
∆ =

e

= 05(125 + 1 + 125 + 2 + 325 + 5)

rS

(b) 3 = 1 · (−1) + 1 · (0) + 1 ·  (1) = 1 · 2 + 1 · 1 + 1 · 2 = 5

al

= 05(1375) = 6875

6 = 05[ (−1) +  (−05) +  (0) +  (05) +  (1) + (15)]
= 05(2 + 125 + 1 + 125 + 2 + 325)

Fo

= 05(1075) = 5375

(c) 3 = 1 ·  (−05) + 1 ·  (05) + 1 ·  (15)

ot

= 1 · 125 + 1 · 125 + 1 · 325 = 575

6 = 05[(−075) +  (−025) + (025)

N

+  (075) +  (125) +  (175)]

= 05(15625 + 10625 + 10625 + 15625 + 25625 + 40625)
= 05(11875) = 59375

(d) 6 appears to be the best estimate.
7.  () = 2 + sin , 0 ≤  ≤ , ∆ = .

 = 2:

The maximum values of  on both subintervals occur at  =
 
  upper sum =   ·  +   · 
2
2
2
2
=3·


2

+3·


2


,
2

so

= 3 ≈ 942.

The minimum values of  on the subintervals occur at  = 0 and
 = , so lower sum =  (0) ·
=2·


2


2

+ () ·

+2·


2


2

= 2 ≈ 628.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 4.1

 = 4:

 = 8:

AREAS AND DISTANCES

¤

191

 
 
   
   upper sum =   +   +   +  3
4
2
2
4
4
√ 
√   


= 2 + 1 2 + (2 + 1) + (2 + 1) + 2 + 1 2
2
2
4
√  

= 10 + 2  ≈ 896
4


 
  lower sum =  (0) +  4 +  3 +  () 
4
4
√  
√ 


 
1
1
= (2 + 0) + 2 + 2 2 + 2 + 2 2 + (2 + 0) 
4
√   

= 8 + 2 4 ≈ 739
 
 
 
 
   upper sum =   +   +  3 +   +  
8
4
8
2
2
 5 
 3 
 7   
+  8 + 4 + 8
8
≈ 865

 
 
 
  lower sum =  (0) +   +   +  3 +  5
8
4
8
8
 
 
 
+  3 +  7 +  () 
4
8
8

e

≈ 786

al

9. Here is one possible algorithm (ordered sequence of operations) for calculating the sums:

1 Let SUM = 0, X_MIN = 0, X_MAX = 1, N = 10 (depending on which sum we are calculating),

rS

DELTA_X = (X_MAX - X_MIN)/N, and RIGHT_ENDPOINT = X_MIN + DELTA_X.
2 Repeat steps 2a, 2b in sequence until RIGHT_ENDPOINT  X_MAX.
2a Add (RIGHT_ENDPOINT)^4 to SUM.

Fo

Add DELTA_X to RIGHT_ENDPOINT.

At the end of this procedure, (DELTA_X)·(SUM) is equal to the answer we are looking for. We find that

100 =




1 100
100 =1

4

≈ 02533, 30


100

4


10


30
1 
=
30 =1




30

4

≈ 02170, 50

50
1 
=
50 =1




50

4

≈ 02101 and

ot

10

10
1 
=
10 =1

≈ 02050. It appears that the exact area is 02. The following display shows the program

N

SUMRIGHT and its output from a TI-83/4 Plus calculator. To generalize the program, we have input (rather than assign) values for Xmin, Xmax, and N. Also, the function, 4 , is assigned to Y1 , enabling us to evaluate any right sum

merely by changing Y1 and running the program.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

192

¤

CHAPTER 4

INTEGRALS

11. In Maple, we have to perform a number of steps before getting a numerical answer. After loading the student package

[command: with(student);] we use the command left_sum:=leftsum(1/(xˆ2+1),x=0..1,10 [or 30, or 50]); which gives us the expression in summation notation. To get a numerical approximation to the sum, we use evalf(left_sum);. Mathematica does not have a special command for these sums, so we must type them in manually. For example, the first left sum is given by
(1/10)*Sum[1/(((i-1)/10)ˆ2+1)],{i,1,10}], and we use the N command on the resulting output to get a numerical approximation.
In Derive, we use the LEFT_RIEMANN command to get the left sums, but must define the right sums ourselves.
(We can define a new function using LEFT_RIEMANN with  ranging from 1 to  instead of from 0 to  − 1.)
2


1
1
1 
, 0 ≤  ≤ 1, the left sums are of the form  =
. Specifically, 10 ≈ 08100,


+1
 =1 −1 2 + 1





1
1
. Specifically, 10 ≈ 07600,
 
 =1  2 + 1

al

30 ≈ 07937, and 50 ≈ 07904. The right sums are of the form  =

e

(a) With  () =

30 ≈ 07770, and 50 ≈ 07804.



rS

(b) In Maple, we use the leftbox (with the same arguments as left_sum) and rightbox commands to generate the

Fo

graphs.

left endpoints,  = 30

left endpoints,  = 50

N

ot

left endpoints,  = 10

right endpoints,  = 10

right endpoints,  = 30

right endpoints,  = 50

(c) We know that since  = 1(2 + 1) is a decreasing function on (0 1), all of the left sums are larger than the actual area, and all of the right sums are smaller than the actual area. Since the left sum with  = 50 is about 07904  0791 and the right sum with  = 50 is about 07804  0780, we conclude that 0780  50  exact area  50  0791, so the exact area is between 0780 and 0791.
13. Since  is an increasing function, 6 will give us a lower estimate and 6 will give us an upper estimate.

6 = (0 fts)(05 s) + (62)(05) + (108)(05) + (149)(05) + (181)(05) + (194)(05) = 05(694) = 347 ft
6 = 05(62 + 108 + 149 + 181 + 194 + 202) = 05(896) = 448 ft c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 4.1

AREAS AND DISTANCES

¤

193

15. Lower estimate for oil leakage: 5 = (76 + 68 + 62 + 57 + 53)(2) = (316)(2) = 632 L.

Upper estimate for oil leakage: 5 = (87 + 76 + 68 + 62 + 57)(2) = (35)(2) = 70 L.
17. For a decreasing function, using left endpoints gives us an overestimate and using right endpoints results in an underestimate.

We will use 6 to get an estimate. ∆ = 1, so
6 = 1[(05) + (15) + (25) + (35) + (45) + (55)] ≈ 55 + 40 + 28 + 18 + 10 + 4 = 155 ft
For a very rough check on the above calculation, we can draw a line from (0 70) to (6 0) and calculate the area of the triangle: 1 (70)(6) = 210. This is clearly an overestimate, so our midpoint estimate of 155 is reasonable.
2
2
, 1 ≤  ≤ 3. ∆ = (3 − 1) = 2 and  = 1 + ∆ = 1 + 2.
2 + 1

→∞

→∞ =1

( )∆ = lim




→∞ =1

2(1 + 2)
2
· .
(1 + 2)2 + 1 

√ sin , 0 ≤  ≤ . ∆ = ( − 0) =  and  = 0 +  ∆ = .

 = lim  = lim
→∞




→∞ =1

al

21.  () =




e

 = lim  = lim



 sin() · .
→∞ =1


( ) ∆ = lim

rS

19.  () =




 

tan can be interpreted as the area of the region lying under the graph of  = tan  on the interval 0  ,
4
→∞ =1 4
4

23. lim

ot

Fo



4 − 0


=
,  = 0 +  ∆ =
, and ∗ =  , the expression for the area is since for  = tan  on 0  with ∆ =

4

4
4
 






 = lim
. Note that this answer is not unique, since the expression for the area is
 (∗ ) ∆ = lim tan 
→∞ =1
→∞ =1
4 4

 the same for the function  = tan( − ) on the interval   +  , where  is any integer.
4

25. (a) Since  is an increasing function,  is an underestimate of  [lower sum] and  is an overestimate of  [upper sum].

(b)

N

Thus, ,  , and  are related by the inequality      .
 = (1 )∆ +  (2 )∆ + · · · +  ( )∆
 = (0 )∆ +  (1 )∆ + · · · +  (−1 )∆

 −  = ( )∆ − (0 )∆
= ∆[ ( ) −  (0 )]
=

−
[ () −  ()]


In the diagram,  −  is the sum of the areas of the shaded rectangles. By sliding the shaded rectangles to the left so that they stack on top of the leftmost shaded rectangle, we form a rectangle of height () −  () and width
(c)    , so  −    −  ; that is,  −  

−
.


−
[ () −  ()].


c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

194

¤

CHAPTER 4 INTEGRALS

2
2−0
2
=
and  = 0 +  ∆ = .



 5





 2
 325 2
64  5
2
 = lim  = lim
 ( ) ∆ = lim
· = lim
· = lim
 .
→∞
→∞ =1
→∞ =1

 →∞ =1 5  →∞ 6 =1

27. (a)  =  () = 5 . ∆ =

(b)




CAS

5 =

=1



2 ( + 1)2 22 + 2 − 1
12

 2
 2



2
2
 + 2 + 1 22 + 2 − 1
64  ( + 1) 2 + 2 − 1
64
= lim (c) lim 6 ·
→∞ 
12
12 →∞
2 · 2



16
2
2
1
1
=
2+ − 2 = lim 1 + + 2
3 →∞
 


29.  =  () = cos . ∆ =

16
3

·1·2=

32
3


−0

= and  = 0 +  ∆ = .




then  = sin  = 1.
2

rS


2,

If  =

al

e


 


1
 
+1
 sin 





2

 CAS
  CAS
 = sin 
 
·
 ( ) ∆ = lim cos = lim 
 = lim  = lim
−
→∞
→∞ =1
→∞ =1
→∞ 


2 

2 sin
2

4.2 The Definite Integral

14 − 2
−
=
= 2.

6

Fo

1.  () = 3 − 1 , 2 ≤  ≤ 14. ∆ =
2

Since we are using left endpoints, ∗ = −1 .

6

6 =
 (−1 ) ∆
=1

ot

= (∆) [ (0 ) +  (1 ) +  (2 ) +  (3 ) +  (4 ) + (5 )]

= 2[ (2) +  (4) +  (6) + (8) +  (10) +  (12)]

N

= 2[2 + 1 + 0 + (−1) + (−2) + (−3)] = 2(−3) = −6
The Riemann sum represents the sum of the areas of the two rectangles above the -axis minus the sum of the areas of the three rectangles below the -axis; that is, the net area of the rectangles with respect to the -axis.
3. 5 =

5


( ) ∆

=1

[∗ =  = 1 (−1 +  ) is a midpoint and ∆ = 1]

2

= 1 [(15) +  (25) +  (35)
+ (45) +  (55)]

[ () =

√
 − 2]

≈ −0856759
The Riemann sum represents the sum of the areas of the two rectangles above the -axis minus the sum of the areas of the three rectangles below the -axis.

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 4.2 THE DEFINITE INTEGRAL



5. (a)

(b)

 ()  ≈ 5 = [ (2) + (4) +  (6) +  (8) +  (10)] ∆
= [−1 + 0 + (−2) + 2 + 4](2) = 3(2) = 6
10

 ()  ≈ 5 = [ (0) +  (2) + (4) +  (6) +  (8)] ∆

0



(c)

0

195

10

0



¤

= [3 + (−1) + 0 + (−2) + 2](2) = 2(2) = 4
10

 ()  ≈ 5 = [ (1) +  (3) + (5) +  (7) +  (9)] ∆
= [0 + (−1) + (−1) + 0 + 3](2) = 1(2) = 2

7. Since  is increasing, 5 ≤

Lower estimate = 5 =

 30

5


10

 ()  ≤ 5 .

(−1 )∆ = 4[ (10) +  (14) +  (18) +  (22) +  (26)]

=1

5


 ()∆ = 4[ (14) + (18) +  (22) +  (26) +  (30)]

al

Upper estimate = 5 =

e

= 4[−12 + (−6) + (−2) + 1 + 3] = 4(−16) = −64
=1

rS

= 4[−6 + (−2) + 1 + 3 + 8] = 4(4) = 16

9. ∆ = (8 − 0)4 = 2, so the endpoints are 0, 2, 4, 6, and 8, and the midpoints are 1, 3, 5, and 7. The Midpoint Rule gives
0

sin

4
√
√
√ 
 √

√
  ≈
(¯ ) ∆ = 2 sin 1 + sin 3 + sin 5 + sin 7 ≈ 2(30910) = 61820.

=1

11. ∆ = (2 − 0)5 =

gives

0

2

so the endpoints are 0, 2 , 4 , 6 , 8 , and 2, and the midpoints are 1 , 3 , 5 ,
5 5 5 5
5 5 5

5


2
 ≈
 (¯ ) ∆ =

+1
5
=1



1
5

1
5

+1

+

ot



2
,
5

Fo

8

3
5

3
5

+1

+

5
5

5
5

+1

+

7
5
7
5

+1

+

9
5
9
5

+1



=

2
5



7
5

127
56

and 9 . The Midpoint Rule
5


=

127
≈ 09071.
140

N

13. In Maple 14, use the commands with(Student[Calculus1]) and

ReimannSum(x/(x+1),0..2,partition=5,method=midpoint,output=plot). In some older versions of
Maple, use with(student) to load the sum and box commands, then m:=middlesum(x/(x+1),x=0..2), which gives us the sum in summation notation, then M:=evalf(m) to get the numerical approximation, and finally middlebox(x/(x+1),x=0..2) to generate the graph. The values obtained for  = 5, 10, and 20 are 09071, 09029, and
09018, respectively.

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¤

196

CHAPTER 4 INTEGRALS

15. We’ll create the table of values to approximate


0

sin   by using the


program in the solution to Exercise 4.1.9 with Y1 = sin , Xmin = 0,

5

17. On [2 6], lim


 1 − 2

∆ =
4 + 2


→∞ =1

19. On [2 7], lim




2

6

1983524

50

1999342

100

The values of  appear to be approaching 2.

1933766

10

Xmax = , and  = 5, 10, 50, and 100.





1999836

1 − 2
.
4 + 2

[5(∗ )3 − 4∗ ] ∆ =



→∞ =1

7
2

(53 − 4) .

3
5−2
3
= and  = 2 +  ∆ = 2 + .








 5





3 
3 3
3
= lim
(4 − 2)  = lim
( ) ∆ = lim
 2+
4−2 2+
→∞ =1
→∞ =1
  →∞  =1

2





 




3
3
6
18
( + 1)
6
−
= lim
= lim
−
 = lim − 2
→∞  =1
→∞ 
→∞

 =1

2





18
1
+1
= lim −
= −9 lim 1 +
= −9(1) = −9
→∞
→∞
2



23. Note that ∆ =

rS

al

e

21. Note that ∆ =

2
0 − (−2)
2
= and  = −2 +  ∆ = −2 + .






2 



2 
2
2 2
2
−2 +
= lim
( + )  = lim
 ( ) ∆ = lim
 −2 +
+ −2 +
→∞ =1
→∞ =1
  →∞  =1


−2
0




2




Fo









2 
2  42
42
2
6
8
+ 2 −2+
= lim
+2
−
4−
→∞  =1
→∞  =1



2









8 ( + 1)(2 + 1)
2 4  2 6 
12 ( + 1)
2
= lim
− 2
+ · (2)
 −
+
2 = lim
→∞  2 =1
→∞ 3
 =1
6

2

=1






4 ( + 1)(2 + 1)
4  + 1 2 + 1
1
+1
= lim
+ 4 = lim
−6 1+
+4
−6
→∞ 3
→∞ 3
2




 





4
1
1
1
4
2
= lim
1+
2+
−6 1+
+ 4 = (1)(2) − 6(1) + 4 =
→∞ 3



3
3

N

ot

= lim

25. Note that ∆ =



0

1

1−0

1
= and  = 0 +  ∆ = .




 
 
 2 
3



1


∆ = lim
( − 3 )  = lim
 ( ) ∆ = lim

−3
→∞ =1
→∞ =1
→∞ =1




3




2











1  3
1 1  3
32
3  2
− 2 = lim
 − 2

→∞  =1 3
→∞  3 =1

 =1

 

2

1 ( + 1)
1 +1 +1
1  + 1 2 + 1
3 ( + 1)(2 + 1)
= lim
−
− 3
= lim
→∞
→∞ 4
4
2

6


2 






 
1
1
1
1
1
1
1
1
3
1+
1+
−
1+
2+
= (1)(1) − (1)(2) = −
= lim
→∞ 4


2


4
2
4
= lim

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 4.2 THE DEFINITE INTEGRAL

¤

197









( − ) 
− 
( − )2 
−
 = lim
27.
  = lim
1+

+
→∞
→∞
 =1


2
=1
=1






( − )
( − )2
( − )2 ( + 1)
1
+
=  ( − ) + lim
1+
·
→∞
→∞

2
2
2

 2



= ( − ) + 1 ( − )2 = ( − )  + 1  − 1  = ( − ) 1 ( + ) = 1  − 2
2
2
2
2
2

= lim

29.  () =



so

6

2




6−2
4
4
,  = 2,  = 6, and ∆ =
= . Using Theorem 4, we get ∗ =  = 2 +  ∆ = 2 + ,

1 + 5







 = lim  = lim
→∞
→∞ =1
1 + 5

4
2+
4
  5 · .

4
1+ 2+


31. ∆ = ( − 0) =  and ∗ =  = .


33. (a) Think of

so
5
0

2
0

2
0

()  = 1 (1 + 3)2 = 4.
2

 ()  =

5

9
7

 ()  +

1
(1
2

3
2

 ()  +

rectangle

+ 3)2 +

3·1

5
3

triangle

+

1
2

·2·3

= 4 + 3 + 3 = 10
7
5

()  = − 1 · 2 · 3 = −3.
2

 ()  is the negative of the area of a trapezoid with bases 3 and 2 and height 2, so it equals

+ ) = − 1 (3 + 2)2 = −5. Thus,
2
5
7
9
9
 ()  = 0  ()  + 5  ()  + 7  ()  = 10 + (−3) + (−5) = 2.
0

−1

(1 − )  can be interpreted as the difference of the areas of the two

shaded triangles; that is, 1 (2)(2) − 1 (1)(1) = 2 −
2
2

37.

 () 

 ()  is the negative of the area of the triangle with base 2 and height 3.

N

2

0

trapezoid

− 1 (
2

35.

2

ot

(d)

7

e

()  as the area of a trapezoid with bases 1 and 3 and height 2. The area of a trapezoid is  = 1 ( + ),
2

=
(c)

al

0

(b)

 


 
 


2
1
5  CAS
5 CAS
2
(sin 5 )
=  lim
=  sin = lim cot =
→∞ =1
→∞ =1
→∞ 

 
2
5
5



sin 5  = lim

rS



Fo



1
2

= 3.
2

√
0 

1 + 9 − 2  can be interpreted as the area under the graph of
−3
√
 () = 1 + 9 − 2 between  = −3 and  = 0. This is equal to one-quarter the area of the circle with radius 3, plus the area of the rectangle, so
√
0 

1 + 9 − 2  = 1  · 32 + 1 · 3 = 3 + 9 .
4
4
−3

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¤

198

39.

CHAPTER 4 INTEGRALS

2

−1

||  can be interpreted as the sum of the areas of the two shaded

triangles; that is, 1 (1)(1) + 1 (2)(2) =
2
2

1
2

= 5.
2

4
2

+

0

47.

49.
51.

1
0

4
1

sin2  cos4   = 0 since the limits of intergration are equal.

(5 − 62 )  =

3
0

5  − 6
4

1
0

 
2  = 5(1 − 0) − 6 1 = 5 − 2 = 3
3

2  − 3

1

 ()  +

5
2

 ()  −

[2 () + 3()]  = 2

3

3

4

  +

1

 −1
−2

9

2

4
1

1 

= 2 · 1 (4 − 1 ) − 3 · 1 (4 − 12 ) + 1(4 − 1) =
3
2

−2

0

0

(22 − 3 + 1)  = 2

2
9

1

 ()  =
=

0

−2

5

9

 ()  + 3

0

5

−1

 ()  +
 () 

 −2
−1

45
2

 () 

= 225
[by Property 5 and reversing limits]
[Property 5]

e

45.



al

43.



()  = 2(37) + 3(16) = 122

 ()  is clearly less than −1 and has the smallest value. The slope of the tangent line of  at  = 1,  0 (1), has a value

rS

41.

between −1 and 0, so it has the next smallest value. The largest value is

value about 1 unit less than

8
3

2

−4

 ()    0 (1) 

[ () + 2 + 5]  =

2

−4

8
0

 ()  

 ()  + 2

2

−4

8

1 = −3 [area below -axis]

4

 ()  

  +

ot

53.  =

Fo

0

3

 () , followed by

 () . Still positive, but with a smaller value than

quantities from smallest to largest gives us
3

8

2

−4

8
3

8
4

8
4

() , is

 () , which has a

8
0

 () . Ordering these

 ()  or B  E  A  D  C

5  = 1 + 22 + 3

+ 3 − 3 = −3

N

2 = − 1 (4)(4) [area of triangle, see figure]
2

+ 1 (2)(2)
2

= −8 + 2 = −6

3 = 5[2 − (−4)] = 5(6) = 30

Thus,  = −3 + 2(−6) + 30 = 15.
55. 2 − 4 + 4 = ( − 2)2 ≥ 0 on [0 4], so

4
0

(2 − 4 + 4)  ≥ 0 [Property 6].

√
√
1 + 2 ≤ 2 and
√
√
1 √
1 √
1[1 − (−1)] ≤ −1 1 + 2  ≤ 2 [1 − (−1)] [Property 8]; that is, 2 ≤ −1 1 + 2  ≤ 2 2.

57. If −1 ≤  ≤ 1, then 0 ≤ 2 ≤ 1 and 1 ≤ 1 + 2 ≤ 2, so 1 ≤

59. If 1 ≤  ≤ 4, then 1 ≤
61. If


4

≤≤


,
3

4√
4√
√
 ≤ 2 so 1(4 − 1) ≤ 1   ≤ 2(4 − 1); that is, 3 ≤ 1   ≤ 6.

then 1 ≤ tan  ≤

√
√ 
  3


3, so 1  −  ≤ 4 tan   ≤ 3  −  or
3
4
3
4


12

≤

 3
4

tan   ≤

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.


12

√
3.

SECTION 4.3 THE FUNDAMENTAL THEOREM OF CALCULUS

63. For −1 ≤  ≤ 1, 0 ≤ 4 ≤ 1 and 1 ≤

or 2 ≤
65.

√
1 √
1 + 4  ≤ 2 2.
−1

¤

199

√
√
√
1 √
1 + 4 ≤ 2, so 1[1 − (−1)] ≤ −1 1 + 4  ≤ 2 [1 − (−1)]

√
√


3√
3
4 + 1 ≥ 4 = 2 , so 1 4 + 1  ≥ 1 2  = 1 33 − 13 =
3

26
.
3

67. Using right endpoints as in the proof of Property 2, we calculate




 ()  = lim




→∞ =1

 ( ) ∆ = lim 
→∞




 ( ) ∆ =  lim

→∞ =1

=1

69. Suppose that  is integrable on [0 1]  that is, lim







→∞ =1

 ( ) ∆ = 




 () .

(∗ ) ∆ exists for any choice of ∗ in [−1   ]. Let n denote a





 


1
1 2
−1
positive integer and divide the interval [0 1] into n equal subintervals 0
,

,  ,
 1 . If we choose ∗ to be


 

a rational number in the ith subinterval, then we obtain the Riemann sum




→∞ =1



=1

 (∗ ) ·


 (∗ ) ·


lim




→∞ =1

1
= lim 0 = 0. Now suppose we choose ∗ to be an irrational number. Then we get

→∞


al




1
= 0, so






1
1
1
1
1 · =  · = 1 for each , so lim
(∗ ) · = lim 1 = 1. Since the value of
=

→∞ =1
→∞




=1

rS

lim

 (∗ ) ·


e

=1

 (∗ ) ∆ depends on the choice of the sample points ∗ , the limit does not exist, and  is not integrable on [0 1].



 



 4
 4 1
  4 1
. At this point, we need to recognize the limit as being of the form
= lim
· = lim
→∞ =1 5
→∞ =1 4
 →∞ =1  



→∞ =1

is

1
0

 ( ) ∆, where ∆ = (1 − 0) = 1,  = 0 +  ∆ = , and  () = 4 . Thus, the definite integral

4 .


√
73. Choose  = 1 + and ∗ = −1  =






1 
1
1


 = lim 
−1

→∞  =1 1 +
→∞ =1 ( +  − 1)( + )
1+ 




−1



 1
 1
1
1
−
[by the hint] = lim 
−
= lim 
→∞ =1
→∞
+−1
+
=0  + 
=1  + 
 


1
1
1
1
1
1
+
+··· +
−
+ ··· +
+
= lim 
→∞

+1
2 − 1
+1
2 − 1
2




1
1
= lim 
−
= lim 1 − 1 = 1
2
2
→∞
→∞

2

N

2




−1

1+
1+
. Then



ot

lim

Fo

71. lim

1

−2  = lim

4.3 The Fundamental Theorem of Calculus
1. One process undoes what the other one does. The precise version of this statement is given by the Fundamental Theorem of

Calculus. See the statement of this theorem and the paragraph that follows it on page 317. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

200

¤

CHAPTER 4 INTEGRALS



3. (a) () =

 () .

0

0

(0) =

0

1

(1) =

0

2

(2) =

0

 ()  = 0
 ()  = 1 · 2 = 2
[rectangle],
1
2
2
 ()  = 0  ()  + 1  ()  = (1) + 1  () 

=2+1·2+ 1 ·1·2 =5
[rectangle plus triangle],
2
3
3
(3) = 0  ()  = (2) + 2  ()  = 5 + 1 · 1 · 4 = 7,
2
6
(6) = (3) + 3  () 
[the integral is negative since  lies under the -axis]
 1

=7+ − 2 ·2·2+1·2 =7−4 =3

(b)  is increasing on (0 3) because as  increases from 0 to 3, we keep

(d)

adding more area.

e

(c)  has a maximum value when we start subtracting area; that is,

al

at  = 3.

(a) By FTC1 with  () = 2 and  = 1, () =

rS

5.

 0 () =  () = 2 .

7.  () =

1 and () =
3 + 1





1


1

1

2  ⇒

 1 3 
1 3
3 1 = 3 −

2  =

1
3

⇒  0 () = 2 .

Fo

(b) Using FTC2, () =



1
1
, so by FTC1, 0 () =  () = 3
. Note that the lower limit, 1, could be any
3 + 1
 +1

ot

real number greater than −1 and not affect this answer.

11.  () =







13. Let  =

0 () =


5

( − 2 )8 , so by FTC1,  0 () =  () = ( − 2 )8 .

N

9.  () = ( − 2 )8 and () =

√
1 + sec   = −







√

1 + sec   ⇒  0 () = −








√
√
1 + sec   = − 1 + sec 


1
 
1

. Then
= − 2 . Also,
=
, so




 





1

sin4   =

2






2



sin4   ·

− sin4 (1)


= sin4 
=
.


2



 
= sec2 . Also,
=
, so


 

 tan  
 

√
√
√
√ 



= + 
= tan  + tan  sec2 .
 +   =
 +   ·
0 =
 0
 0



15. Let  = tan . Then

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

SECTION 4.3 THE FUNDAMENTAL THEOREM OF CALCULUS



 
= −3. Also,
=
, so


 
 1
 



3(1 − 3)3
3
3
3

3
 =
 ·
 ·
(−3) =
=−
=−
1 + 2
  1 + 2

 1 1 + 2

1 + 2
1 + (1 − 3)2

17. Let  = 1 − 3. Then

0 =

1−3

 4
 4
2
 

 3


 


2
(−1)4
 − 2  =
− 2
− 22 −
− (−1)2 = (4 − 4) − 1 − 1 = 0 − − 3 =
=
4
4
4
4
4
−1
−1



2

1



4

(5 − 2 + 32 )  = 5 − 2 + 3 1 = (20 − 16 + 64) − (5 − 1 + 1) = 68 − 5 = 63

9

√
  =

1



6

27.

( + 2)( − 3)  =



9

1

31.
33.

35.

0

2
1

=

1

2
3


9
32 = 2 (932 − 132 ) = 2 (27 − 1) =
3
3
1

(2 −  − 6)  =

9



1

3
3

1 
− 1 2 − 6 0 = 1 −
2
3


4
sec2   = tan  0 = tan  − tan 0 = 1 − 0 = 1
4

 5 + 3 6
=
4


2
1



2


2 
(1 + 4 + 42 )  =  + 2 2 + 4  3 1 = 2 + 8 +
3

1

37. If  () =

sin 


0

cos 

 ()  =

32
3




2


 + 3 2  = 1  2 +  3 1 = (2 + 8) − 1 + 1 =
2
2

if 0 ≤   2

N

1



(1 + 2)2  =

2

0

1

9

52
3

1
2


− 6 − 0 = − 37
6


 9

9

1
√ −√
(12 − −12 )  = 2 32 − 212
 =
3
1


1
1
2
 2

 4  40
= 3 · 27 − 2 · 3 − 3 − 2 = 12 − − 3 = 3

−1
√  =


 4



6



32
32

 √

√


= − cos  − − cos  = −(−1) − − 32 = 1 + 32
6



sin   = − cos 

1

0

29.



12  =

1





9

ot

25.



e

4

3
4

al

23.

1

rS

21.



Fo

19.




if 2 ≤  ≤ 

 2
0

sin   +







− 1+2+ 4 =
3

62
3

−

13
3

=

49
3

17
2

then

2

= −0 + 1 + 0 − 1 = 0


2 

cos   = − cos  0 + sin  2 = − cos  + cos 0 + sin  − sin 
2
2

Note that  is integrable by Theorem 3 in Section 4.2.
39.  () = −4 is not continuous on the interval [−2 1], so FTC2 cannot be applied. In fact,  has an infinite discontinuity at

 = 0, so

1

−2

201

−4  does not exist.

41.  () = sec  tan  is not continuous on the interval [3 ], so FTC2 cannot be applied. In fact,  has an infinite

discontinuity at  = 2, so



3

sec  tan   does not exist.

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202

¤

CHAPTER 4 INTEGRALS

43. From the graph, it appears that the area is about 60. The actual area is

 27
0

13  =



3 43
4

27

3
4

=

0

· 81 − 0 =

= 6075. This is

243
4

3
4

of the

area of the viewing rectangle.

45. It appears that the area under the graph is about

2
3

of the area of the viewing

rectangle, or about 2  ≈ 21. The actual area is
3
 sin   = [− cos ] = (− cos ) − (− cos 0) = − (−1) + 1 = 2.
0
0

−1

3  =

1
4

4

2

−1

1
4

=

0

2 − 1
 +
2 + 1

=4−

15
4

= 375

e

2

49. () =



3

2

2 − 1
 =
2 + 1



2

2

51. () =



3

0



2 − 1
 = −
2 + 1

2

2

0



2 − 1
 +
2 + 1

2

3

0

2 − 1
 ⇒
2 + 1

2

(3) − 1 
4 − 1
9 − 1
(2) − 1 
·
(2) +
·
(3) = −2 · 2
+3· 2
(2)2 + 1 
(3)2 + 1 
4 + 1
9 + 1

3

√



cos(2 )  =



0

√

Fo

 0 () = −



rS

al

47.

cos(2 )  +





0

3

cos(2 )  = −



√



cos(2 )  +

0



3

0

cos(2 )  ⇒





0

 00 =

2
2
 ⇒ 0 = 2
2 +  + 2
 ++2

N

53.  =

ot

√
  √

1
2
0 () = − cos (  ) ·
(  ) + [cos(3 )2 ] ·
(3 ) = − √ cos  + 32 cos(6 )


2 
⇒

(2 +  + 2)(2) − 2 (2 + 1)
23 + 22 + 4 − 23 − 2
2 + 4
( + 4)
=
= 2
= 2
.
2 +  + 2)2
2 +  + 2)2
(
(
( +  + 2)2
( +  + 2)2

The curve  is concave downward when  00  0; that is, on the interval (−4 0).
55. By FTC2,

4
1

 0 ()  = (4) −  (1), so 17 =  (4) − 12 ⇒ (4) = 17 + 12 = 29.

57. (a) The Fresnel function () =


0





sin  2  has local maximum values where 0 =  0 () = sin  2 and
2
2

 0 changes from positive to negative. For   0, this happens when  2 = (2 − 1) [odd multiples of ] ⇔
2
√
2 = 2(2 − 1) ⇔  = 4 − 2,  any positive integer. For   0,  0 changes from positive to negative where
√
 2
 = 2 [even multiples of ] ⇔ 2 = 4 ⇔  = −2 .  0 does not change sign at  = 0.
2

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SECTION 4.3 THE FUNDAMENTAL THEOREM OF CALCULUS

¤

203

(b)  is concave upward on those intervals where  00 ()  0. Differentiating our expression for  0 (), we get





 00 () = cos  2 2   =  cos  2 . For   0,  00 ()  0 where cos(  2 )  0 ⇔ 0   2   or
2
2
2
2
2
2




√
√
2 − 1    2  2 + 1 ,  any integer ⇔ 0    1 or 4 − 1    4 + 1,  any positive integer.
2
2
2




00
For   0,  ()  0 where cos(  2 )  0 ⇔ 2 − 3    2  2 − 1 ,  any integer ⇔
2
2
2
2
√
√
√
√
4 − 3  2  4 − 1 ⇔
4 − 3  ||  4 − 1 ⇒
4 − 3  −  4 − 1 ⇒
 √

√
√
√
− 4 − 3    − 4 − 1, so the intervals of upward concavity for   0 are − 4 − 1 − 4 − 3 ,  any
√ 
 √
 √ √   √ positive integer. To summarize:  is concave upward on the intervals (0 1), − 3 −1 ,
3 5 , − 7 − 5 ,
√ 
7 3 ,    .
(c) In Maple, we use plot({int(sin(Pi*tˆ2/2),t=0..x),0.2},x=0..2);. Note that
Maple recognizes the Fresnel function, calling it FresnelS(x). In Mathematica, we use

Fo

rS

al

e

Plot[{Integrate[Sin[Pi*tˆ2/2],{t,0,x}],0.2},{x,0,2}]. In Derive, we load the utility file

 
FRESNEL and plot FRESNEL_SIN(x). From the graphs, we see that 0 sin  2  = 02 at  ≈ 074.
2

59. (a) By FTC1, 0 () =  (). So 0 () =  () = 0 at  = 1 3 5 7, and 9.  has local maxima at  = 1 and 5 (since  = 0

changes from positive to negative there) and local minima at  = 3 and 7. There is no local maximum or minimum at
 = 9, since  is not defined for   9.

5
0


 


 

9
 3
  5

 7
  9

  = (1) −  1   +  3  , and (9) = 0   = (5) −  5   +  7  . Thus,

N

(5) =

ot

 
 
 
 




  3
  5
  7
  9

 1

 1
(b) We can see from the graph that  0     1     3     5     7  . So (1) =  0  ,
(1)  (5)  (9), and so the absolute maximum of () occurs at  = 9.
(c)  is concave downward on those intervals where  00  0. But  0 () =  (),

 so  00 () =  0 (), which is negative on (approximately) 1  2 , (4 6) and
2

(d)

(8 9). So  is concave downward on these intervals.



 3
1−0 
= lim
→∞ =1 4
→∞
 =1

61. lim

 3  1
 4 1


1
=
3  =
=

4 0
4
0

63. Suppose   0. Since  is continuous on [ +  ], the Extreme Value Theorem says that there are numbers  and  in

[ +  ] such that  () =  and  () =  , where  and  are the absolute minimum and maximum values of  on

c
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¤

204

CHAPTER 4 INTEGRALS

[ +  ]. By Property 8 of integrals, (−) ≤



+

()  ≤ (−); that is, ()(−) ≤ −

Since −  0, we can divide this inequality by −:  () ≤
( + ) − ()
1
=

 case where   0.



1




+



 ()  for  6= 0, and hence  () ≤



 ()  ≤  ()(−).

 ()  ≤  (). By Equation 2,



+

 +

( + ) − ()
≤  (), which is Equation 3 in the


√
√
 ⇒  0 () = 1(2  )  0 for   0 ⇒  is increasing on (0 ∞). If  ≥ 0, then 3 ≥ 0, so
√


1 + 3 ≥ 1 and since  is increasing, this means that  1 + 3 ≥  (1) ⇒
1 + 3 ≥ 1 for  ≥ 0. Next let

65. (a) Let  () =

0

1  ≤


1
 1  1 √
 0 ≤ 0 1 + 3  ≤  + 1 4 0
4

1√
1
1 + 3  ≤ 0 (1 + 3 )  ⇔
0
⇔ 1≤

al

1

1√
1 + 3  ≤ 1 +
0

1
4

= 125.

rS

(b) From part (a) and Property 7:

e

() = 2 −  ⇒  0 () = 2 − 1 ⇒  0 ()  0 when  ≥ 1. Thus,  is increasing on (1 ∞). And since (1) = 0,
√
√
() ≥ 0 when  ≥ 1. Now let  = 1 + 3 , where  ≥ 0. 1 + 3 ≥ 1 (from above) ⇒  ≥ 1 ⇒ () ≥ 0 ⇒
√

 √
1 + 3 − 1 + 3 ≥ 0 for  ≥ 0. Therefore, 1 ≤ 1 + 3 ≤ 1 + 3 for  ≥ 0.

2
1
2
 4 = 2 on [5 10], so
2 +1
+



10


 10
 10
1
1
2
1
1
1
0≤
 = −
=− − −
 
=
= 01
4 + 2 + 1
2
 5
10
5
10
5
5
4

Fo

67. 0 



√
 ()
 ()
1
 = 2  to get 2 = 2 √
⇒ () = 32 .
2

2 

 
√
√
()
To find , we substitute  =  in the original equation to obtain 6 +
 = 2  ⇒ 6 + 0 = 2  ⇒
2


√
3 =  ⇒  = 9.


 () . Then, by FTC1,  0 () = () = rate of depreciation, so  () represents the loss in value over the

N

71. (a) Let  () =



ot

69. Using FTC1, we differentiate both sides of 6 +

0

interval [0 ].
(b) () =



 
 +  ()
1
 ()  =
+
represents the average expenditure per unit of  during the interval [0 ],


0

assuming that there has been only one overhaul during that time period. The company wants to minimize average expenditure. 



 
 
1
1
1
+
 ()  . Using FTC1, we have  0 () = − 2  +
 ()  +  ().



0
0


 
 
1
 0 () = 0 ⇒   () =  +
()  ⇒ () =
 ()  = ().
+

0
0

(c) () =

73.



1

9

1
1
 =
2
2



1

9


9
1
 = 1 ln || = 1 (ln 9 − ln 1) =
2
2

1

1
2

ln 9 − 0 = ln 912 = ln 3

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 4.4

75.



√
32

12

6
√
 = 6
1 − 2



√

32

12

INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM

¤

√ 

 


 

√32
1
√
 = 6 sin−1  12 = 6 sin−1 23 − sin−1 1 = 6  −  = 6  = 
2
3
6
6
1 − 2

1

77. −1 +1  = +1 −1 = 2 − 0 = 2 − 1 [or start with +1 =  1 ]
1

4.4 Indefinite Integrals and the Net Change Theorem



  sin  − 1 (sin )3 +  = cos  − sin3  +  =
3




7.



9.



( + 4)(2 + 1)  =

11.



3 − 2



 4 1 3 1
1 4
1 2
5
−
+
− 2 +  = 1 5 − 1 4 + 1 2 − 2 + 
 − 2  + 4  − 2  =
5
8
8
5
2 4
4 2

√


 

 =



(22 + 9 + 4)  = 2




2
2
3
9
+9
+ 4 +  = 3 + 2 + 4 + 
3
2
3
2

212
3
−



 =

(2 − 2−12 )  =

√
3
12
−2
+  = 1 3 − 4  + 
3
3
12

( − csc  cot )  = 1 2 + csc  + 
2
(1 + tan2 )  =



sec2   = tan  + 



cos  + 1   = sin  + 1 2 + . The members of the family
2
4

N

17.



−1
1
3
1
+
+  = 3 − + 
3
−1
3


(2 + −2 )  =

ot

15.



· 3(sin )2 (cos ) + 0

= cos (1 − sin2 ) = cos (cos2 ) = cos3 

5.

13.

1
3

e

1
3

al

  sin  −


rS

3.

 √



 · 1 (1 + 2 )−12 (2) − (1 + 2 )12 · 1

1 + 2

(1 + 2 )12
2
−
+ =
−
+ =−
+0




()2


(1 + 2 )−12 2 − (1 + 2 )
−1
1
=−
= √
=−
2
(1 + 2 )12 2
2 1 + 2

Fo

1.

in the figure correspond to  = −5, 0, 5, and 10.

19.
21.
23.
25.

3

−2

(2 − 3)  =

 0 1
−2

2
0

3

3


3 − 3 −2 = (9 − 9) − − 8 + 6 =
3


1
+ 1 3 −   = 10 5 +
4

(2 − 3)(42 + 1)  =


0

4
2

1

2
0

205

1 4
16 

− 1 2
2

0

−2

=0−



8
3

−

18
3

1
10 (−32)

= − 10
3
+

1
16 (16)




− 1 (4) = − − 16 + 1 − 2 =
2
5

21
5


2
(83 − 122 + 2 − 3)  = 24 − 43 + 2 − 3 0 = (32 − 32 + 4 − 6) − 0 = −2



(4 sin  − 3 cos )  = − 4 cos  − 3 sin  0 = (4 − 0) − (−4 − 0) = 8

c
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¤

206

27.



1

29.

CHAPTER 4

4

4 + 6
√




INTEGRALS

 =



4

1



4
6
√ +√





 =



4

1


4
(4−12 + 612 )  = 812 + 432 = (16 + 32) − (8 + 4) = 36
1

√ 4
√  √ 4 √
√
4
5  = 5 1 −12  = 5 2  = 5 (2 · 2 − 2 · 1) = 2 5
1
1

31.


4 
4
4√
 (1 + )  = 1 (12 + 32 )  = 2 32 + 2 52 = 16 +
3
5
3
1
1

33.



4

0

35.



64

1

64
5

4 



−

2


 4
1
cos2 
+
 =
(sec2  + 1) 
2
2 cos cos
0
0


4 
= tan  +  0 = tan  +  − (0 + 0) = 1 + 
4
4
4

1 + cos2 
 = cos2 



3

+

2
5



=

14
3

256
15

if  − 3 ≥ 0

−3
−( − 3)

Thus,

e



=

−3

if  ≥ 3

+

5
9

−0 =1

rS



4
9

if  − 3  0
3 −  if   3
3 
5
3
5

5
| − 3|  = 2 (3 − )  + 3 ( − 3)  = 3 − 1 2 2 + 1 2 − 3 3
2
2
2



 

= 9 − 9 − (6 − 2) + 25 − 15 − 9 − 9 = 5
2
2
2
2

Fo

39. | − 3| =

−1

=

al

1 
 94
1
√ 
 1 54
 1 √
95

4
45
5 + 5 4
 = 0 (
+
37. 0

+  )  =
= 4 94 + 5 95 =
9
9
94
95 0
0

2

62
5


√
 64 
 64
 64 

13
1+ 3
1
√
+ 12  =
(−12 + −16 ) 
−12 + (13) − (12)  =
 =
12


1
1
1
64 

 

= 16 + 192 − 2 + 6 = 14 + 186 = 256
= 212 + 6 56
5
5
5
5
5
1

41.

+

0
2
2
0
2


[ − 2(−)]  + 0 [ − 2()]  = −1 3  + 0 (−)  = 3 1 2 −1 − 1 2 0
2
2


= 3 0 − 1 − (2 − 0) = − 7 = −35
2
2

( − 2 ||)  =

0

−1

ot

43. The graph shows that  = 1 − 2 − 54 has -intercepts at

 =  ≈ −086 and at  =  ≈ 042. So the area of the region that

N

lies under the curve and above the -axis is



(1 − 2 − 54 )  =  − 2 − 5 


= ( − 2 − 5 ) − ( − 2 − 5 )

≈ 136
45.  =



2 

 2
2 − 2  =  2 − 1  3 0 = 4 − 8 − 0 =
3
3
0

4
3

47. If 0 () is the rate of change of weight in pounds per year, then () represents the weight in pounds of the child at age . We

know from the Net Change Theorem that

 10
5

0 ()  = (10) − (5), so the integral represents the increase in the child’s

weight (in pounds) between the ages of 5 and 10.
49. Since () is the rate at which oil leaks, we can write () = − 0 (), where  () is the volume of oil at time . [Note that the

minus sign is needed because  is decreasing, so  0 () is negative, but () is positive.] Thus, by the Net Change Theorem,

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 4.4

 120
0

()  = −

 120
0

INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM

¤

207

 0 ()  = − [ (120) −  (0)] =  (0) −  (120), which is the number of gallons of oil that leaked

from the tank in the first two hours (120 minutes).
 5000

51. By the Net Change Theorem,

1000

0 ()  = (5000) − (1000), so it represents the increase in revenue when

production is increased from 1000 units to 5000 units.
53. In general, the unit of measurement for



 ()  is the product of the unit for  () and the unit for . Since  () is



measured in newtons and  is measured in meters, the units for abbreviated N·m.)
55. (a) Displacement =

3
0

(3 − 5)  =
3

3
2

3
2 − 5 0 =

27
2

 100
0

 ()  are newton-meters. (A newton-meter is

− 15 = − 3 m
2

 53
3
|3 − 5|  = 0 (5 − 3)  + 53 (3 − 5) 
53 
3


= 5 − 3 2 0 + 3 2 − 5 53 = 25 − 3 · 25 + 27 − 15 − 3 ·
2
2
3
2
9
2
2

(b) Distance traveled =
=

⇒ () = 1 2 + 4 + 
2

 10
0

500
3

0

25
3



=

41
6

m


 10  1 2
 



  + 4 + 5  = 10 1 2 + 4 + 5  = 1 3 + 22 + 5 10
0

+ 200 + 50 =

4

−

⇒ (0) =  = 5 ⇒ () = 1 2 + 4 + 5 ms
2

()  =

2

0

416 2
3

2

6

0

m


4
√ 
 4
9 + 2   = 9 + 4 32 = 36 +
3
0

Fo

59. Since 0 () = (),  =

|()|  =

25
9

al

57. (a)  0 () = () =  + 4

e

0

rS

(b) Distance traveled =

0

61. Let  be the position of the car. We know from Equation 2 that (100) − (0) =

 100
0

32
3

−0 =

ot

1
180 (38

= 46 2 kg.
3

() . We use the Midpoint Rule for

0 ≤  ≤ 100 with  = 5. Note that the length of each of the five time intervals is 20 seconds =
So the distance traveled is
 100
1
()  ≈ 180 [(10) + (30) + (50) + (70) + (90)] =
0

140
3

20
3600

+ 58 + 51 + 53 + 47) =

hour =

247
180

1
180

hour.

≈ 14 miles.

N

63. By the Net Change Theorem, the amount of water after four days is

25,000 +

4
0

()  ≈ 25,000 + 4 = 25,000 +

4−0
4

[(05) + (15) + (25) + (35)]

≈ 25,000 + [1500 + 1770 + 740 + (−690)] = 28,320 liters

65. Power is the rate of change of energy with respect to time; that is,  () =  0 (). By the Net Change Theorem and the

Midpoint Rule,
(24) − (0) =



24
0

 ()  ≈

24 − 0
[ (1) +  (3) +  (5) + · · · +  (21) +  (23)]
12

≈ 2(16,900 + 16,400 + 17,000 + 19,800 + 20,700 + 21,200
+ 20,500 + 20,500 + 21,700 + 22,300 + 21,700 + 18,900)
= 2(237,600) = 475,200
Thus, the energy used on that day was approximately 4.75 × 105 megawatt-hours. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

208

67.



CHAPTER 4 INTEGRALS

(sin  + sinh )  = − cos  + cosh  + 

 
69.
2 + 1 +

1
2 + 1





0

√
1 3

2 − 1
 =
4 − 1



=

71.


6

3
+  + tan−1  + 
3

 =

1

√

3

0



2 − 1
 =
2 + 1)(2 − 1)
(

1

√

3

2

0


6

−0 =

 √ 

1√3
1
 = arctan  0
= arctan 1 3 − arctan 0
+1

4.5 The Substitution Rule
1. Let  = . Then  =   and

1


 = , so

, so



sin 


 =

1
(− cos )


1
+  = −  cos  + .



3 (−) = −

4
+  = − 1 cos4  + .
4
4
, so



 sin(2 )  =

9. Let  = 1 − 2. Then  = −2  and  = − 1 , so
2

(1 − 2)9  =





Fo

1
2

rS

cos3  sin   =

7. Let  = 2 . Then  = 2  and   =





 √


 1 32
1 2
2 3 + 1  =
+  = · 32 +  = 2 (3 + 1)32 + .
 1  =
3
9
3 32
3 3

5. Let  = cos . Then  = − sin   and sin   = −, so



1

al



1
3

sin   =

e

3. Let  = 3 + 1. Then  = 32  and 2  =





9 − 1  = − 1 ·
2
2

1 10

10



sin  1  = − 1 cos  +  = − 1 cos(2 ) + .
2
2
2

1
+  = − 20 (1 − 2)10 + .

11. Let  = 2 + 2 . Then  = (2 + 2)  = 2(1 + )  and ( + 1)  =



1
2

, so

N

ot

 √


32

 1 32
( + 1) 2 + 2  =
+  = 1 2 + 2
 1  =
+ .
2
3
2 32
√
Or: Let  = 2 + 2 . Then 2 = 2 + 2 ⇒ 2  = (2 + 2)  ⇒   = (1 + ) , so
√



( + 1) 2 + 2  =  ·   = 2  = 1 3 +  = 1 (2 + 2 )32 + .
3
3

13. Let  = 3. Then  = 3  and  =

1
3





, so sec 3 tan 3  = sec  tan  1  =
3

1
3

sec  +  =

1
3

sec 3 + .

15. Let  = 3 + 3 . Then  = (3 + 32 )  = 3( + 2 ) , so



 + 2
√
 =
3 + 3



1
3



12

=

1
3



17. Let  = tan . Then  = sec2  , so

−12  =


(2 + 1)(3 + 3)4  =



· 212 +  =

sec2  tan3   =

19. Let  = 3 + 3. Then  = (32 + 3)  and



1
3

4 ( 1 ) =
3

1
3

1
3



2
3


3 + 3 + .

3  = 1 4 +  =
4

1
4

tan4  + .

 = (2 + 1) , so

· 1 5 +  =
5

1
(3
15

+ 3)5 + .

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 4.5 THE SUBSTITUTION RULE



21. Let  = sin . Then  = cos  , so

cos 
 = sin2 



1
 =
2



−2  =

[or −csc  +  ].
23. Let  = 1 +  3 . Then  = 3 2  and  2  =



2
√
 =
3
1 + 3



−13

1
3


 =

1
3

1
3

1
1
−1
+ =− + =−
+
−1

sin 

, so

· 3 23 +  = 1 (1 +  3 )23 + .
2
2

25. Let  = cot . Then  = − csc2   and csc2   = −, so

 √
 √
32
+  = − 2 (cot )32 + . cot  csc2   =
 (−) = −
3
32

27. Let  = sec . Then  = sec  tan  , so



sec2  (sec  tan )  =

29. Let  = 2 + 5. Then  = 2  and  =



(2 + 5)8  =

− 5), so
 9
1
( − 5)8 ( 1 ) = 1 ( − 58 ) 
2
2
4
1
(
2

1
= 1 ( 10 10 − 5 9 ) +  =
4
9

31.  () = (2 − 1)3 .



2  = 1 3 +  =
3

1
40 (2

1
3

sec3  + .

+ 5)10 −

5
36 (2

+ 5)9 + 

rS





e

sec3  tan   =

al



 = 2 − 1 ⇒  = 2  so

(2 − 1)3  =



3

1
2


 = 1 4 +  = 1 (2 − 1)4 + 
8
8

Fo

Where  is positive (negative),  is increasing (decreasing). Where 

changes from negative to positive (positive to negative),  has a local minimum (maximum).



⇒  = cos  , so

ot

33.  () = sin3  cos .  = sin 

sin3  cos   =

2,

3  = 1 4 +  =
4

1
4

sin4  + 

 changes from positive to negative and  has a local

N

Note that at  =



maximum. Also, both  and  are periodic with period , so at  = 0 and at  = ,  changes from negative to positive and  has local minima.
35. Let  =


,
2

so  =
1
0


2

. When  = 0,  = 0; when  = 1,  =

cos(2)  =

 2

cos 

0

2




 =

2



.
2

Thus,

[sin ]2 =
0

2




sin  − sin 0 =
2

2
 (1

− 0) =

2


3
28 (16

− 1) =

45
28

37. Let  = 1 + 7, so  = 7 . When  = 0,  = 1; when  = 1,  = 8. Thus,



1

√
3
1 + 7  =

0

1

39. Let  = 4, so  =


0

sec2 (4)  =



1
4

 4
0

8

13 ( 1 ) =
7

1
7



3 43
4

8
1

=

43
3
28 (8

¤

− 143 ) =

. When  = 0,  = 0; when  = ,  = 4. Thus,

4

 sec2  (4 ) = 4 tan  0 = 4 tan  − tan 0 = 4(1 − 0) = 4.
4

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

209

¤

210

41.

CHAPTER 4 INTEGRALS

 4

−4

(3 + 4 tan )  = 0 by Theorem 6(b), since  () = 3 + 4 tan  is an odd function.

43. Let  = 1 + 2, so  = 2 . When  = 0,  = 1; when  = 13,  = 27. Thus,



13



=
3
(1 + 2)2

0



27

−23

1

1
2

27
 
 = 1 · 313
= 3 (3 − 1) = 3.
2
2
1

45. Let  = 2 + 2 , so  = 2  and   =









2 + 2  =

22

12

2

0

1
2

1
2

. When  = 0,  = 2 ; when  = ,  = 22 . Thus,


22 
22


 √


 = 1 2 32
= 1 32
= 1 (22 )32 − (2 )32 = 1 2 2 − 1 3
2 3
3
3
3
2

2

47. Let  =  − 1, so  + 1 =  and  = . When  = 1,  = 0; when  = 2,  = 1. Thus,
1

√
 − 1  =



1

0

 1
1

√
( + 1)   =
(32 + 12 )  = 2 52 + 2 32 =
5
3
0

0

49. Let  = −2 , so  = −2−3 . When  =



1

12

cos(−2 )
 =
3

51. Let  = 1 +





1

cos 

4

√
, so  =

2


−2



=

1
2



1
,
2

2
3

=

16
15 .

cos   =

1

4
1
1 sin  1 = (sin 4 − sin 1).
2
2

√
1
√  ⇒ 2   =  ⇒ 2( − 1)  = . When  = 0,  = 1; when  = 1,



2
 2
1
1
1
1
1
· [2( − 1) ] = 2
− 4  = 2 − 2 + 3
4
3

2
3 1
1 
1
 1
  1

1 1
1
= 2 − 8 + 24 − − 2 + 1 = 2 12 = 6
3


√
=
(1 +  )4



2

Fo

0

1

+

 = 4; when  = 1,  = 1. Thus,

4

 = 2. Thus,


2
5

e



al

2

rS



53. From the graph, it appears that the area under the curve is about

ot



1 + a little more than 1 · 1 · 07 , or about 14. The exact area is given by
2
1√
 = 0 2 + 1 . Let  = 2 + 1, so  = 2 . The limits change to

N

2 · 0 + 1 = 1 and 2 · 1 + 1 = 3, and

3
 √

 √
3√ 
 = 1  1  = 1 2 32 = 1 3 3 − 1 = 3 −
2
2 3
3
1

1
3

≈ 1399.

55. First write the integral as a sum of two integrals:

2

√
2 √
2 √
( + 3) 4 − 2  = 1 + 2 = −2  4 − 2  + −2 3 4 − 2 . 1 = 0 by Theorem 6(b), since
√
 () =  4 − 2 is an odd function and we are integrating from  = −2 to  = 2. We interpret 2 as three times the area of

 a semicircle with radius 2, so  = 0 + 3 · 1  · 22 = 6.
2
=

−2

57. The volume of inhaled air in the lungs at time  is

 () =



0



 ()  =





0

25
5 
=
− cos  0
4





 25

1
1
2
5
sin
  = sin 

substitute  =
2
5
2
2
0







5
2
5
2
=
− cos
 +1 =
1 − cos

liters
4
5
4
5

2
,
5

 =

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

2
5




SECTION 4.5 THE SUBSTITUTION RULE

59. Let  = 2. Then  = 2 , so

2
0

(2)  =

61. Let  = −. Then  = −, so




 (−)  =

 −
−

 ()(−) =

 −
−

4
0



 () 1  =
2

 ()  =

 −
−

1
2

4
0

¤

211

 ()  = 1 (10) = 5.
2

 () 

From the diagram, we see that the equality follows from the fact that we are reflecting the graph of  , and the limits of integration, about the -axis.
63. Let  = 1 − . Then  = 1 −  and  = −, so

1
0

65.

 (1 − )  =

 2
0

0
1

(1 − )  (−) =

1
0

 (1 − )  =

1
0

 (1 − ) .

 2   



 sin 2 −  
 =  − ,  = −
2
0
 2
 2
0
= 2 (sin )(−) = 0  (sin )  = 0  (sin ) 

 (cos )  =


=
5 − 3



1 1 
− 3  = − 1 ln || +  = − 1 ln |5 − 3| + .
3
3


69. Let  = ln . Then  =


, so




rS



al

67. Let  = 5 − 3. Then  = −3  and  = − 1 , so
3

e

Continuity of  is needed in order to apply the substitution rule for definite integrals.


(ln )2
 = 2  = 1 3 +  = 1 (ln )3 + .
3
3


71. Let  = 1 +  . Then  =  , so



√
√
 1 +   =
  = 2 32 +  = 2 (1 +  )32 + .
3
3

Fo

√
Or: Let  = 1 +  . Then 2 = 1 +  and 2  =  , so

 √
 1 +   =  · 2  = 2 3 +  = 2 (1 +  )32 + .
3
3

73. Let  = tan . Then  = sec2  , so

77.





  =  +  = tan  + .


 1

1

2
 +
 = tan−1  +
= tan−1  + 1 ln|| + 
2
1 + 2
1 + 2





= tan−1  + 1 ln1 + 2  +  = tan−1  + 1 ln 1 + 2 +  [since 1 + 2  0].

1+
 =
1 + 2



N



tan  sec2   =

ot

75. Let  = 1 + 2 . Then  = 2 , so



2

2



sin 2 sin  cos 
 = 2
 = 2. Let  = cos . Then  = − sin  , so
1 + cos2 
1 + cos2 

 
= −2 · 1 ln(1 + 2 ) +  = − ln(1 + 2 ) +  = − ln(1 + cos2 ) + .
2 = −2
2
1 + 2

Or: Let  = 1 + cos2 .




1
cos 
. Let  = sin . Then  = cos  , so cot   =
 = ln || +  = ln |sin | + . cot   =
79.
sin 


. When  = ,  = 1; when  = 4 ;  = 4. Thus,

 4
4

=
−12  = 2 12 = 2(2 − 1) = 2.

81. Let  = ln , so  =





4




√
ln 

1

1

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¤

212

CHAPTER 4 INTEGRALS

83. Let  =  + , so  = ( + 1) . When  = 0,  = 1; when  = 1,  =  + 1. Thus,



1

0

85.

 + 1
 =
 + 



+1

1


+1
1
= ln | + 1| − ln |1| = ln( + 1).
 = ln ||

1

 sin 

sin 
. By Exercise 64,
=   (sin ), where () =
=·
1 + cos2 
2 − 2
2 − sin2 
 


 
 sin  sin 
 
 
 =

  (sin )  =
 (sin )  =
2
2 0
2 0 1 + cos2 
0 1 + cos 
0
Let  = cos . Then  = − sin  . When  = ,  = −1 and when  = 0,  = 1. So



1
 
 1 

sin 
 −1 
=
=
 = − tan−1  −1
2 0 1 + cos2 
2 1 1 + 2
2 −1 1 + 2
2

      2
[tan−1 1 − tan−1 (−1)] =
− −
=
2
2 4
4
4

e

=



=1

rS

1. (a)

al

4 Review

 (∗ ) ∆ is an expression for a Riemann sum of a function  .


∗ is a point in the th subinterval [−1   ] and ∆ is the length of the subintervals.


Fo

(b) See Figure 1 in Section 4.2.

(c) In Section 4.2, see Figure 3 and the paragraph beside it.
2. (a) See Definition 4.2.2.

ot

(b) See Figure 2 in Section 4.2.

(c) In Section 4.2, see Figure 4 and the paragraph by it (contains “net area”).

N

3. See the Fundamental Theorem of Calculus on page 317.
4. (a) See the Net Change Theorem on page 324.

(b)
5. (a)

(b)
(c)
6. (a)

 2
1

()  represents the change in the amount of water in the reservoir between time 1 and time 2 .

 120
60

 120
60

 120
60



()  represents the change in position of the particle from  = 60 to  = 120 seconds.
|()|  represents the total distance traveled by the particle from  = 60 to 120 seconds.
()  represents the change in the velocity of the particle from  = 60 to  = 120 seconds.

 ()  is the family of functions { |  0 =  }. Any two such functions differ by a constant.

(b) The connection is given by the Net Change Theorem:




()  =




 ()   if  is continuous.

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CHAPTER 4 REVIEW

¤

213

7. The precise version of this statement is given by the Fundamental Theorem of Calculus. See the statement of this theorem and

the paragraph that follows it at the end of Section 4.3.
8. See the Substitution Rule (4.5.4). This says that it is permissible to operate with the  after an integral sign as if it were a

differential.

1. True by Property 2 of the Integral in Section 4.2.
3. True by Property 3 of the Integral in Section 4.2.

For example, let  () = 2 . Then



1√
1
1 2
2  = 0   = 1 , but
  = 1 =
2
3
0
0

7. True by Comparison Property 7 of the Integral in Section 4.2.

The integrand is an odd function that is continuous on [−1 1], so the result follows from Theorem 4.5.6(b).

al

9. True.

1
√ .
3

e

5. False.

11. False.

For example, the function  = || is continuous on R, but has no derivative at  = 0.

13. True.

By Property 5 in Section 4.2,



sin 
 =




3



[by reversing limits].
15. False.




rS

2

3

Fo



 2
 3 sin  sin  sin 
 =
 +
 ⇒





2
 3
 2
 3
 2 sin  sin  sin  sin  sin 
 −
 ⇒
 =
 +






2


3


 ()  is a constant, so


  
 ()  = 0, not  () [unless  () = 0]. Compare the given statement



17. False.

ot

carefully with FTC1, in which the upper limit in the integral is .

The function  () = 14 is not bounded on the interval [−2 1]. It has an infinite discontinuity at  = 0, so it is

N

not integrable on the interval. (If the integral were to exist, a positive value would be expected, by Comparison
Property 6 of Integrals.)

1. (a)

6 =

6


=1

 (−1 ) ∆

[∆ =

6−0
6

= 1]

=  (0 ) · 1 + (1 ) · 1 + (2 ) · 1 + (3 ) · 1 + (4 ) · 1 +  (5 ) · 1
≈ 2 + 35 + 4 + 2 + (−1) + (−25) = 8
The Riemann sum represents the sum of the areas of the four rectangles above the -axis minus the sum of the areas of the two rectangles below the
-axis.

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¤

214

CHAPTER 4 INTEGRALS

(b)

6 =

6


[∆ =

 ( ) ∆

=1

6−0
6

= 1]

=  (1 ) · 1 +  (2 ) · 1 +  (3 ) · 1 +  (4 ) · 1 +  (5 ) · 1 +  (6 ) · 1
=  (05) + (15) +  (25) +  (35) +  (45) +  (55)
≈ 3 + 39 + 34 + 03 + (−2) + (−29) = 57
The Riemann sum represents the sum of the areas of the four rectangles above the -axis minus the sum of the areas of the two rectangles below the
-axis.
√
1√
 1

1
3. 0  + 1 − 2  = 0   + 0 1 − 2  = 1 + 2 .

1 can be interpreted as the area of the triangle shown in the figure

6
0

 ()  =

4
0

 ()  +

6
4

+ .
4

 ()  ⇒ 10 = 7 +

7. First note that either  or  must be the graph of

13.

0

 () , since

6
4

()  = 10 − 7 = 3

0

 ()  = 0, and (0) 6= 0. Now notice that   0 when 

is increasing, and that   0 when  is increasing. It follows that  is the graph of  (),  is the graph of  0 (), and  is the

graph of 0  () .


2 
2 

 2 3
8 + 32  = 8 · 1 4 + 3 · 1 3 1 = 24 + 3 1 = 2 · 24 + 23 − (2 + 1) = 40 − 3 = 37
4
3
1
 1


1 − 9  =  −
0


9

1


1 10 1
10 
0


= 1−

1
10



0

N

1



−0 =

9
10

√
 9
9

 − 22
 =
(−12 − 2)  = 212 − 2 = (6 − 81) − (2 − 1) = −76

1
1

15. Let  =  2 + 1, so  = 2  and   =

17.

 ()  ⇒

Fo

11.

0

4

ot

9.



6

rS

5.

1
2

al

Area = 1 (1)(1) + 1 ()(1)2 =
2
4

e

and 2 can be interpreted as the area of the quarter-circle.

( 2 + 1)5  =

5

1

2
1

1
2

. When  = 0,  = 1; when  = 1,  = 2. Thus,




2
5 1  = 1 1 6 1 =
2
2 6

1
(64
12

− 1) =

63
12

=

21
.
4

1

does not exist because the function  () = has an infinite discontinuity at  = 4;
( − 4)2
( − 4)2

that is,  is discontinuous on the interval [1 5].
19. Let  =  3 , so  = 3 2 . When  = 0,  = 0; when  = 1,  = 1. Thus,

1
0

21.



 2 cos( 3 )  =

4

−4

1
0

cos 

1
3


1

 = 1 sin  0 = 1 (sin 1 − 0) =
3
3

1
3

sin 1.

4 tan 
4 tan 
 = 0 by Theorem 4.5.6(b), since  () = is an odd function.
2 + cos 
2 + cos 

23. Let  = sin . Then  =  cos  , so



sin  cos   =

 1 
   =

1


· 1 2 +  =
2

1
(sin )2
2

+ .

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CHAPTER 4 REVIEW

¤

215

25. Let  = 2. Then  = 2 , so

 8
0

sec 2 tan 2  =

 4
0

sec  tan 


4

√



 = 1 sec  0 = 1 sec  − sec 0 = 1 2 − 1 =
2
2
4
2

1
2



1
2



√
2 − 1.
2

27. Since 2 − 4  0 for 0 ≤   2 and 2 − 4  0 for 2   ≤ 3, we have 2 − 4 = −(2 − 4) = 4 − 2 for 0 ≤   2 and

 2

 − 4 = 2 − 4 for 2   ≤ 3. Thus,
2  3
3

 2
 3
 3
 2


3
 − 4  =
− 4
(4 − 2 )  +
(2 − 4)  = 4 −
+
3 0
3
0
0
2
2

8

 16
8
16
32
= 8 − 3 − 0 + (9 − 12) − 3 − 8 = 3 − 3 + 3 = 3 − 9 = 23
3
3

29. Let  = 1 + sin . Then  = cos  , so


√
cos  
√
= −12  = 212 +  = 2 1 + sin  + .
1 + sin 

al

e



31. From the graph, it appears that the area under the curve  = 

√
 between  = 0

rS

and  = 4 is somewhat less than half the area of an 8 × 4 rectangle, so perhaps about 13 or 14. To find the exact value, we evaluate

4
4 √
4
   = 0 32  = 2 52 = 2 (4)52 =
5
5
0
33.  () =



0





64
5

= 128.

Fo

0

2

 ⇒  0 () =
1 + 3




0

2
2
 =
3
1+
1 + 3

 


= 43 . Also,
=
, so


 
 4
 




= cos(2 )
= 43 cos(8 ). cos(2 )  = cos(2 )  ·
 0 () =
 0
 0




ot

35. Let  = 4 . Then

 
 1
 
 √ cos  cos  cos  cos  cos 
37.  = √
 =
 + √
 =
 −







1

1
1
√
√
cos  cos  1
2 cos  − cos 
0
√ =
 =
− √

2
 2 

N



⇒

√
√
√
√
√
12 + 3 ≤ 2 + 3 ≤ 32 + 3 ⇒ 2 ≤ 2 + 3 ≤ 2 3, so
√
√
3√
3√
2(3 − 1) ≤ 1 2 + 3  ≤ 2 3(3 − 1); that is, 4 ≤ 1 2 + 3  ≤ 4 3.

39. If 1 ≤  ≤ 3, then

41. 0 ≤  ≤ 1

⇒ 0 ≤ cos  ≤ 1 ⇒ 2 cos  ≤ 2

43. ∆ = (3 − 0)6 =

1
,
2

⇒

1
0

2 cos   ≤

1
0

2  =

1
3

 3 1
 0=

1
3

so the endpoints are 0, 1 , 1, 3 , 2, 5 , and 3, and the midpoints are 1 , 3 , 5 , 7 , 9 , and
2
2
2
4 4 4 4 4

[Property 7].
11
.
4

The Midpoint Rule gives
3
0

sin(3 )  ≈

6


=1

( ) ∆ =

1
2

  
 3
 3
 3
 3
 3 
3
sin 1 + sin 3 + sin 5 + sin 7 + sin 9 + sin 11
≈ 0280981.
4
4
4
4
4
4

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¤

216

CHAPTER 4 INTEGRALS

45. Note that () = 0 (), where () = the number of barrels of oil consumed up to time . So, by the Net Change Theorem,

8
0

()  = (8) − (0) represents the number of barrels of oil consumed from Jan. 1, 2000, through Jan. 1, 2008.

47. We use the Midpoint Rule with  = 6 and ∆ =

 24
0

= 4. The increase in the bee population was

24 − 0
6

()  ≈ 6 = 4[(2) + (6) + (10) + (14) + (18) + (22)]
≈ 4[50 + 1000 + 7000 + 8550 + 1350 + 150] = 4(18,100) = 72,400

49. Let  = 2 sin . Then  = 2 cos   and when  = 0,  = 0; when  =





 ()  =  sin  +

0



0


 () 1 −

1
1 + 2



2
0





 () 1  =
2

1
2

2
0

 ()  =

1
2

2
0

()  = 1 (6) = 3.
2

 ()
()
 ⇒ () =  cos  + sin  +
[by differentiation] ⇒
1 + 2
1 + 2



2
1 + 2
( cos  + sin )
=  cos  + sin  ⇒  ()
=  cos  + sin  ⇒  () =
2
1+
2

53. Let  =  () and  =  0 () . So 2

1
0



 () 0 ()  = 2

 (1 − )  =

0
1

  ()
 ()

  ()
  = 2  () = [ ()]2 − [ ()]2 .

()(−) =

1
0

 ()  =

1
0

 () .

N

ot

Fo

55. Let  = 1 − . Then  = −, so



e

51.

 (2 sin ) cos   =

al

0

 = 2. Thus,

rS

 2


,
2

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PROBLEMS PLUS
1. Differentiating both sides of the equation  sin  =

 2
0

 ()  (using FTC1 and the Chain Rule for the right side) gives

sin  +  cos  = 2 (2 ). Letting  = 2 so that  (2 ) =  (4), we obtain sin 2 + 2 cos 2 = 4 (4), so
 (4) = 1 (0 + 2 · 1) =
4


.
2



3. Differentiating the given equation,

0

 ()  = [ ()]2 , using FTC1 gives  () = 2 ()  0 () ⇒

 ()[2 0 () − 1] = 0, so  () = 0 or  0 () = 1 . Since  () is never 0, we must have  0 () =
2
 () = 1  + . To find , we substitute into the given equation to get
2

0

2



2
 +   = 1  + 
2

1
2

⇒

⇔



()



cos 

[1 + sin(2 )] . Using FTC1 and the Chain Rule (twice) we have

0

rS

0

1
√
, where () =
1 + 3

e

+  = 1 2 +  +  2 . It follows that  2 = 0, so  = 0, and () = 1 .
4
2

5.  () =

 
1
1
0 () = 
[1 + sin(cos2 )](− sin ). Now   =
 0 () = 
2
3
3
1 + [()]
1 + [()]

2



0

[1 + sin(2 )]  = 0, so

0

1
= √
(1 + sin 0)(−1) = 1 · 1 · (−1) = −1.
1+0

Fo

0

and  0 () =

al

1 2

4

  1

1
2

7.  () = 2 +  − 2 = (− + 2)( + 1) = 0

else. The integral




⇔  = 2 or  = −1.  () ≥ 0 for  ∈ [−1 2] and  ()  0 everywhere

(2 +  − 2 )  has a maximum on the interval where the integrand is positive, which is [−1 2]. So

ot

 = −1,  = 2. (Any larger interval gives a smaller integral since  ()  0 outside [−1 2]. Any smaller interval also gives a smaller integral since  () ≥ 0 in [−1 2].)

0


  

[[]]  into the sum

N

9. (a) We can split the integral

=1

−1


[[]]  . But on each of the intervals [ − 1 ) of integration,

[[]] is a constant function, namely  − 1. So the ith integral in the sum is equal to ( − 1)[ − ( − 1)] = ( − 1). So the original integral is equal to




( − 1) =

=1

(b) We can write
Now


0




[[]]  =

[[]]  =

 [[]]
0


0

[[]]  −

[[]]  +



[ ]
[]

−1


=

=1


0

( − 1)
.
2

[[]] .

[[]] . The first of these integrals is equal to 1 ([[]] − 1) [[]],
2

by part (a), and since [[]] = [[]] on [[[]]  ], the second integral is just [[]] ( − [[]]). So


0

[[]]  = 1 ([[]] − 1) [[]] + [[]] ( − [[]]) =
2

Therefore,




[[]]  =

1
2

[[]] (2 − [[]] − 1) −

1
2
1
2

[[]] (2 − [[]] − 1) and similarly
[[]] (2 − [[]] − 1).


0

[[]]  =

1
2

[[]] (2 − [[]] − 1).

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217

218

¤

CHAPTER 4 PROBLEMS PLUS

11. Let () =











 ()  =  + 2 + 3 + 4 =  + 2 + 3 + 4 . Then (0) = 0, and (1) = 0 by the
2
3
4
2
3
4
0



0

given condition,  +

 

+ + = 0. Also, 0 () =  () =  +  + 2 + 3 by FTC1. By Rolle’s Theorem, applied to
2 3 4

 on [0 1], there is a number  in (0 1) such that 0 () = 0, that is, such that  () = 0. Thus, the equation  () = 0 has a root between 0 and 1.
More generally, if  () = 0 + 1  + 2 2 + · · · +   and if 0 +

1
2

+
+ ··· +
= 0 then the equation
2
3
+1

 () = 0 has a root between 0 and 1 The proof is the same as before:
 
1 2 2 3

 ()  = 0  +
Let () =
 +
 + ··· +
 . Then (0) = (1) = 0 and 0 () =  (). By
2
3
+1
0
Rolle’s Theorem applied to  on [0 1], there is a number  in (0 1) such that 0 () = 0, that is, such that  () = 0.


0



0



  
()   =


0

 ()( − )  =

  
 






 ()( − )  =
()  −
() 

 0
0


= 0 ()  + () − () = 0 () 

   
0

0


 ()   + . Setting  = 0 gives  = 0.

Fo


1
1
1
+√ √
+··· + √ √
√ √
 +
 +1
 +2




1



= lim
+
+ ··· +
→∞ 
+1
+2
+


1
1
1
1

+
+··· + √
= lim
→∞ 
1+1
1 + 1
1 + 2

ot

→∞

0



 


1

→∞ 

=1

= lim

N

15. lim





 ()  by FTC1, while

0




Hence,



e



al




rS

13. Note that

=



0

1



1 where  () = √
1+

√

 √
1
1
√
 = 2 1 +  0 = 2 2 − 1
1+

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5

APPLICATIONS OF INTEGRATION

5.1 Areas Between Curves

=0



5.  =

2
3



−
2

−1

=

( −  )  =





2

−1

1
3


+ 1 − (0) =

4

0

=1

=0

=



1

0



(5 − 2 ) −   =



0

4

4 

(4 − 2 )  = 22 − 1 3 0 = 32 −
3

(8 −  − 2 ) 

= 22 − 3 +

1
2

=

1
2

− (0) =

32
3

0



(9 − 2 ) − ( + 1) 

2

3
2
−
= 8 −
2
3 −1
 

8
= 16 − 2 − 3 − −8 −



 1



1
 − ( 2 − 1)  =
( 12 −  2 + 1)  = 2  32 − 1  3 + 
3
3
0

4
3

64
3

e

3.  =

( −  )  =

=4

+

1
3

39
2



⇔ 2 − 4 + 4 =  ⇔ 2 − 5 + 4 = 0 ⇔

Fo

7. The curves intersect when ( − 2)2 = 

al



rS

1.  =

( − 1)( − 4) = 0 ⇔  = 1 or 4.
=



1

4

[ − ( − 2)2 ]  =



4

1

(−2 + 5 − 4) 

=

9
2

N

ot


4
= − 1 3 + 5 2 − 4 1
3
2

 
= − 64 + 40 − 16 − − 1 +
3
3

9. First find the points of intersection:

5
2


−4

√
+3
+3 =
2

⇒

√
2
+3 =



+3
2

2

⇒  + 3 = 1 ( + 3)2
4

⇒

4( + 3) − ( + 3)2 = 0 ⇒ ( + 3)[4 − ( + 3)] = 0 ⇒ ( + 3)(1 − ) = 0 ⇒  = −3 or 1. So

 1 
√
+3
=
+3−

2
−3
1

( + 3)2
= 2 ( + 3)32 −
3
4
−3
 16

4
= 3 − 4 − (0 − 0) = 3

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219

220

¤

CHAPTER 5 APPLICATIONS OF INTEGRATION

11. The curves intersect when 1 −  2 =  2 − 1

=



=

⇔  2 = 1 ⇔  = ±1.



(1 − 2 ) − (2 − 1) 

1

−1



⇔ 2 = 22

1

2(1 −  2 ) 

−1

=2·2



1

0

(1 −  2 ) 

1



= 4  − 1 3 0 = 4 1 − 1 =
3
3
13. 12 − 2 = 2 − 6

8
3

⇔ 22 = 18 ⇔

2 = 9 ⇔  = ±3, so
 3


(12 − 2 ) − (2 − 6) 
=
0



18 − 22 

[by symmetry]


3
= 2 18 − 2 3 0 = 2 [(54 − 18) − 0]
3

= 2(36) = 72

for 0     . By symmetry,
2
 3
(8 cos  − sec2 ) 
= 2

=


3

0

⇒ 8 cos3  = 1 ⇒ cos3  =

Fo

15. The curves intersect when 8 cos  = sec2 

al

3

rS



=2

e

−3

1
8

⇒ cos  =

1
2

⇒

ot


3
= 2 8 sin  − tan  0
 √
√ 
 √ 
= 2 8 · 23 − 3 = 2 3 3

N

√
3

=6

17. 2 2 = 4 +  2

=



2

−2

=2





(4 + 2 ) − 2 2 

2

0

⇔  2 = 4 ⇔  = ±2, so

(4 −  2 ) 

[by symmetry]


2


= 2 4 − 1  3 0 = 2 8 − 8 =
3
3

32
3

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SECTION 5.1 AREAS BETWEEN CURVES

19. By inspection, the curves intersect at  = ± 1 .
2

=



12

[cos  − (42 − 1)] 

−12

=2



12

0

(cos  − 42 + 1)  [by symmetry]

12
 1
1
= 2  sin  − 4 3 +  0 = 2  −
3

1
2
=2  +1 =  +2
3
3

1
6

+

1
2




−0

21. From the graph, we see that the curves intersect at  = 0,  =


,
2

and  = . By symmetry,







 2 
 2 


2
2
cos  − 1 − 2   = 2
 = 2

cos  − 1 − cos  − 1 +

 


0
0
0




2


2
1
1
= 2 sin  −  +  2 0 = 2 1 −  +  ·  − 0 = 2 1 −  +  = 2 − 
2
4
2
4
2




Fo

rS

al

e

=

23. Notice that cos  = sin 2 = 2 sin  cos 

=



6

(cos  − sin 2)  +

0

=

25.

or


.
2



2

6

(sin 2 − cos ) 

6 
2
cos 2 0 + − 1 cos 2 − sin  6
2


 
 
− 0 + 1 · 1 + 1 − 1 − −1 · 1 − 1 =
2
2
2 2
2

N


= sin  +


6

ot

2 sin  = 1 or cos  = 0 ⇔  =

⇔ 2 sin  cos  − cos  = 0 ⇔ cos  (2 sin  − 1) = 0 ⇔

1
2

+

1
2

1
2

1
2

·

1
2

√
=  ⇒ 1 2 =  ⇒ 2 − 4 = 0 ⇒ ( − 4) = 0 ⇒  = 0 or 4, so
4
 4
 9
√
1

√ 
=
 − 1   +
 −  
2
2
1

2

0

=
=
=



4

2 32

3

 16
3

81
4

+

4

+

− 26 =

59
12

− 1 2
4

0



1 2

4

− 2 32
3

9
4


 
 
− 4 − 0 + 81 − 18 − 4 −
4
32
3

16
3



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¤

221

222

¤

CHAPTER 5 APPLICATIONS OF INTEGRATION

27. Graph the three functions  = 12 ,  = , and  =

1
8 ;




 2
1
1
1
−  
 −   +
8
2
8
0
1
1 7
 2  −2 1 
= 0 8   + 1  − 8  
1 
2

1
7 2
1
+ − − 2
=

16

16
0
1
 1
 

7
1
1
= 16 + − 2 − 4 − −1 − 16 = 3
4

=



1

29. An equation of the line through (0 0) and (3 1) is  =



then determine the points of intersection: (0 0), (1 1), and 2 1 .
4

1
;
3

through (0 0) and (1 2) is  = 2;

through (3 1) and (1 2) is  = − 1  + 5 .
2
2
 3
 1

 1



2 − 1   +
− 2  + 5 − 1  
=
3
2
3

=
=

5
 
3

+

1
2 0
6

 5

− 6  + 5 
2

 5
3
+ − 12 2 + 5  1
2
  5


+ − 15 + 15 − − 12 + 5 =
4
2
2

5
5
6

1

3

al

0



5
2

31. The curves intersect when sin  = cos 2

(on [0 2]) ⇔ sin  = 1 − 2 sin2  ⇔ 2 sin2  + sin  − 1 = 0 ⇔



2

|sin  − cos 2| 

0



6

=
=
=

33.

1
2

1 √
3+
4
3
2

6

sin 2 + cos 

√
3−1

1
2

0



⇒ =


6.

2

6

(sin  − cos 2) 

ot

0

(cos 2 − sin )  +


+ − cos  −

1
2

2 sin 2 6

√ 
√

3 − (0 + 1) + (0 − 0) − − 1 3 −
2

N

=

1
2

Fo

(2 sin  − 1)(sin  + 1) = 0 ⇒ sin  =
=

rS

=

1

1

e

0



1
4

√ 
3

From the graph, we see that the curves intersect at  = 0 and  =  ≈ 0896, with
 sin(2 )  4 on (0 ). So the area  of the region bounded by the curves is
 




=
 sin(2 ) − 4  = − 1 cos(2 ) − 1 5 0
2
5
0

= − 1 cos(2 ) − 1 5 +
2
5

1
2

≈ 0037

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 5.1 AREAS BETWEEN CURVES

¤

From the graph, we see that the curves intersect at

35.

 =  ≈ −111  =  ≈ 125 and  =  ≈ 286 with
3 − 3 + 4  32 − 2 on ( ) and 32 − 2  3 − 3 + 4 on ( ). So the area of the region bounded by the curves is

=
=










=

4

(3 − 32 −  + 4)  +













 2

(3 − 2) − (3 − 3 + 4) 

(−3 + 32 +  − 4) 

 

4 − 3 − 1 2 + 4  + − 1 4 + 3 + 1 2 − 4  ≈ 838
2
4
2

Graph Y1 =2/(1+xˆ4) and Y2 =xˆ2. We see that Y1  Y2 on (−1 1), so the

 1 
2
area is given by
− 2 . Evaluate the integral with a
1 + 4
−1

al

e

37.

1



 3

( − 3 + 4) − (32 − 2)  +

command such as fnInt(Y1 -Y2 ,x,-1,1) to get 280123 to five decimal places. rS

Another method: Graph () = Y1 =2/(1+xˆ4)-xˆ2 and from the graph

evaluate ()  from −1 to 1.
The curves intersect at  = 0 and  =  ≈ 0749363.
 
√

 − tan2   ≈ 025142
=

Fo

39.

ot

0

41. As the figure illustrates, the curves  =  and  = 5 − 63 + 4

N

enclose a four-part region symmetric about the origin (since
5 − 63 + 4 and  are odd functions of ). The curves intersect at values of  where 5 − 63 + 4 = ; that is, where
(4 − 62 + 3) = 0. That happens at  = 0 and where
√




√
√
√
√
√
36 − 12
 =
= 3 ± 6; that is, at  = − 3 + 6, − 3 − 6, 0, 3 − 6, and 3 + 6. The exact area is
2
 √3+√6
 √3+√6
 5
 5


( − 63 + 4) −   = 2
 − 63 + 3 
2
2

6±

0

0

=2

 √3−√6
0

 √3+√6
( − 6 + 3)  + 2 √ √ (−5 + 63 − 3) 
5

3

3− 6

√
= 12 6 − 9

CAS

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

223

224

¤

CHAPTER 5 APPLICATIONS OF INTEGRATION

43. 1 second =

∆ =

hour, so 10 s =

1
3600

1360−0
5

=

1
,
1800

1
360

h. With the given data, we can take  = 5 to use the Midpoint Rule.

so

distance Kelly − distance Chris =

 1360
0

≈ 5 =

  −
1
1800

 1360
0

  =

 1360
0

( −  ) 

[( −  )(1) + ( −  )(3) + ( −  )(5)
+ ( −  )(7) + ( −  )(9)]

=

1
1800 [(22

=

1
1800 (2

− 20) + (52 − 46) + (71 − 62) + (86 − 75) + (98 − 86)]

+ 6 + 9 + 11 + 12) =

1
1800 (40)

=

1
45

mile, or 117 1 feet
3

45. Let () denote the height of the wing at  cm from the left end.

200 − 0
[(20) + (60) + (100) + (140) + (180)]
5

e

 ≈ 5 =



 ()  =  (), where  () is the velocity of car A

and A is its displacement. Similarly, the area under curve  between  = 0 and  =  is 0 B ()  = B ().
0

rS

47. We know that the area under curve  between  = 0 and  =  is

al

= 40(203 + 290 + 273 + 205 + 87) = 40(1058) = 4232 cm2

(a) After one minute, the area under curve  is greater than the area under curve . So car A is ahead after one minute.

(b) The area of the shaded region has numerical value A (1) − B (1), which is the distance by which A is ahead of B after

Fo

1 minute.

(c) After two minutes, car B is traveling faster than car A and has gained some ground, but the area under curve  from  = 0 to  = 2 is still greater than the corresponding area for curve , so car A is still ahead.
(d) From the graph, it appears that the area between curves  and  for 0 ≤  ≤ 1 (when car A is going faster), which

ot

corresponds to the distance by which car A is ahead, seems to be about 3 squares. Therefore, the cars will be side by side at the time  where the area between the curves for 1 ≤  ≤  (when car B is going faster) is the same as the area for

49.

N

0 ≤  ≤ 1. From the graph, it appears that this time is  ≈ 22. So the cars are side by side when  ≈ 22 minutes.
To graph this function, we must first express it as a combination of explicit
√
functions of ; namely,  = ±  + 3. We can see from the graph that the loop extends from  = −3 to  = 0, and that by symmetry, the area we seek is just twice the area under the top half of the curve on this interval, the equation of the
0 

√
√
top half being  = −  + 3. So the area is  = 2 −3 −  + 3 . We

substitute  =  + 3, so  =  and the limits change to 0 and 3, and we get
3
3
√
 = −2 0 [( − 3)  ]  = −2 0 (32 − 312 ) 

3
  √ 
 √ 
= −2 2 52 − 232 = −2 2 32 3 − 2 3 3 =
5
5
0

24
5

√
3

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 5.1 AREAS BETWEEN CURVES

¤

By the symmetry of the problem, we consider only the first quadrant, where

 = 2 ⇒  = . We are looking for a number  such that
 
 4



4
  =
  ⇒ 2  32 = 2  32
⇒
3
3

51.

0

32



0



32

=4

32

32

−

⇒ 2



32

=8 ⇒ 

= 4 ⇒  = 423 ≈ 252.

53. We first assume that   0, since  can be replaced by − in both equations without changing the graphs, and if  = 0 the

curves do not enclose a region. We see from the graph that the enclosed area  lies between  = − and  = , and by symmetry, it is equal to four times the area in the first quadrant. The enclosed area is

0







(2 − 2 )  = 4 2  − 1 3 0 = 4 3 − 1 3 = 4 2 3 = 8 3
3
3
3
3

So  = 576 ⇔

8 3

3

= 576 ⇔ 3 = 216 ⇔  =

√
3
216 = 6.

2



= ln 2 −

1
2

≈ 019

rS



Fo



2
1
1
1
− 2  = ln  +
 
 1
1


= ln 2 + 1 − (ln 1 + 1)
2

55.  =

al

Note that  = −6 is another solution, since the graphs are the same.

e

=4

(on [−3 3]) ⇔ sin  = 2 sin  cos  ⇔

ot

57. The curves intersect when tan  = 2 sin 

N

2 sin  cos  − sin  = 0 ⇔ sin  (2 cos  − 1) = 0 ⇔ sin  = 0 or cos  =
 3
(2 sin  − tan ) 
[by symmetry]
= 2

1
2

⇔  = 0 or  = ±  .
3

0


3
= 2 −2 cos  − ln |sec |
0

= 2 [(−1 − ln 2) − (−2 − 0)]
= 2(1 − ln 2) = 2 − 2 ln 2

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

225

226

¤

CHAPTER 5

APPLICATIONS OF INTEGRATION

5.2 Volumes


2

1. A cross-section is a disk with radius 2 − 1  so its area is () =  2 − 1  .
2
2
2

()  =

1

=



2

1



2

1


2
 2 − 1  
2



4 − 2 + 1 2 
4


2
1
=  4 − 2 + 12 3 1
 

8
=  8 − 4 + 12 − 4 − 1 +


7
=  1 + 12 = 19 
12

1
12

3. A cross-section is a disk with radius



 =

5



√
2
√
 − 1, so its area is () =   − 1 = ( − 1).

()  =

5

1

( − 1)  = 

1
2

5

 

2 −  1 =  25 − 5 − 1 − 1 = 8
2
2

  2
() =  2  .




9

()  =

0

9

0

= 4

1

2


2 9
0


  
2
 2   = 4

N

 =


, so its area is

ot

5. A cross-section is a disk with radius 2

Fo

rS

1



e



al

 =

9

 

0

= 2(81) = 162

7. A cross-section is a washer (annulus) with inner

radius 3 and outer radius , so its area is
() = ()2 − (3 )2 = (2 − 6 ).
 =



1

()  =

0

=



0

1
3

3 − 1 7
7

1
0

1

(2 − 6 ) 

=

1
3

−

1
7



=

4

21

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 5.2

VOLUMES

9. A cross-section is a washer with inner radius  2

and outer radius 2, so its area is
() = (2)2 − ( 2 )2 = (4 2 −  4 ).
 2
 2
()  = 
(4 2 −  4 ) 
 =
3

 −

3


1 5 2
 0
5

=

 32
3

−

32
5



=

64

15

11. A cross-section is a washer with inner radius 1 −


√ 2
() =  (1 − 2 )2 − (1 −  )


√
=  (1 − 22 + 4 ) − (1 − 2  + )


√
=  4 − 22 + 2  −  .
1

√
 and outer radius 1 − 2 , so its area is

1

(4 − 22 + 212 − ) 
1

=  1 5 − 2 3 + 4 32 − 1 2
5
3
3
2
0

=

()  =

1

−

5

2
3

+

4
3

0

1
2

−

0



=

11

30

rS

 =

al

=

0

4

e

0

13. A cross-section is a washer with inner radius (1 + sec ) − 1 = sec  and outer radius 3 − 1 = 2, so its area is



() =  22 − (sec )2 = (4 − sec2 ).


3

()  =

−3

= 2



3

−3

3

0



Fo

 =

(4 − sec2 ) 

(4 − sec2 ) 

[by symmetry]

ot


3
√ 


= 2 4 − 3 − 0
= 2 4 − tan 
3
0

3

√ 
− 3

N

= 2

 4

15. A cross-section is a washer with inner radius 2 − 1 and outer radius 2 −

√
3 , so its area is






√
√
() =  (2 − 3  )2 − (2 − 1)2 =  4 − 4 3  + 3  2 − 1 .
 =



0

1

()  =



0

1


1


(3 − 413 +  23 )  =  3 − 3 43 + 3  23 =  3 − 3 + 3 = 3 .
5
5
5
0

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227

228

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CHAPTER 5

APPLICATIONS OF INTEGRATION

17. From the symmetry of the curves, we see they intersect at  =

1
2

and so 2 =


⇔  = ± 1 . A cross-section is a
2

1
2

washer with inner radius 3 − (1 −  2 ) and outer radius 3 −  2 , so its area is


() =  (3 − 2 )2 − (2 + 2 )2


=  (9 − 62 +  4 ) − (4 + 4 2 +  4 )
= (5 − 10 2 ).
 √
12

() 

√

−

12

 √12

=2

0

5(1 − 22 ) 

[by symmetry]

√
√22

= 10 22 −
= 10  − 2  3 0
3
= 10

√ 
2
3

=

10
3

√

2
6

√
2



e

 =



1

()  =

0



1

()2  = 

0

1
3

3

1
0

= 1
3

21. R1 about AB (the line  = 1):



1

()  =

0



1

0

(1 − )2  = 

23. R2 about OA (the line  = 0):



1

()  =

0



1

0

1

0

1

(1 − 2 +  2 )  =   −  2 + 1  3 0 = 1 
3
3

 1


1
 √ 2 


 12 − 4 
(1 − 12 )  =   − 2 32 =  1 − 2 = 1 
 = 
3
3
3
0

0

ot

 =



Fo

 =

rS

 =

al

19. R1 about OA (the line  = 0):

25. R2 about AB (the line  = 1):



1

()  =

0

=





1

0

1

0

[12 − (1 − 4 )2 ]  = 

N

 =

(2 4 −  8 )  = 

2
5

5 − 1 9
9

1
0

=



2
5

1

0

−

[1 − (1 − 24 +  8 )] 
1
9



=

13

45

27. R3 about OA (the line  = 0):

 =



0

1

()  =



0

1



 1

1
 √ 



2
4
 − 2  = 
(12 − 2 )  =  2 32 − 1 3 =  2 − 1 = 1 
3
3
3
3
3
0

0

Note: Let R = R1 ∪ R2 ∪ R3 . If we rotate R about any of the segments , , , or , we obtain a right circular cylinder of height 1 and radius 1. Its volume is 2  = (1)2 · 1 = . As a check for Exercises 19, 23, and 27, we can add the answers, and that sum must equal . Thus, 1  + 1  + 1  = .
3
3
3

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SECTION 5.2

VOLUMES

¤

29. R3 about AB (the line  = 1):

 =



1

()  =

0

=





0

1

0

1

[(1 −  4 )2 − (1 − )2 ]  = 

( 8 − 2 4 −  2 + 2)  = 

1

9

9



1

0

[(1 − 2 4 +  8 ) − (1 − 2 + 2 )] 

− 2 5 − 1 3 + 2
5
3

1
0

=

1
9

−

Note: See the note in Exercise 27. For Exercises 21, 25, and 29, we have 1  +
3

2
5

−

13

45

1
3

+


+1 =

17

45

17
45 

= .

31. (a) About the -axis:

 4
0

(tan )2  = 

0

tan2  

e

≈ 067419

 4

(b) About  = −1:
 4
0

=
=



 [tan  − (−1)]2 − [0 − (−1)]2 

 4
0

 4
0

[(tan  + 1)2 − 12 ] 
(tan2  + 2 tan ) 

33. (a) About  = 2:

Fo

≈ 285178



2



−2


 
2 
2 


2 − − 1 − 2 4
− 2 − 1 − 2 4

N



⇒  2 = 1 − 2 4 ⇒

ot

2 + 4 2 = 4 ⇒ 42 = 4 − 2

 = ± 1 − 2 4
 =

rS

 =

al

 =

= 2

2

8

0


1 − 2 4  ≈ 78.95684

(b) About  = 2:


2 + 4 2 = 4 ⇒ 2 = 4 − 4 2 ⇒  = ± 4 − 4 2
 1 
 
2 
2 

 =
 2 − − 4 − 4 2
− 2 − 4 − 4 2

−1

= 2



0

1


8 4 − 4 2  ≈ 7895684

[Notice that this is the same approximation as in part (a). This can be explained by Pappus’s Theorem in Section 8.3.]

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229

¤

230

CHAPTER 5

APPLICATIONS OF INTEGRATION

 = 2 + 2 cos  and  = 4 +  + 1 intersect at

35.

 =  ≈ −1288 and  =  ≈ 0884
 
[(2 + 2 cos )2 − (4 +  + 1)2 ]  ≈ 23780
 =


37.  = 





0



2 sin2  − (−1) − [0 − (−1)]2 

39. 


0

sin   = 

al

e

CAS 11 2
= 8

2
  √ sin   describes the volume of solid obtained by rotating the region
0

rS



√
R = ( ) | 0 ≤  ≤ , 0 ≤  ≤ sin  of the -plane about the -axis.


1
( 4 −  8 )  =  0 ( 2 )2 − ( 4 )2  describes the volume of the solid obtained by rotating the region


R = ( ) | 0 ≤  ≤ 1  4 ≤  ≤  2 of the -plane about the -axis.
1
0

Fo

41. 

43. There are 10 subintervals over the 15-cm length, so we’ll use  = 102 = 5 for the Midpoint Rule.

 =

 15
0

()  ≈ 5 =

15−0
[(15)
5

+ (45) + (75) + (105) + (135)]

N

ot

= 3(18 + 79 + 106 + 128 + 39) = 3 · 370 = 1110 cm3


 10
−
45. (a)  = 2  [ ()]2  ≈  10 4 2 [(3)]2 + [ (5)]2 + [ (7)]2 + [(9)]2


≈ 2 (15)2 + (22)2 + (38)2 + (31)2 ≈ 196 units3

4 
(b)  = 0  (outer radius)2 − (inner radius)2 

 
 
 

≈  4 − 0 (99)2 − (22)2 + (97)2 − (30)2 + (93)2 − (56)2 + (87)2 − (65)2
4
≈ 838 units3

47. We’ll form a right circular cone with height  and base radius  by

revolving the line  =   about the -axis.


  2
 
 2
 2
2 1 3
  = 

  =  2
 =
2

 3
0
0 
0


1
2 1 3
 = 2 
= 2
 3
3

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SECTION 5.2

VOLUMES


 +  about the -axis.


  2
 
2
 2 22

∗
 −
−  +   = 
 =
 + 2 

2

0
0
Another solution: Revolve  = −

=



2 3  2 2
 −  + 2 
32




=

0

1
3


2  − 2  + 2  = 1 2 
3



 and  = −  to get


0




 0 

1
 1 3
1


− 3 = 2 .
2 −  = −
= −


 3

3
3


∗

Or use substitution with  =  −

⇔ 2 = 2 −  2



 

 
 2

3
3
( − )3
 =
=  3 −
 −  2  =  2  −
− 2 ( − ) −
3 −
3
3
−

2 3 1
 2
=  3  − 3 ( − ) 3 − ( − )2




= 1  23 − ( − ) 32 − 2 − 2 + 2
3



= 1  23 − ( − ) 22 + 2 − 2
3


= 1  23 − 23 − 22  + 2 + 22  + 22 − 3
3


 1 2


2
3
2
1
= 3  3 −  = 3  (3 − ), or, equivalently,   −
3

rS

al

e

49. 2 +  2 = 2

Fo


2
−

=
, so  =  1 −
.
2




. So
Similarly, for cross-sections having 2 as their base and  replacing ,  = 2 1 −

 
  
  
 
 1−
 =
2 1 −

()  =


0
0

 
 

2
 2
2
=
22 1 −
 = 22
1−
+ 2 



0
0

2
3 




+ 2
= 22  −  + 1 
= 22  −
3

3 0

N

ot

51. For a cross-section at height , we see from similar triangles that

= 2 2  [ = 1  where  is the area of the base, as with any pyramid.]
3
3

53. A cross-section at height  is a triangle similar to the base, so we’ll multiply the legs of the base triangle, 3 and 4, by a

proportionality factor of (5 − )5. Thus, the triangle at height  has area

 


5−
5−
1
 2
() = · 3
·4
=6 1−
, so
2
5
5
5


 5
 5
 0
 = 1 − 5,
 2
 =
1−
()  = 6
 = 6
2 (−5 )
5
 = − 1 
0
0
1
5
 1 3 0
 1
= −30 3  1 = −30 − 3 = 10 cm3

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¤

231

232

¤

CHAPTER 5

APPLICATIONS OF INTEGRATION

55. If  is a leg of the isosceles right triangle and 2 is the hypotenuse,

then 2 + 2 = (2)2
2

⇒ 2 = 2 2 .

⇒ 22 = 4 2

2
2
()  = 2 0 1 ()()  = 2 0  2 
2
2
2
= 2 0 1 (36 − 92 )  = 9 0 (4 − 2 ) 
4
2

2


= 9 4 − 1 3 0 = 9 8 − 8 = 24
2
3
2
3

 =

−2

()  = 2

2
0

57. The cross-section of the base corresponding to the coordinate  has length

 = 1 − . The corresponding square with side  has area
() = 2 = (1 − )2 = 1 − 2 + 2 . Therefore,
1

()  =

0



0

1

(1 − 2 + 2 ) 

1 


=  − 2 + 1 3 0 = 1 − 1 + 1 − 0 = 1
3
3
3
 1
 0
1

Or:
(1 − )2  =
2 (−) [ = 1 − ] = 1 3 0 =
3
1

1
3

rS

0

e



al

 =

59. The cross-section of the base  corresponding to the coordinate  has length 1 − 2 . The height  also has length 1 − 2 ,

−1

=2·

1
2



0

1

(1 − 22 + 4 ) 

[by symmetry]
2
3

+

1
5



−0 =

8
15

ot

1 

=  − 2 3 + 1 5 0 = 1 −
3
5

Fo

so the corresponding isosceles triangle has area () = 1  = 1 (1 − 2 )2 . Therefore,
2
2
 1
1
 =
(1 − 2 )2 
2

61. (a) The torus is obtained by rotating the circle ( − )2 +  2 = 2 about

N

the -axis. Solving for , we see that the right half of the circle is given by


 =  + 2 −  2 =  () and the left half by  =  − 2 −  2 = ().
So

 =

 

[ ()]2 − [()]2 
−

= 2
= 2

 



   2
 + 2 2 −  2 + 2 −  2 − 2 − 2 2 −  2 + 2 −  2 
0


0

4



2 −  2  = 8 0 2 −  2 

(b) Observe that the integral represents a quarter of the area of a circle with radius , so

8 0 2 −  2  = 8 · 1 2 = 2 2 2 .
4

63. (a) Volume(1 ) =


0

()  = Volume(2 ) since the cross-sectional area () at height  is the same for both solids.

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SECTION 5.2

VOLUMES

¤

233

(b) By Cavalieri’s Principle, the volume of the cylinder in the figure is the same as that of a right circular cylinder with radius  and height , that is, 2 .
65. The volume is obtained by rotating the area common to two circles of radius , as

shown. The volume of the right half is
 2
 2  2  1
2 
 − 2  +  
right =  0  2  =  0


3 2


 
=  2  − 1 1  + 
=  1  3 − 1 3 − 0 −
3 2
2
3
0

So by symmetry, the total volume is twice this, or

1 3

24

3
5
12  .



=

5
3
24

67. Take the -axis to be the axis of the cylindrical hole of radius .

al

A quarter of the cross-section through  perpendicular to the

e

Another solution: We observe that the volume is the twice the volume of a cap of a sphere, so we can use the formula from
 2 

5
Exercise 49 with  = 1 :  = 2 · 1 2 (3 − ) = 2  1  3 − 1  = 12 3 .
2
3
3
2
2

-axis, is the rectangle shown. Using the Pythagorean Theorem



−







2 −  2 2 −  2  = 8 0 2 −  2 2 −  2 

Fo

 =

rS

twice, we see that the dimensions of this rectangle are


 = 2 −  2 and  = 2 −  2 , so


1
2 −  2 2 −  2 , and
4 () =  =
()  =

−

4

69. (a) The radius of the barrel is the same at each end by symmetry, since the

function  =  − 2 is even. Since the barrel is obtained by rotating

ot

the graph of the function  about the -axis, this radius is equal to the

N

 2 value of  at  = 1 , which is  −  1  =  −  = .
2
2
(b) The barrel is symmetric about the -axis, so its volume is twice the volume of that part of the barrel for   0. Also, the barrel is a volume of rotation, so

 2
 2  = 2
 =2
0

= 2

0

1
2

2

 −

1
3
12

+

2


2
2

 − 2  = 2 2  − 2 3 + 1 2 5 0
3
5

1 2 5
 
160





Trying to make this look more like the expression we want, we rewrite it as  = 1  22 + 2 − 1 2 +
3
2




2
2
2
2 2
2 2
2
1
3 2 4
1
1 2 4
2 1
2 2
− 40   = ( − ) − 5 4 
=  − 5 .
But  − 2  + 80   =  − 4 
 2

Substituting this back into  , we see that  = 1  2 + 2 − 2  2 , as required.
3
5

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3 2 4
 
80


.

234

¤

CHAPTER 5

APPLICATIONS OF INTEGRATION

5.3 Volumes by Cylindrical Shells
If we were to use the “washer” method, we would first have to locate the

1.

local maximum point ( ) of  = ( − 1)2 using the methods of
Chapter 3. Then we would have to solve the equation  = ( − 1)2 for  in terms of  to obtain the functions  = 1 () and  = 2 () shown in the first figure. This step would be difficult because it involves the cubic formula. Finally we would find the volume using


 =  0 [1 ()]2 − [2 ()]2 .

Using shells, we find that a typical approximating shell has radius , so its circumference is 2. Its height is , that is,



1

√
2 3   = 2

0



1

43 

0



3 73
7

1
0

= 2

3
7

= 6
7

Fo

= 2

rS

3.  =

0

 5
1
 4


4
3

 − 23 + 2  = 2
−2
+
=
5
4
3 0
15

e

0

1

al

( − 1)2 . So the total volume is

 1


2 ( − 1)2  = 2
 =

2
2(4 − 2 )  = 2 0 (4 − 3 ) 

2
= 2 22 − 1 4 0 = 2(8 − 4)
4

5.  =

2
0

7. 2 = 6 − 22

 =



2

0

= 2

N

ot

= 8

⇔ 32 − 6 = 0 ⇔ 3( − 2) = 0 ⇔  = 0 or 2.

2[(6 − 22 ) − 2 ] 



2

(−33 + 62 ) 

0

2

= 2 − 3 4 + 23 0
4

= 2 (−12 + 16) = 8

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SECTION 5.3

VOLUMES BY CYLINDRICAL SHELLS

1
⇒  =  , so
 3  
1
 = 2



1
 3
 3
 = 2  1
= 2

9.  = 1

1

= 2(3 − 1) = 4

8

0

= 2



8

 

( 3  − 0) 
 43  = 2

0

3 73

7

8
0

6 73
6 7
768
(8 ) =
(2 ) =

7
7
7

Fo

rS

=



e



al

11.  = 2









13. The height of the shell is 2 − 1 + ( − 2)2 = 1 − ( − 2)2 = 1 −  2 − 4 + 4 = − 2 + 4 − 3.

= 2

3
1

3
1

(− 2 + 4 − 3) 

ot

 = 2

(− 3 + 4 2 − 3) 

N

3

= 2 − 1  4 + 4  3 − 3  2 1
4
3
2


= 2 − 81 + 36 −
4
 8  16
= 2 3 = 3 

27
2




− −1 +
4

4
3

−

3
2



15. The shell has radius 2 − , circumference 2(2 − ), and height 4 .

1

2(2 − )4 
1
= 2 0 (24 − 5 ) 

1
= 2 2 5 − 1 6 0
5
6


7

= 2 2 − 1 − 0 = 2 30 =
5
6

 =

0

7

15

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¤

235

236

¤

CHAPTER 5

APPLICATIONS OF INTEGRATION

17. The shell has radius  − 1, circumference 2( − 1), and height (4 − 2 ) − 3 = −2 + 4 − 3.

3

2( − 1)(−2 + 4 − 3) 
3
= 2 1 (−3 + 52 − 7 + 3) 
3

= 2 − 1 4 + 5 3 − 7 2 + 3 1
4
3
2

 
= 2 − 81 + 45 − 63 + 9 − − 1 +
4
2
4
4 8
= 2 3 = 3 

=

1

5
3

−

7
2


+3

19. The shell has radius 1 − , circumference 2(1 − ), and height 1 −

1

2(1 − )(1 −  13 ) 
1
= 2 0 (1 −  −  13 +  43 ) 
1

= 2  − 1  2 − 3  43 + 3  73
2
4
7


 = 3

⇔

=

 
3
 .

0

e

=


3


0

 3
2

rS

21. (a)  =

al




= 2 1 − 1 − 3 + 3 − 0
2
4
7
5
5
= 2 28 = 14 
2 sin  



2

−2

= 4



2

−2

( − )[cos4  − (− cos4 )] 
( − ) cos4  


0

N

(b)  ≈ 4650942
25. (a)  =

ot

23. (a)  = 2

Fo

(b) 9869604

2(4 − )

√ sin  

(b)  ≈ 3657476

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SECTION 5.3

27.  =

1
0

2

VOLUMES BY CYLINDRICAL SHELLS

¤

√
√
1 + 3 . Let  () =  1 + 3 .

Then the Midpoint Rule with  = 5 gives
1
0

 ()  ≈

1−0
5

[(01) +  (03) +  (05) +  (07) +  (09)]

≈ 02(29290)

Multiplying by 2 gives  ≈ 368.
29.

3
0

25  = 2

cylindrical shells.
31.

1
0

3
0

(4 ) . The solid is obtained by rotating the region 0 ≤  ≤ 4 , 0 ≤  ≤ 3 about the y-axis using

2(3 − )(1 − 2 ) . The solid is obtained by rotating the region bounded by (i)  = 1 −  2 ,  = 0, and  = 0 or

(ii)  =  2 ,  = 1, and  = 0 about the line  = 3 using cylindrical shells.
From the graph, the curves intersect at  = 0 and at  =  ≈ 132, with

e

33.

al

 + 2 − 4  0 on the interval (0 ). So the volume of the solid obtained

2

 
2

0



−  sin2  − sin4  

ot

CAS 1 3
= 32 

rS



Fo

35.  = 2

by rotating the region about the -axis is


 = 2 0 [( + 2 − 4 )]  = 2 0 (2 + 3 − 5 ) 


= 2 1 3 + 1 4 − 1 6 0 ≈ 405
3
4
6

37. Use shells:

4

2(−2 + 6 − 8)  = 2

4
= 2 − 1 4 + 23 − 42 2
4
2

N

 =

4
2

(−3 + 62 − 8) 

= 2[(−64 + 128 − 64) − (−4 + 16 − 16)]
= 2(4) = 8
√
 = ± 2 ± 1


2 

 (2 − 0)2 −
2 + 1 − 0

39. Use washers:  2 − 2 = 1

 =



√

3

√
− 3

= 2



√

3

0

= 2



0

√

3

⇒

[4 − (2 + 1)] 

[by symmetry]

√3

(3 − 2 )  = 2 3 − 1 3 0
3

√ 
√
 √
= 2 3 3 − 3 = 4 3 

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237

238

¤

CHAPTER 5

APPLICATIONS OF INTEGRATION


⇔  = ± 1 − ( − 1)2
 2 
 2
2
 =
1 − ( − 1)2  = 
(2 −  2 ) 

41. Use disks: 2 + ( − 1)2 = 1
0

0

2



=  2 − 1 3 0 =  4 − 8 = 4 
3
3
3

43.  + 1 = ( − 1)2

⇔  + 1 = 2 − 2 + 1 ⇔ 0 = 2 − 3

⇔

0 = ( − 3) ⇔  = 0 or 3.

Use disks:
 3


 =
[( + 1) − (−1)]2 − [( − 1)2 − (−1)]2 
0

0

=



3

0

[( + 2)2 − ( 2 − 2 + 2)2 ] 
[( 2 + 4 + 4) − ( 4 − 4 3 + 8 2 − 8 + 4)]  = 



0

(−4 + 4 3 − 7 2 + 12) 

117
5 

rS

3



=  − 1  5 +  4 − 7  3 + 6 2 0 =  − 243 + 81 − 63 + 54 =
5
3
5

3

e

3

al

=



45. Use shells:

0

Fo

√


 = 2 0 2 2 − 2  = −2 0 (2 − 2 )12 (−2) 


= −2 · 2 (2 − 2 )32 = − 4 (0 − 3 ) = 4 3
3
3
3



N



ot




  2


−
+  
 −  +   = 2


0
0

 3
2
2 

2
+
=
= 2 −
= 2
3
2 0
6
3

47.  = 2

5.4 Work
1. (a) The work done by the gorilla in lifting its weight of 360 pounds to a height of 20 feet

is  =   = (360 lb)(20 ft) = 7200 ft-lb.
(b) The amount of time it takes the gorilla to climb the tree doesn’t change the amount of work done, so the work done is still 7200 ft-lb.
10

 1


 10
3.  =   ()  = 1 5−2  = 5 −−1
= 5 − 10 + 1 = 45 ft-lb
1

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SECTION 5.4 WORK

¤

239

5. The force function is given by  () (in newtons) and the work (in joules) is the area under the curve, given by

8
0

 ()  =

4
0

 ()  +

8
4

 ()  = 1 (4)(30) + (4)(30) = 180 J.
2

7. According to Hooke’s Law, the force required to maintain a spring stretched  units beyond its natural length is proportional

to , that is,  () = . Here, the amount stretched is 4 in. =

1
3

ft and the force is 10 lb. Thus, 10 = 

 = 30 lbft, and  () = 30. The work done in stretching the spring from its natural length to 6 in. = length is  =
9. (a) If

 012
0

 12
0


12
30  = 152 0 =

  = 2 J, then 2 =

15
4


2 012
2  0

1

1
3

1
2

⇒

ft beyond its natural

ft-lb.
= 1 (00144) = 00072 and  =
2

2
00072

=

2500
9

≈ 27778 Nm.

Thus, the work needed to stretch the spring from 35 cm to 40 cm is
110
 1


 010 2500
1
  = 1250 2 120 = 1250 100 − 400 = 25 ≈ 104 J.
9
9
9
24
005
2500
9 

and  =

270
2500

m = 108 cm

e

(b)  () = , so 30 =

 02
01

  = 


2 02
2  01

1

=



4
200

−

1
200



  = 

3
200 .

=

0

Thus, 2 = 31 .

rS

Now 2 =

 01

al

11. The distance from 20 cm to 30 cm is 01 m, so with  () = , we get 1 =


2 01
2 0

1

=

1
200 .

In Exercises 13 – 20,  is the number of subintervals of length ∆, and ∗ is a sample point in the th subinterval [−1   ].

13. (a) The portion of the rope from  ft to ( + ∆) ft below the top of the building weighs

1
2

∆ lb and must be lifted ∗ ft,


Fo

so its contribution to the total work is 1 ∗ ∆ ft-lb. The total work is
2 

 1 ∗

→∞ =1 2

 = lim

∆ =

 50
0

1
 
2

=

1
4

2

50
0

=

2500
4

= 625 ft-lb

Notice that the exact height of the building does not matter (as long as it is more than 50 ft).

N

ot

(b) When half the rope is pulled to the top of the building, the work to lift the top half of the rope is
25

 25
1 = 0 1   = 1 2 0 = 625 ft-lb. The bottom half of the rope is lifted 25 ft and the work needed to accomplish
2
4
4
 50
 50 1 that is 2 = 25 2 · 25  = 25  25 = 625 ft-lb. The total work done in pulling half the rope to the top of the building
2
2 is  = 1 + 2 =

625
2

+

625
4

=

3
4

15. The work needed to lift the cable is lim

→∞

· 625 =



=1

1875
4

ft-lb.

2∗ ∆ =


 500
0

 500
2  = 2 0 = 250,000 ft-lb. The work needed to lift

the coal is 800 lb · 500 ft = 400,000 ft-lb. Thus, the total work required is 250,000 + 400,000 = 650,000 ft-lb.

17. At a height of  meters (0 ≤  ≤ 12), the mass of the rope is (08 kgm)(12 −  m) = (96 − 08) kg and the mass of the

water is

 36
12

 kgm (12 −  m) = (36 − 3) kg. The mass of the bucket is 10 kg, so the total mass is

(96 − 08) + (36 − 3) + 10 = (556 − 38) kg, and hence, the total force is 98(556 − 38) N. The work needed to lift the bucket ∆ m through the th subinterval of [0 12] is 98(556 − 38∗ )∆, so the total work is

 = lim




→∞ =1

98(556 − 38∗ ) ∆ =


 12
0


12
(98)(556 − 38)  = 98 556 − 192
= 98(3936) ≈ 3857 J
0

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

240

¤

CHAPTER 5

APPLICATIONS OF INTEGRATION

19. A “slice” of water ∆ m thick and lying at a depth of ∗ m (where 0 ≤ ∗ ≤



1
2)

has volume (2 × 1 × ∆) m3 , a mass of

2000 ∆ kg, weighs about (98)(2000 ∆) = 19,600 ∆ N, and thus requires about 19,600∗ ∆ J of work for its removal.

So  = lim




→∞ =1

19,600∗ ∆ =


 12
0


12
19,600  = 98002 0 = 2450 J.

21. A rectangular “slice” of water ∆ m thick and lying  m above the bottom has width  m and volume 8 ∆ m3 . It weighs

about (98 × 1000)(8 ∆) N, and must be lifted (5 − ) m by the pump, so the work needed is about
(98 × 103 )(5 − )(8 ∆) J. The total work required is
3

3
3
 ≈ 0 (98 × 103 )(5 − )8  = (98 × 103 ) 0 (40 − 82 )  = (98 × 103 ) 202 − 8 3 0
3
= (98 × 103 )(180 − 72) = (98 × 103 )(108) = 10584 × 103 ≈ 106 × 106 J

23. Let  measure depth (in feet) below the spout at the top of the tank. A horizontal

− ) ft () and volume 2 ∆ =  ·

9
(16
64

− )2 ∆ ft3 . It weighs

al

3
(16
8

e

disk-shaped “slice” of water ∆ ft thick and lying at coordinate  has radius

about (625) 9 (16 − )2 ∆ lb and must be lifted  ft by the pump, so the
64

rS

work needed to pump it out is about (625) 9 (16 − )2 ∆ ft-lb. The total
64
work required is
8
8
 ≈ 0 (625) 9 (16 − )2  = (625) 9 0 (256 − 32 + 2 ) 
64
64

= (625) 9 0 (256 − 322 + 3 )  = (625) 9 1282 −
64
64


9 11 264
= 33,000 ≈ 104 × 105 ft-lb
= (625)
64
3


32 3
1 4 8
3  + 4 0

Fo

8

() From similar triangles,
So  = 3 +  = 3 +


3
= .
8−
8
3
8 (8

− )

=

3
3(8)
+ (8 − )
8
8

=

3
8 (16

− )

25. If only 47 × 105 J of work is done, then only the water above a certain level (call

ot

it ) will be pumped out. So we use the same formula as in Exercise 21, except that

N

the work is fixed, and we are trying to find the lower limit of integration:

3

3
47 × 105 ≈  (98 × 103 )(5 − )8  = 98 × 103 202 − 8 3 
3
 


47
⇔
× 102 ≈ 48 = 20 · 32 − 8 · 33 − 202 − 8 3
98
3
3

⇔

23 − 152 + 45 = 0. To find the solution of this equation, we plot 23 − 152 + 45 between  = 0 and  = 3.
We see that the equation is satisfied for  ≈ 20. So the depth of water remaining in the tank is about 20 m.
27.  = 2 , so  is a function of  and  can also be regarded as a function of . If 1 = 2 1 and 2 = 2 2 , then

 =



2

 ()  =

1

=





2

2  ( ())  =

1

2

 ( ) 



2

 ( ())  ()

[Let  () = 2 , so  () = 2 .]

1

by the Substitution Rule.

1

29. (a)  =







 ()  =













1 2
−1
1
1
 = 1 2
= 1 2
−
2
 



c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 5.5

AVERAGE VALUE OF A FUNCTION

¤



1
1
(b) By part (a),  = 
−
where  = mass of the earth in kg,  = radius of the earth in m,

 + 1,000,000 and  = mass of satellite in kg. (Note that 1000 km = 1,000,000 m.) Thus,


1
1
−11
24
 = (667 × 10 )(598 × 10 )(1000) ×
−
≈ 850 × 109 J
637 × 106
737 × 106

5.5 Average Value of a Function
1. ave =

1
−

3. ave =

1
−

5. ave =

1
−









9

 ()  =

1
4−0

()  =

1
8−1

 ()  =

1
2−0

4
0

(4 − 2 )  =

1
4

 2 1 3 4

2 − 3  0 = 1 32 −
4


8
8 √
3
  = 1 3 43 =
7 4
1

3
(16
28

1

2
0

2 (1 + 3 )4  =

9

1
2

1

1

4

3

− 1) =




45
28

64
3




 
− 0 = 1 32 =
4 3

8
3

[ = 1 + 3 ,  = 32 ]

5

1

=

1
(95
30

− 1) =

al

5

e

29,524
= 196826
15

 −1
1
1
=  − 0 0 cos4  sin   =  1 4 (−) [ = cos ,  = − sin  ]

1
1
1
1
1
2
2
=  −1 4  =  · 2 0 4  [by Theorem 4.5.6(a)] =  1 5 0 = 5
5
1
6

7. ave

1



rS

5

 5
1
1 1
( − 3)2  =
( − 3)3
5−2 2
3 3
2
 3
 1
= 1 2 − (−1)3 = 9 (8 + 1) = 1
9

(c)

Fo

9. (a) ave =

(b)  () = ave

⇔ ( − 3)2 = 1 ⇔

 − 3 = ±1 ⇔  = 2 or 4

ot

 
1
(2 sin  − sin 2) 
 −0 0


1
=  −2 cos  + 1 cos 2 0
2

 

1
4
=  2 + 1 − −2 + 1 = 
2
2

(c)

N

11. (a) ave =

(b)  () = ave

⇔ 2 sin  − sin 2 =

4


1 ≈ 1238 or 2 ≈ 2808

⇔

13.  is continuous on [1 3], so by the Mean Value Theorem for Integrals there exists a number  in [1 3] such that

3
1

 ()  = ()(3 − 1) ⇒ 8 = 2 (); that is, there is a number  such that  () =

8
2

= 4.

15. Use geometric interpretations to find the values of the integrals.

8
0

 ()  =
=

1
0

 ()  +

−1
2

+

1
2

+

1
2

2
1

 ()  +

+1+4+

3
2

3
2

 ()  +

+2=9

Thus, the average value of  on [0 8] = ave =

1
8−0

8
0

4
3

 ()  +

6
4

 ()  +

7
6

 ()  +

8
7

() 

()  = 1 (9) = 9 .
8
8

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

241

242

¤

CHAPTER 5 APPLICATIONS OF INTEGRATION

17. Let  = 0 and  = 12 correspond to 9 AM and 9 PM , respectively.



12
 12 
1
1
1
50 + 14 sin 12   = 12 50 − 14 · 12 cos 12  0

0

 

1
= 12 50 · 12 + 14 · 12 + 14 · 12 = 50 + 28 ◦ F ≈ 59 ◦ F



1
12 − 0

ave =

19. ave

1
=
8

21. ave =

=



0


1 5
5

0

1
4

8

12
3
√
 =
2
+1

 ()  =


−

23. Let  () =




5
2

1
5

5

5
0 4



0

8

 √
8
( + 1)−12  = 3  + 1 0 = 9 − 3 = 6 kgm




1 − cos 2   =
5


5
sin 2  0 =
5

1
4

[(5 − 0) − 0] =

1
4
5
4



5
1 − cos 2  
5
0

≈ 04 L

()  for  in [ ]. Then  is continuous on [ ] and differentiable on ( ), so by the Mean Value

e

Theorem there is a number  in ( ) such that  () −  () =  0 ()( − ). But  0 () =  () by the Fundamental

Theorem of Calculus. Therefore,   ()  − 0 =  ()( − ).

rS

al

5 Review

1. (a) See Section 5.1, Figure 2 and Equations 5.1.1 and 5.1.2.

(b) Instead of using “top minus bottom” and integrating from left to right, we use “right minus left” and integrate from bottom

Fo

to top. See Figures 11 and 12 in Section 5.1.

2. The numerical value of the area represents the number of meters by which Sue is ahead of Kathy after 1 minute.
3. (a) See the discussion in Section 5.2, near Figures 2 and 3, ending in the Definition of Volume.

(b) See the discussion between Examples 5 and 6 in Section 5.2. If the cross-section is a disk, find the radius in terms of  or 

ot

and use  = (radius)2 . If the cross-section is a washer, find the inner radius in and outer radius out and use
 2 
 2
 =  out −  in .

N

4. (a)  = 2 ∆ = (circumference)(height)(thickness)

(b) For a typical shell, find the circumference and height in terms of  or  and calculate

 =  (circumference)(height)( or ), where  and  are the limits on  or .

(c) Sometimes slicing produces washers or disks whose radii are difficult (or impossible) to find explicitly. On other occasions, the cylindrical shell method leads to an easier integral than slicing does.

5.

6
0

 ()  represents the amount of work done. Its units are newton-meters, or joules.

6. (a) The average value of a function  on an interval [ ] is ave =

1
−





() .



(b) The Mean Value Theorem for Integrals says that there is a number  at which the value of  is exactly equal to the average value of the function, that is,  () = ave . For a geometric interpretation of the Mean Value Theorem for Integrals, see
Figure 2 in Section 5.5 and the discussion that accompanies it. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 5 REVIEW

1. The curves intersect when 2 = 4 − 2

¤

243

⇔ 22 − 4 = 0 ⇔

2( − 2) = 0 ⇔  = 0 or 2.
2

2
 = 0 (4 − 2 ) − 2  = 0 (4 − 22 ) 
2 

= 22 − 2 3 0 = 8 −
3

16
3




−0 =

8
3

3. If  ≥ 0, then |  | = , and the graphs intersect when  = 1 − 22

 12 

 12
(1 − 22 ) −   = 2 0 (−22 −  + 1) 
0


12
 1
= 2 − 2 3 − 1 2 +  0 = 2 − 12 −
3
2
7
7
= 2 24 = 12


   

sin
− (2 − 2) 
2
0

2
   1
2
= − cos
− 3 + 2

2
3
0
2
  2

8
=  − 3 + 4 − − − 0 + 0 =
2

+

1
2




−0

Fo

5.  =

1
8

al

= 2

e

or −1, but −1  0. By symmetry, we can double the area from  = 0 to  = 1 .
2

1
2

rS

=

⇔ 22 +  − 1 = 0 ⇔ (2 − 1)( + 1) = 0 ⇔

4
3

+

4


ot

7. Using washers with inner radius 2 and outer radius 2, we have

2

2
(2)2 − (2 )2  =  0 (42 − 4 ) 
0
2



=  4 3 − 1 5 0 =  32 − 32
3
5
3
5

N

 =

= 32 ·

9.  = 

2
15

=

64

15


 3 
2
(9 −  2 ) − (−1) − [0 − (−1)]2 
−3

= 2

3

3
(10 − 2 )2 − 1  = 2 0 (100 − 20 2 +  4 − 1) 
0
3


(99 − 20 2 +  4 )  = 2 99 −


= 2 297 − 180 + 243 = 1656 
5
5
= 2

0

20 3

3

+ 1 5
5

3
0

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

244

¤

CHAPTER 5 APPLICATIONS OF INTEGRATION

11. The graph of 2 −  2 = 2 is a hyperbola with right and left branches.

√
⇒  = ± 2 − 2 .

Solving for  gives us  2 = 2 − 2

We’ll use shells and the height of each shell is
√
√

 √
2 − 2 − − 2 − 2 = 2 2 − 2 .
√
 +
2 · 2 2 − 2 . To evaluate, let  = 2 − 2 ,
The volume is  = 

so  = 2  and   =

1
2

. When  = ,  = 0, and when  =  + ,

 = ( + )2 − 2 = 2 + 2 + 2 − 2 = 2 + 2 .

13. A shell has radius


2

− , circumference 2

 = cos2  intersects  =
±1
2

cos  =
 =



when cos2  =

[ || ≤ 2] ⇔  =

3

2

−3

1
4


2

1
4


−  , and height cos2  − 1 .
4
⇔

e

0

2+2

√ 1 
32
2 32
4 
 = 2 

=  2 + 2
.
2
3
3
0

±.
3



1
−  cos2  −

2
4



al

2+2

rS

Thus,  = 4



15. (a) A cross-section is a washer with inner radius 2 and outer radius .

 =



1  2

1
1

 () − (2 )2  = 0 (2 − 4 )  =  1 3 − 1 5 0 =  1 − 1 =
3
5
3
5
0

Fo


(b) A cross-section is a washer with inner radius  and outer radius .


1
1

 1  2


2
 = 0 
 −   = 0 ( −  2 )  =  1  2 − 1  3 0 =  1 − 1 =
2
3
2
3

2

15


6

ot

(c) A cross-section is a washer with inner radius 2 −  and outer radius 2 − 2 .


1 

1
1
 = 0  (2 − 2 )2 − (2 − )2  = 0 (4 − 52 + 4)  =  1 5 − 5 3 + 22 0 =  1 −
5
3
5

17. (a) Using the Midpoint Rule on [0 1] with  () = tan(2 ) and  = 4, we estimate

5
3


+2 =

8

15

   
  
  
  
2
2
2
2 tan 1
+ tan 3
+ tan 5
+ tan 7
≈ 1 (153) ≈ 038
8
8
8
8
4

N

=

1
0

tan(2 )  ≈

1
4

(b) Using the Midpoint Rule on [0 1] with  () =  tan2 (2 ) (for disks) and  = 4, we estimate
 =
19.

21.

 2
0

1
0

  
  
  
  

2
2
2
2
+ tan2 3
+ tan2 5
+ tan2 7
≈
 ()  ≈ 1  tan2 1
4
8
8
8
8

2 cos   =

 2
0

0

(2 − sin )2 

≈ 087

(2) cos  


The solid is obtained by rotating the region R = ( ) | 0 ≤  ≤



4 (1114)


20


≤  ≤ cos  about the -axis.

The solid is obtained by rotating the region R = {( ) | 0 ≤  ≤  0 ≤  ≤ 2 − sin } about the -axis.

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 5 REVIEW

23. Take the base to be the disk 2 +  2 ≤ 9. Then  =

3

−3

() , where (0 ) is the area of the isosceles right triangle

whose hypotenuse lies along the line  = 0 in the -plane. The length of the hypotenuse is 2 each leg is

√ √
√ √
2
2 9 − 2 . () = 1 2 9 − 2 = 9 − 2 , so
2
 =2

3
0

()  = 2

3
0

· 1 ·
4

27.  () = 

 =

 008
0

√
3
8 

=

√

3 2
64  .

 =

 20
0

()  =

√

3
64

 20
0

2  =

√

3
64

√

3
.
8

1


3 20
3 0

Therefore,
=

√
8000 3
64 · 3

=

√
125 3
3

m3 .

⇒ 30 N = (15 − 12) cm ⇒  = 10 Ncm = 1000 Nm. 20 cm − 12 cm = 008 m ⇒
  = 1000

 008
0

 008
  = 500 2 0 = 500(008)2 = 32 N·m = 32 J.

e

1
2

√
9 − 2 and the length of


3
(9 − 2 )  = 2 9 − 1 3 0 = 2(27 − 9) = 36
3

25. Equilateral triangles with sides measuring 1  meters have height 1  sin 60◦ =
4
4

() =

¤

29. (a) The parabola has equation  = 2 with vertex at the origin and passing through
1
4

⇒  = 1 2
4

⇒ 2 = 4

⇒



. Each circular disk has radius 2  and is moved 4 −  ft.
 4   2
4
 = 0  2  625(4 − )  = 250 0 (4 − ) 

rS

=2

⇒ =

al

(4 4). 4 =  · 42

64
3



=

8000
3

≈ 8378 ft-lb

Fo



4
= 250 2 2 − 1  3 0 = 250 32 −
3

(b) In part (a) we knew the final water level (0) but not the amount of work done. Here we use the same equation, except with the work fixed, and the lower limit of

ot

integration (that is, the final water level — call it ) unknown:  = 4000 ⇔

4
 


250 22 − 1  3  = 4000 ⇔ 16 = 32 − 64 − 22 − 1 3
⇔
3

3
3

3 − 62 + 32 −

48


= 0. We graph the function  () = 3 − 62 + 32 −

48


N

on the interval [0 4] to see where it is 0. From the graph,  () = 0 for  ≈ 21.
So the depth of water remaining is about 21 ft.
31. lim ave = lim
→0

→0

1
( + ) − 



+



 ()  = lim

→0


 ( + ) −  ()
, where  () =   () . But we recognize this


limit as being  0 () by the definition of a derivative. Therefore, lim ave =  0 () =  () by FTC1.
→0

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245

N ot e

al

rS

Fo

PROBLEMS PLUS
1. (a) The area under the graph of  from 0 to  is equal to



 () , so the requirement is that

0


0

 ()  = 3 for all . We

differentiate both sides of this equation with respect to  (with the help of FTC1) to get  () = 32 . This function is positive and continuous, as required.
(b) The volume generated from  = 0 to  =  is


0

[ ()]2 . Hence, we are given that 2 =


0

[ ()]2  for all

  0. Differentiating both sides of this equation with respect to  using the Fundamental Theorem of Calculus gives


2 = [()]2 ⇒  () = 2, since  is positive. Therefore,  () = 2.

3. Let  and  be the -coordinates of the points where the line intersects the

curve. From the figure, 1 = 2

e

⇒

 − 42 +

27 4

4


= 42 −

= 42 −

27 4

4

27 4

4


−  

rS


27 4 
 0
4


 − 42 +

al




  


 − 8 − 273  =  8 − 273 −  
0
 
−  − 42 −

27 4

4

−  = 42 −

= 42 −

27 4

4

− 82 + 274 =

= 2
16
81

 81
4


− 



−  8 − 273

81 4

4

− 42


2 − 4

 
 64 
⇒  = 4 . Thus,  = 8 − 273 = 8 4 − 27 729 =
9
9

32
9

−

64
27

=

32
.
27

ot

So for   0, 2 =

27 4

4

Fo

0 = 42 −

27 4

4

5. (a)  = 2 ( − 3) =

1
2 (3
3

− ). See the solution to Exercise 5.2.49.

N

(b) The smaller segment has height  = 1 −  and so by part (a) its volume is
 = 1 (1 − )2 [3(1) − (1 − )] = 1 ( − 1)2 ( + 2). This volume must be 1 of the total volume of the sphere,
3
3
3
4  which is 4 (1)3 . So 1 ( − 1)2 ( + 2) = 1 3  ⇒ (2 − 2 + 1)( + 2) = 4 ⇒ 3 − 3 + 2 = 4 ⇒
3
3
3
3
3
33 − 9 + 2 = 0. Using Newton’s method with () = 33 − 9 + 2,  0 () = 92 − 9, we get

+1 =  −

33 − 9 + 2

. Taking 1 = 0, we get 2 ≈ 02222, and 3 ≈ 02261 ≈ 4 , so, correct to four decimal
92 − 9


places,  ≈ 02261.

(c) With  = 05 and  = 075, the equation 3 − 32 + 43  = 0 becomes 3 − 3(05)2 + 4(05)3 (075) = 0 ⇒
 
3 − 3 2 + 4 1 3 = 0 ⇒ 83 − 122 + 3 = 0. We use Newton’s method with  () = 83 − 122 + 3,
2
8 4
 0 () = 242 − 24, so +1 =  −

83 − 122 + 3


. Take 1 = 05. Then 2 ≈ 06667, and 3 ≈ 06736 ≈ 4 .
242 − 24


So to four decimal places the depth is 06736 m.

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247

248

¤

CHAPTER 5 PROBLEMS PLUS

(d) (i) From part (a) with  = 5 in., the volume of water in the bowl is
 = 1 2 (3 − ) = 1 2 (15 − ) = 52 − 1 3 . We are given that
3
3
3
when  = 3. Now



= 02 in3s and we want to find







02
= 10
− 2 , so
=
. When  = 3, we have




(10 − 2 )


02
1
=
=
≈ 0003 ins.

(10 · 3 − 32 )
105
(ii) From part (a), the volume of water required to fill the bowl from the instant that the water is 4 in. deep is
 =

1
2

· 4 (5)3 − 1 (4)2 (15 − 4) =
3
3
743
02

this volume by the rate: Time =

2
3

· 125 −

=

370
3

16
3

· 11 =

74
3 .

To find the time required to fill the bowl we divide

≈ 387 s ≈ 65 min.

7. We are given that the rate of change of the volume of water is


= −(), where  is some positive constant and () is



 

. But by the Chain Rule,
=
, so the first equation can be written


 

al

with respect to time, that is,

e

the area of the surface when the water has depth . Now we are concerned with the rate of change of the depth of the water

rS


 
= −() (). Also, we know that the total volume of water up to a depth  is  () = 0 () , where () is
 

the area of a cross-section of the water at a depth . Differentiating this equation with respect to , we get  = ().
Substituting this into equation , we get ()() = −() ⇒  = −, a constant.
9. We must find expressions for the areas  and , and then set them equal and see what this says about the curve . If

Fo





 =  22 , then area  is just 0 (22 − 2 )  = 0 2  = 1 3 . To find area , we use  as the variable of
3
 integration. So we find the equation of the middle curve as a function of :  = 22 ⇔  = 2, since we are

concerned with the first quadrant only. We can express area  as

0

22


22  22


 22
4
4
(2)32
2 − ()  =
−
()  = 3 −
() 
3
3
0
0
0

ot



N

where () is the function with graph . Setting  = , we get 1 3 = 4 3 −
3
3

 22
0

() 

⇔

 22
0

()  = 3 .

Now we differentiate this equation with respect to  using the Chain Rule and the Fundamental Theorem:


(22 )(4) = 32 ⇒ () = 3 2, where  = 22 . Now we can solve for :  = 3 2 ⇒
4
4

2 =

9
16 (2)

⇒ =

32 2
9  .

11. (a) Stacking disks along the -axis gives us  =

(b) Using the Chain Rule,


0

 [ ()]2 .


 

=
·
=  [ ()]2
.

 



√
√

 14

 √
. Set
= : [ ()]2  =   ⇒ [ ()]2 =
 ; that
 = [ ()]2
 ⇒ () =





 14

 . The advantage of having
=  is that the markings on the container are equally spaced. is, () =



(c) 

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 5 PROBLEMS PLUS

13. The cubic polynomial passes through the origin, so let its equation be

 = 3 + 2 + . The curves intersect when 3 + 2 +  = 2

⇔

3 + ( − 1)2 +  = 0. Call the left side  (). Since  () =  () = 0 another form of  is
 () = ( − )( − ) = [2 − ( + ) + ]
= [3 − ( + )2 + ]
Since the two areas are equal, we must have


0

 ()  = −




()  ⇒

e

[ ()] = [ ()] ⇒  () −  (0) =  () −  () ⇒  (0) =  (), where  is an antiderivative of  .
0





Now  () =  ()  = [3 − ( + )2 + ]  =  1 4 − 1 ( + )3 + 1 2 + , so
4
3
2




 (0) =  () ⇒  =  1 4 − 1 ( + )3 + 1 3 +  ⇒ 0 =  1 4 − 1 ( + )3 + 1 3 ⇒
4
3
2
4
3
2
0 = 3 − 4( + ) + 6 [multiply by 12(3 ),  6= 0] ⇒ 0 = 3 − 4 − 4 + 6 ⇒  = 2.

N

ot

Fo

rS

al

Hence,  is twice the value of .

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¤

249

N ot e

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rS

Fo

6

INVERSE FUNCTIONS:
Exponential, Logarithmic, and Inverse Trigonometric Functions

6.1 Inverse Functions
1. (a) See Definition 1.

(b) It must pass the Horizontal Line Test.
3.  is not one-to-one because 2 6= 6, but  (2) = 20 =  (6).
5. We could draw a horizontal line that intersects the graph in more than one point. Thus, by the Horizontal Line Test, the

function is not one-to-one.

e

7. No horizontal line intersects the graph more than once. Thus, by the Horizontal Line Test, the function is one-to-one.

−2

=−
= 1. Pick any -values equidistant
2
2(1)

al

9. The graph of  () = 2 − 2 is a parabola with axis of symmetry  = −

from 1 to find two equal function values. For example,  (0) = 0 and  (2) = 0, so  is not one-to-one.
1 6= 2

⇒ 11 6= 12

⇒  (1 ) 6=  (2 ), so  is one-to-one.

rS

11. () = 1.

Geometric solution: The graph of  is the hyperbola shown in Figure 14 in Section 1.2. It passes the Horizontal Line Test, so  is one-to-one.

Fo

13. () = 1 + cos  is not one-to-one since ( + 2) = () for any real number  and any integer .
15. A football will attain every height  up to its maximum height twice: once on the way up, and again on the way down.

Thus, even if 1 does not equal 2 ,  (1 ) may equal  (2 ), so  is not 1-1.
⇔  −1 (17) = 6.

ot

17. (a) Since  is 1-1,  (6) = 17

(b) Since  is 1-1,  −1 (3) = 2 ⇔  (2) = 3.

√
√
 ⇒ 0 () = 1 + 1(2  )  0 on (0 ∞). So  is increasing and hence, 1-1. By inspection,
√
(4) = 4 + 4 = 6, so −1 (6) = 4.

N

19. () =  +

21. We solve  =

5
(
9

− 32) for  : 9  =  − 32 ⇒  = 9  + 32. This gives us a formula for the inverse function, that
5
5

is, the Fahrenheit temperature  as a function of the Celsius temperature .  ≥ −45967 ⇒
9

5

9

5

+ 32 ≥ −45967 ⇒

≥ −49167 ⇒  ≥ −27315, the domain of the inverse function.

23.  =  () = 3 − 2
25.  = () = 1 +

⇒ 2 = 3 − 

√
2 + 3

( ≥ 1)

⇒ =

3−
3−
3−
. Interchange  and :  =
. So  −1 () =
.
2
2
2

⇒ −1=

√
2 + 3

⇒ ( − 1)2 = 2 + 3 ⇒ ( − 1)2 − 2 = 3 ⇒

 = 1 ( − 1)2 − 2 . Interchange  and :  = 1 ( − 1)2 − 2 . So  −1 () = 1 ( − 1)2 − 2 . Note that the domain of  −1
3
3
3
3
3
3 is  ≥ 1.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

251

252

¤

CHAPTER 6

INVERSE FUNCTIONS

√
1− 
√ , the domain is  ≥ 0.  (0) = 1 and as  increases,  decreases. As  → ∞,
27. For  () =
1+ 
√
√
√
1 −  1 
1  − 1
−1
√ · √ = √
= −1, so the range of  is −1   ≤ 1. Thus, the domain of  −1 is −1   ≤ 1.
→
1
1 +  1 
1  + 1
√
√
√
√
√
√
√
1− 
√
=
+ = 1− ⇒
⇒ (1 +  ) = 1 −  ⇒  +   = 1 −  ⇒
1+ 


2
2
√
√
1−
1−
1−
(1 + ) = 1 −  ⇒
=
. Interchange  and :  =
. So
⇒ =
1+
1+
1+
2

1− with −1   ≤ 1.
 −1 () =
1+
√
⇒  − 1 = 4 ⇒  = 4  − 1 [not ± since
√
√
 ≥ 0]. Interchange  and :  = 4  − 1. So  −1 () = 4  − 1. The
√
√ graph of  = 4  − 1 is just the graph of  = 4  shifted right one unit.

rS

al

From the graph, we see that  and  −1 are reflections about the line  = .

e

29.  =  () = 4 + 1

31. Reflect the graph of  about the line  = . The points (−1 −2), (1 −1),

(2 2), and (3 3) on  are reflected to (−2 −1), (−1 1), (2 2), and (3 3)

33. (a)  =  () =
2

2

Fo

on  −1 .

√
1 − 2

(0 ≤  ≤ 1 and note that  ≥ 0) ⇒

⇒  = 1 −  2 ⇒  = 1 −  2 . So
2

ot

 =1−
√
 −1 () = 1 − 2 , 0 ≤  ≤ 1. We see that  −1 and  are the same function. N

(b) The graph of  is the portion of the circle 2 +  2 = 1 with 0 ≤  ≤ 1 and
0 ≤  ≤ 1 (quarter-circle in the first quadrant). The graph of  is symmetric with respect to the line  = , so its reflection about  =  is itself, that is,
 −1 =  .
35. (a) 1 6= 2

⇒ 3 6= 3
1
2

⇒  (1 ) 6=  (2 ), so  is one-to-one.

(b)  () = 3 and (2) = 8 ⇒  −1 (8) = 2, so ( −1 )0 (8) = 1 0 ( −1 (8)) = 1 0 (2) =
0

2

(c)  = 3

⇒  =  13 . Interchanging  and  gives  = 13 ,

 so  −1 () = 13 . Domain  −1 = range() = R.

1
.
12

(e)

Range( −1 ) = domain( ) = R.

(d)  −1 () = 13 ⇒ ( −1 )0 () = 1 −23
3
 
−1 0
1 1
1
( ) (8) = 3 4 = 12 as in part (b).

⇒

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 6.1

37. (a) Since  ≥ 0, 1 6= 2

⇒ 2 6= 2
1
2

⇒ 9 − 2 6= 9 − 2
1
2

(b)  0 () = −2 and  (1) = 8 ⇒  −1 (8) = 1, so ( −1 )0 (8) =
√
⇒ 2 = 9 −  ⇒  = 9 − .
√
√
Interchange  and :  = 9 − , so  −1 () = 9 − .

INVERSE FUNCTIONS

¤

253

⇒  (1 ) 6= (2 ), so  is 1-1.

1
1
1
1
= 0
=
=− .
 0 ( −1 (8))
 (1)
−2
2

(c)  = 9 − 2

(e)

Domain( −1 ) = range ( ) = [0 9].
Range( −1 ) = domain ( ) = [0 3].
 √

(d) ( −1 )0 () = −1 2 9 − 
⇒ ( −1 )0 (8) = − 1 as in part (b).
2
⇒  −1 (4) = 0, and () = 23 + 32 + 7 + 4 ⇒  0 () = 62 + 6 + 7 and  0 (0) = 7.

 0 (0) =


2

43.  (4) = 5

 (3) =

⇒  −1 (5) = 4. Thus, ( −1 )0 (5) =


2

sec2 (2) and

1
1
1
3
= 0
=
= .
 0 ( −1 (5))
 (4)
23
2

√
√
1 + 3  ⇒  0 () = 1 + 3  0, so  is an increasing function and it has an inverse. Since
3

3√
1 + 3  = 0,  −1 (0) = 3. Thus, ( −1 )0 (0) =
3

Fo

45.  () =

⇒  −1 (3) = 0, and  () = 3 + 2 + tan(2) ⇒  0 () = 2 +
0



· 1 =  . Thus,  −1 (3) = 1 0  −1 (3) = 1 0 (0) = 2.
2

rS

41.  (0) = 3

1
1
1
= 0
= .
 0 ( −1 (4))
 (0)
7

e

Thus, ( −1 )0 (4) =

al

39.  (0) = 4

1
1
1
1
= √ .
= 0
= √
3
 0 ( −1 (0))
 (3)
1+3
28

√
3 + 2 +  + 1 is increasing, so  is 1-1.

(We could verify this by showing that  0 ()  0.) Enter  =  3 +  2 +  + 1

47. We see that the graph of  =  () =

ot

and use your CAS to solve the equation for . Using Derive, we get two

(irrelevant) solutions involving imaginary expressions, as well as one which can be

N

simplified to the following:
√ √
√ 
√
3
 =  −1 () = − 64 3  − 272 + 20 − 3  + 272 − 20 + 3 2
√ √ where  = 3 3 274 − 402 + 16.

Maple and Mathematica each give two complex expressions and one real expression, and the real expression is equivalent

to that given by Derive. For example, Maple’s expression simplifies to
 = 1082 + 12

1  23 − 8 − 2 13
, where
6
2 13

√
48 − 1202 + 814 − 80.

49. (a) If the point ( ) is on the graph of  = (), then the point ( −  ) is that point shifted  units to the left. Since 

is 1-1, the point ( ) is on the graph of  =  −1 () and the point corresponding to ( −  ) on the graph of  is
(  − ) on the graph of  −1 . Thus, the curve’s reflection is shifted down the same number of units as the curve itself is shifted to the left. So an expression for the inverse function is  −1 () =  −1 () − . c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

254

¤

CHAPTER 6 INVERSE FUNCTIONS

(b) If we compress (or stretch) a curve horizontally, the curve’s reflection in the line  =  is compressed (or stretched) vertically by the same factor. Using this geometric principle, we see that the inverse of () = () can be expressed as
−1 () = (1)  −1 ().

6.2 Exponential Functions and Their Derivatives
1. (a)  () =  ,   0

(b) R

(c) (0 ∞)

(d) See Figures 6(c), 6(b), and 6(a), respectively.

3. All of these graphs approach 0 as  → −∞, all of them pass through the point

(0 1), and all of them are increasing and approach ∞ as  → ∞. The larger the base, the faster the function increases for   0, and the faster it approaches 0 as

with bases less than 1

 1 
3

and



 
1 
10

are decreasing. The graph of



1 
10

 1 
3

is the

is the reflection of

rS

reflection of that of 3 about the -axis, and the graph of

al

5. The functions with bases greater than 1 (3 and 10 ) are increasing, while those

e

 → −∞.

that of 10 about the -axis. The graph of 10 increases more quickly than that of

7. We start with the graph of  = 10

Fo

3 for   0, and approaches 0 faster as  → −∞.
(Figure 3) and shift it 2 units to the left to

N

ot

obtain the graph of  = 10+2 .

9. We start with the graph of  = 2 (Figure 2),

reflect it about the -axis, and then about the
-axis (or just rotate 180◦ to handle both reflections) to obtain the graph of  = −2− .

In each graph,  = 0 is the horizontal asymptote.  = 2

 = 2−

 = −2−

11. We start with the graph of  =  (Figure 12) and reflect about the -axis to get the graph of  = − . Then we compress

the graph vertically by a factor of 2 to obtain the graph of  = 1 − and then reflect about the -axis to get the graph of
2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 6.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES

¤

255

 = − 1 − . Finally, we shift the graph upward one unit to get the graph of  = 1 − 1 − .
2
2

13. (a) To find the equation of the graph that results from shifting the graph of  =  2 units downward, we subtract 2 from the

original function to get  =  − 2.
(b) To find the equation of the graph that results from shifting the graph of  =  2 units to the right, we replace  with  − 2 in the original function to get  = (−2) .

e

(c) To find the equation of the graph that results from reflecting the graph of  =  about the -axis, we multiply the original

al

function by −1 to get  = − . the original function to get  = − .

rS

(d) To find the equation of the graph that results from reflecting the graph of  =  about the -axis, we replace  with − in
(e) To find the equation of the graph that results from reflecting the graph of  =  about the -axis and then about the

get  = −− .

Fo

-axis, we first multiply the original function by −1 (to get  = − ) and then replace  with − in this equation to
2

15. (a) The denominator is zero when 1 − 1− = 0

⇔

2

2

1− = 1

⇔

1 − 2 = 0

⇔

 = ±1. Thus,

1− has domain { |  6= ±1} = (−∞ −1) ∪ (−1 1) ∪ (1 ∞).
1 − 1−2
1+
(b) The denominator is never equal to zero, so the function  () = cos  has domain R, or (−∞ ∞).

 


6 3
6
=  and 24 = 3 ⇒ 24 =

17. Use  =  with the points (1 6) and (3 24). 6 = 1


N

ot

the function () =

4 = 2

⇒  = 2 [since   0] and  =

19. 2 ft = 24 in,  (24) = 242 in = 576 in = 48 ft.

6
2

= 3. The function is  () = 3 · 2 .

(24) = 224 in = 224 (12 · 5280) mi ≈ 265 mi

21. The graph of  finally surpasses that of  at  ≈ 358.

23. lim (1001) = ∞ by (3), since 1001  1.
→∞

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

⇒

256

¤

CHAPTER 6 INVERSE FUNCTIONS

25. Divide numerator and denominator by 3 : lim

→∞

3 − −3
1 − −6
1−0
=1
= lim
=
3 + −3
→∞ 1 + −6

1+0

27. Let  = 3(2 − ). As  → 2+ ,  → −∞. So lim 3(2−) = lim  = 0 by (10).
→−∞

→2+

29. Since −1 ≤ cos  ≤ 1 and −2  0, we have −−2 ≤ −2 cos  ≤ −2 . We know that lim (−−2 ) = 0 and
→∞



lim −2 = 0, so by the Squeeze Theorem, lim (−2 cos ) = 0.

→∞

→∞

31.  () = 5 is a constant function, so its derivative is 0, that is,  0 () = 0.
33. By the Product Rule,  () = (3 + 2)

⇒

 0 () = (3 + 2)( )0 +  (3 + 2)0 = (3 + 2) +  (32 + 2)

39.  () = 

0

1

⇒  () = 

√
1 + 23




 

 
−1
1
1 −1
=
1
=

2
2

⇒  0 () =  sin 2 ( sin 2)0 =  sin 2 ( · 2 cos 2 + sin 2 · 1) =  sin 2 (2 cos 2 + sin 2)

⇒ 0 =


⇒  0 =  ·

1
1
33

(23 · 3) = √
(1 + 23 )−12
(1 + 23 ) = √
3
2

2 1 + 2
1 + 23



 
( ) =  ·  or  +


 + 
 + 

⇒

N

47. By the Quotient Rule,  =

0 =


·


ot

45.  = 

3

(3 ) = 32  .






⇒  0 =  − (−) + − · 1 = − (− + 1) or (1 − )−

41. By (9),  () =  sin 2
43.  =

3

rS

1

⇒  0 = 

Fo

37.  = −

3

al

35. By (9),  = 

e

=  [(3 + 2) + (32 + 2)] =  (3 + 32 + 2 + 2)

( + )( ) − ( + )( )
( +  −  − )
( − )
=
=
.
( + )2
( + )2
( + )2


49.  = cos

1 − 2
1 + 2



⇒







1 − 2
 1 − 2
1 − 2
(1 + 2 )(−22 ) − (1 − 2 )(22 )
 = − sin
·
= − sin
·
2
2
2
1+
 1 + 
1+
(1 + 2 )2








−22 (1 + 2 ) + (1 − 2 )
−22 (2)
1 − 2
1 − 2
42
1 − 2
·
·
= − sin
= − sin
=
· sin
1 + 2
(1 + 2 )2
1 + 2
(1 + 2 )2
(1 + 2 )2
1 + 2
0

51.  = 2 cos 

⇒  0 = 2 (− sin ) + (cos )(22 ) = 2 (2 cos  −  sin ).

At (0 1),  0 = 1(2 − 0) = 2, so an equation of the tangent line is  − 1 = 2( − 0), or  = 2 + 1. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 6.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES

53.


 
( ) =
( − ) ⇒ 


 ·

 · 1 −  · 0
= 1 − 0
2

 
 
·
= 1 − 0
 

⇒

 ·

¤

257

⇒

 0
1
−
·  = 1 − 0

2

0 −

⇒

 0

· =1−
2


⇒

 − 
( −  )

⇒ 0 = 2
= 2

 − 
 − 
2





 − 
0 1 −
=
2


55.  =  + −2

⇒  0 =  − 1 −2 ⇒  00 =  + 1 −2 , so
2
4

 
 

200 −  0 −  = 2  + 1 −2 −  − 1 −2 −  + −2 = 0.
4
2

57.  = 

⇒  0 = 

⇒

 00 = 2  , so if  =  satisfies the differential equation  00 + 6 0 + 8 = 0,

e

then 2  + 6 + 8 = 0; that is,  (2 + 6 + 8) = 0. Since   0 for all , we must have 2 + 6 + 8 = 0,

 000 () = 22 · 22 = 23 2

⇒  00 () = 2 · 22 = 22 2

⇒ ···

⇒

⇒  () () = 2 2

rS

⇒  0 () = 22

59.  () = 2

al

or ( + 2)( + 4) = 0, so  = −2 or −4.

61. (a)  () =  +  is continuous on R and (−1) = −1 − 1  0  1 =  (0), so by the Intermediate Value Theorem,

 +  = 0 has a root in (−1 0).

Fo

(b)  () =  +  ⇒  0 () =  + 1, so +1 =  −

 + 
. Using 1 = −05, we get 2 ≈ −0566311,
 + 1

3 ≈ −0567143 ≈ 4 , so the root is −0567143 to six decimal places.
63. (a) lim () = lim

→∞

1
1
= 1, since   0 ⇒ − → −∞ ⇒ − → 0.
=
1 + −
1+·0
⇒

N

(b) () = (1 + − )−1
(c)


−
= −(1 + − )−2 (−− ) =

(1 + − )2

ot

→∞

From the graph of () = (1 + 10−05 )−1 , it seems that () = 08
(indicating that 80% of the population has heard the rumor) when
 ≈ 74 hours.

65.  () =  − 

⇒  0 () = 1 −  = 0 ⇔  = 1 ⇔  = 0. Now  0 ()  0 for all   0 and  0 ()  0 for all

  0, so the absolute maximum value is  (0) = 0 − 1 = −1.
2 8

67.  () = −

, [−1 4].  0 () =  · −

2 8

· (−  ) + −
4

2 8

2 8

· 1 = −

2

2 8

(−  + 1). Since −
4

is never 0,

 0 () = 0 ⇒ −2 4 + 1 = 0 ⇒ 1 = 2 4 ⇒ 2 = 4 ⇒  = ±2, but −2 is not in the given interval, [−1 4].
√
 (−1) = −−18 ≈ −088,  (2) = 2−12 ≈ 121, and (4) = 4−2 ≈ 054. So  (2) = 2−12 = 2  is the absolute
√
maximum value and  (−1) = −−18 = −1 8  is the absolute minimum value. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

258

¤

CHAPTER 6 INVERSE FUNCTIONS

69. (a)  () = (1 − )−

⇒  0 () = (1 − )(−− ) + − (−1) = − ( − 2)  0 ⇒   2, so  is increasing on

(2 ∞) and decreasing on (−∞ 2).
(b)  00 () = − (1) + ( − 2)(−− ) = − (3 − )  0 ⇔   3, so  is CU on (−∞ 3) and CD on (3 ∞).


(c)  00 changes sign at  = 3, so there is an IP at 3 −2−3 .

71.  = () = −1(+1)

C. No symmetry D.
−1 ( + 1) → −∞,

A.  = { |  6= −1} = (−∞ −1) ∪ (−1, ∞) B. No -intercept; -intercept =  (0) = −1

lim −1(+1) = 1 since −1( + 1) → 0, so  = 1 is a HA.

→±∞

lim −1(+1) = ∞ since −1( + 1) → ∞, so  = −1 is a VA.

→−1−

E.  0 () = −1(+1) ( + 1)2

⇒  0 ()  0 for all  except 1, so

 is increasing on (−∞ −1) and (−1 ∞). F. No extreme values
−1(+1)

−1(+1)

H.

−1(+1)

(−2)



+
=−
( + 1)4
( + 1)3

(2 + 1)
( + 1)4

e

G.  00 () =

lim −1(+1) = 0 since

→−1+

⇒

73.  = 1(1 + − )

rS

al

 00 ()  0 ⇔ 2 + 1  0 ⇔   − 1 , so  is CU on (−∞ −1)
2






and −1 − 1 , and CD on − 1 , ∞ .  has an IP at − 1 , −2 .
2
2
2

A.  = R B. No -intercept; -intercept =  (0) = 1  C. No symmetry
2

D. lim 1(1 + − ) =
→∞

1
1+0

= 1 and lim 1(1 + − ) = 0 since lim − = ∞, so  has horizontal asymptotes
→−∞

→−∞

F. No extreme values G.  00 () =

Fo

 = 0 and  = 1. E.  0 () = −(1 + − )−2 (−− ) = − (1 + − )2 . This is positive for all , so  is increasing on R.
(1 + − )2 (−− ) − − (2)(1 + − )(−− )
− (− − 1)
=
(1 + − )4
(1 + − )3

The second factor in the numerator is negative for   0 and positive for   0,

H.

N

ot

and the other factors are always positive, so  is CU on (−∞, 0) and CD

 on (0 ∞). IP at 0, 1
2
75. () =  − with  = 001,  = 4, and  = 007. We will find the

zeros of  00 for () =  − .
 0 () =  (−− ) + − (−1 ) = − (− + −1 )
 00 () = − (−−1 + ( − 1)−2 ) + (− + −1 )(−− )
= −2 − [− + ( − 1) + 2 2 − ]
= −2 − (2 2 − 2 + 2 − )
Using the given values of  and  gives us  00 () = 2 −007 (000492 − 056 + 12). So  00 () = 001 00 () and its zeros are  = 0 and the solutions of 000492 − 056 + 12 = 0, which are 1 =

200
7

≈ 2857 and 2 =

600
7

≈ 8571.

At 1 minutes, the rate of increase of the level of medication in the bloodstream is at its greatest and at 2 minutes, the rate of decrease is the greatest. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 6.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES
3

77.  () = 

−

¤

259

→ 0 as  → −∞, and

 () → ∞ as  → ∞. From the graph,

it appears that  has a local minimum of about  (058) = 068, and a local maximum of about  (−058) = 147.
To find the exact values, we calculate

 3
1
 0 () = 32 − 1  − , which is 0 when 32 − 1 = 0 ⇔  = ± √3 . The negative root corresponds to the local


√ 3
√
√
1
maximum  − √3 = (−1 3) − (−1 3) = 2 39 , and the positive root corresponds to the local minimum




1
√
3



= (1

3)3 − (1

√

3)

√

= −2

39

. To estimate the inflection points, we calculate and graph


 3  
 3 


3
3
  2
3 − 1  − = 32 − 1  − 32 − 1 +  − (6) =  − 94 − 62 + 6 + 1 .


e

 00 () =

√

From the graph, it appears that  00 () changes sign (and thus  has inflection points) at  ≈ −015 and  ≈ −109. From the


1

( +  )  =

0

81.



2



2

0

+1
+ 
+1

1

=

0


1
1
+  − (0 + 1) =
+−1
+1
+1


2
1
1
1
1
−  = − − = − −2 + 0 = (1 − −2 )




0

83. Let  = 1 +  . Then  =  , so
85.





√
√
 1 +   =
  = 2 32 +  = 2 (1 +  )32 + .
3
3

Fo

0


=




rS

79.

al

graph of , we see that these -values correspond to inflection points at about (−015 115) and (−109 082).




( + − )2  = (2 + 2 + −2 )  = 1 2 + 2 − 1 −2 + 
2
2


tan  sec2   =

ot

87. Let  = tan . Then  = sec2  , so



  =  +  = tan  + .

89. Let  = 1, so  = −12 . When  = 1,  = 1; when  = 2,  =
2

1

1
 =
2

91. Area =

1

93.  =

0







12

√
 12
 (−) = −  1 = −(12 − ) =  − .

0

−2



1

 1  3
 



1 
 −   = 1 3 −  0 = 1 3 −  − 1 − 1 = 1 3 −  +
3
3
3
3
0
( )2  = 
2


95. erf() = √





Thus,

N



1
2.

 =

2





1
0

 1
2  = 1  2 0 =
2

2

− 

0



−  =

0



0

−2




 +
2



⇒


−  −





0

2

−  =

0

−2








2

2
3

≈ 4644

 2

 −1

√

erf() By Property 5 of definite integrals in Section 4.2,
2

, so
2

−  =

√
√


erf() − erf() =
2
2

1
2

√
 [erf() − erf()].

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

260

¤

CHAPTER 6 INVERSE FUNCTIONS

97. The rate is measured in liters per minute. Integrating from  = 0 minutes to  = 60 minutes will give us the total amount of oil

that leaks out (in liters) during the first hour.
 60
 60
()  = 0 100−001 
[ = −001,  = −001]
0
 −06 
 −06
= 100 0
 (−100 ) = −10,000  0
= −10,000(−06 − 1) ≈ 45119 ≈ 4512 liters

99. We use Theorem 6.1.7. Note that  (0) = 3 + 0 + 0 = 4, so  −1 (4) = 0. Also  0 () = 1 +  . Therefore,

 −1 0
(4) =


1
1
1
1
= .
= 0
=
 0 ( −1 (4))
 (0)
1 + 0
2
From the graph, it appears that  is an odd function ( is undefined for  = 0).

101.

To prove this, we must show that  (−) = −().

al

e

1
1 − 1 1
1 − 1(−)
1 − (−1)
1 − 1

 (−) =
=
=
· 1 = 1
1(−)
(−1)
1
1+
1+


+1
1 + 1

1
1−
= −()
=−
1 + 1

rS

so  is an odd function.

103. (a) Let  () =  − 1 − . Now  (0) = 0 − 1 = 0, and for  ≥ 0, we have  0 () =  − 1 ≥ 0. Now, since  (0) = 0 and

 is increasing on [0 ∞), () ≥ 0 for  ≥ 0 ⇒  − 1 −  ≥ 0 ⇒  ≥ 1 + .
2

2

Fo

(b) For 0 ≤  ≤ 1, 2 ≤ , so  ≤  [since  is increasing]. Hence [from (a)] 1 + 2 ≤  ≤  .

1 2
1
1
1 2
So 4 = 0 1 + 2  ≤ 0   ≤ 0   =  − 1   ⇒ 4 ≤ 0   ≤ .
3
3

2

+ ··· + for  ≥ 0.
2!
!

ot

105. (a) By Exercise 103(a), the result holds for  = 1. Suppose that  ≥ 1 +  +

2


+1
− ··· −
−
. Then  0 () =  − 1 −  − · · · −
≥ 0 by assumption. Hence
2!
!
( + 1)!
!

N

Let  () =  − 1 −  −

 () is increasing on (0 ∞). So 0 ≤  implies that 0 =  (0) ≤  () =  − 1 −  − · · · −
 ≥ 1 +  + · · · +

+1

−
, and hence
!
( + 1)!


2
+1

+ for  ≥ 0. Therefore, for  ≥ 0,  ≥ 1 +  +
+ ··· + for every positive
!
( + 1)!
2!
!

integer , by mathematical induction.
(b) Taking  = 4 and  = 1 in (a), we have  = 1 ≥ 1 +
(c)  ≥ 1 +  + · · · +

+1

+
!
( + 1)!

⇒

1
2

+

1
6

+

1
24

= 27083  27.




1
1
1
+
≥
.
≥  + −1 + · · · +



!
( + 1)!
( + 1)!



= ∞, so lim  = ∞.
→∞ ( + 1)!
→∞ 

But lim

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SECTION 6.3

LOGARITHMIC FUNCTIONS

¤

6.3 Logarithmic Functions
1. (a) It is defined as the inverse of the exponential function with base , that is, log  = 

(b) (0 ∞)

(c) R

⇔  = .

(d) See Figure 1.
1
1
1
= −3 since 3−3 = 3 =
.
27
3
27

3. (a) log5 125 = 3 since 53 = 125.

(b) log3

5. (a) ln 45 = 45 since ln  =  for   0.

(b) log10 00001 = log10 10−4 = −4 by (2).

6
7. (a) log2 6 − log2 15 + log2 20 = log2 ( 15 ) + log2 20

[by Law 2]
[by Law 1]

6
= log2 ( 15 · 20)

= log2 8, and log2 8 = 3 since 23 = 8.

11. ln

1
2

ln() = 1 (ln  + ln ) =
2

1
2

al

√
 = ln()12 =

18

ln  +

1
2

ln  [assuming that the variables are positive]

rS

9. ln

 100 

e

 100 
− log3 50 = log3 18·50
 
= log3 ( 1 ), and log3 1 = −2 since 3−2 = 1 
9
9
9

(b) log3 100 − log3 18 − log3 50 = log3

2
= ln 2 − ln( 3  4 ) = 2 ln  − (ln  3 + ln  4 ) = 2 ln  − 3 ln  − 4 ln 
3 4

13. 2 ln  + 3 ln  − ln  = ln 2 + ln  3 − ln 
2 3

= ln
15. ln 5 + 5 ln 3 = ln 5 + ln 35

[by Law 1]

Fo

= ln(  ) − ln 

[by Law 3]

2  3


[by Law 3]

[by Law 1]

ot

5

[by Law 2]

= ln(5 · 3 )
= ln 1215
1
3



ln  − ln(2 + 3 + 2)2 = ln[( + 2)3 ]13 +

N

17.

ln( + 2)3 +

1
2

= ln( + 2) + ln

1
2

ln


(2 + 3 + 2)2

√

2 + 3 + 2

√
( + 2) 
( + 1)( + 2)
√

= ln
+1

= ln

[by Laws 3, 2]
[by Law 3]
[by Law 1]

Note that since ln  is defined for   0, we have  + 1,  + 2, and 2 + 3 + 2 all positive, and hence their logarithms are defined.
1
ln 
=
≈ 0402430 ln 12 ln 12 ln 
(c) log2  =
≈ 1651496 ln 2

19. (a) log12  =

(b) log6 1354 =

ln 1354
≈ 1454240 ln 6

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261

262

¤

CHAPTER 6

INVERSE FUNCTIONS

21. To graph these functions, we use log15  =

ln  ln  and log50  =
.
ln 15 ln 50

These graphs all approach −∞ as  → 0+ , and they all pass through the

point (1 0). Also, they are all increasing, and all approach ∞ as  → ∞.

The functions with larger bases increase extremely slowly, and the ones with smaller bases do so somewhat more quickly. The functions with large bases approach the -axis more closely as  → 0+ .
23. (a) Shift the graph of  = log10  five units to the left to

(b) Reflect the graph of  = ln  about the -axis to obtain the graph of  = − ln .

obtain the graph of  = log10 ( + 5). Note the vertical

 = log10 ( + 5)

 = ln 

 = − ln 

rS

 = log10 

al

e

asymptote of  = −5.

25. (a) The domain of  () = ln  + 2 is   0 and the range is R.

(b)  = 0 ⇒ ln  + 2 = 0 ⇒ ln  = −2 ⇒  = −2

⇔

7 − 4 = ln 6

(b) ln(3 − 10) = 2
29. (a) 2−5 = 3

⇔

⇔

3 − 10 = 2

7 − ln 6 = 4
⇔

⇔

3 = 2 + 10

ot

27. (a) 7−4 = 6

Fo

(c) We shift the graph of  = ln  two units upward.

 = 1 (7 − ln 6)
4
⇔

 = 1 (2 + 10)
3

⇔ log2 3 =  − 5 ⇔  = 5 + log2 3.

N



ln 3
Or: 2−5 = 3 ⇔ ln 2−5 = ln 3 ⇔ ( − 5) ln 2 = ln 3 ⇔  − 5 = ln 2

⇔ =5+

ln 3 ln 2

(b) ln  + ln( − 1) = ln(( − 1)) = 1 ⇔ ( − 1) = 1 ⇔ 2 −  −  = 0. The quadratic formula (with  = 1,


√
 = −1, and  = −) gives  = 1 1 ± 1 + 4 , but we reject the negative root since the natural logarithm is not
2


√ defined for   0. So  = 1 1 + 1 + 4 .
2

31.  − −2 = 1
33. ln(ln ) = 1

⇔  − 1 = −2

⇔ ln(ln ) = 1

35. 2 −  − 6 = 0
37. (a) 2+5 = 100

⇔ ln( − 1) = ln −2

⇔ −2 = ln( − 1) ⇔  = − 1 ln( − 1)
2

⇔ ln  = 1 =  ⇔ ln  = 

⇔  = 

⇔ ( − 3)( + 2) = 0 ⇔  = 3 or −2 ⇒  = ln 3 since   0.



⇒ ln 2+5 = ln 100 ⇒ 2 + 5 = ln 100 ⇒ 5 = ln 100 − 2 ⇒

 = 1 (ln 100 − 2) ≈ 05210
5

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SECTION 6.3

(b) ln( − 2) = 3 ⇒  − 2 = 3
39. (a) ln   0

⇒   0

is 0    1.

LOGARITHMIC FUNCTIONS

¤

263

⇒  = 3 + 2 ⇒  = ln(3 + 2) ≈ 30949

⇒   1. Since the domain of  () = ln  is   0, the solution of the original inequality

(b)   5 ⇒ ln   ln 5 ⇒   ln 5
41. 3 ft = 36 in, so we need  such that log2  = 36

68,719,476,736 in ·

⇔  = 236 = 68,719,476,736. In miles, this is

1 ft
1 mi
·
≈ 1,084,5877 mi.
12 in 5280 ft

43. If  is the intensity of the 1989 San Francisco earthquake, then log10 () = 71

log10 (16) = log10 16 + log10 () = log10 16 + 71 ≈ 83.
  


45. (a)  =  () = 100 · 23 ⇒
= 23 ⇒ log2
=
100
100
3

⇒  = 3 log2

  
. Using formula (7), we can
100

ln(100)
. This function tells us how long it will take to obtain  bacteria (given the ln 2

e

write this as  =  −1 () = 3 ·

⇒





ln 50,000 ln 500
100
=3
≈ 269 hours ln 2 ln 2

rS

(b)  = 50,000 ⇒  =  −1 (50,000) = 3 ·

al

number ).

47. Let  = 2 − 9. Then as  → 3+ ,  → 0+ , and lim ln(2 − 9) = lim ln  = −∞ by (8).
→3+

→0+

49. lim ln(cos ) = ln 1 = 0. [ln(cos ) is continuous at  = 0 since it is the composite of two continuous functions.]

Fo

→0

51. lim [ln(1 + 2 ) − ln(1 + )] = lim ln
→∞

→∞

parentheses is ∞.

1 + 2
→∞ 1 +  lim 

= ln



1

→∞ 1


lim

+
+1



  

 =  | 2 − 9  0 =   ||  3 = (−∞ −3) ∪ (3 ∞)

√
3 − 2 , we must have 3 − 2 ≥ 0 ⇒ 2 ≤ 3

N

55. (a) For  () =



ot

53.  () = log10 (2 − 9).

1 + 2
= ln
1+

⇒



= ∞, since the limit in

2 ≤ ln 3 ⇒  ≤

1
2

ln 3.

Thus, the domain of  is (−∞ 1 ln 3].
2
(b)  =  () =

√
3 − 2

[note that  ≥ 0] ⇒  2 = 3 − 2

⇒ 2 = 3 −  2

⇒ 2 = ln(3 −  2 ) ⇒

ln(3 − 2 ). So  −1 () = 1 ln(3 − 2 ). For the domain of  −1 ,
2
√
√
√
√
2
2
we must have 3 −   0 ⇒   3 ⇒ ||  3 ⇒ − 3    3 ⇒ 0 ≤   3 since  ≥ 0. Note
√
that the domain of  −1 , [0 3 ), equals the range of .
=

1
2

ln(3 −  2 ). Interchange  and :  =

57. (a) We must have  − 3  0

1
2

⇒   3 ⇒   ln 3. Thus, the domain of  () = ln( − 3) is (ln 3 ∞).

(b)  = ln( − 3) ⇒  =  − 3 ⇒  =  + 3 ⇒  = ln( + 3), so  −1 () = ln( + 3).
Now  + 3  0 ⇒   −3, which is true for any real , so the domain of  −1 is R.
59.  = ln( + 3)

⇒  = ln(+3) =  + 3 ⇒  =  − 3.

Interchange  and : the inverse function is  =  − 3.

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264

¤

CHAPTER 6 INVERSE FUNCTIONS

61.  =  () = 

3



63.  = log10 1 +

⇒ ln  = 3

1




⇒ 10 = 1 +

Interchange  and :  =

1
3

1


√
√
√
3
ln . Interchange  and :  = 3 ln . So  −1 () = 3 ln .

⇒

1
1
= 10 − 1 ⇒  = 
.

10 − 1

1 is the inverse function.
10 − 1

33

 



1
 1

1
⇔ 2  ln 3 = − ln 3 ⇔   − 2 ln 3, so  is increasing on − 2 ln 3 ∞ .

65.  () = 3 − 

2 

⇒ =

⇒  0 () = 33 −  . Thus,  0 ()  0

⇔

33  

⇔

⇔ 32  1 ⇔

67. (a) We have to show that − () =  (−).


√
√


−1 
− () = − ln  + 2 + 1 = ln  + 2 + 1
= ln

1
√
2 + 1
√
√


√

1
 − 2 + 1
 − 2 + 1
√
√
= ln 2 + 1 −  = (−)
·
= ln 2
= ln
2 +1
2 +1
 − 2 − 1
+ 
− 

e

+

al

Thus,  is an odd function.

2 = 2 − 1 ⇔  =

2 − 1
= 1 ( − − ). Thus, the inverse function is  −1 () = 1 ( − − ).
2
2
2

1
· ln  = ln 2 ⇒ 1 = ln 2, a contradiction, so the given equation has no ln 

⇒ ln(1 ln  ) = ln(2) ⇒

Fo

69. 1 ln  = 2

rS

√
√


(b) Let  = ln  + 2 + 1 . Then  =  + 2 + 1 ⇔ ( − )2 = 2 + 1 ⇔ 2 − 2 + 2 = 2 + 1 ⇔

solution. The function  () = 1 ln  = (ln  )1 ln  = 1 =  for all   0, so the function  () = 1 ln  is the constant function () = .

71. (a) Let   0 be given. We need  such that | − 0|   when   . But   

Then   



ot



→−∞

(b) Let   0 be given. We need  such that    when   . But   
⇒   log 

⇔   log  . Let  = log  .

⇒    , so lim  = ∞.

N

Then   

⇔   log . Let  = log .

⇒   log  ⇒ | − 0| =   , so lim  = 0.


→∞

⇒ 0  2 − 2 − 2 ≤ 1. Now 2 − 2 − 2 ≤ 1 gives 2 − 2 − 3 ≤ 0 and hence
√
√
( − 3)( + 1) ≤ 0. So −1 ≤  ≤ 3. Now 0  2 − 2 − 2 ⇒   1 − 3 or   1 + 3. Therefore,
√
√ ln(2 − 2 − 2) ≤ 0 ⇔ −1 ≤   1 − 3 or 1 + 3   ≤ 3.

73. ln(2 − 2 − 2) ≤ 0

6.4 Derivatives of Logarithmic Functions
1. The differentiation formula for logarithmic functions,

3.  () = sin(ln )

⇒  0 () = cos(ln ) ·


1
(log ) =
, is simplest when  =  because ln  = 1.

 ln 


1
cos(ln ) ln  = cos(ln ) · =




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SECTION 6.4 DERIVATIVES OF LOGARITHMIC FUNCTIONS

5.  () = ln

¤

1


Another solution: () = ln

7.  () = log10 (3 + 1)

9.  () = sin  ln(5)

1
1
= ln 1 − ln  = − ln  ⇒  0 () = − .



⇒  0 () =

(3

 3
1
32
( + 1) = 3
+ 1) ln 10 
( + 1) ln 10

⇒  0 () = sin  ·

sin  · 5 sin 
1 
·
(5) + ln(5) · cos  =
+ cos  ln(5) =
+ cos  ln(5)
5 
5


(2 + 1)5
= ln(2 + 1)5 − ln( 2 + 1)12 = 5 ln(2 + 1) − 1 ln( 2 + 1) ⇒
2
2 + 1


1
1
10

8 2 −  + 10
1
0 () = 5 ·
·2− · 2
· 2 =
− 2 or 2 + 1
2  +1
2 + 1  + 1
(2 + 1)(2 + 1)

1
1
1

2 − 1 +  · 
22 − 1
1
+ · 2
· 2 = + 2
=
=
2 − 1)
 2  −1

 −1
(
(2 − 1) ln 
1 + ln(2)

⇒

1
[1 + ln(2)] ·  − ln  ·
[1 + ln(2)]2

17.  () = 5 + 5

1
2

·2

=

rS

 0 () =

ln(2 − 1) ⇒

1
[1


+ ln(2) − ln ]
1 + (ln 2 + ln ) − ln 
1 + ln 2
=
=
[1 + ln(2)]2
[1 + ln(2)]2
[1 + ln(2)]2

Fo

15.  () =

1
2

al

 √

2 − 1 = ln  + ln(2 − 1)12 = ln  +

13. () = ln 

e

11. () = ln 

 0 () =

⇒  0 () = 54 + 5 ln 5

⇒  0 = sec2 [ln( + )] ·

1

·  = sec2 [ln( + )]
 + 
 + 

ot

19.  = tan [ln( + )]

21.  = ln(− + − ) = ln(− (1 + )) = ln(− ) + ln(1 + ) = − + ln(1 + )

23.  = 2 log10

Note:

⇒

−1 −  + 1

1
=
=−
1+
1+
1+

N

 0 = −1 +

√
 = 2 log10 12 = 2 ·

1
2

log10  =  log10  ⇒  0 =  ·

1
1
+ log10  · 1 =
+ log10 
 ln 10 ln 10

1 ln 
1
=
= log10 , so the answer could be written as
+ log10  = log10  + log10  = log10 . ln 10 ln 10 ln 10
√

25.  () = 10

265


 

1  1
1
1
⇒  0 () =
= − 2 =− .
1  





27.  = 2 ln(2)

 00 = 1 + 2 ·

√

⇒  0 () = 10

⇒  0 = 2 ·

√



ln 10

 √  10  ln 10
√
 =

2 

1
· 2 + ln(2) · (2) =  + 2 ln(2) ⇒
2

1
· 2 + ln(2) · 2 = 1 + 2 + 2 ln(2) = 3 + 2 ln(2)
2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

266

¤

CHAPTER 6 INVERSE FUNCTIONS



29.  = ln  +

√

⇒
1 + 2



√

 
1
1
√
√
1 + 1 (1 + 2 )−12 (2)
 + 1 + 2 =
2
2 
2
+ 1+
+ 1+
√


1 + 2 + 

1
1
1
√
√
1+ √
=
· √
= √
=
2
2
2
2
+ 1+
1+
+ 1+
1+
1 + 2

 00 = − 1 (1 + 2 )−32 (2) =
2
31.  () =

0

 () =

−
(1 + 2 )32

⇒

−1
( − 1)[1 − ln( − 1)] + 
 − 1 − ( − 1) ln( − 1) + 
−1 =
−1
=
[1 − ln( − 1)]2
( − 1)[1 − ln( − 1)]2
[1 − ln( − 1)]2

[1 − ln( − 1)] · 1 −  ·

2 − 1 − ( − 1) ln( − 1)
( − 1)[1 − ln( − 1)]2

al

=


1 − ln( − 1)

⇒

e

0 =

33.  () = ln(2 − 2)

⇒  0 () =

rS

Dom() = { |  − 1  0 and 1 − ln( − 1) 6= 0} = { |   1 and ln( − 1) 6= 1}


=  |   1 and  − 1 6= 1 = { |   1 and  6= 1 + } = (1 1 + ) ∪ (1 +  ∞)
2( − 1)
1
(2 − 2) =
.
2 − 2
( − 2)

Dom( ) = { | ( − 2)  0} = (−∞ 0) ∪ (2 ∞).

37.  = ln(2 − 3 + 1)

⇒

0 =

Fo

ln 
1 + 2

1
· (2 − 3)
2 − 3 + 1

⇒

 0 (3) =

1
1

· 3 = 3, so an equation of a tangent line at

ot

35.  () =

 
1
(1 + 2 )
− (ln )(2)

1
2(1) − 0(2)
2
0
⇒  () =
, so  0 (1) =
= = .
(1 + 2 )2
22
4
2

(3 0) is  − 0 = 3( − 3), or  = 3 − 9.

⇒  0 () = cos  + 1.

N

39.  () = sin  + ln 

This is reasonable, because the graph shows that  increases when  0 is positive, and  0 () = 0 when  has a horizontal tangent.

41.  () =  + ln(cos )

 0(  ) = 6
4

⇒

⇒

 0 () =  +

 − tan  = 6
4

⇒

1
· (− sin ) =  − tan . cos 

−1 =6

⇒

 = 7.

43.  = (2 + 2)2 (4 + 4)4

⇒ ln  = ln[(2 + 2)2 (4 + 4)4 ] ⇒ ln  = 2 ln(2 + 2) + 4 ln(4 + 4) ⇒


1
4
1
163
1 0
 =2· 2
· 2 + 4 · 4
· 43 ⇒  0 =  2
+ 4
⇒

 +2
 +4
 +2
 +4


4
163
+ 4
0 = (2 + 2)2 (4 + 4)4
2 + 2
 +4 c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

SECTION 6.4 DERIVATIVES OF LOGARITHMIC FUNCTIONS

45.  =




12
−1
⇒ ln  = ln 4
 +1

−1
4 + 1

1 0
1 1
1 1
 =
−
· 43

2−1
2 4 + 1
47.  = 

⇒ ln  = ln 

267

1
1
ln( − 1) − ln(4 + 1) ⇒
2
2





1
1
−1
23
23
− 4
⇒ 0 =
− 4
⇒ 0 = 
2( − 1)
 +1
4 + 1 2 − 2
 +1
⇒ ln  =

⇒ ln  =  ln  ⇒  0  = (1) + (ln ) · 1 ⇒  0 = (1 + ln ) ⇒

 0 =  (1 + ln )
49.  =  sin 

0 = 



⇒ ln  = ln  sin 

 sin 
+ ln  cos 


51.  = (cos )

⇒

1
0
= (sin ) · + (ln )(cos ) ⇒



 sin 
 0 =  sin 
+ ln  cos 


⇒

⇒ ln  = ln(cos )

ln  = sin  ln  ⇒

1 0
1
 =·
· (− sin ) + ln cos  · 1 ⇒

cos 

⇒ ln  =  ln cos  ⇒

al

e



 sin 
 0 =  ln cos  −
⇒  0 = (cos ) (ln cos  −  tan ) cos 

1 ln tan  ⇒





sec2 
1 0
1
ln tan 
1
1
 = ·
· sec2  + ln tan  · − 2
−
⇒ 0 = 
⇒

 tan 

 tan 
2




sec2  ln tan  ln tan 
1
or 0 = (tan )1 ·
−
csc  sec  −
 0 = (tan )1
 tan 
2



55.  = ln(2 +  2 )

⇒ 0 =

⇒ ln  =

rS

⇒ ln  = ln(tan )1

Fo

53.  = (tan )1

2


1
2 + 2 0
(2 +  2 ) ⇒  0 = 2
2 
+
 + 2

ot

2  0 +  2  0 − 2 0 = 2 ⇒ (2 +  2 − 2) 0 = 2 ⇒ 0 =
⇒  0 () =

N

57.  () = ln( − 1)

 (4) () = −2 · 3( − 1)−4
59.

1
= ( − 1)−1
( − 1)

⇒ ···

⇒ 2  0 +  2  0 = 2 + 2 0

⇒

2
2 +  2 − 2

⇒  00 () = −( − 1)−2

⇒  000 () = 2( − 1)−3

⇒  () () = (−1)−1 · 2 · 3 · 4 · · · · · ( − 1)( − 1)− = (−1)−1

⇒

( − 1)!
( − 1)

From the graph, it appears that the curves  = ( − 4)2 and  = ln  intersect just to the left of  = 3 and to the right of  = 5, at about  = 53. Let
 () = ln  − ( − 4)2 . Then  0 () = 1 − 2( − 4), so Newton’s Method says that +1 =  −  ( ) 0 ( ) =  −

ln  − ( − 4)2
. Taking
1 − 2( − 4)

0 = 3, we get 1 ≈ 2957738, 2 ≈ 2958516 ≈ 3 , so the first root is 2958516, to six decimal places. Taking 0 = 5, we get 1 ≈ 5290755, 2 ≈ 5290718 ≈ 3 , so the second (and final) root is 5290718, to six decimal places.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

268

¤

CHAPTER 6 INVERSE FUNCTIONS

ln 

61.  () = √



⇒  0 () =

√
√
 (1) − (ln )[1(2  )]
2 − ln 
=

232

⇒

232 (−1) − (2 − ln )(312 )
3 ln  − 8
=
 0 ⇔ ln  
43
452

 and CD on (0 83 ). The inflection point is 83  8 −43 .
3

 00 () =

8
3

⇔   83 , so  is CU on (83  ∞)

63.  =  () = ln(sin )

A.  = { in R | sin   0} =

∞


=−∞

(2 (2 + 1) ) = · · · ∪ (−4 −3) ∪ (−2 −) ∪ (0 ) ∪ (2 3) ∪ · · ·

B. No -intercept; -intercepts:  () = 0 ⇔ ln(sin ) = 0 ⇔ sin  = 0 = 1 ⇔ integer .

C.  is periodic with period 2. D.

lim

→(2)+

() = −∞ and

lim

→[(2+1)]−

 = 2 +


2

for each

 () = −∞, so the lines

cos 
= cot , so  0 ()  0 when 2    2 +  for each
2
sin 


   (2 + 1). Thus,  is increasing on 2 2 +  and
2

al

H.

rS

integer , and  0 ()  0 when 2 + 
2


 decreasing on 2 + 2  (2 + 1) for each integer .


F. Local maximum values  2 +  = 0, no local minimum.
2

e

 =  are VAs for all integers . E.  0 () =

G.  00 () = − csc2   0, so  is CD on (2 (2 + 1)) for

Fo

each integer  No IP

A.  = R B. Both intercepts are 0 C.  (−) =  (), so the curve is symmetric about the

65.  =  () = ln(1 + 2 )

-axis. D.

lim ln(1 + 2 ) = ∞, no asymptotes. E.  0 () =

→±∞

2
0 ⇔
1 + 2
H.

ot

  0, so  is increasing on (0 ∞) and decreasing on (−∞ 0) 
F.  (0) = 0 is a local and absolute minimum.

2(1 + 2 ) − 2(2)
2(1 − 2 )
=
0 ⇔
2 )2
(1 + 
(1 + 2 )2

N

G.  00 () =

||  1, so  is CU on (−1 1), CD on (−∞ −1) and (1 ∞).
IP (1 ln 2) and (−1 ln 2).

67. We use the CAS to calculate  0 () =

 00 () =

2 + sin  +  cos  and 2 +  sin 

22 sin  + 4 sin  − cos2  + 2 + 5
. From the graphs, it
2 (cos2  − 4 sin  − 5)

seems that  0  0 (and so  is increasing) on approximately the intervals
(0 27), (45 82) and (109 143). It seems that  00 changes sign
(indicating inflection points) at  ≈ 38, 57, 100 and 120.
Looking back at the graph of  () = ln(2 +  sin ), this implies that the inflection points have approximate coordinates
(38 17), (57 21), (100 27), and (120 29). c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 6.4 DERIVATIVES OF LOGARITHMIC FUNCTIONS

¤

269

69. (a) Using a calculator or CAS, we obtain the model  =  with  ≈ 1000124369 and  ≈ 0000045145933.

(b) Use 0 () =  ln  (from Formula 7) with the values of  and  from part (a) to get 0 (004) ≈ −67063 A.
The result of Example 2 in Section 1.4 was −670 A.
71.



4

2

73.



2

1

3
 = 3




4

2


4
1
4
 = 3 ln || = 3(ln 4 − ln 2) = 3 ln = 3 ln 2

2
2


2


1

1
1
1
1 5
= − ln |8 − 3| = − ln 2 − − ln 5 = (ln 5 − ln 2) = ln
8 − 3
3
3
3
3
3 2
1

Or: Let  = 8 − 3. Then  = −3 , so

2


 2 1
 2
− 3 

1
1
1
1 5
1
=
= − ln || = − ln 2 − − ln 5 = (ln 5 − ln 2) = ln .

3
3
3
3
3 2
1 8 − 3
5
5


1

1 2

2

=

77. Let  = ln . Then  =

79.

+−




⇒

1
2



(ln )2
 =






2  =

1 3
1
 +  = (ln )3 + .
3
3

 sin 2 sin  cos 
 = 2
 = 2. Let  = cos . Then  = − sin  , so
1 + cos2 
1 + cos2 

 
2 = −2
= −2 · 1 ln(1 + 2 ) +  = − ln(1 + 2 ) +  = − ln(1 + cos2 ) + .
2
1 + 2

Or: Let  = 1 + cos2 .


2

10  =

1



10 ln 10

2

=

1

101
100 − 10
90
102
−
=
=
ln 10 ln 10 ln 10 ln 10

ot

81.




 
 

1
+1+
 = 1 2 +  + ln  1 = 1 2 +  + 1 − 1 + 1 + 0
2
2
2


e



al

1

2 +  + 1
 =


rS



Fo

75.



1

(ln |sin | + ) = cos  = cot 

sin 



cos 

(b) Let  = sin . Then  = cos  , so cot   =
 =
= ln || +  = ln |sin | + . sin 


N

83. (a)



√

85. The cross-sectional area is  1  + 1



0

1

2

= ( + 1). Therefore, the volume is


 = [ln( + 1)]1 = (ln 2 − ln 1) =  ln 2.
0
+1

87.  =

 2
1

  =



1000

600


 = 




1000

600


1000
1
 =  ln | |
= (ln 1000 − ln 600) =  ln 1000 =  ln 5 .
600
3

600

Initially,   = , where  = 150 kPa and  = 600 cm3 , so  = (150)(600) = 90,000. Thus,
 = 90 000 ln 5 ≈ 45 974 kPa · cm3 = 45 974(103 Pa)(10−6 m3 ) = 45 974 Pa·m3 = 45 974 N·m [Pa = Nm2 ]
3
= 45 974 J

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

270

¤

CHAPTER 6 INVERSE FUNCTIONS

⇒  0 () = 2 + 1. If  =  −1 , then  (1) = 2 ⇒ (2) = 1, so

89.  () = 2 + ln 

 0 (2) = 1 0 ((2)) = 1 0 (1) = 1 .
3
91. The curve and the line will determine a region when they intersect at two or

more points. So we solve the equation (2 + 1) =  ⇒  = 0 or


± −4()( − 1)
1
2
=±
− 1.
 +  − 1 = 0 ⇒  = 0 or  =
2

Note that if  = 1, this has only the solution  = 0, and no region is determined. But if 1 − 1  0 ⇔ 1  1 ⇔ 0    1, then there are two solutions. [Another way of seeing this is to observe that the slope of the tangent to  = (2 + 1) at the origin is  0 = 1 and therefore we must have 0    1.]
Note that we cannot just integrate between the positive and negative roots, since the curve and the line cross at the origin.

93. If  () = ln (1 + ), then  0 () =

Thus, lim

1
, so  0 (0) = 1.
1+

ln(1 + )
 ()
() − (0)
= lim
= lim
=  0 (0) = 1.
→0
→0


−0

ot

→0

rS

0


√1−1


−   = 2 1 ln(2 + 1) − 1 2 0
2
2
2 + 1



 
1
1
−1+1 −
−1
− (ln 1 − 0)
= ln


 
1
= ln
+  − 1 =  − ln  − 1


Fo

2

 √1−1 

al

e

Since  and (2 + 1) are both odd functions, the total area is twice the area between the curves on the interval

 
0 1 − 1 . So the total area enclosed is

N

6.2* The Natural Logarithmic Function
1. ln

√
 = ln()12 =

3. ln

2
= ln 2 − ln( 3  4 ) = 2 ln  − (ln  3 + ln  4 ) = 2 ln  − 3 ln  − 4 ln 
3 4

1
2

ln() = 1 (ln  + ln ) =
2

5. 2 ln  + 3 ln  − ln  = ln 2 + ln  3 − ln 

= ln(2  3 ) − ln 

1
2

ln  +

1
2

ln  [assuming that the variables are positive]

[by Law 3]
[by Law 1]

2 3

= ln
7. ln 5 + 5 ln 3 = ln 5 + ln 35
5

= ln(5 · 3 )

 


[by Law 2]
[by Law 3]
[by Law 1]

= ln 1215

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SECTION 6.2* THE NATURAL LOGARITHMIC FUNCTION

9.

1
3

ln( + 2)3 +

1
2



ln  − ln(2 + 3 + 2)2 = ln[( + 2)3 ]13 +
= ln( + 2) + ln

1
2

ln


(2 + 3 + 2)2

√

2 + 3 + 2

√
( + 2) 
( + 1)( + 2)
√

= ln
+1

¤

271

[by Laws 3, 2]
[by Law 3]
[by Law 1]

= ln

Note that since ln  is defined for   0, we have  + 1,  + 2, and 2 + 3 + 2 all positive, and hence their logarithms are defined.
13.

11. Reflect the graph of  = ln  about the -axis to obtain

al

e

the graph of  = − ln .

 = ln( + 3)

rS

 = ln 

 = ln 

 = − ln 

15. Let  = 2 − 9. Then as  → 3+ ,  → 0+ , and lim ln(2 − 9) = lim ln  = −∞ by (4).
17.  () =

Fo

→3+

√
 ln  ⇒  0 () =

1


⇒  0 () =

1 
1 

N

21.  () = ln

⇒  0 () = cos(ln ) ·

Another solution: () = ln
23.  () = sin  ln(5)
25. () = ln

 0 () =

 
1
ln  + 2
√
=

2 

1 cos(ln )

ln  = cos(ln ) · =




ot

19.  () = sin(ln )

√
1
√ ln  + 
2 

→0+

 


1
1
1
= − 2 =− .




1
1
= ln 1 − ln  = − ln  ⇒  0 () = − .



⇒  0 () = sin  ·

sin  · 5 sin 
1 
·
(5) + ln(5) · cos  =
+ cos  ln(5) =
+ cos  ln(5)
5 
5


−
= ln( − ) − ln( + ) ⇒
+

−( + ) − ( − )
1
1
−2
(−1) −
=
= 2
−
+
( − )( + )
 − 2

(2 + 1)5
= ln(2 + 1)5 − ln( 2 + 1)12 = 5 ln(2 + 1) − 1 ln( 2 + 1) ⇒
2
2 + 1


1
1
10

8 2 −  + 10
1
·2− · 2
· 2 =
− 2 or 0 () = 5 ·
2 + 1
2  +1
2 + 1  + 1
(2 + 1)(2 + 1)

27. () = ln 

c
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¤

CHAPTER 6 INVERSE FUNCTIONS

 √

2 − 1 = ln  + ln(2 − 1)12 = ln  +

29. () = ln 

 0 () =

31.  () =

 0 () =

ln 
1 + ln(2)

⇒

1
[1 + ln(2)] ·  − ln  ·
[1 + ln(2)]2





1
2

·2

=

1
[1


+ ln(2) − ln ]
1 + (ln 2 + ln ) − ln 
1 + ln 2
=
=
[1 + ln(2)]2
[1 + ln(2)]2
[1 + ln(2)]2

1
−10 − 1
10 + 1
· (−1 − 10) = or 2 −  − 52
2 −  − 52
52 +  − 2

⇒ 0 =

35.  = tan [ln( + )]

⇒  0 = sec2 [ln( + )] ·

⇒  0 = 2 ·

37.  = 2 ln(2)

=

1
· 2 + ln(2) · (2) =  + 2 ln(2) ⇒
2


1 − ln( − 1)

al

1
· 2 + ln(2) · 2 = 1 + 2 + 2 ln(2) = 3 + 2 ln(2)
2
⇒

−1
( − 1)[1 − ln( − 1)] + 
 − 1 − ( − 1) ln( − 1) + 
−1 =
−1
=
[1 − ln( − 1)]2
( − 1)[1 − ln( − 1)]2
[1 − ln( − 1)]2

[1 − ln( − 1)] · 1 −  ·

2 − 1 − ( − 1) ln( − 1)
( − 1)[1 − ln( − 1)]2

Fo

 () =

1

·  = sec2 [ln( + )]
 + 
 + 

rS

 00 = 1 + 2 ·

0

ln(2 − 1) ⇒

1
1
1
1

2 − 1 +  · 
22 − 1
+ · 2
· 2 = + 2
=
=
2 − 1)
 2  −1

 −1
(
(2 − 1)

33.  = ln 2 −  − 52 

39.  () =

1
2

e

272

√
1 − ln  is defined ⇔   0 [so that ln  is defined] and 1 − ln  ≥ 0 ⇔

N

41.  () =

ot

Dom() = { |  − 1  0 and 1 − ln( − 1) 6= 0} = { |   1 and ln( − 1) 6= 1}


=  |   1 and  − 1 6= 1 = { |   1 and  6= 1 + } = (1 1 + ) ∪ (1 +  ∞)

  0 and ln  ≤ 1 ⇔ 0   ≤ , so the domain of  is (0 ]. Now


1

1
1
−1
√
 0 () = (1 − ln )−12 ·
· −
.
(1 − ln ) = √
=
2


2 1 − ln 
2 1 − ln 

43.  () =

ln 
1 + 2

 
1
− (ln )(2)
(1 + 2 )

2(1) − 0(2)
2
1
0
⇒  () =
, so  0 (1) =
= = .
(1 + 2 )2
22
4
2

45.  () = sin  + ln 

⇒  0 () = cos  + 1.

This is reasonable, because the graph shows that  increases when  0 is positive, and  0 () = 0 when  has a horizontal tangent.

c
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¤

SECTION 6.2* THE NATURAL LOGARITHMIC FUNCTION

⇒ 0 = cos(2 ln ) ·

47.  = sin(2 ln )

273

2
2
. At (1 0), 0 = cos 0 · = 2, so an equation of the tangent line is

1

 − 0 = 2 · ( − 1), or  = 2 − 2.
49.  = ln(2 +  2 )

⇒ 0 =

2

1
2 + 2 0

(2 +  2 ) ⇒  0 = 2
2 
+
 + 2

⇒  0 () =

 (4) () = −2 · 3( − 1)−4

1
= ( − 1)−1
( − 1)

⇒ ···

⇒

2
2 +  2 − 2

2  0 +  2  0 − 2 0 = 2 ⇒ (2 +  2 − 2) 0 = 2 ⇒ 0 =
51.  () = ln( − 1)

⇒ 2  0 +  2  0 = 2 + 2 0

⇒  00 () = −( − 1)−2

⇒  000 () = 2( − 1)−3

⇒  () () = (−1)−1 · 2 · 3 · 4 · · · · · ( − 1)( − 1)− = (−1)−1

⇒

( − 1)!
( − 1)

From the graph, it appears that the curves  = ( − 4)2 and  = ln  intersect

e

53.

just to the left of  = 3 and to the right of  = 5, at about  = 53. Let

al

 () = ln  − ( − 4)2 . Then  0 () = 1 − 2( − 4), so Newton’s Method

rS

says that +1 =  −  ( ) 0 ( ) =  −

ln  − ( − 4)2
. Taking
1 − 2( − 4)

0 = 3, we get 1 ≈ 2957738, 2 ≈ 2958516 ≈ 3 , so the first root is 2958516, to six decimal places. Taking 0 = 5, we

55.  =  () = ln(sin )

Fo

get 1 ≈ 5290755, 2 ≈ 5290718 ≈ 3 , so the second (and final) root is 5290718, to six decimal places.

A.  = { in R | sin   0} =

∞


=−∞

(2 (2 + 1) ) = · · · ∪ (−4 −3) ∪ (−2 −) ∪ (0 ) ∪ (2 3) ∪ · · ·

integer .

ot

B. No -intercept; -intercepts:  () = 0 ⇔ ln(sin ) = 0 ⇔ sin  = 0 = 1 ⇔
C.  is periodic with period 2. D.

lim

→(2)+

() = −∞ and

lim

→[(2+1)]−

 = 2 +


2

for each

 () = −∞, so the lines

cos 
= cot , so  0 ()  0 when 2    2 +  for each
2
sin 


   (2 + 1). Thus,  is increasing on 2 2 +  and
2

N

 =  are VAs for all integers . E.  0 () =

integer , and  0 ()  0 when 2 + 
2


decreasing on 2 +   (2 + 1) for each integer .
2


F. Local maximum values  2 +  = 0, no local minimum.
2

H.

G.  00 () = − csc2   0, so  is CD on (2 (2 + 1)) for each integer  No IP

c
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¤

274

CHAPTER 6 INVERSE FUNCTIONS

A.  = R B. Both intercepts are 0 C.  (−) =  (), so the curve is symmetric about the

57.  =  () = ln(1 + 2 )

-axis. D.

lim ln(1 + 2 ) = ∞, no asymptotes. E.  0 () =

→±∞

2
0 ⇔
1 + 2
H.

  0, so  is increasing on (0 ∞) and decreasing on (−∞ 0) 
F.  (0) = 0 is a local and absolute minimum.
G.  00 () =

2(1 + 2 ) − 2(2)
2(1 − 2 )
=
0 ⇔
2 )2
(1 + 
(1 + 2 )2

||  1, so  is CU on (−1 1), CD on (−∞ −1) and (1 ∞).
IP (1 ln 2) and (−1 ln 2).

 00 () =

2 + sin  +  cos  and 2 +  sin 

22 sin  + 4 sin  − cos2  + 2 + 5
. From the graphs, it
2 (cos2  − 4 sin  − 5)

(0 27), (45 82) and (109 143). It seems that  00 changes sign

rS

(indicating inflection points) at  ≈ 38, 57, 100 and 120.

al

seems that  0  0 (and so  is increasing) on approximately the intervals

e

59. We use the CAS to calculate  0 () =

Looking back at the graph of  () = ln(2 +  sin ), this implies that the inflection points have approximate coordinates
(38 17), (57 21), (100 27), and (120 29).

⇒ ln  = ln[(2 + 2)2 (4 + 4)4 ] ⇒ ln  = 2 ln(2 + 2) + 4 ln(4 + 4) ⇒


4
1
163
1
1 0
 =2· 2
· 2 + 4 · 4
· 43 ⇒  0 =  2
+ 4
⇒

 +2
 +4
 +2
 +4


4
163
0
2
2
4
4
+ 4
 = ( + 2) ( + 4)
2 + 2
 +4

ot


12
−1
⇒ ln  = ln 4
 +1

−1
4 + 1

1 0
1 1
1 1
 =
−
· 43

2−1
2 4 + 1
65.



4

2

67.



1

2

1
1
ln( − 1) − ln(4 + 1) ⇒
2
2





−1
23
1
23
1
− 4
⇒ 0 =
− 4
⇒ 0 = 
2( − 1)
 +1
4 + 1 2 − 2
 +1
⇒ ln  =

N

63.  =



Fo

61.  = (2 + 2)2 (4 + 4)4

3
 = 3




2

4


4
1
4
 = 3 ln || = 3(ln 4 − ln 2) = 3 ln = 3 ln 2

2
2


2



1
1
1
1 5
1
= − ln |8 − 3| = − ln 2 − − ln 5 = (ln 5 − ln 2) = ln
8 − 3
3
3
3
3
3 2
1

Or: Let  = 8 − 3. Then  = −3 , so

2


 2
 2 1
− 3 

1
1
1
1 5
1
=
= − ln || = − ln 2 − − ln 5 = (ln 5 − ln 2) = ln .
8 − 3

3
3
3
3
3 2
1
5
5

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 6.2* THE NATURAL LOGARITHMIC FUNCTION

69.





1

= 1 2 +  −
2

71. Let  = ln . Then  =

73.

¤

275




 
 

2 +  + 1
1
+1+
 =
 = 1 2 +  + ln  1 = 1 2 +  + 1 − 1 + 1 + 0
2
2
2


1





1
2

⇒





(ln )2
 =




2  =

1 3
1
 +  = (ln )3 + .
3
3

 sin 2 sin  cos 
 = 2
 = 2. Let  = cos . Then  = − sin  , so
1 + cos2 
1 + cos2 

 
= −2 · 1 ln(1 + 2 ) +  = − ln(1 + 2 ) +  = − ln(1 + cos2 ) + .
2 = −2
2
1 + 2

Or: Let  = 1 + cos2 .
75. (a)

1

(ln |sin | + ) = cos  = cot 

sin 



√

77. The cross-sectional area is  1  + 1
1

0

2

= ( + 1). Therefore, the volume is


 = [ln( + 1)]1 = (ln 2 − ln 1) =  ln 2.
0
+1

79.  =

 2
1

  =



1000

600

Fo



rS

al

e

(b) Let  = 2 + 2 . Then  = (2 + 2)  = 2(1 + )  and ( + 1)  = 1 , so
2

 √


32

 1 32
( + 1) 2 + 2  =
+  = 1 2 + 2
 1  =
+ .
2
3
2 32
√
Or: Let  = 2 + 2 . Then 2 = 2 + 2 ⇒ 2  = (2 + 2)  ⇒   = (1 + ) , so
√



( + 1) 2 + 2  =  ·   = 2  = 1 3 +  = 1 (2 + 2 )32 + .
3
3


 = 




1000

600


1000
1
 =  ln | |
= (ln 1000 − ln 600) =  ln 1000 =  ln 5 .
600
3

600

ot

Initially,   = , where  = 150 kPa and  = 600 cm3 , so  = (150)(600) = 90,000. Thus,
 = 90 000 ln 5 ≈ 45 974 kPa · cm3 = 45 974(103 Pa)(10−6 m3 ) = 45 974 Pa·m3 = 45 974 N·m [Pa = Nm2 ]
3

N

= 45 974 J

81.  () = 2 + ln 

⇒  0 () = 2 + 1. If  =  −1 , then  (1) = 2 ⇒ (2) = 1, so

 0 (2) = 1 0 ((2)) = 1 0 (1) = 1 .
3
83. (a)

We interpret ln 15 as the area under the curve  = 1 from  = 1 to
 = 15. The area of the rectangle  is 1 · 2 = 1 . The area of the
2 3
3


1
1
2
5
trapezoid  is 2 · 2 1 + 3 = 12 . Thus, by comparing areas, we observe that

1
3

 ln 15 

5
.
12

(b) With  () = 1,  = 10, and ∆ = 005, we have ln 15 =

 15

(1)  ≈ (005)[(1025) +  (1075) + · · · +  (1475)]

 1
1
1
= (005) 1025 + 1075 + · · · + 1475 ≈ 04054
1

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

276

¤

CHAPTER 6 INVERSE FUNCTIONS

The area of  is

85.

The area of  is

1
1
1
1
and so + + · · · + 
+1
2
3




1
1
1 and so 1 + + · · · +


2
−1





1

1



1
 = ln .


1
 = ln .


Note that if  = 1, this has only the solution  = 0, and no region is

al

more points. So we solve the equation (2 + 1) =  ⇒  = 0 or


± −4()( − 1)
1
2
=±
− 1.
 +  − 1 = 0 ⇒  = 0 or  =
2


e

87. The curve and the line will determine a region when they intersect at two or

rS

determined. But if 1 − 1  0 ⇔ 1  1 ⇔ 0    1, then there are two solutions. [Another way of seeing this is to observe that the slope of the tangent to  = (2 + 1) at the origin is  0 = 1 and therefore we must have 0    1.]
Note that we cannot just integrate between the positive and negative roots, since the curve and the line cross at the origin.

2

 √1−1 


√1−1


−   = 2 1 ln(2 + 1) − 1 2 0
2
2
2 +1

 



1
1
= ln
−1+1 −
−1
− (ln 1 − 0)


 
1
+  − 1 =  − ln  − 1
= ln


N

ot

0

Fo

Since  and (2 + 1) are both odd functions, the total area is twice the area between the curves on the interval

 
0 1 − 1 . So the total area enclosed is

89. If  () = ln (1 + ), then  0 () =

Thus, lim

→0

1
, so  0 (0) = 1.
1+

ln(1 + )
 ()
() − (0)
= lim
= lim
=  0 (0) = 1.
→0
→0


−0

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 6.3* THE NATURAL EXPONENTIAL FUNCTION

¤

277

6.3* The Natural Exponential Function
The function value at  = 0 is 1 and the slope at  = 0 is 1.

1.

⇔

⇔

 (5)

10
(5)
= ln(10 ) = 10
(b) ln ln 

1
1
=
52
25

7 − 4 = ln 6

(b) ln(3 − 10) = 2
7. (a) 3+1 = 

(4)

= 5−2 =

⇔

7 − ln 6 = 4

3 − 10 = 2

⇔ 3 + 1 = ln 

⇔

⇔

3 = 2 + 10

⇔  = 1 (ln  − 1)
3

 = 1 (7 − ln 6)
4
⇔

 = 1 (2 + 10)
3

e

5. (a) 7−4 = 6

−2

al



3. (a) −2 ln 5 = ln 5

⇔  − 1 = −2

11. 2 −  − 6 = 0
13. (a) 2+5 = 100

⇔ ln( − 1) = ln −2

⇔ −2 = ln( − 1) ⇔  = − 1 ln( − 1)
2

⇔ ( − 3)( + 2) = 0 ⇔  = 3 or −2 ⇒  = ln 3 since   0.

Fo

9.  − −2 = 1

rS

(b) ln  + ln( − 1) = ln(( − 1)) = 1 ⇔ ( − 1) = 1 ⇔ 2 −  −  = 0. The quadratic formula (with  = 1,


√
 = −1, and  = −) gives  = 1 1 ± 1 + 4 , but we reject the negative root since the natural logarithm is not
2


√ defined for   0. So  = 1 1 + 1 + 4 .
2



⇒ ln 2+5 = ln 100 ⇒ 2 + 5 = ln 100 ⇒ 5 = ln 100 − 2 ⇒

 = 1 (ln 100 − 2) ≈ 05210
5

15. (a) ln   0

⇒  = 3 + 2 ⇒  = ln(3 + 2) ≈ 30949

ot

(b) ln( − 2) = 3 ⇒  − 2 = 3
⇒   0

N

is 0    1.

⇒   1. Since the domain of  () = ln  is   0, the solution of the original inequality

(b)   5 ⇒ ln   ln 5 ⇒   ln 5
17.

 = 

 = −

19. We start with the graph of  =  (Figure 2) and reflect about the -axis to get the graph of  = − . Then we compress

the graph vertically by a factor of 2 to obtain the graph of  = 1 − and then reflect about the -axis to get the graph of
2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

278

¤

CHAPTER 6 INVERSE FUNCTIONS

 = − 1 − . Finally, we shift the graph upward one unit to get the graph of  = 1 − 1 − .
2
2

√
3 − 2 , we must have 3 − 2 ≥ 0 ⇒ 2 ≤ 3

21. (a) For  () =

⇒

2 ≤ ln 3 ⇒  ≤

1
2

ln 3.

Thus, the domain of  is (−∞ 1 ln 3].
2
(b)  =  () =

√
3 − 2

[note that  ≥ 0] ⇒  2 = 3 − 2

⇒ 2 = ln(3 −  2 ) ⇒

ln(3 − 2 ). So  −1 () = 1 ln(3 − 2 ). For the domain of  −1 ,
2
√
√
√
√
2
2
we must have 3 −   0 ⇒   3 ⇒ ||  3 ⇒ − 3    3 ⇒ 0 ≤   3 since  ≥ 0. Note
√
that the domain of  −1 , [0 3 ), equals the range of .
1
2

1
2

⇒  = ln(+3) =  + 3 ⇒  =  − 3.

23.  = ln( + 3)

25.  =  () = 

3

⇒ ln  = 3

⇒ =

rS

Interchange  and : the inverse function is  =  − 3.

al

e

=

ln(3 −  2 ). Interchange  and :  =

⇒ 2 = 3 −  2

√
√
√
3
ln . Interchange  and :  = 3 ln . So  −1 () = 3 ln .
3 − −3
1 − −6
1−0
=1
= lim
=
→∞ 1 + −6
3 + −3
1+0

Fo

27. Divide numerator and denominator by 3 : lim

→∞

29. Let  = 3(2 − ). As  → 2+ ,  → −∞. So lim 3(2−) = lim  = 0 by (6).
→2+

→−∞

31. Since −1 ≤ cos  ≤ 1 and −2  0, we have −−2 ≤ −2 cos  ≤ −2 . We know that lim (−−2 ) = 0 and

ot

→∞



lim −2 = 0, so by the Squeeze Theorem, lim (−2 cos ) = 0.

→∞

→∞

N

33.  () = 5 is a constant function, so its derivative is 0, that is,  0 () = 0.
35. By the Product Rule,  () = (3 + 2)

⇒

 0 () = (3 + 2)( )0 +  (3 + 2)0 = (3 + 2) +  (32 + 2)
=  [(3 + 2) + (32 + 2)] =  (3 + 32 + 2 + 2)
37. By (9),  = 
39.  = −
41.  () = 1

3

⇒  0 = 

3

3

(3 ) = 32  .






⇒  0 =  − (−) + − · 1 = − (− + 1) or (1 − )−
⇒  0 () = 1 ·

43. By (9),  () =  sin 2




 

 

1
−1
−1
= 1
=
1

2
2

⇒  0 () =  sin 2 ( sin 2)0 =  sin 2 ( · 2 cos 2 + sin 2 · 1) =  sin 2 (2 cos 2 + sin 2)

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 6.3* THE NATURAL EXPONENTIAL FUNCTION

45.  =

√
1 + 23

47.  = 





49. By the Quotient Rule,  =

0 =

279


1
1
33
(1 + 23 )−12
(1 + 23 ) = √
(23 · 3) = √
3
2

2 1 + 2
1 + 23

⇒ 0 =

⇒  0 =  ·

¤



 
( ) =  ·  or  +


 + 
 + 

⇒

( + )( ) − ( + )( )
( +  −  − )
( − )
=
=
.
 + )2
 + )2
(
(
( + )2


51.  = cos

1 − 2
1 + 2



⇒

53.  = 2 cos 

al

e







 1 − 2
1 − 2
(1 + 2 )(−22 ) − (1 − 2 )(22 )
1 − 2
·
= − sin
·
 0 = − sin
2
2
2
1+
 1 + 
1+
(1 + 2 )2








−22 (1 + 2 ) + (1 − 2 )
1 − 2
1 − 2
1 − 2
42
−22 (2)
= − sin
= − sin
=
· sin
·
·
1 + 2
(1 + 2 )2
1 + 2
(1 + 2 )2
(1 + 2 )2
1 + 2
⇒  0 = 2 (− sin ) + (cos )(22 ) = 2 (2 cos  −  sin ).



 
( ) =
( − ) ⇒  ·



 ·

 · 1 −  · 0
= 1 − 0
2

 ·

⇒

 0
1
−
·  = 1 − 0

2

⇒

0 −

 0

· =1−
2


⇒

 − 
( −  )

⇒ 0 = 2
= 2

 − 
 − 
2


ot




 − 
0 1 −
=
2



⇒

 

= 1 − 0


Fo

55.

rS

At (0 1),  0 = 1(2 − 0) = 2, so an equation of the tangent line is  − 1 = 2( − 0), or  = 2 + 1.

⇒  0 =  − 1 −2 ⇒  00 =  + 1 −2 , so
2
4

 
 

200 −  0 −  = 2  + 1 −2 −  − 1 −2 −  + −2 = 0.
4
2

N

57.  =  + −2

59.  = 

⇒  0 = 

⇒

 00 = 2  , so if  =  satisfies the differential equation  00 + 6 0 + 8 = 0,

then 2  + 6 + 8 = 0; that is,  (2 + 6 + 8) = 0. Since   0 for all , we must have 2 + 6 + 8 = 0, or ( + 2)( + 4) = 0, so  = −2 or −4.
61.  () = 2

⇒  0 () = 22

 000 () = 22 · 22 = 23 2

⇒  00 () = 2 · 22 = 22 2

⇒ ···

⇒

⇒  () () = 2 2

63. (a)  () =  +  is continuous on R and (−1) = −1 − 1  0  1 =  (0), so by the Intermediate Value Theorem,

 +  = 0 has a root in (−1 0).
(b)  () =  +  ⇒  0 () =  + 1, so +1 =  −

 + 
. Using 1 = −05, we get 2 ≈ −0566311,
 + 1

3 ≈ −0567143 ≈ 4 , so the root is −0567143 to six decimal places. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

280

¤

CHAPTER 6 INVERSE FUNCTIONS

65. (a) lim () = lim
→∞

→∞

1
1
= 1, since   0 ⇒ − → −∞ ⇒ − → 0.
=
1 + −
1+·0

(b) () = (1 + − )−1

⇒


−
= −(1 + − )−2 (−− ) =

(1 + − )2

(c)
From the graph of () = (1 + 10−05 )−1 , it seems that () = 08
(indicating that 80% of the population has heard the rumor) when
 ≈ 74 hours.

67.  () =  − 

⇒  0 () = 1 −  = 0 ⇔  = 1 ⇔  = 0. Now  0 ()  0 for all   0 and  0 ()  0 for all

  0, so the absolute maximum value is  (0) = 0 − 1 = −1.
, [−1 4].  0 () =  · −

2 8

· (−  ) + −
4

2 8

2 8

· 1 = −

2 8

2

(−  + 1). Since −
4

e

2 8

69.  () = −

is never 0,

 () = 0 ⇒ − 4 + 1 = 0 ⇒ 1 =  4 ⇒  = 4 ⇒  = ±2, but −2 is not in the given interval, [−1 4].
√
 (−1) = −−18 ≈ −088,  (2) = 2−12 ≈ 121, and (4) = 4−2 ≈ 054. So  (2) = 2−12 = 2  is the absolute
√
maximum value and  (−1) = −−18 = −1 8  is the absolute minimum value.
0

2

2

71. (a)  () = (1 − )−

rS

al

2

⇒  0 () = (1 − )(−− ) + − (−1) = − ( − 2)  0 ⇒   2, so  is increasing on

(2 ∞) and decreasing on (−∞ 2).

Fo

(b)  00 () = − (1) + ( − 2)(−− ) = − (3 − )  0 ⇔   3, so  is CU on (−∞ 3) and CD on (3 ∞).


(c)  00 changes sign at  = 3, so there is an IP at 3 −2−3 .

73.  = () = −1(+1)

→±∞

lim −1(+1) = ∞ since −1( + 1) → ∞, so  = −1 is a VA.

E.  0 () = −1(+1) ( + 1)2

⇒  0 ()  0 for all  except 1, so

 is increasing on (−∞ −1) and (−1 ∞). F. No extreme values
G.  00 () =

lim −1(+1) = 0 since

→−1+

→−1−

N

−1 ( + 1) → −∞,

lim −1(+1) = 1 since −1( + 1) → 0, so  = 1 is a HA.

ot

C. No symmetry D.

A.  = { |  6= −1} = (−∞ −1) ∪ (−1, ∞) B. No -intercept; -intercept =  (0) = −1

H.

−1(+1)
−1(+1) (−2)
−1(+1) (2 + 1)
+
=−
( + 1)4
( + 1)3
( + 1)4

⇒

 00 ()  0 ⇔ 2 + 1  0 ⇔   − 1 , so  is CU on (−∞ −1)
2




 1

1 and −1 − 2 , and CD on − 2 , ∞ .  has an IP at − 1 , −2 .
2
75.  = 1(1 + − )

A.  = R B. No -intercept; -intercept =  (0) = 1  C. No symmetry
2

D. lim 1(1 + − ) =
→∞

1
1+0

= 1 and lim 1(1 + − ) = 0 since lim − = ∞, so  has horizontal asymptotes

0

→−∞

− −2

 = 0 and  = 1. E.  () = −(1 + 
F. No extreme values G.  00 () =

)

→−∞

−

(−

−

)=

− 2

(1 + 

) . This is positive for all , so  is increasing on R.

(1 + − )2 (−− ) − − (2)(1 + − )(−− )
− (− − 1)
=
− )4
(1 + 
(1 + − )3

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 6.3* THE NATURAL EXPONENTIAL FUNCTION

The second factor in the numerator is negative for   0 and positive for   0,

¤

281

H.

and the other factors are always positive, so  is CU on (−∞, 0) and CD

 on (0 ∞). IP at 0, 1
2
77. () =  − with  = 001,  = 4, and  = 007. We will find the

zeros of  00 for () =  − .
 0 () =  (−− ) + − (−1 ) = − (− + −1 )
 00 () = − (−−1 + ( − 1)−2 ) + (− + −1 )(−− )
= −2 − [− + ( − 1) + 2 2 − ]

e

= −2 − (2 2 − 2 + 2 − )
Using the given values of  and  gives us  00 () = 2 −007 (000492 − 056 + 12). So  00 () = 001 00 () and its zeros
200
7

≈ 2857 and 2 =

al

are  = 0 and the solutions of 000492 − 056 + 12 = 0, which are 1 =

600
7

≈ 8571.

At 1 minutes, the rate of increase of the level of medication in the bloodstream is at its greatest and at 2 minutes, the rate of

3 −

79.  () = 

rS

decrease is the greatest.
→ 0 as  → −∞, and

 () → ∞ as  → ∞. From the graph,

Fo

it appears that  has a local minimum of about  (058) = 068, and a local

maximum of about  (−058) = 147.

1
√
3



√

√

√

N





ot

To find the exact values, we calculate

 3
1
 0 () = 32 − 1  − , which is 0 when 32 − 1 = 0 ⇔  = ± √3 . The negative root corresponds to the local


√ 3
√
√
1
maximum  − √3 = (−1 3) − (−1 3) = 2 39 , and the positive root corresponds to the local minimum
= (1

 00 () =

3)3 − (1

3)

= −2

39

. To estimate the inflection points, we calculate and graph

 3  
 3 



3
3
  2
3 − 1  − = 32 − 1  − 32 − 1 +  − (6) =  − 94 − 62 + 6 + 1 .


From the graph, it appears that  00 () changes sign (and thus  has inflection points) at  ≈ −015 and  ≈ −109. From the graph of , we see that these -values correspond to inflection points at about (−015 115) and (−109 082).
81.



1

( +  )  =

0

83.



0

2


=




0

2



+1
+ 
+1

1

=

0




1
1
+  − (0 + 1) =
+−1
+1
+1


2
1
1
1
1
−  = − − = − −2 + 0 = (1 − −2 )




0

85. Let  = 1 +  . Then  =  , so



√
√
 1 +   =
  = 2 32 +  = 2 (1 +  )32 + .
3
3

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

282




( + − )2  = (2 + 2 + −2 )  = 1 2 + 2 − 1 −2 + 
2
2

89. Let  = tan . Then  = sec2  , so



tan  sec2   =



  =  +  = tan  + .

91. Let  = 1, so  = −12 . When  = 1,  = 1; when  = 2,  =
2

1

1


 =
2

93. Area =
95.  =

1
0

0





12

1

Thus,

√
 12
 (−) = −  1 = −(12 − ) =  − .



1 
 1  3
 

 −   = 1 3 −  0 = 1 3 −  − 1 − 1 = 1 3 −  +
3
3
3
3
0
( )2  = 
2


97. erf() = √





2

−  =
2



2

0

 1
2  = 1  2 0 =
2

− 

0



0

2

−  +


2



⇒

0



−  =





1



−  −



2

−  =

0


2

2
3

≈ 4644

 2

 −1

√

erf() By Property 5 of definite integrals in Section 4.2,
2

2

− , so







2

−  =

√
√


erf() − erf() =
2
2

1
2

√
 [erf() − erf()].

al



1
.
2

e

87.

CHAPTER 6 INVERSE FUNCTIONS

0

99. The rate is measured in liters per minute. Integrating from  = 0 minutes to  = 60 minutes will give us the total amount of oil

rS

that leaks out (in liters) during the first hour.
 60
 60
()  = 0 100−001 
[ = −001,  = −001]
0
 −06
 −06 
 (−100 ) = −10,000  0
= −10,000(−06 − 1) ≈ 45119 ≈ 4512 liters
= 100 0
 −1 0

(4) =

Fo

101. We use Theorem 6.1.7. Note that  (0) = 3 + 0 + 0 = 4, so  −1 (4) = 0. Also  0 () = 1 +  . Therefore,

1
1
1
1
= 0
=
= .
 0 ( −1 (4))
 (0)
1 + 0
2

From the graph, it appears that  is an odd function ( is undefined for  = 0).

ot

103.

N

To prove this, we must show that  (−) = −().

1
1 − 1 1
1 − 1(−)
1 − (−1)
1 − 1

=
=
· 1 = 1
1(−)
(−1)
1
1+
1+


+1
1 + 1

1 − 1
= −()
=−
1 + 1

 (−) =

so  is an odd function.
105. Using the second law of logarithms and Equation 5, we have ln( ) = ln  − ln  =  −  = ln( −  ). Since ln is a

one-to-one function, it follows that  =  −  .
107. (a) Let  () =  − 1 − . Now  (0) = 0 − 1 = 0, and for  ≥ 0, we have  0 () =  − 1 ≥ 0. Now, since  (0) = 0 and

 is increasing on [0 ∞), () ≥ 0 for  ≥ 0 ⇒  − 1 −  ≥ 0 ⇒  ≥ 1 + .
2

2

(b) For 0 ≤  ≤ 1, 2 ≤ , so  ≤  [since  is increasing]. Hence [from (a)] 1 + 2 ≤  ≤  .
1
1
1 2

1 2
So 4 = 0 1 + 2  ≤ 0   ≤ 0   =  − 1   ⇒ 4 ≤ 0   ≤ .
3
3 c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 6.4* GENERAL LOGARITHMIC AND EXPONENTIAL FUNCTIONS

109. (a) By Exercise 107(a), the result holds for  = 1. Suppose that  ≥ 1 +  +


2
+ ··· + for  ≥ 0.
2!
!

 () is increasing on (0 ∞). So 0 ≤  implies that 0 =  (0) ≤  () =  − 1 −  − · · · −


+1
−
, and hence
!
( + 1)!

+1


2
+ for  ≥ 0. Therefore, for  ≥ 0,  ≥ 1 +  +
+ ··· + for every positive
!
( + 1)!
2!
!

integer , by mathematical induction.
(b) Taking  = 4 and  = 1 in (a), we have  = 1 ≥ 1 +
(c)  ≥ 1 +  + · · · +

+1

+
!
( + 1)!

⇒

1
2

+

1
6

+

1
24

= 27083  27.




1
1
1
+
≥
.
≥  + −1 + · · · +



!
( + 1)!
( + 1)!

e



= ∞, so lim  = ∞.
→∞ ( + 1)!
→∞ 

rS

6.4* General Logarithmic and Exponential Functions

al

But lim

1. (a)  =  ln 

(b) The domain of  () =  is R.

(c) The range of  () =  [ 6= 1] is (0 ∞).
(ii) See Figure 3.

(iii) See Figure 2.

Fo

(d) (i) See Figure 1.

3. Since  =  ln  , 4− = − ln 4 .
2

2

5. Since  =  ln  , 10 = 

ln 10

.

ot

7. (a) log5 125 = 3 since 53 = 125.

N

6
9. (a) log2 6 − log2 15 + log2 20 = log2 ( 15 ) + log2 20
6
= log2 ( 15 · 20)

(b) log3

1
1
1
= −3 since 3−3 = 3 =
.
27
3
27
[by Law 2]
[by Law 1]

= log2 8, and log2 8 = 3 since 23 = 8.
 
 100 
(b) log3 100 − log3 18 − log3 50 = log3 100 − log3 50 = log3 18·50
18
 
= log3 ( 1 ), and log3 1 = −2 since 3−2 = 1 
9
9
9

11. All of these graphs approach 0 as  → −∞, all of them pass through the point

(0 1), and all of them are increasing and approach ∞ as  → ∞. The larger the base, the faster the function increases for   0, and the faster it approaches 0 as
 → −∞.

13. (a) log12  =

283


+1
2

− ··· −
−
. Then  0 () =  − 1 −  − · · · −
≥ 0 by assumption. Hence
2!
!
( + 1)!
!

Let  () =  − 1 −  −

 ≥ 1 +  + · · · +

¤

1 ln 
=
≈ 0402430 ln 12 ln 12

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

284

¤

CHAPTER 6 INVERSE FUNCTIONS

(b) log6 1354 =
(c) log2  =

ln 1354
≈ 1454240 ln 6

ln 
≈ 1651496 ln 2

15. To graph these functions, we use log15  =

ln  ln  and log50  =
.
ln 15 ln 50

These graphs all approach −∞ as  → 0+ , and they all pass through the

point (1 0). Also, they are all increasing, and all approach ∞ as  → ∞.

The functions with larger bases increase extremely slowly, and the ones with smaller bases do so somewhat more quickly. The functions with large bases approach the -axis more closely as  → 0+ .

⇒  = 2 [since   0] and  =

6
2



6
=  and 24 = 3

⇒ 24 =

 
6 3



⇒

= 3. The function is  () = 3 · 2 .

19. (a) 2 ft = 24 in,  (24) = 242 in = 576 in = 48 ft.

e

4 = 2

6 = 1

(24) = 224 in = 224 (12 · 5280) mi ≈ 265 mi

al

17. Use  =  with the points (1 6) and (3 24).

(b) 3 ft = 36 in, so we need  such that log2  = 36 ⇔  = 236 = 68,719,476,736. In miles, this is
1 mi
1 ft
·
≈ 1,084,5877 mi.
12 in 5280 ft

21. lim (1001) = ∞ by Figure 1, since 1001  1.
→∞

→−∞

⇒  0 () = 54 + 5 ln 5

25.  () = 5 + 5
√

27.  () = 10



[where  = −2 ] = 0

√

0

⇒  () = 10
2

 √  10  ln 10
√
ln 10
 =

2 

 2   2
 2
 2
2
2
 2
4 = sec2 4 · 4 ln 4
( ) = 2 ln 4 sec2 4 · 4
⇒ 0 () = sec 2 4



N

29. () = tan(4 )

√



ot

→∞

Fo

2

23. lim 2− = lim 2

rS

68,719,476,736 in ·

31.  () = log2 (1 − 3)
33.  = 2 log10

Note:

⇒  0 () =

1

−3
3
(1 − 3) = or (1 − 3) ln 2 
(1 − 3) ln 2
(3 − 1) ln 2

√
 = 2 log10 12 = 2 ·

1
2

log10  =  log10  ⇒  0 =  ·

1
1
+ log10  · 1 =
+ log10 
 ln 10 ln 10

ln 
1
1
=
= log10 , so the answer could be written as
+ log10  = log10  + log10  = log10 . ln 10 ln 10 ln 10

35.  = 

⇒ ln  = ln 

⇒ ln  =  ln  ⇒  0  = (1) + (ln ) · 1 ⇒  0 = (1 + ln ) ⇒

 0 =  (1 + ln )
37.  =  sin 

0 = 



⇒ ln  = ln  sin 

 sin 
+ ln  cos 


⇒

0
1
= (sin ) · + (ln )(cos ) ⇒



 sin 
 0 =  sin 
+ ln  cos 


⇒

ln  = sin  ln  ⇒

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 6.4* GENERAL LOGARITHMIC AND EXPONENTIAL FUNCTIONS

39.  = (cos )

⇒ ln  = ln(cos )

¤

285

1 0
1
 =·
· (− sin ) + ln cos  · 1 ⇒

cos 

⇒ ln  =  ln cos  ⇒



 sin 
⇒  0 = (cos ) (ln cos  −  tan )
 0 =  ln cos  − cos 
1
ln tan  ⇒





sec2 
1 0
1
ln tan 
1
1
 = ·
· sec2  + ln tan  · − 2
−
⇒ 0 = 
⇒

 tan 

 tan 
2




sec2  ln tan  ln tan 
0
1
0
1 1
−
csc  sec  −
·
or  = (tan )
 = (tan )
 tan 
2



41.  = (tan )1

43.  = 10

⇒ ln  = ln(tan )1

⇒ ln  =

⇒  0 = 10 ln 10, so at (1 10), the slope of the tangent line is 101 ln 10 = 10 ln 10, and its equation is

45.



47.





1

2

10
10  = ln 10


2

=

1

101
100 − 10
90
102
−
=
=
ln 10 ln 10 ln 10 ln 10


(ln )(ln 10)
1
ln 
1
log10 
 =
 =
. Now put  = ln , so  = , and the expression becomes

 ln 10




1
1
1 1 2
=
  =
(ln )2 + .
2  + 1 ln 10 ln 10
2 ln 10

log10 

and we get
 = 1 ln 10(log10 )2 + .
Or: The substitution  = log10  gives  =
2
 ln 10



Fo



rS

al



e

 − 10 = 10 ln 10( − 1), or  = (10 ln 10) + 10(1 − ln 10).

49. Let  = sin . Then  = cos   and

3sin  cos   =

3  =

1 sin 
3
+ =
3
+ . ln 3 ln 3

0
1
 
2
5
5
2
−
−
+
ln 2 ln 5 −1 ln 5 ln 2 0
−1
0
 
 
 


1
12 15
5
2
1
1
1
−
−
−
+
−
−
−
=
ln 2 ln 5 ln 2 ln 5 ln 5 ln 2 ln 5 ln 2
0

(2 − 5 )  +



1

(5 − 2 )  =



ot

=



16
1
−
5 ln 5
2 ln 2

N

51.  =

53. We see that the graphs of  = 2 and  = 1 + 3− intersect at  ≈ 06. We

let  () = 2 − 1 − 3− and calculate  0 () = 2 ln 2 + 3− ln 3, and using the formula +1 =  −  ( )  0 ( ) (Newton’s Method), we get
1 = 06, 2 ≈ 3 ≈ 0600967. So, correct to six decimal places, the root

occurs at  = 0600967.


55.  = log10 1 +

1




⇒ 10 = 1 +

Interchange  and :  =

1


⇒

1
1
= 10 − 1 ⇒  = 
.

10 − 1

1 is the inverse function.
10 − 1

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

286

¤

CHAPTER 6

INVERSE FUNCTIONS

57. If  is the intensity of the 1989 San Francisco earthquake, then log10 () = 71

⇒

log10 (16) = log10 16 + log10 () = log10 16 + 71 ≈ 83.
59. We find  with the loudness formula from Exercise 58, substituting 0 = 10−12 and  = 50:

50 = 10 log10


10−12

respect to :  = 10 log10
0 (50) =

10 ln 10



1
10−7




0
=

⇒


10−12


⇔  = 10−7 wattm2 . Now we differentiate  with
10−12
 
 
10
1
1
1

=
= 10
. Substituting  = 10−7 , we get

(0 ) ln 10 0 ln 10 

⇔ 5 = log10

⇔ 105 =

dB
108
≈ 434 × 107
.
ln 10 wattm2 61. (a) Using a calculator or CAS, we obtain the model  =  with  ≈ 1000124369 and  ≈ 0000045145933.

(b) Use 0 () =  ln  (from Formula 4) with the values of  and  from part (a) to get 0 (004) ≈ −67063 A.

e

The result of Example 2 in Section 1.4 was −670 A.

(a)  =   =  + 

⇒ log () =  +  = log  + log 





=  =  − 



(c)  = ( ) = 

⇒ log


=  −  = log  − log 


⇒ log ( ) =  =  log 

6.5 Exponential Growth and Decay

1 

= 07944, so
= 07944 and, by Theorem 2,  () =  (0)07944 = 207944 .
 


ot

1. The relative growth rate is

Fo

(b)

 ln 

= .
 ln 


rS

65. Let log  =  and log  = . Then  =  and  = .

al

63. Using Definition 1 and the second law of exponents for  , we have  −  = ( − ) ln  =  ln  −  ln  =

N

Thus,  (6) = 207944(6) ≈ 23499 or about 235 members.
3. (a) By Theorem 2,  () =  (0) = 100 . Now  (1) = 100(1) = 420

⇒  =

420
100

⇒  = ln 42.

So  () = 100(ln 42) = 100(42) .
(b)  (3) = 100(42)3 = 74088 ≈ 7409 bacteria
(c)  = 



⇒  0 (3) =  ·  (3) = (ln 42) 100(42)3 [from part (a)] ≈ 10,632 bacteriah

(d)  () = 100(42) = 10,000 ⇒ (42) = 100 ⇒  = (ln 100)(ln 42) ≈ 32 hours

5. (a) Let the population (in millions) in the year  be  (). Since the initial time is the year 1750, we substitute  − 1750 for  in

Theorem 2, so the exponential model gives  () =  (1750)(−1750) . Then  (1800) = 980 = 790(1800−1750)
980
790

= (50)

⇒ ln 980 = 50
790

⇒ =

1
50

ln 980 ≈ 00043104. So with this model, we have
790

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

⇒

SECTION 6.5

EXPONENTIAL GROWTH AND DECAY

¤

287

 (1900) = 790(1900−1750) ≈ 1508 million, and  (1950) = 790(1950−1750) ≈ 1871 million. Both of these estimates are much too low.
(b) In this case, the exponential model gives  () =  (1850)(−1850) ln 1650 = (50) ⇒  =
1260

1
50

⇒  (1900) = 1650 = 1260(1900−1850)

⇒

ln 1650 ≈ 0005393. So with this model, we estimate
1260

 (1950) = 1260(1950−1850) ≈ 2161 million. This is still too low, but closer than the estimate of  (1950) in part (a).
(c) The exponential model gives  () =  (1900)(−1900) ln 2560 = (50) ⇒  =
1650

1
50

⇒  (1950) = 2560 = 1650(1950−1900)

⇒

ln 2560 ≈ 0008785. With this model, we estimate
1650

 (2000) = 1650(2000−1900) ≈ 3972 million. This is much too low. The discrepancy is explained by the fact that the world birth rate (average yearly number of births per person) is about the same as always, whereas the mortality rate

e

(especially the infant mortality rate) is much lower, owing mostly to advances in medical science and to the wars in the first part of the twentieth century. The exponential model assumes, among other things, that the birth and mortality rates will

(b) () = −00005 = 09


= −00005


⇒ () = (0)−00005 = −00005 .

rS

7. (a) If  = [N2 O5 ] then by Theorem 2,

al

remain constant.

⇒ −00005 = 09 ⇒ −00005 = ln 09 ⇒  = −2000 ln 09 ≈ 211 s

Fo

9. (a) If () is the mass (in mg) remaining after  years, then () = (0) = 100 .

(30) = 10030 = 1 (100) ⇒ 30 =
2

⇒  = −(ln 2)30 ⇒ () = 100−(ln 2)30 = 100 · 2−30

1
2

(b) (100) = 100 · 2−10030 ≈ 992 mg

001
⇒  = −30 lnln 2 ≈ 1993 years

ot

1
(c) 100−(ln 2)30 = 1 ⇒ −(ln 2)30 = ln 100

11. Let () be the level of radioactivity. Thus, () = (0)− and  is determined by using the half-life:

N

(5730) = 1 (0) ⇒ (0)−(5730) = 1 (0) ⇒ −5730 =
2
2

1
2

⇒ −5730 = ln 1
2

If 74% of the 14 C remains, then we know that () = 074(0) ⇒ 074 = −(ln 2)5730
=−

⇒ =−

ln 1 ln 2
2
=
.
5730
5730

⇒ ln 074 = −

 ln 2
5730

⇒

5730(ln 074)
≈ 2489 ≈ 2500 years. ln 2

13. (a) Using Newton’s Law of Cooling,



= ( −  ), we have
= ( − 75). Now let  =  − 75, so



(0) =  (0) − 75 = 185 − 75 = 110, so  is a solution of the initial-value problem  =  with (0) = 110 and by
Theorem 2 we have () = (0) = 110 .
(30) = 11030 = 150 − 75 ⇒ 30 =
45

75
110

=

15
22

⇒ =

1
30

1

15

ln 15 , so () = 110 30  ln( 22 ) and
22

15

(45) = 110 30 ln( 22 ) ≈ 62◦ F. Thus,  (45) ≈ 62 + 75 = 137◦ F. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

288

¤

CHAPTER 6 INVERSE FUNCTIONS
1

15

1

15

(b)  () = 100 ⇒ () = 25. () = 110 30  ln( 22 ) = 25 ⇒  30  ln( 22 ) =
=

15.

25
110

1
15
30  ln 22

⇒

25
= ln 110

⇒

25
30 ln 110
≈ 116 min.
15
ln 22



= ( − 20). Letting  =  − 20, we get
= , so () = (0) . (0) =  (0) − 20 = 5 − 20 = −15, so


(25) = (0)25 = −1525 , and (25) =  (25) − 20 = 10 − 20 = −10, so −1525 = −10 ⇒ 25 = 2 . Thus,
3

 2 25

  ln 2 , so () = (0) = −15(125) ln(23) . More simply, 25 =
3

⇒ () = −15 ·

3

 2 25
3

(a)  (50) = 20 + (50) = 20 − 15 ·

(25) ln

2
3

3

= 20 − 15 ·

 2 25
3

⇒ 15 ·

 2 2
3

= 20 −

 2 25
3

17. (a) Let  () be the pressure at altitude . Then  = 

20
3

= 13¯ ◦ C
3

=5 ⇒

 
   
= ln 1
⇒  = 25 ln 1 ln 2 ≈ 6774 min.
3
3
3

 2 25

8714

8714

=

1
3

 8714 

rS

8714

1
1000

 () = 1013  1000  ln( 1013 ) , so  (3000) = 10133 ln( 1013 ) ≈ 645 kPa.
6187

3

⇒

⇒  () =  (0) = 1013 .



⇒ =
 (1000) = 10131000 = 8714 ⇒ 1000 = ln 8714
1013
1

⇒  =

.

 2 5025

(b) 15 =  () = 20 + () = 20 − 15 ·

2
3

e

 =

1
25

al

 
25 = ln 2 and  =
3

ln

1013

⇒

N

ot

Fo

(b)  (6187) = 1013  1000 ln( 1013 ) ≈ 399 kPa

 
19. (a) Using  = 0 1 + with 0 = 3000,  = 005, and  = 5, we have:

1·5

= $382884
(i) Annually:  = 1;
 = 3000 1 + 005
1


005 2·5
= $384025
(ii) Semiannually:  = 2;
 = 3000 1 + 2
12·5

(iii) Monthly:  = 12;
 = 3000 1 + 005
= $385008
12


005 52·5
= $385161
(iv) Weekly:  = 52;
 = 3000 1 + 52

365·5
= $385201
(v) Daily:  = 365;
 = 3000 1 + 005
365
(vi) Continuously:

 = 3000(005)5 = $385208

(b)  = 005 and (0) = 3000.

6.6 Inverse Trigonometric Functions
1. (a) sin−1 (05) =


6

since sin  = 05.
6

(b) cos−1 (−1) =  since cos  = −1 and  is in [0 ].
3. (a) csc−1

(b) sin−1

√
2=


4

since csc  =
4

√
2.

1
√
2


4

since sin  =
4

1
√
2

=

and


4



is in −    .
2 2

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

 2 125
3

⇒

SECTION 6.6

INVERSE TRIGONOMETRIC FUNCTIONS

5. (a) In general, tan(arctan ) =  for any real number . Thus, tan(arctan 10) = 10.





(b) sin−1 sin 7 = sin−1 sin  = sin−1
3
3

[Recall that 7 =
3
 
−1 2
.
7. Let  = sin
3


3

√
3
2

=


3

since sin  =
3

√

3
2

and


3

+ 2 and the sine function is periodic with period 2.]



is in −    .
2 2


 
2
Then tan sin−1 2 = tan  = √ .
3
5

9. Let  = tan−1

√
2. Then

11. Let  = sin−1 . Then −  ≤  ≤
2


2

al

e

√

 √ 
√ 

1
2 2
2
√
=
sin 2 tan−1 2 = sin(2) = 2 sin = 2 √
3
3
3

⇒ cos  ≥ 0, so cos(sin−1 ) = cos  =

15.

Fo


.
sin(tan−1 ) = sin  = √
1 + 2

rS

13. Let  = tan−1 . Then tan  = , so from the triangle we see that


√
1 − sin2  = 1 − 2 .

The graph of sin−1  is the reflection of the graph of

N

ot

sin  about the line  = .

17. Let  = cos−1 . Then cos  =  and 0 ≤  ≤ 

⇒ − sin 


=1 ⇒



1
1
1
=−
= −
. [Note that sin  ≥ 0 for 0 ≤  ≤ .]
= −√

sin 
1 − 2
1 − cos2 
19. Let  = cot−1 . Then cot  = 

21. Let  = csc−1 . Then csc  = 

⇒ − csc2 


=1 ⇒


⇒ − csc  cot 


1
1
1
=− 2 =−
=−
.

csc 
1 + cot2 
1 + 2


=1 ⇒


1
1

1

=−
=−
. Note that cot  ≥ 0 on the domain of csc−1 .
=− √
2−1
 csc  cot 
 2 − 1 csc  csc c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

289

290

¤

CHAPTER 6 INVERSE FUNCTIONS

23.  = (tan−1 )2

⇒  0 = 2(tan−1 )1 ·

25.  = sin−1 (2 + 1)


1
2 tan−1 
(tan−1 ) = 2 tan−1  ·
=
2

1+
1 + 2

⇒

1
1

2
1
= √
·
0 = 
·2= √
(2 + 1) = 
−42 − 4
−2 − 
1 − (2 + 1)2 
1 − (42 + 4 + 1)

27.  =  sin−1  +

√
1 − 2

⇒

1
1


+ (sin−1 )(1) + (1 − 2 )−12 (−2) = √
+ sin−1  − √
= sin−1 
0 =  · √
2
2
2
1−
1−
1 − 2

⇒ 0 =

31.  = arctan(cos )

1 sin 
(− sin ) = −
1 + (cos )2
1 + cos2 

33. () = cot−1 () + cot−1 (1)



1
1
 1
2
1
1
1
1
−
·
− 2
+ 2
=−
· − 2 =−
= 0.
1 + 2
1 + (1)2  
1 + 2
 +1

1 + 2
 +1



 +  cos 
⇒
35.  = arccos
 +  cos 
0 = − 

1

1−



 +  cos 
 +  cos 

2

3

for   0 and () = for   0.
2
2

Fo

Note that this makes sense because () =

rS

0 () = −

⇒

e

1
 2
22
( ) = − √
⇒ 0 = − 
·
1 − 4
1 − (2 )2 

al

29.  = cos−1 (2 )

( +  cos )(− sin ) − ( +  cos )(− sin )
( +  cos )2

ot

(2 − 2 ) sin 
1
= √
2 + 2 cos2  − 2 − 2 cos2  | +  cos |

N

√
(2 − 2 ) sin 
1
2 − 2 sin 
√
= √
=
| +  cos | |sin |
2 − 2 1 − cos2  | +  cos |
But 0 ≤  ≤ , so |sin | = sin . Also     0 ⇒  cos  ≥ −  −, so  +  cos   0. Thus 0 =
37. () = cos−1 (3 − 2)

√
2 − 2
.
 +  cos 

1
2
⇒ 0 () = − 
(−2) = 
.
2
1 − (3 − 2)2
1 − (3 − 2)

Domain() = { | −1 ≤ 3 − 2 ≤ 1} = { | −4 ≤ −2 ≤ −2} = { | 2 ≥  ≥ 1} = [1 2].

 

Domain(0 ) =  | 1 − (3 − 2)2  0 =  | (3 − 2)2  1 = { | |3 − 2|  1}

= { | −1  3 − 2  1} = { | −4  −2  −2} = { | 2    1} = (1 2)

39. () =  sin−1



4
 
 0 (2) = sin−1 1 =
2

+

6

√
16 − 2

⇒  0 () = sin−1


4

+





⇒
−√
= sin−1
4
16 − 2
4 1 − (4)2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 6.6

41.  () =

INVERSE TRIGONOMETRIC FUNCTIONS

¤

√
√
−12
1
1
 arcsin 
1 − 2
1 − 2 arcsin  ⇒  0 () = 1 − 2 · √
+ arcsin  ·
(−2) = 1 − √
2
2
1−
1 − 2
Note that  0 = 0 where the graph of  has a horizontal tangent. Also note that  0 is negative when  is decreasing and  0 is positive when  is increasing. 43.

lim sin−1  = sin−1 (−1) = − 
2

→−1+

45. Let  =  . As  → ∞,  → ∞. lim arctan( ) = lim arctan  =
→∞

→∞


2

by (8).

2
5
and tan  =
. Since

3−
 


5
2
− tan−1
⇒
 +  +  = 180◦ = ,  =  − tan−1

3−





5
1
2
1
=−
2
 2 − 2 −



(3 − )2
5
2
1+
1+

3−

From the figure, tan  =

rS

al

e

47.

=

=0


⇒

5
2
= 2
2 + 25
 − 6 + 13

⇒ 22 + 50 = 52 − 30 + 65 ⇒

Fo

Now

2
(3 − )2
5
2
.
· 2 −
·
2 + 25 
(3 − )2 + 4 (3 − )2

√
2 − 10 + 5 = 0 ⇒  = 5 ± 2 5. We reject the root with the + sign, since it is
√
√ larger than 3.   0 for   5 − 2 5 and   0 for   5 − 2 5, so  is maximized when
√
| | =  = 5 − 2 5 ≈ 053.
   
110
,
,
= 
10 
1 − (10)2


210

 
110
(2) rads,
=  rads =
=
= 

 
  = 6
1 − (10)2
1 − (610)2


= 2 fts, sin  =

10

N

49.

⇒

ot

32 − 30 + 15 = 0

⇒  = sin−1

1
4

rads

51.  = () = sin−1 (( + 1))

A.  = { | −1 ≤ ( + 1) ≤ 1}. For   −1 we have − − 1 ≤  ≤  + 1 ⇔


2 ≥ −1 ⇔  ≥ − 1 , so  = − 1  ∞  B. Intercepts are 0 C. No symmetry
2
2





1
= lim sin−1
= sin−1 1 =  , so  =  is a HA.
D. lim sin−1
2
2
→∞
→∞
+1
1 + 1

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

291

¤

292

CHAPTER 6 INVERSE FUNCTIONS

1
1
( + 1) − 
√
 0,
E.  0 () = 
=
2
( + 1)2
( + 1) 2 + 1
1 − [( + 1)]

 so  is increasing on − 1  ∞  F. No local maximum or minimum,
2
 
 − 1 = sin−1 (−1) = −  is an absolute minimum
2
2

H.

√
√
2 + 1 + ( + 1)/ 2 + 1
( + 1)2 (2 + 1)


3 + 2
 0 on , so  is CD on − 1  ∞ .
=−
2
( + 1)2 (2 + 1)32

G.  00 () = −

53.  =  () =  − tan−1 

A.  = R B. Intercepts are 0 C.  (−) = − (), so the curve is symmetric about the

origin. D. lim ( − tan−1 ) = ∞ and lim ( − tan−1 ) = −∞, no HA.
→−∞

H.

2

e

→∞



But  () −  −  = − tan−1  +  → 0 as  → ∞, and
2
2


−1


 () −  + 2 = − tan  − 2 → 0 as  → −∞, so  =  ±

are

slant asymptotes. E.  0 () = 1 −

1

= 2
 0, so  is
2 + 1
 +1

G.  00 () =

rS

increasing on R. F. No extrema

al

2

(1 + 2 )(2) − 2 (2)
2
=
 0 ⇔   0, so
(1 + 2 )2
(1 + 2 )2

 is CU on (0, ∞), CD on (−∞, 0). IP at (0, 0).

N

ot

Fo

55.  () = arctan(cos(3 arcsin )). We use a CAS to compute  0 and  00 , and to graph  ,  0 , and  00 :

From the graph of  0 , it appears that the only maximum occurs at  = 0 and there are minima at  = ±087. From the graph of  00 , it appears that there are inflection points at  = ±052.
57.  () =

59.



√
3

1

√

3

2 + 2
1 + (1 + 2 )
1
=
=
+ 1 ⇒  () = tan−1  +  + 
2
1+
1 + 2
1 + 2

   4

√3

8

−
=8
=
 = 8 arctan  √ = 8
2
1+
3
6
6
3
1 3

. When  = 0,  = 0; when  = 1 ,  =
2
1 − 2

61. Let  = sin−1 , so  = √



0

12

sin−1 
√
 =
1 − 2



0

6

  =



2
2

6
0

=


.
6

Thus,

2
.
72

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 6.6

INVERSE TRIGONOMETRIC FUNCTIONS

¤

293

63. Let  = 1 + 2 . Then  = 2 , so




 1

1

2
 +
 = tan−1  +
= tan−1  + 1 ln|| + 
2
1 + 2
1 + 2



= tan−1  + 1 ln1 + 2  +  = tan−1  + 1 ln(1 + 2 ) +  [since 1 + 2  0].
2
2

1+
 =
1 + 2



1
, so
1 − 2

65. Let  = sin−1 . Then  = √




√
=
1 − 2 sin−1 





1
 = ln || +  = ln sin−1  + .




2
√
 =
1 − 6



71. Let  = . Then  = , so




√
=
2 − 2



73.

The integral represents the area below the curve  = sin−1  on the interval

67. Let  = 3 . Then  = 32  and

1
3

sin−1  +  =

1
3

sin−1 (3 ) + .



√
√


2 
√
. Then  = √ and
= 2 tan−1  +  = 2 tan−1  + .
=
2
1+
2 
 (1 + )




√
= sin−1  +  = sin−1 + .

1 − 2

e



=
 1 − ()2

al

69. Let  =

1

√3
=
1 − 2

 ∈ [0 1]. The bounding curves are  = sin−1  ⇔  = sin ,  = 0 and

rS

 = 1. We see that  ranges between sin−1 0 = 0 and sin−1 1 = integrate the function  = 1 − sin  between  = 0 and  =

1
0

1
2


+ arctan 1 = arctan
3

1
2

+1
3
1− 1 · 1
2 3



 2
0

(1 − sin )  =

Fo

75. (a) arctan

sin−1   =

= arctan 1 =


4


2




(b) 2 arctan 1 + arctan 1 = arctan 1 + arctan 1 + arctan 1 = arctan
3
7
3
3
7

ot


= arctan 3 + arctan 1 = arctan
4
7

3
4

+1
7
1− 3 · 1
4
7




.
2

So we have to


:
2


+ cos  − (0 + cos 0) =
2

1
3

+1
3
1− 1 · 1
3
3



= arctan 1 =


2

− 1.

+ arctan 1
7

4

2
4
2
4
√
−
= √
−
=0
2
1 − 2
1 − 2
2 1 − 2
1 − (1 − 22 )

N

77. Let  () = 2 sin−1  − cos−1 (1 − 22 ). Then  0 () = √

[since  ≥ 0] Thus  0 () = 0 for all  ∈ [0 1). Thus  () = . To find  let  = 0. Thus

2 sin−1 (0) − cos−1 (1) = 0 = . Therefore we see that  () = 2 sin−1  − cos−1 (1 − 22 ) = 0 ⇒
2 sin−1  = cos−1 (1 − 22 ).
79.  = sec−1 

⇒ sec  =  ⇒ sec  tan 


=1 ⇒



1
=
. Now tan2  = sec2  − 1 = 2 − 1, so

sec  tan 

√

 tan  = ± 2 − 1. For  ∈ 0  ,  ≥ 1, so sec  =  = || and tan  ≥ 0 ⇒
2

For  ∈


2

√

  ,  ≤ −1, so || = − and tan  = − 2 − 1 ⇒


1
1
√
= √
=
.

 2 − 1
|| 2 − 1


1
1
1
1
√
√
 =
=
=  √
=
2 −1
2 −1

sec  tan 
 − 
(−) 
|| 2 − 1

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

294

¤

CHAPTER 6

INVERSE FUNCTIONS

6.7 Hyperbolic Functions
1. (a) sinh 0 =

0
1
2 (

(b) cosh 0 = 1 (0 + 0 ) = 1 (1 + 1) = 1
2
2

− 0 ) = 0

2−
ln 2 − (ln 2 )−1
2 − 2−1
ln 2 − −ln 2
=
=
=
2
2
2
2

3. (a) sinh(ln 2) =

1
2

=

3
4

(b) sinh 2 = 1 (2 − −2 ) ≈ 362686
2
5. (a) sech 0 =

1
1
= =1 cosh 0
1

(b) cosh−1 1 = 0 because cosh 0 = 1.

7. sinh(−) = 1 [− − −(−) ] =
2

−
1
2 (

−  ) = − 1 (− −  ) = − sinh 
2

9. cosh  + sinh  = 1 ( + − ) + 1 ( − − ) =
2
2
11. sinh  cosh  + cosh  sinh  =

1
2

1
(2 )
2

= 


 


( − − ) 1 ( + − ) + 1 ( + − ) 1 ( − − )
2
2
2

e

= 1 [(+ + − − −+ − −− ) + (+ − − + −+ − −− )]
4

cosh2  sinh2 
1
−
=
sinh2  sinh2  sinh2 

rS

13. Divide both sides of the identity cosh2  − sinh2  = 1 by sinh2 :

al

= 1 (2+ − 2−− ) = 1 [+ − −(+) ] = sinh( + )
4
2

⇔ coth2  − 1 = csch2 .

15. Putting  =  in the result from Exercise 11, we have

17. tanh(ln ) =

Fo

sinh 2 = sinh( + ) = sinh  cosh  + cosh  sinh  = 2 sinh  cosh .

sinh(ln )
 − 1
(ln  − − ln  )2
 − (ln  )−1
(2 − 1)
 − −1
2 − 1
=
=
= ln 
=
= 2
= 2 cosh(ln )
(
+ − ln  )2
 + (ln  )−1
 + −1
 + 1
( + 1)
 +1

1 cosh 

1
3
= .
53
5

⇒ sech  =

N

21. sech  =

ot

19. By Exercise 9, (cosh  + sinh ) = ( ) =  = cosh  + sinh .

cosh2  − sinh2  = 1 ⇒ sinh2  = cosh2  − 1 = csch  =

1 sinh 

tanh  =

sinh  cosh 

⇒ tanh  =

1 tanh 

⇒ coth  =

3

−1 =

16
9

⇒ sinh  =

4
3

[because   0].

43
4
= .
53
5

coth  =

 5 2

1
5
= .
45
4

⇒ csch  =

23. (a) lim tanh  = lim
→∞

→∞

1
3
= .
43
4

1−0
 − − −
1 − −2
·
= lim
=
=1
→∞ 1 + −2
 + − −
1+0

 − − 
2 − 1
0−1
·  = lim 2
=
= −1
→−∞  + −
→−∞ 

+1
0+1

(b) lim tanh  = lim
→−∞

 − −
=∞
→∞
2

(c) lim sinh  = lim
→∞

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 6.7

(d) lim sinh  = lim
→−∞

→−∞

(e) lim sech  = lim
→∞

→∞

HYPERBOLIC FUNCTIONS

¤

295

 − −
= −∞
2

2
=0
 + −

 + − −
1 + −2
1+0
= 1 [Or: Use part (a)]
· − = lim
=
 − −
→∞ 
→∞ 1 − −2

1−0

(f ) lim coth  = lim
→∞

(g) lim coth  = lim

cosh 
= ∞, since sinh  → 0 through positive values and cosh  → 1. sinh 

(h) lim coth  = lim

cosh 
= −∞, since sinh  → 0 through negative values and cosh  → 1. sinh 

→0+

→0−

→0+

→0−

(i) lim csch  = lim
→−∞

→−∞

2
=0
 − −
⇒ [with cosh   0]

e

25. Let  = sinh−1 . Then sinh  =  and, by Example 1(a), cosh2  − sinh2  = 1

1 +  = 2 − 2

sinh 
2 − 1
( − − )2 
= 
·  = 2 cosh 
( + − )2 
 +1

√


⇒  = ln  + 1 + 2 .

⇒ 2 +  = 2 − 1 ⇒



1+
⇒ 2 = ln
⇒ =
1−

rS

27. (a) Let  = tanh−1 . Then  = tanh  =

al


√
√ cosh  = 1 + sinh2  = 1 + 2 . So by Exercise 9,  = sinh  + cosh  =  + 1 + 2

1+
1−

⇒ 1 +  = 2 (1 − ) ⇒ 2 =

1
2



1+
.
ln
1−

Fo

(b) Let  = tanh−1 . Then  = tanh , so from Exercise 18 we have




1 + tanh 
1+
1+
1+
=
⇒ 2 = ln
⇒  = 1 ln
.
2 =
2
1 − tanh 
1−
1−
1−
29. (a) Let  = cosh−1 . Then cosh  =  and  ≥ 0

ot


1
1
1
=
= 
= √
2
2 −1

sinh 

cosh  − 1

⇒ sinh 

[since sinh  ≥ 0 for  ≥ 0]. Or: Use Formula 4.

(b) Let  = tanh−1 . Then tanh  =  ⇒ sech2 

N


=1 ⇒



=1 ⇒


Or: Use Formula 5.


1
1
1
=
.
=
=

1 − 2 sech2 
1 − tanh2 


=1 ⇒



1
=−
. By Exercise 13,

csch  coth 

√
√
coth  = ± csch2  + 1 = ± 2 + 1. If   0, then coth   0, so coth  = 2 + 1. If   0, then coth   0,

(c) Let  = csch−1 . Then csch  =  ⇒ − csch  coth 

√

1
1
. so coth  = − 2 + 1. In either case we have
=−
=− √

csch  coth 
|| 2 + 1
(d) Let  = sech−1 . Then sech  =  ⇒ − sech  tanh 


=1 ⇒



1
1
1

=−
=−
. [Note that   0 and so tanh   0.]
=− √
2
 sech  tanh 
 1 − 2 sech  1 − sech 

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

296

¤

CHAPTER 6

INVERSE FUNCTIONS

(e) Let  = coth−1 . Then coth  =  ⇒ − csch2 


=1 ⇒


by Exercise 13.
31.  () =  sinh  − cosh 
33. () = ln(cosh )

1
1

1
=
=
=−

1 − 2 csch2 
1 − coth2 

⇒  0 () =  (sinh )0 + sinh  · 1 − sinh  =  cosh 

⇒ 0 () =

1 sinh 
(cosh )0 =
= tanh  cosh  cosh 

⇒  0 = cosh 3 · sinh 3 · 3 = 3cosh 3 sinh 3

35.  = cosh 3

37.  () = sech2 ( ) = [sech( )]2

⇒



 0 () = 2[sech( )] [sech( )]0 = 2 sech( ) − sech( ) tanh( ) ·  = −2 sech2 ( ) tanh( )

√

1
1
1
 √
1
√ = 
( ) = √
⇒ 0 = 
√ 2

−1 2 
2 ( − 1)
( ) − 1

43.  =  sinh−1 (3) −

 0 = sinh−1

√
9 + 2



⇒



13

2

+ 
+√
− √
= sinh−1
−√
= sinh−1
3
3
3
2 9 + 2
9 + 2
9 + 2
1 + (3)2

45.  = coth−1 (sec )

⇒

 sec  tan  sec  tan  sec  tan 
1
(sec ) =
=
=
1 − (sec )2 
1 − sec2 
1 − (tan2  + 1)
− tan2 
1 cos  sec 
1
=−
=−
= − csc  tan  sin  cos  sin 

N

=−
47.

e

−2 sinh 
(1 + cosh )2

41.  = cosh−1 

0 =

al

(1 + cosh )(− sinh ) − (1 − cosh )(sinh )
− sinh  − sinh  cosh  − sinh  + sinh  cosh 
=
(1 + cosh )2
(1 + cosh )2

rS

=

⇒

Fo

0 () =

1 − cosh 
1 + cosh 

ot

39. () =



1 sech2 
1 cosh2  arctan(tanh ) =
(tanh ) =
=

1 + (tanh )2 
1 + tanh2 
1 + (sinh2 ) cosh2 
=

1
1
=
[by Exercise 16] = sech 2
2
cosh 2 cosh  + sinh 
2



2
2
gets large, and from Figure 3 or Exercise 23(a), tanh





 
2


 approaches 1. Thus,  = tanh ≈
(1) =
.
2

2
2

49. As the depth  of the water gets large, the fraction

51. (a)  = 20 cosh(20) − 15

⇒  0 = 20 sinh(20) ·

1
20

= sinh(20). Since the right pole is positioned at  = 7,

7 we have  0 (7) = sinh 20 ≈ 03572. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 6.7

HYPERBOLIC FUNCTIONS

¤

297

7
(b) If  is the angle between the tangent line and the -axis, then tan  = slope of the line = sinh 20 , so


7
 = tan−1 sinh 20 ≈ 0343 rad ≈ 1966◦ . Thus, the angle between the line and the pole is  = 90◦ −  ≈ 7034◦ .

53. (a) From Exercise 52, the shape of the cable is given by  =  () =

  

cosh
. The shape is symmetric about the





 and the poles are at  = ±100. We want to find  when the lowest
-axis, so the lowest point is (0  (0)) = 0

kg-m

= 60 ⇒  = 60 = (60 m)(2 kgm)(98 ms2 ) = 1176
, or 1176 N (newtons).

s2




 · 100
100
 cosh = 60 cosh
≈ 16450 m.
The height of each pole is (100) =


60

point is 60 m, so

(b) If the tension is doubled from  to 2 , then the low point is doubled since

2
= 120. The height of the


e





 · 100
100
2 cosh = 120 cosh
≈ 16413 m, just a slight decrease.

2
120

al

poles is now  (100) =


= 60 ⇒


55. (a)  =  sinh  +  cosh 

⇒ 0 =  cosh  +  sinh  ⇒

rS

00 = 2  sinh  + 2  cosh  = 2 ( sinh  +  cosh ) = 2 

(b) From part (a), a solution of  00 = 9 is () =  sinh 3 +  cosh 3. So −4 = (0) =  sinh 0 +  cosh 0 = , so
 = −4. Now  0 () = 3 cosh 3 − 12 sinh 3 ⇒ 6 =  0 (0) = 3 ⇒  = 2, so  = 2 sinh 3 − 4 cosh 3.

Fo

57. The tangent to  = cosh  has slope 1 when  0 = sinh  = 1

Since sinh  = 1 and  = cosh  =

√ 

⇒  = sinh−1 1 = ln 1 + 2 , by Equation 3.


√
√  √ 
 
1 + sinh2 , we have cosh  = 2. The point is ln 1 + 2 , 2 .

59. Let  = cosh . Then  = sinh  , so



sinh  cosh2   =



2  = 1 3 +  =
3

1
3

cosh3  + .



cosh 
 = cosh2  − 1



cosh 
 = sinh2 

N

63.

ot

√


√
√

sinh 
√
√ and
61. Let  = . Then  =
 = sinh  · 2  = 2 cosh  +  = 2 cosh  + .
2 



1 cosh 
·
 = sinh  sinh 



coth  csch   = − csch  + 

65. Let  = 3. Then  = 3  and



4

6

1
√
 =
2 − 9



2

43


2
= cosh−1 

43

67. Let  =  . Then  =   and





1
1 +  or ln
+ .
2
1 − 



2

 

√ or = cosh−1 2 − cosh−1 4
= cosh−1 
3
43
2 − 1
43
√ 
√ 


 
√ 
√
 2

4+ 7
6+3 3
√
= ln  + 2 − 1
= ln 2 + 3 − ln
= ln
3
43
4+ 7

1
√
3  =
92 − 9



2


 =
1 − 2




= tanh−1  +  = tanh−1 ( ) + 
1 − 2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

298

¤

CHAPTER 6

INVERSE FUNCTIONS

69. (a) From the graphs, we estimate

that the two curves  = cosh 2 and  = 1 + sinh  intersect at
 = 0 and at  =  ≈ 0481.

(b) We have found the two roots of the equation cosh 2 = 1 + sinh  to be  = 0 and  =  ≈ 0481. Note from the first graph that 1 + sinh   cosh 2 on the interval (0 ), so the area between the two curves is
=


0


(1 + sinh  − cosh 2)  =  + cosh  −

= [ + cosh  −

1
2

sinh 2] − [0 + cosh 0 −

1
2

1
2

 sinh 2 0

sinh 0] ≈ 00402

gives us

= ±2

(1) and  = ±  − (2). We need to find  and . Dividing equation (1) by equation (2)
2

 
⇒ =
⇒ () 2 = ln ± 


2

2

and  = ± , so
=±

2

2

() If




1
2

  ln ±  . Solving equations (1) and (2) for  gives us


⇒ 2 = ±4 ⇒  = 2

 0, we use the + sign and obtain a cosh function, whereas if

function.

Fo

 =




 
2

al

and − , we have  =

rS

 + −

[or  sinh( + )], then

 +
   
 


=  
± −− = 2   ± − − =    ±  − − . Comparing coefficients of 
2
2
2

e

71. If  + − =  cosh( + )

In summary, if  and  have the same sign, we have  + − = 2
√

− sinh  +

1
2

  ln −  .


ot

opposite sign, then  + − = 2

√
±.



 0, we use the − sign and obtain a sinh

√

 cosh  +

1
2

 ln  , whereas, if  and  have the


6.8 Indeterminate Forms and l'Hospital's Rule

H

1. (a) lim

→

N

Note: The use of l’Hospital’s Rule is indicated by an H above the equal sign: =

 ()
0
is an indeterminate form of type .
()
0

(b) lim

 ()
= 0 because the numerator approaches 0 while the denominator becomes large.
()

(c) lim

()
= 0 because the numerator approaches a finite number while the denominator becomes large.
()

→

→

(d) If lim () = ∞ and  () → 0 through positive values, then lim
→

→

and () = 2 .] If  () → 0 through negative values, then lim

→

()
= ∞. [For example, take  = 0, () = 12 ,
 ()

()
= −∞. [For example, take  = 0, () = 12 ,
()

and  () = −2 .] If  () → 0 through both positive and negative values, then the limit might not exist. [For example, take  = 0, () = 12 , and  () = .] c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 6.8

(e) lim

→

INDETERMINATE FORMS AND L’HOSPITAL’S RULE

¤

299

()
∞
is an indeterminate form of type
.
()
∞

3. (a) When  is near , () is near 0 and () is large, so  () − () is large negative. Thus, lim [ () − ()] = −∞.
→

(b) lim [ () − ()] is an indeterminate form of type ∞ − ∞.
→

(c) When  is near , () and () are both large, so () + () is large. Thus, lim [ () + ()] = ∞.
→

5. From the graphs of  and , we see that lim  () = 0 and lim () = 0, so l’Hospital’s Rule applies.
→2

→2

0

lim

→2

lim  ()
 ()
 0 ()
 0 (2)
18
9
→2
= lim 0
=
= 0
= 4 =
→2  ()
()
lim  0 ()
 (2)
4
5
→2

→1

2 − 1
( + 1)( − 1)
+1
1+1
= lim
= lim
=
=2
→1
2 −  →1
( − 1)

1

9. This limit has the form 0 .
0

lim

→1

3 − 22 + 1 H
32 − 4
1
= lim
=−
→1
3 − 1
32
3

lim

→(2)+

rS

Note: Alternatively, we could factor and simplify.

cos  H
− sin 
= lim
= lim tan  = −∞.
1 − sin  →(2)+ − cos  →(2)+

2 − 1 H
22
2(1)
=
=2
= lim
→0 cos  sin 
1

Fo

11. This limit has the form 0 .
0

al

lim

e

7. This limit has the form 0 . We can simply factor and simplify to evaluate the limit.
0

13. This limit has the form 0 .
0

→0

15. This limit has the form 0 .
0

→2

lim

1 − sin  H
− cos  H sin 
1
= lim
= lim
=
→2 −2 sin 2
→2 −4 cos 2
1 + cos 2
4

ot

lim

17. This limit has the form

∞
.
∞

ln  H
1
2 lim √ = lim 1 −12 = lim √ = 0
→∞ 
→∞


2

→∞

→0+

N

19. lim [(ln )] = −∞ since ln  → −∞ as  → 0+ and dividing by small values of  just increases the magnitude of the

quotient (ln ). L’Hospital’s Rule does not apply.
21. This limit has the form 0 .
0

lim

→1

8 − 1 H
87
8
8
8
= lim 4 = lim 3 = (1) =
5 −1
→1 5

5 →1
5
5

23. This limit has the form 0 .
0

lim

→0

√
√
1
(1 + 2)−12 · 2 − 1 (1 − 4)−12 (−4)
1 + 2 − 1 − 4 H
2
= lim 2
→0

1


1
2
1
2
+√
= √ +√ =3
= lim √
→0
1 + 2
1 − 4
1
1

25. This limit has the form 0 . lim
0

→0

 − 1 −  H
 − 1 H

1
=
= lim
= lim
2
→0
→0 2

2
2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

300

¤

CHAPTER 6

INVERSE FUNCTIONS

27. This limit has the form 0 . lim
0

tanh  H sech 2  sech2 0
1
=
= =1
= lim
→0 sec2  tan  sec2 0
1

29. This limit has the form 0 .
0

lim

→0

√
1
sin−1  H
1 1 − 2
1
= lim
= =1
= lim √
→0
→0

1
1
1 − 2

31. This limit has the form 0 .
0

→0

lim

3 H
3 ln 3 + 3
3 ( ln 3 + 1)
 ln 3 + 1
1
= lim
= lim
=
= lim
→0
→0
3 − 1 →0
3 ln 3
3 ln 3 ln 3 ln 3

→0

33. lim

→0

 + sin 
0+0
0
=
= = 0. L’Hospital’s Rule does not apply.
 + cos 
0+1
1 lim →1

1 −  + ln  H
−1 + 1 H
−12
−1
1
=
=− 2
= lim
= lim
→1 − sin 
→1 − 2 cos 
1 + cos 
− 2 (−1)


37. This limit has the form 0 .
0

lim

 −  +  − 1 H
−1 −  H
( − 1)−2
( − 1)
=
= lim
= lim
2
→1 2( − 1)
→1
( − 1)
2
2

→1

e

35. This limit has the form 0 .
0

rS

41. This limit has the form ∞ · 0. We’ll change it to the form 0 .
0

al

cos  − 1 + 1 2 H
− sin  +  H
− cos  + 1 H sin  H cos 
1
2
= lim
= lim
= lim
= lim
=
→0
→0
→0
→0 24
→0 24
4
43
122
24

39. This limit has the form 0 . lim
0

sin() H cos()(−2 )
= lim
=  lim cos() = (1) = 
→∞
→∞
→∞
1
−12

lim  sin() = lim

→∞

lim cot 2 sin 6 = lim

→0

→0

Fo

43. This limit has the form ∞ · 0. We’ll change it to the form 0 .
0

sin 6 H
6 cos 6
6(1)
=
= lim
=3
tan 2 →0 2 sec2 2
2(1)2
2

3 H
32
3 H
3
2 = lim
2 = lim
2 = lim
2 = 0
→∞ 
→∞ 2
→∞ 2
→∞ 4

45. This limit has the form ∞ · 0. lim 3 − = lim

ot

→∞

47. This limit has the form 0 · (−∞).

1
2
ln 
1
H
= lim
=−
= cot(2) →1+ (−2) csc2 (2)
(−2)(1)2


N

lim ln  tan(2) = lim

→1+

→1+

49. This limit has the form ∞ − ∞.

lim



→1


1
−
−1
ln 



= lim

→1

H

 ln  − ( − 1) H
(1) + ln  − 1 ln 
= lim
= lim
→1 ( − 1)(1) + ln 
→1 1 − (1) + ln 
( − 1) ln 

1

2
1
1
· 2 = lim
=
=
→1 12 + 1
→1 1 + 

1+1
2

= lim
51. This limit has the form ∞ − ∞.

lim

→0+



1
1
− 

 −1



= lim

→0+

 − 1 −  H
 − 1

1
1
H
=
= lim
= lim
=
 − 1)
+  +  − 1
+  +  + 
(
0+1+1
2
→0
→0

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 6.8

INDETERMINATE FORMS AND L’HOSPITAL’S RULE

53. The limit has the form ∞ − ∞ and we will change the form to a product by factoring out .



1 ln  ln  H
= lim lim ( − ln ) = lim  1 −
= ∞ since lim
= 0.
→∞
→∞
→∞ 
→∞ 1

√

55.  = 



⇒ ln  =

lim ln  = lim

→0+

→0+

√


lim 

→0+

√
 ln , so

√
 ln  = lim

→0+

√ ln  H
1
= lim
= −2 lim  = 0 ⇒
1
−12
→0+ − −32
→0+
2

= lim ln  = 0 = 1.
→0+

57.  = (1 − 2)1

⇒ ln  =

ln(1 − 2) H
−2(1 − 2)
1
ln(1 − 2), so lim ln  = lim
= −2 ⇒
= lim
→0
→0
→0


1

lim (1 − 2)1 = lim ln  = −2 .
→0

1 ln  H
1
1 ln , so lim ln  = lim ln  = lim
= −1 ⇒
= lim
+
+ 1−
+ 1−
+ −1
1−
→1
→1
→1
→1

e

⇒ ln  =

lim 1(1−) = lim ln  = −1 =

→1+

→1+

61.  = 1

al

59.  = 1(1−)

1
.


⇒ ln  = (1) ln  ⇒
→∞

63.  = (4 + 1)

→∞

4 ln(4 + 1) H
4 + 1 = 4 ⇒
⇒ ln  = cot  ln(4 + 1), so lim ln  = lim
= lim
2
tan 
→0+
→0+
→0+ sec 

Fo

cot 

lim ln  = lim

→∞

lim 1 = lim ln  = 0 = 1

→∞

1 ln  H
= lim
=0 ⇒
→∞ 1


rS

→0

lim (4 + 1)cot  = lim ln  = 4 .
→0+

65.  = (cos )1

1 ln cos  ⇒
2

ot

→0+

2

⇒ ln  =

ln cos  H
− tan  H
− sec2 
1
= lim
= lim
=−
2

2
2
2
→0+
→0+
→0+
→0+
√
2 lim (cos )1 = lim ln  = −12 = 1 
→0+

67.

N

lim ln  = lim

⇒

→0+

From the graph, if  = 500,  ≈ 736. The limit has the form 1∞ .




2
2
⇒
Now  = 1 +
⇒ ln  =  ln 1 +




1
2
− 2
1 + 2

ln(1 + 2) H lim ln  = lim
= lim
→∞
→∞
→∞
1
−12
= 2 lim

→∞

1
= 2(1) = 2
1 + 2

⇒



2
1+
= lim ln  = 2 [≈ 739]
→∞
→∞

lim

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

301

302

¤

CHAPTER 6

INVERSE FUNCTIONS

From the graph, it appears that lim

69.

→0

We calculate lim

→0

71. lim

→∞

 ()
 0 ()
= lim 0
= 025
→0  ()
()

 ()
 − 1 H

1
= lim 3
= .
= lim
→0  + 4
→0 32 + 4
()
4

 H



H
H
H
= lim
= lim
= · · · = lim
=∞
→∞ −1
→∞ ( − 1)−2
→∞ !


1
H
= lim
= lim 1 2
2 +1
→∞ ( + 1)−12 (2)
→∞

2

73. lim √
→∞

√
2 + 1
. Repeated applications of l’Hospital’s Rule result in the




1
1
by : lim √
= lim 
= lim 
= =1
2 +1
2 2 + 12
2
→∞
→∞
→∞
1


1 + 1
D. lim − = lim
→∞

→∞

 H
1
= lim  = 0, so  = 0 is a HA.
→∞ 


al

A.  = R B. Intercepts are 0 C. No symmetry

75.  =  () = −

e

original limit or the limit of the reciprocal of the function. Another method is to try dividing the numerator and denominator

lim − = −∞

→−∞

H.

rS

E.  0 () = − − − = − (1 − )  0 ⇔   1, so  is increasing on (−∞ 1) and decreasing on (1 ∞)  F. Absolute and local maximum

2

about the origin. D.
2

A.  = R B. Intercepts are 0 C.  (−) = − (), so the curve is symmetric
2

lim − = lim

→±∞

→±∞

 H
1
= lim
2 = 0, so  = 0 is a HA.
→±∞ 2
2

ot

77.  =  () = −

Fo

value  (1) = 1 G.  00 () = − ( − 2)  0 ⇔   2, so  is CU

 on (2 ∞) and CD on (−∞ 2). IP at 2 22

2

2

E.  0 () = − − 22 − = − (1 − 22 )  0 ⇔ 2 

1
2

⇔ || 

1
√ ,
2



1
1
so  is increasing on − √2  √2

N


 



√
1
1
1
and decreasing on −∞ − √2 and √2  ∞ . F. Local maximum value  √2 = 1 2, local minimum



√
2
2
2
1 value  − √2 = −1 2 G.  00 () = −2− (1 − 22 ) − 4− = 2− (22 − 3)  0 ⇔




3
2




3
or − 3    0, so  is CU on
2
2  ∞ and

H.

 


  
 
− 3  0 and CD on −∞ − 3 and 0 3 .
2
2
2


 
IP are (0, 0) and ± 3  ± 3 −32 .
2
2

A.  = { |   −1} = (−1 ∞) B. Intercepts are 0 C. No symmetry

 ln(1 + )
D. lim [ − ln(1 + )] = ∞, so  = −1 is a VA. lim [ − ln(1 + )] = lim  1 −
= ∞,
→∞
→∞

→−1+

79.  =  () =  − ln(1 + )

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 6.8

since lim

→∞

INDETERMINATE FORMS AND L’HOSPITAL’S RULE

ln(1 + ) H
1(1 + )
= lim
= 0
→∞

1

E.  0 () = 1 −

¤

303

H.


1
=
 0 ⇔   0 since  + 1  0.
1+
1+

So  is increasing on (0 ∞) and decreasing on (−1 0).
F.  (0) = 0 is an absolute minimum.
G.  00 () = 1(1 + )2  0, so  is CU on (−1, ∞).

e

81. (a)  () = −

H

lim ln  () = lim −

→0+

→0+

ln 
, so
1

al

(b)  =  () = − . We note that ln  () = ln − = − ln  = −

1
= lim  = 0. Thus lim  () = lim ln  () = 0 = 1.
−−2
→0+
→0+
→0+

rS

(c) From the graph, it appears that there is a local and absolute maximum of about  (037) ≈ 144. To find the exact value, we
  

1
+ ln (−1) = −− (1 + ln ). This is 0 only differentiate:  () = − = − ln  ⇒  0 () = − ln  −


Fo

when 1 + ln  = 0 ⇔  = −1 . Also  0 () changes from positive to negative at −1 . So the maximum value is
 (1) = (1)−1 = 1 .

We differentiate again to get

(d)

ot

 00 () = −− (1) + (1 + ln )2 (− ) = − [(1 + ln )2 − 1]

From the graph of  00 (), it seems that  00 () changes from negative to

N

positive at  = 1, so we estimate that  has an IP at  = 1.

83. (a)  () = 1

(b) Recall that  =  ln  . lim 1 = lim (1) ln  . As  → 0+ ,
→0+

→0+

ln 
→ −∞, so 1 = (1) ln  → 0. This


indicates that there is a hole at (0 0). As  → ∞, we have the indeterminate form ∞0 . lim 1 = lim (1) ln  ,
→∞

but lim

→∞

→∞

ln  H
1
= lim
= 0, so lim 1 = 0 = 1. This indicates that  = 1 is a HA.
→∞ 1
→∞


c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

304

¤

CHAPTER 6

INVERSE FUNCTIONS

(c) Estimated maximum: (272 145). No estimated minimum. We use logarithmic differentiation to find any critical




1 − ln 
0
1 1
1
1
⇒ 0 = 1
=0 ⇒
= · + (ln ) − 2 numbers.  = 1 ⇒ ln  = lnx ⇒


 

2 ln  = 1 ⇒  = . For 0    ,  0  0 and for   , 0  0, so  () = 1 is a local maximum value. This point is approximately (27183 14447), which agrees with our estimate.
From the graph, we see that  00 () = 0 at  ≈ 058 and  ≈ 437. Since  00

(d)

changes sign at these values, they are -coordinates of inflection points.

⇒  0 () =  −  = 0 ⇔  =  ⇔  = ln ,   0.  00 () =   0, so  is CU on
  

 H


= lim
(−∞ ∞). lim ( − ) = lim 
−
= 1 . Now lim
= ∞, so 1 = ∞, regardless
→∞
→∞
→∞ 
→∞ 1


al

e

85.  () =  − 

of the value of . For  = lim ( − ),  → 0, so  is determined
→−∞

rS

by −. If   0, − → ∞, and  = ∞. If   0, − → −∞, and

 = −∞. Thus,  has an absolute minimum for   0. As  increases, the

Fo

minimum points (ln   −  ln ), get farther away from the origin.




 

 
, which is of the form 1∞ .  = 1 +
⇒ ln  =  ln 1 +
1+
, so
→∞






−2 ln(1 + ) H


=  lim
=  lim
= 
=  lim lim ln  = lim  ln 1 +
→∞
→∞
→∞
→∞ (1 + )(−12 )
→∞ 1 + 

1

 
→ 0  . lim  =  . Thus, as  → ∞,  = 0 1 +
→∞


 
 + −
1
89. lim  () = lim
−
 − −

→0+
→0+



 


  + − − 1  − −
 + − −  + − form is 0
= lim
= lim
0
( − − ) 
 − −
→0+
→0+




 +  · 1 +  −− + − · 1 −  + −−
H
= lim
 +  · 1 − [(−− ) + − · 1]
→0+

N

ot

87. First we will find lim

= lim

→0+ 

=

0
 − −
, where  = lim
2+

→0+

Thus, lim  () =
→0+

 − −
= lim
+  + − − −
→0+

0
= 0.
2+2

 − −

−
 +
+ − −





form is 0
0

H

= lim

→0+

[divide by ]

 + −
1+1
=
=2
1
1

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

⇒

SECTION 6.8

INDETERMINATE FORMS AND L’HOSPITAL’S RULE

¤

305

91. Both numerator and denominator approach 0 as  → 0, so we use l’Hospital’s Rule (and FTC1):

()
= lim lim →0 3
→0


0

sin(22) H sin(22) H
 cos(22)


= · cos 0 =
= lim
= lim
→0
→0
3
32
6
6
6

93. We see that both numerator and denominator approach 0, so we can use l’Hospital’s Rule:

lim

→

 
√
√
1
(23  − 4 )−12 (23 − 43 ) −  1 ()−23 2
23  − 4 −  3  H
3
√
= lim 2
4
→
− 1 (3 )−34 (32 )
 − 3
4
=

=

1
(23 
2

− 4 )−12 (23 − 43 ) − 1 3 (2 )−23
3
− 1 (3 )−34 (32 )
4

 
− − 1 
(4 )−12 (−3 ) − 1 3 (3 )−23
3
3
=
= 4 4 =
3 3 4 −34
3 3
− 4  ( )
−3
4

16

9







1+
1
 − 2 ln
= lim  − 2 ln
+ 1 . Let  = 1, so as  → ∞,  → 0+ .
→∞
→∞



→0+

1

1−
 − ln( + 1) H
1
1
1
1
 + 1 = lim ( + 1) = lim
= lim
− 2 ln( + 1) = lim
=


2
2
2
2
→0+
→0+
→0+
→0+ 2 ( + 1)

al

 = lim



e

95. The limit,  = lim

rS

Note: Starting the solution by factoring out  or 2 leads to a more complicated solution.
97. Since  (2) = 0, the given limit has the form 0 .
0

lim

→0

 (2 + 3) +  (2 + 5) H
 0 (2 + 3) · 3 +  0 (2 + 5) · 5
= lim
=  0 (2) · 3 +  0 (2) · 5 = 8 0 (2) = 8 · 7 = 56
→0

1

Fo

99. Since lim [ ( + ) − ( − )] =  () −  () = 0 ( is differentiable and hence continuous) and lim 2 = 0, we use
→0

l’Hospital’s Rule:

→0

 ( + ) − ( − ) H
 0 ( + )(1) −  0 ( − )(−1)
 0 () +  0 ()
2 0 ()
=
=
=  0 ()
= lim
→0
→0
2
2
2
2

ot

lim

 ( + ) −  ( − ) is the slope of the secant line between
2

N

( −   ( − )) and ( +   ( + )). As  → 0, this line gets closer to the tangent line and its slope approaches  0 ().

101. (a) We show that lim

→0

 ()
1
= 0 for every integer  ≥ 0. Let  = 2 . Then


2

 ()
−1
 H
 −1 H
!
H
= lim
= lim  = lim
= · · · = lim  = 0 ⇒
2
2 )
→0 
→0 (
→∞ 
→∞
→∞ 

lim

lim

→0

 ()
 ()
 ()
 () −  (0)
 ()
= lim  2 = lim  lim 2 = 0. Thus,  0 (0) = lim
= lim
= 0.
→0
→0
→0 
→0
→0


−0


(b) Using the Chain Rule and the Quotient Rule we see that  () () exists for  6= 0. In fact, we prove by induction that for each  ≥ 0, there is a polynomial  and a non-negative integer  with  () () =  () () for  6= 0. This is

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

306

¤

CHAPTER 6 INVERSE FUNCTIONS

true for  = 0; suppose it is true for the th derivative. Then  0 () =  ()(23 ), so


 (+1) () =  [0 () () +  ()  0 ()] −   −1  ()  () −2



=  0 () +  ()(23 ) −   −1  ()  ()−2



=  +3 0 () + 2 () −   +2  ()  ()−(2 +3)


which has the desired form.

Now we show by induction that  () (0) = 0 for all . By part (a),  0 (0) = 0. Suppose that  () (0) = 0. Then
 () () −  () (0)
 () ()
 ()  ()
 ()  ()
= lim
= lim
= lim
→0
→0
→0
→0
−0


 +1

 (+1) (0) = lim

= lim  () lim
→0

→0

 ()
=  (0) · 0 = 0
 +1

al

e

6 Review

1. (a) A function  is called a one-to-one function if it never takes on the same value twice; that is, if  (1 ) 6=  (2 ) whenever

rS

1 6= 2 . (Or,  is 1-1 if each output corresponds to only one input.)

Use the Horizontal Line Test: A function is one-to-one if and only if no horizontal line intersects its graph more

than once. defined by

Fo

(b) If  is a one-to-one function with domain  and range , then its inverse function  −1 has domain  and range  and is
 −1 () =  ⇔  () = 

for any  in . The graph of  −1 is obtained by reflecting the graph of  about the line  = .
1
 0 ( −1 ())

ot

(c) ( −1 )0 () =

2. (a) The function  () =  has domain R and range (0 ∞).

N

(b) The function  () = ln  has domain (0 ∞) and range R.
(c) The graphs are reflections of one another about the line  = . See Figure 6.3.3 or Figure 6.3*.1.
(d) log  =

ln  ln 

3. (a) The inverse sine function  () = sin−1  is defined as follows:

sin−1  = 
Its domain is −1 ≤  ≤ 1 and its range is −

⇔

sin  = 

and

−



≤≤
2
2



≤≤ .
2
2

(b) The inverse cosine function  () = cos−1  is defined as follows: cos−1  = 

⇔

Its domain is −1 ≤  ≤ 1 and its range is 0 ≤  ≤ .

cos  = 

and

0≤≤

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 6 REVIEW

¤

307

(c) The inverse tangent function  () = tan−1  is defined as follows: tan−1  = 
Its domain is R and its range is −
4. sinh  =

⇔

tan  = 

and

−




2
2



 .
2
2

 + − sinh 
 − −
 − −
, cosh  =
, tanh  =
= 
2
2 cosh 
 + −

5. (a)  = 

(b)  = 

⇒  0 = 

⇒  0 =  ln 

√
(e)  = sin−1  ⇒  0 = 1 1 − 2

(d)  = log  ⇒  0 = 1( ln )
√
(f )  = cos−1  ⇒  0 = −1 1 − 2

(g)  = tan−1  ⇒ 0 = 1(1 + 2 )

(h)  = sinh  ⇒  0 = cosh 

(i)  = cosh  ⇒ 0 = sinh 
√
(k)  = sinh−1  ⇒ 0 = 1 1 + 2

(j)  = tanh  ⇒ 0 = sech2 
√
(l)  = cosh−1  ⇒ 0 = 1 2 − 1

e

(c)  = ln  ⇒ 0 = 1

al

(m)  = tanh−1  ⇒ 0 = 1(1 − 2 )
 − 1
= 1.
→0


(b)  = lim (1 + )1
→0

(c) The differentiation formula for  = 

rS

6. (a)  is the number such that lim

[ 0 =  ln ] is simplest when  =  because ln  = 1.

7. (a)


= 


Fo

(d) The differentiation formula for  = log  [ 0 = 1( ln )] is simplest when  =  because ln  = 1.

(b) The equation in part (a) is an appropriate model for population growth, assuming that there is enough room and nutrition to support the growth.

ot

(c) If (0) = 0 , then the solution is () = 0  .

N

8. (a) See l’Hospital’s Rule and the three notes that follow it in Section 6.8.

(b) Write   as



or
.
1
1

(c) Convert the difference into a quotient using a common denominator, rationalizing, factoring, or some other method.
(d) Convert the power to a product by taking the natural logarithm of both sides of  =   or by writing   as  ln  .

1. True.
3. False.
5. True.
7. True.

If  is one-to-one, with domain R, then 

−1

( (6)) = 6 by the first cancellation equation [see (4) in Section 6.1].



For example, cos  = cos −  , so cos  is not 1-1.
2
2

The function  = ln  is increasing on (0 ∞), so if 0    , then ln   ln .

We can divide by  since  6= 0 for every . c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

308

¤

CHAPTER 6 INVERSE FUNCTIONS

9. False.

Let  = . Then (ln )6 = (ln )6 = 16 = 1, but 6 ln  = 6 ln  = 6 · 1 = 6 6= 1 = (ln )6 . What is true, however,

is that ln(6 ) = 6 ln  for   0.


(ln 10), is 0, not


11. False.

ln 10 is a constant, so its derivative,

13. False.

The “−1” is not an exponent; it is an indication of an inverse function.

15. True.

See Figure 2 in Section 6.7.

17. True.



16

2

1
.
10

16
 
16
= ln || 2 = ln 16 − ln 2 = ln
= ln 8 = ln 23 = 3 ln 2

2

(b) ( −1 )0 (3) =

1
1
1
= 0
=
 0 ( −1 (3))
 (7)
8

al

3. (a)  −1 (3) = 7 since  (7) = 3.

e

1. No.  is not 1-1 because the graph of  fails the Horizontal Line Test.

Fo

rS

5.

 = 5 − 1

7. Reflect the graph of  = ln  about the -axis to obtain

N

ot

the graph of  = − ln .

9.

 = ln 

 = 2 arctan 

 = − ln 

11. (a) 2 ln 3 = (ln 3 )2 = 32 = 9

(b) log10 25 + log10 4 = log10 (25 · 4) = log10 100 = log10 102 = 2
13. ln  =


1
3

⇔ log  =

15.  = 17

1
3

⇒  = 13



⇒ ln  = ln 17 ⇒  = ln 17 ⇒ ln  = ln(ln 17) ⇒  = ln ln 17

⇒ ln [( + 1)( − 1)] = 1 ⇒ ln(2 − 1) = ln  ⇒ 2 − 1 =  ⇒
√
2 =  + 1 ⇒  =  + 1 since ln( − 1) is defined only when   1.

17. ln( + 1) + ln( − 1) = 1

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 6 REVIEW

19. tan−1  = 1

309

⇒ tan tan−1  = tan 1 ⇒  = tan 1 (≈ 15574)
⇒  0 () = 2 ·

21.  () = 2 ln 

1
+ (ln )(2) =  + 2 ln  or (1 + 2 ln )


⇒ 0 () = tan 2 · sec2 2 · 2 = 2 sec2 (2) tan 2

23. () = tan 2

25.  = ln |sec 5 + tan 5|

⇒

1
5 sec 5 (tan 5 + sec 5)
(sec 5 tan 5 · 5 + sec2 5 · 5) =
= 5 sec 5 sec 5 + tan 5 sec 5 + tan 5

27.  =  tan−1 (4)

⇒ 0 =  ·

1
4
· 4 + tan−1 (4) · 1 =
+ tan−1 (4)
1 + (4)2
1 + 162

⇒ 0 = (2 sec )(sec  tan ) = 2 tan 

29.  = ln(sec2 ) = 2 ln |sec |

e

 0
2 (1 )0 − 1 2
2 (1 )(−12 ) − 1 (2)
−1 (1 + 2)
⇒  =
=
=
(2 )2
4
4

1
31.  = 2


0

35. () =  tan−1 

⇒  0 () =  ·

1

+ tan−1  · 1 =
+ tan−1 
1 + 2
1 + 2

1
2

Fo

⇒ 0 =  cosh(2 ) · 2 + sinh(2 ) · 1 = 22 cosh(2 ) + sinh(2 )

37.  =  sinh(2 )
39.  = ln sin  −




1
( ln ) = 3 ln  (ln 3)  · + ln  · 1 = 3 ln  (ln 3)(1 + ln )



al

⇒  0 = 3 ln  (ln 3)

33.  = 3 ln 

rS

0 =

¤

sin2  ⇒  0 =

1
· cos  − sin 

1
2

· 2 sin  · cos  = cot  − sin  cos 

 
1
1
+
= ln −1 + (ln )−1 = − ln  + (ln )−1

ln 

⇒  0 = −1 ·

1
1
1
1
+ (−1)(ln )−2 · = − −


  (ln )2

ot

41.  = ln

43.  = ln(cosh 3)

⇒  0 = (1 cosh 3)(sinh 3)(3) = 3 tanh 3

⇒ 0 = (cosh ) sinh2  − 1

N

45.  = cosh−1 (sinh )




√
tan 3

47.  = cos 

⇒

  √
0
 √
 √
 √
 0 = − sin  tan 3 ·  tan 3 = − sin  tan 3  tan 3 · 1 (tan 3)−12 · sec2 (3) · 3
2
 √
 √
−3 sin  tan 3  tan 3 sec2 (3)
√
=
2 tan 3

49.  () = ()

⇒  0 () = ()  0 ()

51.  () = ln |()|
53.  () = 2

⇒  0 () =

1 0
 0 ()
 () =
()
()

⇒  0 () = 2 ln 2 ⇒  00 () = 2 (ln 2)2

⇒ ···

⇒  () () = 2 (ln 2)

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

310

¤

CHAPTER 6 INVERSE FUNCTIONS

55. We first show it is true for  = 1:  0 () =  +  = ( + 1) . We now assume it is true for  = :

 () () = ( + ) . With this assumption, we must show it is true for  =  + 1:
  () 

 () =
[( + ) ] =  + ( + ) = [ + ( + 1)]  .
 (+1) () =



Therefore,  () () = ( + ) by mathematical induction.
57.  = (2 + )−

⇒  0 = (2 + )(−− ) + − · 1 = − [−(2 + ) + 1] = − (− − 1). At (0 2),  0 = 1(−1) = −1,

so an equation of the tangent line is  − 2 = −1( − 0), or  = − + 2.
2

59.  = [ln( + 4)]

 + 4 = 0

⇒

⇒  0 = 2[ln( + 4)]1 ·

ln( + 4)
1
·1=2 and 0 = 0 ⇔ ln( + 4) = 0 ⇔
+4
+4

 + 4 = 1 ⇔  = −3, so the tangent is horizontal at the point (−3 0).

1
4

= 1 ( + ln 4) or  = 1  + 1 (ln 4 + 1).
4
4
4

(b) The slope of the tangent at the point (  ) is


 
=  . Thus, an equation of the tangent line is
 
  = 

rS

is  −

1
4

al

Since  =  , the -coordinate is

1 when  0 =  = 1 ⇒  = ln 1 = − ln 4.
4
4
4


1 and the point of tangency is − ln 4 4 . Thus, an equation of the tangent line

e

61. (a) The line  − 4 = 1 has slope 1 . A tangent to  =  has slope
4

 −  =  ( − ). We substitute  = 0,  = 0 into this equation, since we want the line to pass through the origin: is  −  = ( − 1) or  = .

Fo

0 −  =  (0 − ) ⇔ − =  (−) ⇔  = 1. So an equation of the tangent line at the point (  ) = (1 )

63. lim −3 = 0 since −3 → −∞ as  → ∞ and lim  = 0.
→∞

→−∞

lim 2(−3) = lim  = 0

→3−

ot

65. Let  = 2( − 3). As  → 3− ,  → −∞.

69. lim

→∞

lim ln(sinh ) = lim ln  = −∞

→0+

→0+

N

67. Let  = sinh . As  → 0+ ,  → 0+ .

→−∞

(1 + 2 )2
12 + 1
0+1
=
= −1
= lim
 )2
→∞ 12 − 1
(1 − 2
0−1
lim

 − 1 H

1
= =1
= lim
→0 sec2  tan 
1

lim

4 − 1 − 4 H
44 − 4 H
164
= lim 84 = 8 · 1 = 8
= lim
= lim
2
→0
→0
→0

2
2

71. This limit has the form 0 .
0

→0

73. This limit has the form 0 .
0

→0

75. This limit has the form ∞ · 0.

 H

2 − 3  ∞
2 − 32  ∞ form = lim form ∞
∞
−2
→−∞ 
→−∞ −2−2
 H
2 − 6  ∞
−6
H
= lim form = lim
=0
∞
−2
→−∞ 4
→−∞ −8−2

lim (2 − 3 )2 = lim

→−∞

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 6 REVIEW

77. This limit has the form ∞ − ∞.

lim

→1+




1
−
−1
ln 



= lim

→1+

H

= lim

→1+



 ln  −  + 1
( − 1) ln 



H

= lim

→1+

¤

 · (1) + ln  − 1 ln 
= lim
( − 1) · (1) + ln  →1+ 1 − 1 + ln 

1
1
1
=
=
12 + 1
1+1
2

79.  =  () =  sin , − ≤  ≤ 

A.  = [− ] B. -intercept:  (0) = 0;  () = 0 ⇔ sin  = 0 ⇒

 = − 0 . C. No symmetry D. No asymptote E.  0 () =  cos  + sin  ·  =  (cos  + sin ).

e

 0 () = 0 ⇔ − cos  = sin  ⇔ −1 = tan  ⇒  = −   3 .  0 ()  0 for −     3 and  0 ()  0
4
4
4
4



  3 


3
 for −    − 4 and 4    , so  is increasing on − 4  4 and  is decreasing on − − 4 and 3   .
4
√


H.
F. Local minimum value  −  = (− 22)−4 ≈ −032 and
4
 3  √
 34 local maximum value  4 =
22 
≈ 746

al

G.  00 () =  (− sin  + cos ) + (cos  + sin ) =  (2 cos )  0 ⇒

A.  = (0 ∞) B. No -intercept; -intercept 1. C. No symmetry D. No asymptote

Fo

81.  =  () =  ln 

rS

−      and  00 ()  0 ⇒ −    −  and     , so  is
2
2
2
2


 
  

CU on − 2  2 , and  is CD on − − 2 and 2   . There are inflection



 points at −   −−2 and   2 .
2
2

[Note that the graph approaches the point (0 0) as  → 0+ .]
0

0

H.

E.  () = (1) + (ln )(1) = 1 + ln , so  () → −∞ as  → 0 and
+

 0 () → ∞ as  → ∞.  0 () = 0 ⇔ ln  = −1 ⇔  = −1 = 1.

ot

 0 ()  0 for   1, so  is decreasing on (0 1) and increasing on
(1 ∞). F. Local minimum:  (1) = −1. No local maximum.

N

G.  00 () = 1, so  00 ()  0 for   0. The graph is CU on (0 ∞) and

there is no IP.

83.  =  () = ( − 2)−

A.  = R B. -intercept:  (0) = −2; -intercept:  () = 0 ⇔  = 2

C. No symmetry D. lim

→∞

−2 H
1
= lim  = 0, so  = 0 is a HA. No VA
→∞ 


E.  0 () = ( − 2)(−− ) + − (1) = − [−( − 2) + 1] = (3 − )− .

H.

0

 ()  0 for   3, so  is increasing on (−∞ 3) and decreasing on (3 ∞).
F. Local maximum value  (3) = −3 , no local minimum value
G.  00 () = (3 − )(−− ) + − (−1) = − [−(3 − ) + (−1)]
= ( − 4)−  0

for   4, so  is CU on (4 ∞) and CD on (−∞ 4). IP at (4 2−4 ) c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

311

¤

312

CHAPTER 6 INVERSE FUNCTIONS

85. If   0, then lim  () = lim − = lim
→−∞

→−∞

→−∞

 H
1
= lim
= 0, and lim () = ∞.
→−∞ 
→∞


H

If   0, then lim  () = −∞, and lim  () = lim
→−∞

→∞

→∞

1
= 0.


If  = 0, then  () = , so lim  () = ±∞, respectively.
→±∞

So we see that  = 0 is a transitional value. We now exclude the case  = 0, since we know how the function behaves in that case. To find the maxima and minima of  , we differentiate:  () = −
0

−

 () = (−

−

)+

−

= (1 − )

⇒

. This is 0 when 1 −  = 0 ⇔  = 1. If   0 then this

represents a minimum value of  (1) = 1(), since  0 () changes from negative to positive at  = 1; and if   0, it represents a maximum value. As || increases, the maximum or minimum point gets closer to the origin. To find the inflection points, we differentiate again:  0 () = − (1 − )

 00 () = − (−) + (1 − )(−− ) = ( − 2)− . This changes sign
⇔

 = 2. So as || increases, the points of inflection get

al

when  − 2 = 0 closer to the origin.

−

() =  () = {

⇒

rS

87. () = − cos( + )
0

e

⇒

[− sin( + )] + cos( + )(−− )} = −− [ sin( + ) +  cos( + )] ⇒

() =  0 () = −{− [2 cos( + ) −  sin( + )] + [ sin( + ) +  cos( + )](−− )}

Fo

= −− [2 cos( + ) −  sin( + ) −  sin( + ) − 2 cos( + )]
= −− [(2 − 2 ) cos( + ) − 2 sin( + )] = − [(2 − 2 ) cos( + ) + 2 sin( + )]
89. (a) () = (0) = 200

⇒ (05) = 20005 = 360 ⇒ 05 = 18 ⇒ 05 = ln 18 ⇒
⇒

() = 200(ln 324) = 200(324)

ot

 = 2 ln 18 = ln(18)2 = ln 324

(b) (4) = 200(324)4 ≈ 22,040 bacteria

N

(c) 0 () = 200(324) · ln 324, so  0 (4) = 200(324)4 · ln 324 ≈ 25,910 bacteria per hour
(d) 200(324) = 10,000 ⇒
91. Let  () =

(324) = 50 ⇒  ln 324 = ln 50 ⇒  = ln 50 ln 324 ≈ 333 hours

64

=
= (1 +  )−1 , where  = 64,  = 31, and  = −07944.
1 + 31−07944
1 + 
 0 () = −(1 +  )−2 ( ) = − (1 +  )−2
 00 () = − [ − 2(1 +  )−3 ( )] + (1 +  )−2 ( − 2  )
= −2  (1 +  )−3 [−2 + (1 +  )] = −

2  (1 −  )
(1 +  )3

The population is increasing most rapidly when its graph changes from CU to CD; that is, when  00 () = 0 in this case.
 00 () = 0 ⇒  = 1 ⇒  =




1
1
ln
 



=

1


⇒  = ln

1


⇒ =

ln(1) ln(131) =
≈ 432 days. Note that

−07944






=
= , one-half the limit of  as  → ∞.
=
=
1 + (1)
1+1
2
1 + (1) ln(1)
1 + ln(1)

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CHAPTER 6 REVIEW

93. Let  = −2 2 . Then  = −4  and

1
0

313


 −2

 − 1  = − 1  0 = − 1 (−2 − 1) = 1 (1 − −2 ).
4
4
4
4

 −2

2

−2  =

¤

0

95. Let  =  , so  =   When  = 0  = 1; when  = 1  = . Thus,



1

0


 =
1 + 2





1



1
 = arctan  1 = arctan  − arctan 1 = arctan  −  .
4
2
1+

√

97. Let  = . Then  = √
2 

⇒



√

 
√  = 2




√

  = 2 +  = 2



+ .

99. Let  = 2 + 2. Then  = (2 + 2)  = 2( + 1)  and
1
2


=


1
2

101. Let  = ln(cos ). Then  =



tan  ln(cos )  = −



ln || +  =

  = − 1 2 +  = − 1 [ ln(cos )]2 + .
2
2

1−




√


1

1
4−1

113.  =



1

0



1

0

⇒

1



 ⇒  0 () =




4

1

2  =

2
2tan 
+ =
+ . ln 2 ln 2

 cos   ≤

0



√



1
0

  = 

√

1
0

=−1

√

√


   √
  1
 
√ =
 = √
= √

2
 
2 

1

4
1
1
1
1
 = ln || 1 = [ln 4 − ln 1] = ln 4

3
3
3

2
 by cylindrical shells. Let  = 2
1 + 4

⇒  = 2 . Then

   2

1



 =  tan−1  0 =  tan−1 1 − tan−1 0 = 
=
.
1 + 2
4
4

115.  () = ln  + tan−1 

⇒ (1) = ln 1 + tan−1 1 =

1
1
1
1
2
, so  0
 0 () = +
= 0
=
= .
2
 1+
4
 (1)
32
3

117.



2

 
2
1
1
1
− + 1  = − − 2 ln || +  + 
− 1  =

2



N

 =

 

2tan  sec2   =

ot

111. ave =

 =

⇒  cos  ≤ 

107. cos  ≤ 1
109.  () =

2



Fo

 



ln 2 + 2 + .

− sin 
 = − tan   ⇒ cos 

103. Let  = tan . Then  = sec2   and

105.

1
2

e



al

+1
 =
2 + 2

rS




4

⇒ 


4

= 1 [where  =  −1 ].

We find the equation of a tangent to the curve  = − , so that we can find the
- and -intercepts of this tangent, and then we can find the area of the triangle.



 −
The slope of the tangent at the point  − is given by
= −− ,


=

and so the equation of the tangent is  − − = −− ( − ) ⇔
 = − ( −  + 1).
The -intercept of this line is  = − ( − 0 + 1) = − ( + 1). To find the -intercept we set  = 0 ⇒

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314

¤

CHAPTER 6 INVERSE FUNCTIONS



− ( −  + 1) = 0 ⇒  =  + 1. So the area of the triangle is () = 1 − ( + 1) ( + 1) = 1 − ( + 1)2 . We
2
2
 −

 1 − 
0
2 −
2
1 differentiate this with respect to :  () = 2  (2)( + 1) + ( + 1)  (−1) = 2 
1 −  . This is 0 at  = ±1, and the root  = 1 gives a maximum, by the First Derivative Test. So the maximum area of the triangle is

(1) = 1 −1 (1 + 1)2 = 2−1 = 2.
2
 + 1 −  + 1 H
 + 1 ln  −  + 1 ln 
= ln  − ln  =  (−1), so  is continuous at −1.
= lim
→−1
→−1
+1
1

119. lim  () = lim
→−1

121. Using FTC1, we differentiate both sides of the given equation,


0

()  = 2 +

0

⇒ () =

−  () , and get
2 (1 + 2)
.
1 − −

N

ot

Fo

rS

al

e



 () = 2 + 22 + −  () ⇒ () 1 − − = 2 + 22



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PROBLEMS PLUS
2

2

1. Let  =  () = − . The area of the rectangle under the curve from − to  is () = 2− where  ≥ 0. We maximize


2
2
2
(): 0 () = 2− − 42 − = 2− 1 − 22 = 0 ⇒  =

for 0 ≤  

1
√
2

and 0 ()  0 for  

1
√ .
2

1
√ .
2

This gives a maximum since 0 ()  0

We next determine the points of inflection of  (). Notice that

2

1
 0 () = −2− = −(). So  00 () = −0 () and hence,  00 ()  0 for − √2   

and  

1
√ .
2

1
√
2

1 and  00 ()  0 for   − √2

1
So () changes concavity at  = ± √2 , and the two vertices of the rectangle of largest area are at the inflection

points.

 () =



10( − 2) − 2

if  − 2  0

10[−( − 2)] −  if  − 2  0
2

−2 + 10 if   2

−2 − 10 if   2

=



−2 + 10 − 20 if   2
−2 − 10 + 20 if   2

⇒

rS

0



al

() = 10| − 2| − 2 =

e

3.  () has the form () , so it will have an absolute maximum (minimum) where  has an absolute maximum (minimum).

 0 () = 0 if  = −5 or  = 5, and 0 (2) does not exist, so the critical numbers of  are −5, 2, and 5. Since 00 () = −2 for all  6= 2,  is concave downward on (−∞ 2) and (2 ∞), and  will attain its absolute maximum at one of the critical

Fo

numbers. Since (−5) = 45, (2) = −4, and (5) = 5, we see that  (−5) = 45 is the absolute maximum value of  . Also, lim () = −∞, so lim  () = lim () = 0 But  ()  0 for all , so there is no absolute minimum value of  .

→∞

→∞

5. Consider the statement that

→∞

 
( sin ) =   sin( + ). For  = 1,


ot

 
( sin ) =  sin  +  cos , and


N

 sin( + ) =  [sin  cos  + cos  sin ] =  since tan  =




⇒ sin  =





 sin  + cos  =  sin  +  cos 





and cos  = . So the statement is true for  = 1.



Assume it is true for  = . Then

+1 
   
  sin( + ) =   sin( + ) +    cos( + )
( sin ) =
+1

=   [ sin( + ) +  cos( + )]

But sin[ + ( + 1)] = sin[( + ) + ] = sin( + ) cos  + sin  cos( + ) =




sin( + ) +




cos( + ).

Hence,  sin( + ) +  cos( + ) =  sin[ + ( + 1)]. So
+1 
( sin ) =   [ sin(+ )+  cos(+)] =   [ sin( +( + 1))] = +1  [sin(+ ( + 1))].
+1
Therefore, the statement is true for all  by mathematical induction.

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315

316

¤

CHAPTER 6 PROBLEMS PLUS

7. We first show that

 0 () =



 tan−1  for   0. Let  () = tan−1  −
. Then
1 + 2
1 + 2

1
1(1 + 2 ) − (2)
(1 + 2 ) − (1 − 2 )
22
−
=
=
 0 for   0. So  () is increasing
1 + 2
(1 + 2 )2
(1 + 2 )2
(1 + 2 )2

on (0 ∞). Hence, 0   ⇒ 0 =  (0)   () = tan−1  −



. So
 tan−1  for 0  . We next show
1 + 2
1 + 2

that tan−1    for   0. Let () =  − tan−1 . Then 0 () = 1 −

1
2
=
 0. Hence, () is increasing
2
1+
1 + 2

on (0 ∞). So for 0  , 0 = (0)  () =  − tan−1 . Hence, tan−1    for   0, and we conclude that

 tan−1    for   0.
1 + 2
√
√
1 + 3  ⇒  0 () = 1 + 3  0 for   −1.
1

e

9. By the Fundamental Theorem of Calculus,  () =



→∞

+
−



= lim

11. If  = lim



1
√ .
2

rS

( −1 )0(0) = 1 0 (1) =

al

So  is increasing on (−1 ∞) and hence is one-to-one. Note that  (1) = 0, so  −1 (1) = 0 ⇒

, then  has the indeterminate form 1∞ , so

→∞

Fo

1
1




− ln( + ) − ln( − ) H
+
+
+
−
= lim ln  = lim ln
= lim  ln
= lim
→∞
→∞
→∞
→∞
−
−
1
−12
( − ) − ( + ) −2
·
( + )( − )
1



= lim

→∞

22
2
= lim
= 2
→∞ 1 − 22
2 − 2

ot

Hence, ln  = 2, so  = 2 . From the original equation, we want  = 1

⇒ 2 = 1 ⇒  = 1 .
2

N

13. Both sides of the inequality are positive, so cosh(sinh )  sinh(cosh )

⇔ cosh2 (sinh )  sinh2 (cosh ) ⇔ sinh2 (sinh ) + 1  sinh2 (cosh )
⇔ 1  [sinh(cosh ) − sinh(sinh )][sinh(cosh ) + sinh(sinh )]


 

 
 
 

 

 + −
 − −
 + −
 − −
⇔ 1  sinh
− sinh sinh + sinh
2
2
2
2

⇔ 1  [2 cosh(2) sinh(2)][2 sinh(2) cosh(2)]

[use the addition formulas and cancel]

⇔ 1  [2 sinh(2) cosh(2)][2 sinh(2) cosh(2)] ⇔ 1  sinh  sinh − , by the half-angle formula. Now both  and − are positive, and sinh    for   0, since sinh 0 = 0 and
(sinh  − )0 = cosh  − 1  0 for   0, so 1 =  −  sinh  sinh − . So, following this chain of reasoning backward, we arrive at the desired result.

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CHAPTER 6 PROBLEMS PLUS

Let  () = 2 and () = 

15.

¤

317

√
 [  0]. From the graphs of  and ,

we see that  will intersect  exactly once when  and  share a tangent line. Thus, we must have  =  and  0 =  0 at  = .
 () = () ⇒ 2 =  and So we must have 
 = 212 = 2

√
=


√
4 

⇒

√
 ≈ 3297.

 0 () =  0 () ⇒ 22 =

√ 2


=
4

√
 ()


√
2 

⇒  = 1 . From (), 2(14) = 
4

⇒ 2 =


√ .
4 


14 ⇒

10

. We find the

e

17. Suppose that the curve  =  intersects the line  = . Then 0 = 0 for some 0  0, and hence  = 0

maximum value of () = 1 ,   0, because if  is larger than the maximum value of this function, then the curve  = 


 
1
1 1
1
1
=
− 2 ln  + ·
(1 − ln ). This is 0 only where

 
2

rS

(1) ln 

al

does not intersect the line  = .  () = 
0

 = , and for 0    ,  0 ()  0, while for   ,  0 ()  0, so  has an absolute maximum of () = 1 . So if
 =  intersects  = , we must have 0   ≤ 1 . Conversely, suppose that 0   ≤ 1 . Then  ≤ , so the graph of

Fo

 =  lies below or touches the graph of  =  at  = . Also 0 = 1  0, so the graph of  =  lies above that of  =  at  = 0. Therefore, by the Intermediate Value Theorem, the graphs of  =  and  =  must intersect somewhere between

N

ot

 = 0 and  = .

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

N ot e

al

rS

Fo

7

TECHNIQUES OF INTEGRATION

7.1 Integration by Parts
1. Let  = ln ,  = 2 

1
⇒  =  ,  = 1 3 . Then by Equation 2,
3



 1 3    1 3  1 
 2
 = 1 3 ln  − 1 2  = 1 3 ln  − 1 1 3 + 
 ln   = (ln ) 3  −
3

3
3
3
3 3




= 1 3 ln  − 1 3 +  or 1 3 ln  − 1 + 
3
9
3
3

Note: A mnemonic device which is helpful for selecting  when using integration by parts is the LIATE principle of precedence for :

Logarithmic
Inverse trigonometric

e

Algebraic
Exponential

al

Trigonometric

If the integrand has several factors, then we try to choose among them a  which appears as high as possible on the list. For example, in



2 

rS

the integrand is 2 , which is the product of an algebraic function () and an exponential function (2 ). Since Algebraic appears before Exponential, we choose  = . Sometimes the integration turns out to be similar regardless of the selection of  and  , but it is advisable to refer to LIATE when in doubt. ⇒  = ,  = 1 sin 5. Then by Equation 2,
5
 1

1
1
1
 cos 5  = 5  sin 5 − 5 sin 5  = 5  sin 5 + 25 cos 5 + .

 = ,  = − 1 −3 . Then by Equation 2,
3
 1 −3

 −3
1 −3
1 −3
− − 3   = − 3 
+ 1 −3  = − 1 −3 − 1 −3 + .
  = − 3 
3
3
9
⇒

ot

5. Let  = ,  = −3 

Fo

3. Let  = ,  = cos 5 

⇒  = (2 + 2) ,  = sin . Then by Equation 2,

 = (2 + 2) cos   = (2 + 2) sin  − (2 + 2) sin  . Next let  = 2 + 2,  = sin   ⇒  = 2 ,


 = − cos , so (2 + 2) sin   = −(2 + 2) cos  − −2 cos   = −(2 + 2) cos  + 2 sin . Thus,


N

7. First let  = 2 + 2,  = cos  

 = (2 + 2) sin  + (2 + 2) cos  − 2 sin  + .



√
1 −23
1
1
3

,  = . Then
,  =  ⇒  = √
 =
3
3
 3


√
√
√
1
1
ln 3   =  ln 3  −  ·
 =  ln 3  −  + .
3
3

 √
1
Second solution: Rewrite ln 3   = 3 ln  , and apply Example 2.
 √

√
Third solution: Substitute  = 3 , to obtain ln 3   = 3 2 ln  , and apply Exercise 1.

9. Let  = ln

11. Let  = arctan 4,  = 



⇒  =

arctan 4  =  arctan 4 −



4
4
 =
,  = . Then
1 + (4)2
1 + 162

1
4
 =  arctan 4 −
1 + 162
8



32
 =  arctan 4 −
1 + 162

1
8

ln(1 + 162 ) + .

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

319

¤

320

CHAPTER 7

TECHNIQUES OF INTEGRATION

13. Let  = ,  = sec2 2 



 sec2 2  = 1  tan 2 −
2


1
2

15. First let  = (ln )2 ,  = 

tan 2  = 1  tan 2 −
2
1


⇒  = 2 ln  ·



 = (ln )2  = (ln )2 − 2  ln  ·

 = 1 ,  =  to get



tan 2. Then

1
2

⇒  = ,  =

1


1
4

ln |sec 2| + .

,  = . Then by Equation 2,

 = (ln )2 − 2

ln   =  ln  −





ln  . Next let  = ln ,  =  ⇒

 · (1)  =  ln  −



 =  ln  −  + 1 . Thus,

 = (ln )2 − 2( ln  −  + 1 ) = (ln )2 − 2 ln  + 2 + , where  = −21 .

 = 1 2 to get
2



2 cos 3 . Next let  = cos 3,  = 2 

2 cos 3  = 1 2 cos 3 +
2

 = 1 2 sin 3 − 3 2 cos 3 −
2
4

9
4





3
2

2 sin 3  = 1 2 sin 3 − 3 2 cos 3 − 9 
2
4
4

= 1 2 sin 3 − 3 2 cos 3 + 1 . Hence,  =
2
4

19. First let  =  3 ,  =  

1 =  

1 2
 (2 sin 3
13

⇒  = 3 2 ,  =  . Then 1 =

⇒ 1 = 2 , 1 =  . Then 2 =  2  − 2

2 =  . Then



  =  −



⇒

 = −3 sin 3 ,

2 sin 3 . Substituting in the previous formula gives



⇒

− 3 cos 3) + , where  =


 3   =  3  − 3

rS

13

4

2

Then

e

2 sin 3  = 1 2 sin 3 −
2


3

1 2
 .
2



4
 .
13 1

 2  . Next let 1 =  2 ,

 . Finally, let 2 = , 2 =  

⇒ 2 = ,

  =  −  + 1 . Substituting in the expression for 2 , we get

Fo

=



⇒  = 3 cos 3 ,  =

al

17. First let  = sin 3,  = 2 

2 =  2  − 2( −  + 1 ) =  2  − 2 + 2 − 21 . Substituting the last expression for 2 into 1 gives
1 =  3  − 3( 2  − 2 + 2 − 21 ) =  3  − 3 2  + 6 − 6 + , where  = 61 .
1
1
.
 ⇒  = ( · 22 + 2 · 1)  = 2 (2 + 1) ,  = −
(1 + 2)2
2(1 + 2)

ot

21. Let  = 2 ,  =

Then by Equation 2,




N

1
2
2
+
 = −
(1 + 2)2
2(1 + 2)
2

The answer could be written as
23. Let  = ,  = cos  



12

 cos   =

0

=



2 (2 + 1)
2
1
 = −
+
1 + 2
2(1 + 2)
2



2  = −

1
2
+ 2 + 
2(1 + 2)
4

2
+ .
4(2 + 1)

⇒  = ,  =

1


sin . Then

12  12

12
1
1
1
1
1
 sin  sin   =
−0−
− cos 
−


2


0
0
0

1
1
−2
1
1
+ 2 (0 − 1) =
− 2 or
2

2

22

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 7.1

INTEGRATION BY PARTS

¤

321

25. Let  = ,  = cosh  

1
0

⇒  =   = sinh . Then

1  1

1
 cosh   =  sinh  0 − 0 sinh   = (sinh 1 − sinh 0) − cosh  0 = sinh 1 − (cosh 1 − cosh 0)
= sinh 1 − cosh 1 + 1.

We can use the definitions of sinh and cosh to write the answer in terms of : sinh 1 − cosh 1 + 1 = 1 (1 − −1 ) − 1 (1 + −1 ) + 1 = −−1 + 1 = 1 − 1.
2
2
27. Let  = ln ,  = 3 

3
1

3 ln   =

1
4

4 ln 

3
1

⇒  =
−

1


,  =

3

1 3
 
1 4

=

81
4

1 4
 .
4

ln 3 − 0 −

Then
1
4

1
4

4

3

=

1

81
4

ln 3 −

1
(81
16

− 1) =

81
4


= −2  ⇒  = ,  = − 1 −2 . Then
2
2
 1
1

1




 = − 1 −2 + 1
−2  = − 1 −2 + 0 − 1 −2 = − 1 −2 − 1 −2 +
2
2
2
4
2
4
2

0
0
0

ln 3 − 5.

29. Let  = ,  =
1

0

0

+

1
2

1

34



1
4

− 3 −2 .
4

−12  =

⇒  =


6

+

1

√ 1
 34 =


6



−12 − 1  , where  = 1 − 2
2

+1−

√

3
2

cos 
,  = sin . Then sin 

=

1
6

⇒

√ 

+6−3 3 .

Fo

33. Let  = ln (sin ),  = cos  

=

1−

34

rS

0


6

=

al


⇒  = − √
,  = . Then
1 − 2
 12
 12


12

√  = 1 ·  +
=
cos−1   =  cos−1  0 +
2
3
2

31. Let  = cos−1 ,  = 

 = −2 . Thus,  =

1
4

e



cos  ln(sin )  = sin  ln(sin ) −



cos   = sin  ln(sin ) − sin  + .

Another method: Substitute  = sin , so  = cos  . Then  = ln   =  ln  −  +  (see Example 2) and so

ot

 = sin  (ln sin  − 1) + .

35. Let  = (ln )2 ,  = 4 

1

2



5
(ln )2
5

N



4 (ln )2  =

⇒  = 2
2
1

−2



1

2

ln 
5
,  =
. By (6),

5

4 ln   =
5

2
32
5 (ln 2)

−0−2



1

2

4 ln  .
5

4
1
5
 ⇒  = ,  =
.
5

25
 5
2  2 4
 5 2
 2 4
 32




32
Then ln   = ln  −
 = 25 ln 2 − 0 −
= 32 ln 2 − 125 −
25
5
25
25
125 1
1
1
1

 32
2 4
2
2
32
31
62
So 1  (ln )  = 5 (ln 2) − 2 25 ln 2 − 125 = 32 (ln 2)2 − 64 ln 2 + 125 .
5
25
Let  = ln ,  =

1
125


.

√
, so that  = 1 −12  =
2

√



1
1
√  =
. Thus, cos   = cos  (2 ) = 2  cos  . Now
2
2 

 use parts with  = ,  = cos  ,  = ,  = sin  to get  cos   =  sin  − sin   =  sin  + cos  + 1 ,
√
√
√
√

so cos   = 2 sin  + 2 cos  +  = 2  sin  + 2 cos  + .

37. Let  =

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

CHAPTER 7

TECHNIQUES OF INTEGRATION



39. Let  = 2 , so that  = 2 . Thus,

√



√
/2

 
3 cos 2  =



/2

= 1 ( sin  + cos ) −
2

41. Let  = 1 +  so that  = . Thus,

 =

1


,  = 1  2 −  to get
2


( − 1) ln   =



1
2


2



√
/2

parts with  = ,  = cos  ,  = ,  = sin  to get


 
 


1
 cos   = 1  sin  /2 − sin   =
2
2
/2

√

1
2

 
2 cos 2 · 1 (2 ) =
2

1
2





 cos  . Now use

/2



 sin  + cos  /2



 sin  + cos  = 1 ( · 0 − 1) − 1  · 1 + 0 = − 1 −
2
2
2
2 2
2


4


 ln(1 + )  = ( − 1) ln  . Now use parts with  = ln   = ( − 1) ,

1
2


 1

 2 −  ln  −
 − 1  = 1 ( − 2) ln  − 1  2 +  + 
2
2
4

= 1 (1 + )( − 1) ln(1 + ) − 1 (1 + )2 + 1 +  + ,
2
4

which can be written as 1 (2 − 1) ln(1 + ) − 1 2 + 1  +
2
4
2

3
4

+ .

⇒  = ,  = − 1 −2 . Then
2
 −2
 1 −2
−2
−2
1
1

 = − 2 
+ 2
 = − 2 
− 1 −2 + . We
4

al

43. Let  = ,  = −2 

e

322

rS

see from the graph that this is reasonable, since  has a minimum where 

changes from negative to positive. Also,  increases where  is positive and
 decreases where  is negative.
1 2
 ,
2

 = 2

√
1 + 2  ⇒  =  ,  = 2 (1 + 2 )32 .
3

Fo

45. Let  =

Then


 3√
 1 + 2  = 1 2 2 (1 + 2 )32 −
2
3

2
3

= 1 2 (1 + 2 )32 −
3

2
3

= 1 2 (1 + 2 )32 −
3

2
(1
15

2
5

(1 + 2 )32 

· 1 (1 + 2 )52 + 
2

ot

·



+ 2 )52 + 

N

We see from the graph that this is reasonable, since  increases where  is positive and  decreases where  is negative.
Note also that  is an odd function and  is an even function.
Another method: Use substitution with  = 1 + 2 to get 1 (1 + 2 )52 − 1 (1 + 2 )32 + .
5
3


1
 sin 2
1
1  = −
+ .
47. (a) Take  = 2 in Example 6 to get sin2   = − cos  sin  +
2
2
2
4


3
(b) sin4   = − 1 cos  sin3  + 3 sin2   = − 1 cos  sin3  + 3  − 16 sin 2 + .
4
4
4
8


1
−1 sin−2  . Using (6),
49. (a) From Example 6, sin   = − cos  sin−1  +


2

 2

cos  sin−1 
 − 1 2 −2 sin   = −
+
sin
 


0
0
0


 − 1 2 −2
 − 1 2 −2 sin   = sin  
= (0 − 0) +


0
0

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 7.1

(b) Using  = 3 in part (a), we have
Using  = 5 in part (a), we have

 2
0

 2
0

sin3   =

2
3

sin5   =

4
5

 2
0

 2
0

INTEGRATION BY PARTS


2
sin   = − 2 cos  0 = 2 .
3
3 sin3   =

4
5

·

2
3

=

8
.
15

(c) The formula holds for  = 1 (that is, 2 + 1 = 3) by (b). Assume it holds for some  ≥ 1. Then
 2
2 · 4 · 6 · · · · · (2)
. By Example 6, sin2+1   =
3 · 5 · 7 · · · · · (2 + 1)
0


2

sin2+3   =

0

2 + 2
2 + 3



2

sin2+1   =

0

2 + 2
2 · 4 · 6 · · · · · (2)
·
2 + 3 3 · 5 · 7 · · · · · (2 + 1)

2 · 4 · 6 · · · · · (2)[2 ( + 1)]
,
=
3 · 5 · 7 · · · · · (2 + 1)[2 ( + 1) + 1] so the formula holds for  =  + 1. By induction, the formula holds for all  ≥ 1.

tan   =



tan−2  tan2   =

=  − tan−2  .





al



tan−2  (sec2  − 1)  =

tan−2  sec2   −



tan−2  

rS

53.

⇒  = (ln )−1 (),  = . By Equation 2,



(ln )  = (ln ) − (ln )−1 () = (ln ) −  (ln )−1 .

e

51. Let  = (ln ) ,  = 

Let  = tan−2 ,  = sec2   ⇒  = ( − 2) tan−3  sec2  ,  = tan . Then, by Equation 2,
 = tan−1  − ( − 2)



tan−2  sec2  

Fo

1 = tan−1  − ( − 2)

( − 1) = tan−1 
=



tan   =

ot

Returning to the original integral,

tan−1 
−1

tan−1  
− tan−2  .
−1

55. By repeated applications of the reduction formula in Exercise 51,





(ln )3  =  (ln )3 − 3 (ln )2  = (ln )3 − 3 (ln )2 − 2 (ln )1 



=  (ln )3 − 3(ln )2 + 6 (ln )1 − 1 (ln )0 

=  (ln )3 − 3(ln )2 + 6 ln  − 6 1  =  (ln )3 − 3(ln )2 + 6 ln  − 6 + 

N



57. The curves  = 2 ln  and  = 4 ln  intersect when 2 ln  = 4 ln 
2

⇔

2

 ln  − 4 ln  = 0 ⇔ ( − 4) ln  = 0 ⇔
 = 1 or 2 [since   0]. For 1    2, 4 ln   2 ln  Thus,
2
2 area = 1 (4 ln  − 2 ln )  = 1 [(4 − 2 ) ln ] . Let  = ln ,

 = (4 − 2 )  ⇒  =

1


,  = 4 − 1 3 . Then
3

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

323

324

¤

CHAPTER 7

TECHNIQUES OF INTEGRATION

2

 area = (ln ) 4 − 1 3 1 −
3



2

= 16 ln 2 − 4 − 1 3 1 =
3
9

2

1


 2

1


 
4 − 1 3
4 − 1 2 
 = (ln 2) 16 − 0 −
3
3
3

1
 16
 64 ln 2 − 9 − 35 = 3 ln 2 − 29
9
9



16
3

1 
 and  = 2 − 2 intersect at
2

59. The curves  = arcsin

 =  ≈ −175119 and  =  ≈ 117210. From the figure, the area

bounded by the curves is given by
 

 
  

 =  [(2 − 2 ) − arcsin 1  ]  = 2 − 1 3  −  arcsin 1  .
2
3
2

 
1
1
Let  = arcsin 1  ,  =  ⇒  = 
2
 1 2 · 2 ,  = .
1 − 2
Then

e



 
   



1
1 3


−
 arcsin
−

 = 2 − 

3
2
 2


1 − 1 2 
4

al




 
= 2 − 1 3 −  arcsin 1  − 2 1 − 1 2 ≈ 399926
3
2
4


1
0

2 cos(2) . Let  = ,  = cos(2)  ⇒  = ,  =

2


sin(2).

rS

61.  =





  1
  1
2
2
8
2 1   
8
2
 = 2
− 0 − 4 − cos
− 2 · sin = 4 + (0 − 1) = 4 − .
 = 2  sin

2
 0
2


2


0
0


0

−1

2(1 − )− . Let  = 1 − ,  = −  ⇒  = − ,  = −− .

Fo

63. Volume =

0
0

0
0


 = 2 (1 − )(−− ) −1 − 2 −1 −  = 2 ( − 1)(− ) + − −1 = 2 − −1 = 2(0 + ) = 2


 
 4
1
1
 = ,  = sec2  
 ()  =
 sec2  
 = ,  = tan 
− 
4 − 0 0



4  4
4 
√ 
4 
4  
4 
 tan 
− ln |sec |
− ln 2
=
− tan   =
=

 4
 4
0
0
0

ot

65. ave =

4


ln

√
2 or 1 −

N

=1−

2


ln 2

67. Since ()  0 for all , the desired distance is () =

First let  = 2 ,  = − 
Next let  = ,  = − 


0

()  =


0

2 − .




⇒  = 2 ,  = −− . Then () = −2 − 0 + 2 0 − .
⇒  = ,  = −− . Then





  
 
() = −2 − + 2 −− 0 + 0 −  = −2 − + 2 −− + 0 + −− 0

= −2 − + 2(−− − − + 1) = −2 − − 2− − 2− + 2 = 2 − − (2 + 2 + 2) meters

69. For  =

4
1

 00 () , let  = ,  =  00 ()  ⇒  = ,  =  0 (). Then


4  4
 =  0 () 1 − 1  0 ()  = 4 0 (4) − 1 ·  0 (1) − [ (4) −  (1)] = 4 · 3 − 1 · 5 − (7 − 2) = 12 − 5 − 5 = 2.
We used the fact that  00 is continuous to guarantee that  exists.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 7.2 TRIGONOMETRIC INTEGRALS

¤

71. Using the formula for volumes of rotation and the figure, we see that

Volume =


0



2  −

0



2  −



[()]2  = 2  − 2  −



which gives  =  0 ()  and () = , so that  = 2  − 2  − 



[()]2 . Let  = (),




2  0 () .

Now integrate by parts with  = 2 , and  =  0 ()  ⇒  = 2 ,  =  (), and

  

 2 0
  ()  = 2  ()  −  2 ()  = 2  () − 2 () −  2  () , but  () =  and  () =  ⇒


 


 = 2  − 2  −  2  − 2  −  2  ()  =  2  () .

7.2 Trigonometric Integrals s c

The symbols = and = indicate the use of the substitutions { = sin   = cos  } and { = cos   = − sin  }, respectively.

 2
0



e

 sin2  cos2  cos   = sin2  (1 − sin2 ) cos  

s 
= 2 (1 − 2 )  = (2 − 4 )  = 1 3 − 1 5 +  = 1 sin3  −
3
5
3

sin2  cos3   =

0

5. Let  = , so  =   and

sin2 () cos5 ()  =

cos2   =
=

 2
0

1
2


+

11.

13.


0

0



1




sin3 () −

1
2

sin 2

2
0

=

2
5

sin5 () +

1
7

sin7 () + 

[half-angle identity]

1
2

 
2



+ 0 − (0 + 0) =


4

2

[cos2 (2)]2  = 0 1 (1 + cos(2 · 2)) 
[half-angle identity]
2


2
1 
[1 + 2 cos 4 + cos (4)]  = 4 0 [1 + 2 cos 4 + 1 (1 + cos 8)] 
2
0






  3
1
1 3
1
1
+ 2 cos 4 + 2 cos 8  = 4 2  + 2 sin 4 + 16 sin 8 0 = 1 3  + 0 + 0 − 0 = 3 
2
4
2
8
0

=

1
4

cos4 (2)  =

 2

1
3

+ cos 2) 



=

9.

1
(1
2

N

0

sin2  cos5   =

ot

=

 2



sin2  cos4  cos  

 s 1 
1
1
=  sin2  (1 − sin2 )2 cos   =  2 (1 − 2 )2  =  (2 − 24 + 6 ) 


1
1
2
1
=  1 3 − 2 5 + 1 7 +  = 3 sin3  − 5 sin5  + 7 sin7  + 
3
5
7
1


Fo



7.

sin5  + 

 2 sin7  cos4  cos   = 0 sin7  (1 − sin2 )2 cos  
1
1 s 1
= 0 7 (1 − 2 )2  = 0 7 (1 − 22 + 4 )  = 0 (7 − 29 + 11 ) 
1 


1 8 1 10
1
1
15 − 24 + 10
1
1 12
1
=
 −  + 
− +
−0 =
=
=
8
5
12
8
5
12
120
120
0

sin7  cos5   =

 2

1
5

al

3.



rS

1.

0

1
4

sin2  cos2   =

=

 2
0

1
4

1
(4 sin2
4

 2
0

1
2 (1

 cos2 )  =

− cos 4)  =

 1

 2 (1 − cos 2)  =

0

 2
0

1
(2 sin 
4

cos )2  =

(1 − cos 4)  =

1
8

1
4


−

 2
0

1
4



( −  cos 2)  = 1   − 1  cos 2 
2
2


 



 = ,  = cos 2 
= 1 1 2 − 1 1  sin 2 − 1 sin 2 
1
2 2
2 2
2

 sin2   =

1
2



1
8

 2

 = ,

=



= 1 2 − 1  sin 2 + 1 − 1 cos 2 +  = 1 2 − 1  sin 2 −
4
4
2
4
4
4

2

 
2
sin 4 0 = 1  =
8 2

sin 2

1
8

sin2 2 

cos 2 + 

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.


16

325

¤

326

CHAPTER 7

TECHNIQUES OF INTEGRATION

2
 

1 − sin2 
(1 − 2 )2 s √
√
cos   =

sin 



1 − 22 + 4
 = (−12 − 232 + 72 )  = 212 − 4 52 + 2 92 + 
=
5
9
12
√
2
2
= 45 12 (45 − 182 + 54 ) +  = 45 sin  (45 − 18 sin2  + 5 sin4 ) + 


15.



cos5 
√
 = sin 

17.



cos2  tan3   =

23.

sin3  c  = cos 



(1 − 2 )(−)
=


= − ln || + 1 2 +  =
2



cos  + sin 2
 = sin 



1
2

 


−1
+  


cos2  − ln |cos | + 

cos  + 2 sin  cos 
 = sin 



cos 
 + sin 

s

2 cos   =





tan  sec3   =
=



tan  sec  sec2   =

1 3

3

+ =

1
3

3

sec  + 



2 

 tan2   = (sec2  − 1)  = tan  −  + 

 tan4  sec4  (sec2  ) = tan4 (1 + tan2 )2 (sec2  )


= 4 (1 + 2 )2  = (8 + 26 + 4 ) 

tan4  sec6   =



= 1 9 + 2 7 + 1 5 +  =
9
7
5
 3

tan5  sec4   =
=

 3
0

 √3
0

tan5  (tan2  + 1) sec2   =

(7 + 5 )  =

1

8
8

+ 1 6
6

ot

0

1
 + 2 sin 


[ = sec ,  = sec  tan  ]

Fo





e

= ln || + 2 sin  +  = ln |sin | + 2 sin  + 

Or: Use the formula cot   = ln |sin | + .

25. Let  = tan . Then  = sec2  , so

27.



al

21.



rS

19.

cos4 
√
cos   = sin 

√3
0

1
9

 √3
0

=

81
8

tan9  +

2
7

tan7  +

[ = tan ,  = sec2  ]

5 (2 + 1) 
+

27
6

=

81
8

+

tan5  + 

1
5

9
2

=

81
8

+

36
8

=

117
8

N

Alternate solution:
 3
 3
 3 tan5  sec4   = 0 tan4  sec3  sec  tan   = 0 (sec2  − 1)2 sec3  sec  tan  
0
2
2
= 1 (2 − 1)2 3  [ = sec ,  = sec  tan  ] = 1 (4 − 22 + 1)3 

 
2
2 

= 1 (7 − 25 + 3 )  = 1 8 − 1 6 + 1 4 1 = 32 − 64 + 4 − 1 − 1 + 1 =
8
3
4
3
8
3
4

29.

31.




tan3  sec   =



 tan2  sec  tan   = (sec2  − 1) sec  tan  


= (2 − 1)  [ = sec ,  = sec  tan  ] = 1 3 −  +  =
3

sec3  − sec  + 





tan5   = (sec2  − 1)2 tan   = sec4  tan   − 2 sec2  tan   + tan  



= sec3  sec  tan   − 2 tan  sec2   + tan  
=

1
4

sec4  − tan2  + ln |sec | + 

33. Let  =   = sec  tan  



1
3

 sec  tan   =  sec  −



[or

1
4

117
8

sec4  − sec2  + ln |sec | +  ]

⇒  =   = sec . Then sec   =  sec  − ln |sec  + tan | + .

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

SECTION 7.2 TRIGONOMETRIC INTEGRALS

35.
37.

 2
6

 2
4

cot2   =

 2
6

  √
 √

2 
(csc2  − 1)  = − cot  −  6 = 0 −  − − 3 −  = 3 −
2
6

cot5  csc3   =
=

=

 2
4



1

√



2
√

 2

cot4  csc2  csc  cot   =

4


3

(csc2  − 1)2 csc2  csc  cot  

(2 − 1)2 2 (−)

[ = csc ,  = − csc  cot  ]

2

1

1

(6 − 24 + 2 )  =

7 − 2 5 + 1 3
7
5
3

√2
1

=

 √
√
√  
8
2− 8 2+ 2 2 − 1 −
7
5
3
7

2
5

+

22 √
15 − 42 + 35
8
120 − 168 + 70 √
=
2−
2−
=
105
105
105
105


− csc  cot  + csc2  csc  (csc  − cot )
 =
. Let  = csc  − cot  ⇒ csc  − cot  csc  − cot 

 = (− csc  cot  + csc2 ) . Then  =  = ln || = ln |csc  − cot | + .

45.





1
2 [sin(8

− 5) + sin(8 + 5)]  =

= − 1 cos 3 −
6
2b

sin 5 sin   =



1
2 [cos(5

1
26

cos 13 + 

− ) − cos(5 + )]  =

1
2





e

2a

sin 8 cos 5  =

sin 3  +

1
2



cos 4  −

1
2

rS

43.





csc   =

1
2

sin 13 



cos 6  =

1
8

sin 4 −

1
12

sin 6 + 

√  6 √
 6 
 6 √
 6 √
1 + cos 2  = 0
1 + (2 cos2  − 1)  = 0
2 cos2   = 2 0 cos2  
0
√  6
√  6
= 2 0 |cos |  = 2 0 cos  
[since cos   0 for 0 ≤  ≤ 6]

6 √ 
√
√

= 2 1 −0 = 1 2
= 2 sin 
2
2

Fo

41.



al

39.  =

0

49.





1 − tan2 
 = sec2 

 2

cos  − sin2   =



cos 2  =

1
2

sin 2 + 


 sec2   −  



 = ,  = sec2  
=  tan  − tan   − 1 2
2

 tan2   =





(sec2  − 1)  =



ot

47.

N

=  tan  − ln |sec | −

1 2

2

 = ,

 = tan 

+

In Exercises 51–54, let  () denote the integrand and  () its antiderivative (with  = 0).
51. Let  = 2 , so that  = 2 . Then





 sin2  1  = 1 1 (1 − cos 2) 
2
2
2




1
1
= 4  − 2 sin 2 +  = 1  − 1 1 · 2 sin  cos  + 
4
4 2

 sin2 (2 )  =



= 1 2 −
4

1
4

327

sin(2 ) cos(2 ) + 

We see from the graph that this is reasonable, since  increases where  is positive and  decreases where  is negative.
Note also that  is an odd function and  is an even function.

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

1
3



¤

328

TECHNIQUES OF INTEGRATION

=



=

53.



CHAPTER 7

1
6

sin 3 sin 6  =

1
2

1
2 [cos(3



− 6) − cos(3 + 6)] 

(cos 3 − cos 9) 

sin 3 −

1
18

sin 9 + 

Notice that  () = 0 whenever  has a horizontal tangent.

1
2

=

1
2

57.  =





−

0
0

4

−4

=2



sin2  cos3   =

1
2

0

−

sin2  (1 − sin2 ) cos  

2 (1 − 2 )  [where  = sin ] = 0

(cos2  − sin2 )  =

4





4

cos 2 

−4


4 
4
cos 2  = 2 1 sin 2 0 = sin 2 0
2

e

55. ave =

al

=1−0=1

It seems from the graph that

59.

 2
0

cos3   = 0, since the area below the

rS

-axis and above the graph looks about equal to the area above the axis and

2 below the graph. By Example 1, the integral is sin  − 1 sin3  0 = 0.
3

Note that due to symmetry, the integral of any odd power of sin  or cos 

61. Using disks,  =
63. Using washers,



2

 sin2   = 

Fo

between limits which differ by 2 ( any integer) is 0.



1
(1
2 2

− cos 2)  = 

1
2

1
4



sin 2 2 =   − 0 −
2


4


+0 =

2
4

ot


 4 
 (1 − sin )2 − (1 − cos )2 
0

 4 
= 0
(1 − 2 sin  + sin2 ) − (1 − 2 cos  + cos2 ) 
 4
=  0 (2 cos  − 2 sin  + sin2  − cos2 ) 

 4
=  0 (2 cos  − 2 sin  − cos 2)  =  2 sin  + 2 cos  −
√


 √

√
=
2 + 2 − 1 − (0 + 2 − 0) =  2 2 − 5
2
2

−

N

 =

1
2

4 sin 2 0



sin  cos2  . Let  = cos  ⇒  = − sin  . Then

cos 
 cos  2
1
1
1
  = −  1  3 1
= 3 (1 − cos3 ).
 = − 1
3

65.  = () =

0

67. Just note that the integrand is odd [ (−) = − ()].

Or: If  6= , calculate




−

sin  cos   =





−

1
2 [sin(

− ) + sin( + )]  =

1
2



cos( − ) cos( + )
−
−
=0
−
+
−

If  = , then the first term in each set of brackets is zero.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 7.3

69.



−

cos  cos   =

If  6= , this is equal to
If  = , we get





1
[cos(
− 2

TRIGONOMETRIC SUBSTITUTION

¤

− ) + cos( + )] .



1 sin( − ) sin( + )
+
= 0.
2
−
+
−

1
[1
− 2

+ cos( + )]  =



 1  sin( + )
 − +
=  + 0 = .
2
2( + ) −

7.3 Trigonometric Substitution
1. Let  = 2 sin , where −2 ≤  ≤ 2. Then  = 2 cos   and

3. Let  = 2 sec , where 0 ≤  


2

or  ≤  

3
2 .

al

e


√
√
4 − 2 = 4 − 4 sin2  = 4 cos2  = 2 |cos | = 2 cos .



2 cos 

1
√
csc2  
=
Thus,
 =
4
4 sin2 (2 cos )
2 4 − 2
√
1
4 − 2
= − cot  +  = −
+  [see figure]
4
4

Then  = 2 sec  tan   and

rS


√
√
2 − 4 = 4 sec2  − 4 = 4(sec2  − 1)
√
= 4 tan2  = 2 |tan | = 2 tan  for the relevant values of 

ot

Fo

 √ 2


 −4
2 tan 
 =
2 sec  tan   = 2 tan2  

2 sec 
√ 2

  
 −4
= 2 (sec2  − 1)  = 2 (tan  − ) +  = 2
− sec−1
+
2
2

√
+
= 2 − 4 − 2 sec−1
2
5. Let  = sec , so  = sec  tan  ,  =
2

1
√
 =
3 2 − 1

N



√

2



√
2 ⇒ =

3

4

 3

0


=
(2 + 2 )32



4



3

4


.
3

Then

1
 = sec2 



3

+ cos 2)  = 1  + 1 sin 2 4
2
2



√ 
√



= 1  + 1 23 −  + 1 · 1 = 1 12 + 43 − 1 =
2
3
2
4
2
2
2

=

4


2.

 sec2  
2

3

cos2  

4

1
(1
2

32

[2 (1 + tan )]
√

2
1
1
.
= 2
−0 = √

2
2 2
0

and  = 2 ⇒  =

1 sec  tan   = sec3  tan 

7. Let  =  tan , where   0 and −    
2

Thus,
 


,
4


24

+

√

3
8

−

1
4

Then  =  sec2  ,  = 0 ⇒  = 0, and  =  ⇒  =

=



0

4

 sec2  
1
= 2
3 sec3 




0

4

cos   =

4
1  sin 
2

0

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
4.

329

¤

330

CHAPTER 7

TECHNIQUES OF INTEGRATION

9. Let  = 4 tan , where −    
2

Then  = 4 sec2   and

√
√
2 + 16 = 16 tan2  + 16 = 16(tan2  + 1)
√
= 16 sec2  = 4 |sec | for the relevant values of .

= 4 sec 



√
=
2 + 16

(Since




.
2

4 sec2  
=
4 sec 



√

 2 + 16

sec   = ln |sec  + tan | + 1 = ln 
+  + 1

4
4

√

√

= ln  2 + 16 +  − ln |4| + 1 = ln 2 + 16 +  + , where  = 1 − ln 4.

√
2 + 16 +   0, we don’t need the absolute value.)


. Then  = 1 sin ,  = 1 cos  ,
2
2
2

√
2 =
2 = cos . and 1 − 4
1 − (2)



√

1 − 42  = cos  1 cos   = 1 (1 + cos 2) 
2
4


= 1  + 1 sin 2 +  = 1 ( + sin  cos ) + 
4
2
4
√
 −1

1
= 4 sin (2) + 2 1 − 42 + 

13. Let  = 3 sec , where 0 ≤  


2

or  ≤  

3
2 .

rS

al

e

11. Let 2 = sin , where −  ≤  ≤
2

Then

Fo

√
 = 3 sec  tan   and 2 − 9 = 3 tan , so

 √ 2


 −9
1
tan2 
3 tan 
3 sec  tan   =

 =
3
3

27 sec
3
sec2 


1
= 1 sin2   = 1 1 (1 − cos 2)  = 1  − 12 sin 2 +  = 1  −
3
3
2
6
6
√
   √2 − 9
2 − 9 3
1
+
+  = sec−1
−


6
3
22

15. Let  =  sin ,  =  cos  ,  = 0


0

2

sin  cos  + 

ot

 1
1
sec−1
−
6
3
6

N

=

1
6

⇒  = 0 and  = 

⇒ =


.
2

Then

√
 2
 2
2 − 2  = 0 2 sin2  ( cos )  cos   = 4 0 sin2  cos2  


 2
2
1
4 2 2
4 2 1
(2 sin  cos )  = sin 2  =
(1 − cos 4) 
= 4
2
2
4 0
4 0
0
=

4 
−
8

17. Let  = 2 − 7, so  = 2 . Then

1
4



sin 4

2
0

=



 4
4  
−0 −0 =

8
2
16


1
√
 =
2
2 − 7



1
√  =


1
2

·2


√
 +  = 2 − 7 + .

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SECTION 7.3

TRIGONOMETRIC SUBSTITUTION

19. Let  = tan , where −    
2

2

2 . Then  = sec  
√
and 1 + 2 = sec , so
 √


1 + 2 sec  sec 
 = sec2   =
(1 + tan2 ) 

tan  tan 

= (csc  + sec  tan ) 

= ln |csc  − cot | + sec  + 
[by Exercise 7.2.39]
 √
√

√
2
2

 √
 1 + 2
1
 + 1 +  +  = ln  1 +  − 1  + 1 + 2 + 

− 
= ln 




1




06

sin , so  =

cos  ,  = 0 ⇒  = 0, and  = 06 ⇒  =

3
5



 3 2

 sin2   3
9
5 cos   = 125
3 cos 
0
 2

9
9
= 125 0 1 (1 − cos 2)  = 250  −
2



9
9
= 250  − 0 − 0 = 500 
2

2
√
 =
9 − 252

2

5



2

0

1
2


.
2

Then

sin2  

sin 2

2
0

rS

al

0

3
5

e

21. Let  =

23. 5 + 4 − 2 = −(2 − 4 + 4) + 9 = −( − 2)2 + 9. Let

= 9 +
2
9
2

=

9
2

9
4

sin 2 +  = 9  + 9 (2 sin  cos ) + 
2
4
√


5 + 4 − 2
−2
9 −2 sin−1 + ·
·
+
3
2
3
3


√
−2
1
sin−1
+ ( − 2) 5 + 4 − 2 + 
3
2

N

ot
=

Fo

 − 2 = 3 sin , −  ≤  ≤  , so  = 3 cos  . Then
2
2


√
5 + 4 − 2  =
9 − ( − 2)2  =
9 − 9 sin2  3 cos  

√
9 cos2  3 cos   = 9 cos2  
=



= 9 (1 + cos 2)  = 9  + 1 sin 2 + 
2
2
2



25. 2 +  + 1 = 2 +  +

+

1
2

=

√

3
2

1
4



tan , so  =

+
√

3
2

3
4


2  √ 2
=  + 1 + 23 . Let
2

sec2   and

√
2 +  + 1 =

√

3
2

sec .

Then

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¤

331

¤

332



CHAPTER 7

TECHNIQUES OF INTEGRATION


√
 =
2 ++1

=
=

√
3
sec2  
2

√
3
tan  − 1
2
2
√
3 sec 
2



  √3
2

√
3
2

tan  −

sec  −

1
2

1
2



sec   =



√
3
2

tan  sec   −

ln |sec  + tan | + 1
 √
 2
1
ln  √3 2 +  + 1 +
2



1
2

sec  




 + 1  + 1
2
 √
√


 2

= 2 +  + 1 − 1 ln  √3
2 +  + 1 +  + 1  + 1
2
2
√
= 2 +  + 1 −

√
2 +  + 1 −
√
= 2 +  + 1 −
=

2
√
3


√
2 ln √3 − 1 ln 2 +  + 1 +  + 1 + 1
2
2
√

1
ln 2 +  + 1 +  + 1 + , where  = 1 −
2
2
1
2

1
2

2 ln √3

27. 2 + 2 = (2 + 2 + 1) − 1 = ( + 1)2 − 1. Let  + 1 = 1 sec ,




= (sec2  − 1) sec   = sec3   − sec  
1
2

sec  tan  +

1
2

=

1
2

sec  tan  −

1
2

ln |sec  + tan | − ln |sec  + tan | + 

rS

=

√ ln |sec  + tan | +  = 1 ( + 1) 2 + 2 −
2

29. Let  = 2 ,  = 2 . Then



1
2 (1

1
2

cos  · cos  



√ ln  + 1 + 2 + 2  + 



where  = sin ,  = cos  ,
√
and 1 − 2 = cos 



+ cos 2)  = 1  + 1 sin 2 +  = 1  + 1 sin  cos  + 
4
8
4
4
√
√
−1
−1
2
2
= 1 sin  + 1  1 − 2 +  = 1 sin ( ) + 1  1 − 4 + 
4
4
4
4
=

1
2



Fo

√
√


1 − 2 1  =
 1 − 4  =
2

1
2

ot



al

e

√ so  = sec  tan   and 2 + 2 = tan . Then


√
2 + 2  = tan  (sec  tan  ) = tan2  sec  

31. (a) Let  =  tan , where −    
2

Then

√
2 + 2 =  sec  and


√
 2 + 2

  sec2  



√
= sec   = ln|sec  + tan | + 1 = ln
+  + 1
=
 sec 


2 + 2
√


= ln  + 2 + 2 +  where  = 1 − ln ||

N




2.

√
(b) Let  =  sinh , so that  =  cosh   and 2 + 2 =  cosh . Then




 cosh  
√
=  +  = sinh−1 + .
=
2 + 2
 cosh 



√
2 − 1 on the interval [1 7] is
√



2 − 1
1  tan  where  = sec ,  = sec  tan  ,
√
 =
· sec  tan  
2 − 1 = tan , and  = sec−1 7


6 0 sec 




= 1 0 tan2   = 1 0 (sec2  − 1)  = 1 tan  −  0
6
6
6
√

= 1 (tan  − ) = 1 48 − sec−1 7
6
6

33. The average value of () =

1
7−1



1

7

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 7.3

35. Area of 4  =

1
2 ( cos )( sin )

= 1 2 sin  cos . Area of region   =
2

TRIGONOMETRIC SUBSTITUTION



 cos 

¤

333

√
2 − 2 .

Let  =  cos  ⇒  = − sin   for  ≤  ≤  . Then we obtain
2
√


2 − 2  =  sin  (− sin )  = −2 sin2   = − 1 2 ( − sin  cos ) + 
2
√
= − 1 2 cos−1 () + 1  2 − 2 + 
2
2
√
 2

− cos−1 () +  2 − 2  cos 


= 1 0 − (−2  +  cos   sin ) = 1 2  − 1 2 sin  cos 
2
2
2

so

area of region   =

1
2

and thus, (area of sector  ) = (area of 4 ) + (area of region  ) = 1 2 .
2
37. Use disks about the -axis:

 =



0

3





9
2 + 9

2

 = 81



0

3

1

(2 + 9)2

Thus,
 4
 = 81


4.

al

 4
 4
1
1
(1 + cos 2) 
3 sec2   = 3 cos2   = 3
2 )2
(9 sec
2
0
0
0




4
= 3  + 1 sin 2 0 = 3  + 1 − 0 = 3 2 + 3 
2
2
2
4
2
8
4

39. (a) Let  =  sin ,  =  cos  ,  = 0

rS

=3 ⇒ =

e

Let  = 3 tan , so  = 3 sec2  ,  = 0 ⇒  = 0 and

⇒  = 0 and  =  ⇒

0

0



Fo

 = sin−1 (). Then
 sin−1 ()

 
2 − 2  =
 cos  ( cos  ) = 2

sin−1 ()

cos2  

0

sin−1 ()
sin−1 ()
2 
2 
 + 1 sin 2
 + sin  cos 
=
2
2
2
0
0
0
√



 
2
2 − 2
√


= sin−1 + ·
− 0 = 1 2 sin−1 () + 1  2 − 2
2
2
2



2
2

sin−1 ()

(1 + cos 2)  =

ot

=

√
√
2 − 2  represents the area under the curve  = 2 − 2 between the vertical lines  = 0 and  = .
0

N

(b) The integral

The figure shows that this area consists of a triangular region and a sector of the circle 2 +  2 = 2 . The triangular region
√
√ has base  and height 2 − 2 , so its area is 1  2 − 2 . The sector has area 1 2  = 1 2 sin−1 ().
2
2
2

2
2
2
2 − ( − )2 ,
41. We use cylindrical shells and assume that   .  =  − ( − )
⇒ =± 

so () = 2 2 − ( − )2 and

√

 +
[where  =  − ]
 = − 2 · 2 2 − ( − )2  = − 4( + ) 2 − 2 


 √
 √ where  =  sin  ,  =  cos  
= 4 −  2 − 2  + 4 − 2 − 2  in the second integral



 2
 2
+ 4 −2 2 cos2   = − 4 (0 − 0) + 42 −2 cos2  
= 4 − 1 (2 − 2 )32
3
3
= 2

2

 2

−2

−


(1 + cos 2)  = 22  +

1
2

sin 2

2

−2

= 2 2 2

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

[continued]

¤

334

CHAPTER 7

TECHNIQUES OF INTEGRATION

Another method: Use washers instead of shells, so  = 8 integral using  =  sin .


2 −  2  as in Exercise 5.2.61(a), but evaluate the
0

43. Let the equation of the large circle be 2 +  2 = 2 . Then the equation of

the small circle is 2 + ( − )2 = 2 , where  =

√
2 − 2 is the distance

between the centers of the circles. The desired area is
√
  
 √

 = −  + 2 − 2 − 2 − 2 
√
√

 
= 2 0  + 2 − 2 − 2 − 2 

√
√
= 2 0   + 2 0 2 − 2  − 2 0 2 − 2 

The first integral is just 2 = 2

√
2 − 2 . The second integral represents the area of a quarter-circle of radius , so its value

is 1 2 . To evaluate the other integral, note that
4

   2    √2 − 2
  √
2
2 arcsin +
+ = arcsin +
2 − 2 + 
2

2 

2

2

rS

=

al

e

√



2 − 2  = 2 cos2   [ =  sin ,  =  cos  ] = 1 2 (1 + cos 2) 
2


= 1 2  + 1 sin 2 +  = 1 2 ( + sin  cos ) + 
2
2
2
Thus, the desired area is
√
√

 

 = 2 2 − 2 + 2 1 2 − 2 arcsin() +  2 − 2 0
4

√
√
√


2 − 2 + 1 2 − 2 arcsin() +  2 − 2 =  2 − 2 +  2 − 2 arcsin()
2
2

Fo

= 2

7.4 Integration of Rational Functions by Partial Fractions

3. (a)

(b)

5. (a)

ot

(b)



1 + 6
=
+
(4 − 3)(2 + 5)
4 − 3
2 + 5



10

10
=
+ 2 +
= 2
52 − 23
 (5 − 2)


5 − 2

N

1. (a)



4 + 1
4 + 1

 + 
=
+ 2 + 3 + 2
= 3 2
5 + 43
 ( + 4)



 +4


1
1


1
+
+
=
=
=
+
(2 − 9)2
[( + 3)( − 3)]2
( + 3)2 ( − 3)2
+3
( + 3)2
−3
( − 3)2
6
64
= 4 + 42 + 16 +
−4
( + 2)( − 2)

2

= 4 + 42 + 16 +

[by long division]



+
+2
−2

(b)

7.

 + 
 + 
4
 + 
+ 2
+ 2
= 2
(2 −  + 1)(2 + 2)2
 −+1
 +2
( + 2)2



4
 =
−1

 
3 + 2 +  + 1 +

1
−1



 [by division] =

1 4 1 3 1 2
 +  +  +  + ln | − 1| + 
4
3
2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 7.4

9.

INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS

¤

335

5 + 1


=
+
. Multiply both sides by (2 + 1)( − 1) to get 5 + 1 = ( − 1) + (2 + 1) ⇒
(2 + 1)( − 1)
2 + 1  − 1
5 + 1 =  −  + 2 + 

⇒ 5 + 1 = ( + 2) + (− + ).

The coefficients of  must be equal and the constant terms are also equal, so  + 2 = 5 and
− +  = 1. Adding these equations gives us 3 = 6 ⇔  = 2, and hence,  = 1. Thus,


 
5 + 1
1
2
 =
+
 = 1 ln |2 + 1| + 2 ln | − 1| + .
2
(2 + 1)( − 1)
2 + 1
−1
Another method: Substituting 1 for  in the equation 5 + 1 = ( − 1) + (2 + 1) gives 6 = 3

⇔  = 2.

Substituting − 1 for  gives − 3 = − 3  ⇔  = 1.
2
2
2
11.

2
2


=
=
+
. Multiply both sides by (2 + 1)( + 1) to get
22 + 3 + 1
(2 + 1)( + 1)
2 + 1
+1

e

2 = ( + 1) + (2 + 1). The coefficients of  must be equal and the constant terms are also equal, so  + 2 = 0 and
 +  = 2. Subtracting the second equation from the first gives  = −2, and hence,  = 4. Thus,
2
 =
22 + 3 + 1



0

1



4
2
−
2 + 1
+1



 =



1

al

0

1

4 ln |2 + 1| − 2 ln | + 1|
2

0

3
= (2 ln 3 − 2 ln 2) − 0 = 2 ln .
2

rS



Another method: Substituting −1 for  in the equation 2 = ( + 1) + (2 + 1) gives 2 = −

⇔  = −2.

Substituting − 1 for  gives 2 = 1  ⇔  = 4.
2
2
13.



15.

3 − 22 − 4
−4


−4

. Write 2
=
+ 2 +
. Multiplying both sides by 2 ( − 2) gives
=1+ 2
3 − 22
 ( − 2)
 ( − 2)


−2




 =
( − )




 =  ln | − | + 
−

Fo


 =
2 − 

ot

−4 = ( − 2) + ( − 2) + 2 . Substituting 0 for  gives −4 = −2

⇔  = −1. Equating coefficients of 2 , we get 0 =  + , so  = 1. Thus,

−4 = 4

4




4
1
2
2
1
1+ + 2 −
 =  + ln || − − ln | − 2|
 
−2

3
3

 
 7
1
2
2
= 4 + ln 4 − 2 − ln 2 − 3 + ln 3 − 3 − 0 = 6 + ln 3

3 − 22 − 4
 =
3 − 22

N



3

17.

⇔  = 2. Substituting 2 for  gives



4 2 − 7 − 12



=
+
+
( + 2)( − 3)

+2
−3

4

⇒ 42 − 7 − 12 = ( + 2)( − 3) + ( − 3) + ( + 2). Setting

 = 0 gives −12 = −6, so  = 2. Setting  = −2 gives 18 = 10, so  = 9 . Setting  = 3 gives 3 = 15, so  = 1 .
5
5
Now


2

1

4 2 − 7 − 12
 =
( + 2)( − 3)

 2
1

2
95
15
+
+

+2 −3

= 2 ln 2 +

9
5

= 2 ln 2 +

18
5

ln 4 +

1
5

ln 2 −




 = 2 ln || +

ln 1 − 2 ln 1 −

1
5

ln 2 −

9
5

9
5

ln 3 =

ln 3 −

27
5

1
5

9
5

ln | + 2| +

ln 2

ln 2 −

9
5

1
5

ln | − 3|

2
1

ln 3 = 9 (3 ln 2 − ln 3) =
5

9
5

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

ln 8
3

336

19.

¤

CHAPTER 7

TECHNIQUES OF INTEGRATION

2 + 1



+
+
=
. Multiply both sides by ( − 3)( − 2)2 to get
( − 3)( − 2)2
−3
−2
( − 2)2
2 + 1 = ( − 2)2 + ( − 3)( − 2) + ( − 3). Setting  = 2 gives 5 = −

⇔  = −5. Setting  = 3

gives 10 = . Equating coefficients of 2 gives 1 =  + , so  = −9. Thus,


 
2 + 1
5
9
5
10
 =
 = 10 ln | − 3| − 9 ln | − 2| +
−
−
+
( − 3)( − 2)2
−3
−2
( − 2)2
−2
21.

2 + 4 3 + 02
3

By long division,


+ 0 + 4

−4 + 4
3 + 4
=+ 2
. Thus,
2 + 4
 +4

+ 4
−4 + 4

e

10 = 9 − 

 = −1 Thus,


rS



10

 + 
=
+ 2
. Multiply both sides by ( − 1) 2 + 9 to get
( − 1)(2 + 9)
−1
 +9


10 =  2 + 9 + ( + )( − 1) (). Substituting 1 for  gives 10 = 10 ⇔  = 1. Substituting 0 for  gives
⇒  = 9(1) − 10 = −1. The coefficients of the 2 -terms in () must be equal, so 0 =  + 
 



Fo

23.



 
 
4
−4 + 4
4
 =
+ 2

+ 2
− 2
 +4
 +4
 + 22



1
1
1 
1
+  = 2 − 2 ln(2 + 4) + 2 tan−1
+
= 2 − 4 · ln 2 + 4 + 4 · tan−1
2
2
2
2
2
2

3 + 4
 =
2 + 4

al



− − 1
1
+ 2
−1
 +9

 


1
1
 =
− 2
− 2
−1
 +9
 +9
 
= ln| − 1| − 1 ln(2 + 9) − 1 tan−1  + 
2
3
3

10
 =
( − 1)(2 + 9)



⇒



4
4
4

 + 
= 2
=
=
+ 2
. Multiply both sides by
3 + 2 +  + 1
 ( + 1) + 1( + 1)
( + 1)(2 + 1)
+1
 +1

N

25.

ot

In the second term we used the substitution  = 2 + 9 and in the last term we used Formula 10.

( + 1)(2 + 1) to get 4 = (2 + 1) + ( + )( + 1) ⇔ 4 = 2 +  + 2 +  +  + 

⇔

4 = ( + )2 + ( + ) + ( + ). Comparing coefficients give us the following system of equations:
 +  = 0 (1)

 +  = 4 (2)

 +  = 0 (3)

Subtracting equation (1) from equation (2) gives us − +  = 4, and adding that equation to equation (3) gives us
2 = 4 ⇔  = 2, and hence  = −2 and  = 2. Thus,


 
 

−2
2 + 2
−2
2
2
4
 =
+ 2
 =
+ 2
+ 2

3 + 2 +  + 1
+1
 +1
+1
 +1
 +1
= −2 ln | + 1| + ln(2 + 1) + 2 tan−1  + .
27.




3 + 2 + 2 + 1
 + 
 + 
= 2
+ 2
. Multiply both sides by 2 + 1 2 + 2 to get
(2 + 1)(2 + 2)
 +1
 +2




3 + 2 + 2 + 1 = ( + ) 2 + 2 + ( + ) 2 + 1 ⇔ c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 7.4

INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS

¤

337


 

3 + 2 + 2 + 1 = 3 + 2 + 2 + 2 + 3 + 2 +  +  ⇔

3 + 2 + 2 + 1 = ( + )3 + ( + )2 + (2 + ) + (2 + ). Comparing coefficients gives us the following system of equations:
 +  = 1 (1)

 +  = 1 (2)

2 +  = 2 (3)

2 +  = 1 (4)

31.



1
(2 + 2) 
3
3 
 =
+
2 + 2 + 5
2
2 + 2 + 5
( + 1)2 + 4




2 
1  where  + 1 = 2,
= ln 2 + 2 + 5 + 3 and  = 2 
2
4(2 + 1)


1
3
1
3
2
−1
2
−1  + 1
+
= ln( + 2 + 5) + tan  +  = ln( + 2 + 5) + tan
2
2
2
2
2

+4
 =
2 + 2 + 5



+1
 +
2 + 2 + 5



rS



Fo

29.



1
1  2 ln  + 1 + √ tan−1 √ + .
2
2
2

al

Thus,  =

e

Subtracting equation (1) from equation (3) gives us  = 1, so  = 0. Subtracting equation (2) from equation (4) gives us

 

 3

1

 + 2 + 2 + 1
 =
+ 2
. For
, let  = 2 + 1
 = 0, so  = 1. Thus,  =
(2 + 1)(2 + 2)
2 + 1
 +2
2 + 1





1
1
1
1  2
1
so  = 2  and then
 =
 = ln || +  = ln  + 1 + . For
, use
2 + 1
2

2
2
2 + 2


√
1
1
1
−1 
Formula 10 with  = 2. So
 =
√ 2  = √ tan √ + .
2 + 2
2 +
2
2

2

1
1

 + 
=
=
+ 2
3 − 1
( − 1)(2 +  + 1)
−1
 ++1



⇒ 1 =  2 +  + 1 + ( + )( − 1).

ot

Take  = 1 to get  = 1 . Equating coefficients of 2 and then comparing the constant terms, we get 0 =
3
so  = − 1 ,  = − 2
3
3



=

1
3

1
3

1
3

N

1
 =
3 − 1

=


+2
1
 +

ln | − 1| −
−1
3
2 +  + 1


 + 12
1
(32) 
1
 − ln | − 1| −
2 ++1
3

3
( + 12)2 + 34




 2
 1 2
+ 1
−1
1
2
√  tan ln | − 1| − 6 ln  +  + 1 − 2 √
+
3
3 2


1
1 ln | − 1| − 1 ln(2 +  + 1) − √3 tan−1 √3 (2 + 1) + 
6
1
3



−1 − 2
3
3
 =
2 +  + 1

1
3

33. Let  = 4 + 42 + 3 so that  = (43 + 8)  = 4(3 + 2) ,  = 0

Then



0

1

+ , 1 =

⇒

=



1
3

3 + 2
 =
4 + 42 + 3



3

8

1




⇒  = 3, and  = 1 ⇒  = 8.


8
1
1
1 8
1
 = ln || 3 = (ln 8 − ln 3) = ln .
4
4
4
4 3

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

1
3

− ,

CHAPTER 7

TECHNIQUES OF INTEGRATION

1
 + 
 + 

+ 2
+ 2
=
(2 + 4)2

 +4
( + 4)2 gives 1 = 16, so  =

1
.
16

Now compare coefficients.
1=

1
(4
16

1=

1 4

16

1=
So  +


1
16

=

1
16

+ 1 2 + 1 + 4 + 3 + 42 + 4 + 2 + 
2



+  4 + 3 + 1 + 4 +  2 + (4 + ) + 1
2

 

1
16



+

1
2

+ 4 +  = 0 ⇒  = − 1 , and 4 +  = 0 ⇒  = 0. Thus,
4

1
− 16 
−1
+ 2 4 2
2 +4

( + 4)



 =



 1
1
1 1  2
1
1 ln || −
· ln  + 4 −
−
+
16
16 2
4
2 2 + 4

1
1
1 ln || − ln(2 + 4) +
+
16
32
8(2 + 4)

2 − 3 + 7
 + 
 + 
+ 2
= 2
(2 − 4 + 6)2
 − 4 + 6
( − 4 + 6)2

⇒ 2 − 3 + 7 = ( + )(2 − 4 + 6) +  + 

2 − 3 + 7 = 3 + (−4 + )2 + (6 − 4 + ) + (6 + ). So  = 0, −4 +  = 1 ⇒  = 1,

rS

6 − 4 +  = −3 ⇒  = 1, 6 +  = 7 ⇒  = 1. Thus,


 
+1
1
2 − 3 + 7
 =

+ 2
(2 − 4 + 6)2
2 − 4 + 6
( − 4 + 6)2



−2
3
1
 +
 +

=
( − 2)2 + 2
(2 − 4 + 6)2
(2 − 4 + 6)2


Fo

=

= 1 + 2 + 3 .



−2
1
√ tan−1 √
+ 1
√ 2  =
2
2
( − 2)2 +
2

1 =



2 =

1
2



3 = 3



ot

1

2 − 4
 =
(2 − 4 + 6)2

1
2



1
 =
2

N

37.



+ 82 + 16) + (2 + )(2 + 4) + 2 + 

1
= 0 ⇒  = − 16 ,  = 0,


=
(2 + 4)2

⇒ 1 = (2 + 4)2 + ( + )(2 + 4) + ( + ). Setting  = 0

e

35.

¤

al

338

1

√ 2 2  = 3
( − 2)2 +
2



1
2



1
1
−
+ 2 = −
+ 2

2(2 − 4 + 6)

√
1
2 sec2  
[2(tan2  + 1)]2




√
 − 2 = 2 tan ,
√
2
 = 2 sec  

√ 
√ 
√ 
2
sec2 
3 2
3 2
1
 = cos2   =
2 (1 + cos 2) 
4
sec4 
4
4
√
√
√




−2
3 2
3 2 1
3 2
 + 1 sin 2 + 3 = tan−1 √
· 2 sin  cos  + 3
+
=
2
2
8
8
8
2
√
√
√


3 2
2
−2
−2
3 2
+
tan−1 √
·√
=
·√
+ 3
8
8
2 − 4 + 6
2 − 4 + 6
2
√


3( − 2)
−2
3 2
+
tan−1 √
+ 3
=
8
4(2 − 4 + 6)
2

=

3

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

⇒

SECTION 7.4

INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS

¤

339

So  = 1 + 2 + 3
[ = 1 + 2 + 3 ]
√




−1
3( − 2)
−2
−2
1
3 2
+
+
= √ tan−1 √
+
tan−1 √
+
2(2 − 4 + 6)
8
4(2 − 4 + 6)
2
2
2
√ 
√


 √


3( − 2) − 2
3 − 8
4 2
3 2
7 2
−1  − 2
−1  − 2
√
√
+
+
=
+ tan + = tan +
8
8
4(2 − 4 + 6)
8
4(2 − 4 + 6)
2
2
√
 + 1, so 2 =  + 1 and 2  = . Then

 √


 
+1

22
2
 =
(2 ) =
 =
.
2+ 2

2 − 1
2 − 1
 −1

39. Let  =

2


2
=
=
+
2 − 1
( + 1)( − 1)
+1
−1

⇒ 2 = ( − 1) + ( + 1). Setting  = 1 gives  = 1.

√
, so 2 =  and 2  = . Then




2
=
+ 2 +
2 ( + 1)


+1




√ =
2 +  



2 
=
4 + 3

rS

41. Let  =

al

e

Setting  = −1 gives  = −1. Thus,


 
 
1
2
1
 =
+
 = 2 − ln | + 1| + ln | − 1| + 
2+ 2
2−
 −1
+1
−1
√

√

√
= 2  + 1 − ln  + 1 + 1 + ln   + 1 − 1 + 


2 
=
3 + 2



2 
.
2 ( + 1)

⇒ 2 = ( + 1) + ( + 1) + 2 . Setting  = 0 gives  = 2. Setting  = −1



√
3
2 + 1. Then 2 = 3 − 1, 2  = 32  ⇒




ot

43. Let  =

Fo

gives  = 2. Equating coefficients of 2 , we get 0 =  + , so  = −2. Thus,


 

√
√
2 
−2
2
2
2
2
=
+ 2 +
 = −2 ln || − + 2 ln | + 1| +  = −2 ln  − √ + 2 ln  + 1 + .
2 ( + 1)



+1



3 
√
=
3
2 + 1

3 5

10

− 3 2 +  =
4

N

=

(3 − 1) 3 2 
3
2
=

2

45. If we were to substitute  =

3
(2
10

(4 − ) 

+ 1)53 − 3 (2 + 1)23 + 
4

√
, then the square root would disappear but a cube root would remain. On the other hand, the

√
3
 would eliminate the cube root but leave a square root. We can eliminate both roots by means of the
√
6 substitution  = . (Note that 6 is the least common multiple of 2 and 3.)
√
√
√
6
3
Let  = . Then  = 6 , so  = 65  and  = 3 ,  = 2 . Thus,

substitution  =





3
5
65 
 = 6

=6
3 − 2
2 ( − 1)
−1

 
1

[by long division]
=6
2 +  + 1 +
−1
√

√
√
√


3
6
6

= 6 1 3 + 1 2 +  + ln | − 1| +  = 2  + 3  + 6  + 6 ln   − 1 + 
3
2


√
√ =
3
− 



c
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340

¤

CHAPTER 7

TECHNIQUES OF INTEGRATION

47. Let  =  . Then  = ln ,  =






2 
=
2 + 3 + 2

⇒


2 ()
=
2 + 3 + 2



 
=
( + 1)( + 2)

= 2 ln | + 2| − ln | + 1| +  = ln
49. Let  = tan , so that  = sec2  . Then

Now

1


=
+
( + 1)( + 2)
+1
+2



sec2 
 =
2
tan  + 3 tan  + 2



 


−1
2
+

+1
+2

( + 2)2
+
 + 1
1
 =
2 + 3 + 2



1
.
( + 1)( + 2)

⇒ 1 = ( + 2) + ( + 1).


. Then





=
1 + 




.
(1 + )

1


=
+
( + 1)

+1

al

51. Let  =  , so that  =   and  =

e

Setting  = −2 gives 1 = −, so  = −1. Setting  = −1 gives 1 = .


 
1
1
1
Thus,
 =
−
 = ln | + 1| − ln | + 2| +  = ln |tan  + 1| − ln |tan  + 2| + .
( + 1)( + 2)
+1
+2
⇒

rS

1 = ( + 1) + . Setting  = −1 gives  = −1. Setting  = 0 gives  = 1. Thus,


 
1

1
=
−
 = ln || − ln | + 1| +  = ln  − ln( + 1) +  =  − ln( + 1) + .
( + 1)

+1
2 − 1
,  = , and (by integration by parts)
2 −  + 2


 

22 − 
−4
 =  ln(2 −  + 2) −

ln(2 −  + 2)  =  ln(2 −  + 2) −
2+ 2
2 −  + 2
 −+2

 1
(2 − 1)
7

2
=  ln(2 −  + 2) − 2 −
 +
2 −+2

2
( − 1 )2 + 7
2
4
√

√
 where  − 1 = 27 ,
7
2
√

1
7
2

 = 27 ,
=  ln(2 −  + 2) − 2 − ln(2 −  + 2) +
7
2
2
(2 + 1)
2
1 2
7
7
4

N

ot

Fo

53. Let  = ln(2 −  + 2),  = . Then  =

√
= ( − 1 ) ln(2 −  + 2) − 2 + 7 tan−1  + 
2
√
2 − 1
= ( − 1 ) ln(2 −  + 2) − 2 + 7 tan−1 √
+
2
7

55.

( −

2)

+

4

=

4 (

+ 1)

From the graph, we see that the integral will be negative, and we guess that the area is about the same as that of a rectangle with width 2 and height 03, so we estimate the integral to be −(2 · 03) = −06. Now
1
1


=
=
+
2 − 2 − 3
( − 3)( + 1)
−3
+1

⇔

1 = ( + ) +  − 3, so  = − and  − 3 = 1 ⇔  = and  = − 1 , so the integral becomes
4

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

1
4




SECTION 7.4



0

2

2





¤

341


 
2
 − 3 2
1

1

=
ln | − 3| − ln | + 1| = ln 
 + 1
+1
4
4
0
0


1 2
−
4 0
0 −3

 1
= 1 ln 3 − ln 3 = − 1 ln 3 ≈ −055
4
2


1
=
− 2 − 3
4

2

INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS




=
2 − 2

(c)

2

= arctan  ⇒  = 2 arctan  ⇒  =

2
1 + 2

al



e




=
[put  =  − 1]
( − 1)2 − 1
2 − 1






1  − 1
 +  [by Equation 6] = 1 ln   − 2  + 
= ln 
2  + 1
2   


, then = tan−1 . The figure gives
59. (a) If  = tan
2
2


1

= √
= √ and sin
.
cos
2
2
2
1+
1 + 2

 
= 2 cos2
−1
(b) cos  = cos 2 ·
2
2
2

1
2
1 − 2
=2 √
−1 =
−1=
2
2
1+
1 + 2
1+
57.





2 
1






=
=2
3(2) − 4(1 − 2 )
22 + 3 − 2
1 − 2 1 + 2
2
−4
3
1 + 2
1 + 2


 
1 1

2 1
=
=
−

[using partial fractions]
(2 − 1)( + 2)
5 2 − 1
5+2








1  2 − 1 
 +  = 1 ln 2 tan (2) − 1  + 
= 1 ln |2 − 1| − ln | + 2| +  = ln
5
 +2 
 tan (2) + 2 
5
5

1
 =
3 sin  − 4 cos 

Fo



rS

61. Let  = tan(2). Then, using the expressions in Exercise 59, we have



ot

63. Let  = tan (2). Then, by Exercise 59,



2

N

sin 2
 =
2 + cos 



=

0

2

0

2 sin  cos 
 =
2 + cos 

1

8 ·

0

If we now let  = 2 , then

(2

(2



1

2·

0

8(1 − 2 )
2
1 − 2
 1
·
(1 + 2 )2
2
1 + 2 1 + 2
 =

2
2
2
2
1+
1−
0 2(1 +  ) + (1 −  )
2+
1 + 2

1 − 2
 = 
+ 3)(2 + 1)2

1 − 2
1−



=
=
+
+
+ 3)(2 + 1)2
( + 3)( + 1)2
+3
+1
( + 1)2

⇒

1 −  = ( + 1)2 + ( + 3)( + 1) + ( + 3). Set  = −1 to get 2 = 2, so  = 1. Set  = −3 to get 4 = 4, so
 = 1. Set  = 0 to get 1 = 1 + 3 + 3, so  = −1. So
=



0

1





1
4
8
8
8
 = 4 ln(2 + 3) − 4 ln(2 + 1) − 2
− 2
+
2 + 3
 +1
 +1 0
(2 + 1)2

= (4 ln 4 − 4 ln 2 − 2) − (4 ln 3 − 0 − 4) = 8 ln 2 − 4 ln 2 − 4 ln 3 + 2 = 4 ln 2 + 2
3

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

342

¤

CHAPTER 7

TECHNIQUES OF INTEGRATION

2 + 1
3 + 1
= −1 +
. Now
3 − 2
3 − 2

65. By long division,

3 + 1
3 + 1


=
=
+
3 − 2
(3 − )

3−

⇒ 3 + 1 = (3 − ) + . Set  = 3 to get 10 = 3, so  =

10
.
3

Set  = 0 to

get 1 = 3, so  = 1 . Thus, the area is
3


2

1

67.

2 + 1
 =
3 − 2



2

1


−1 +


= −2 +

 +


=
+
 [( − 1) − ]

( − 1) − 

1
3

1
3



10
3

+



 = − +

3−
  ln 2 − 0 − −1 + 0 −

10
3

1
3

ln || −

10
3

 ln 2 = −1 +

2

ln |3 − |

11
3

ln 2

1

⇒  +  =  [( − 1) − ] +  = [( − 1) + ]  − 

⇒

( − 1) +  = 1, − = 1 ⇒  = −1,  = . Now

so  = − ln  +
 = − ln  +

 






−1
−1

+
 = −
+


( − 1) − 

−1
( − 1) − 

e

 +
 =
 [( − 1) − ]

al



 ln|( − 1) − | + . Here  = 010 and  = 900, so
−1

01
−09

ln|−09 − 900| +  = − ln  −

When  = 0,  = 10,000, so 0 = − ln 10,000 −

1
9

ln(|−1| |09 + 900|) = − ln  −

ln(9900) + . Thus,  = ln 10,000 +

Fo

equation becomes
 = ln 10,000 − ln  +
= ln

1
9

rS

=

1
9

ln 9900 −

1
9

ln(09 + 900) = ln

1
9

1
9

ln(09 + 900) + .

ln 9900 [≈ 102326], so our

1
9900
10,000
+ ln

9 09 + 900

1100
10,000
1
11,000
10,000 1
+ ln
= ln
+ ln

9 01 + 100

9  + 1000

ot

69. (a) In Maple, we define  (), and then use convert(f,parfrac,x); to obtain

668323
943880,155
(22,098 + 48,935)260,015
24,1104879
−
−
+
5 + 2
2 + 1
3 − 7
2 +  + 5

N

 () =

In Mathematica, we use the command Apart, and in Derive, we use Expand.
(b)



 ()  =

24,110
4879

·

1
5

=

24,110
4879

·

1
5

=

ln|5 + 2| −

668
323

·

1
2

9438 ln|2 + 1| − 80,155 · 1 ln |3 − 7|
3



22,098  + 1 + 37,886
1
2
 + 
+

2
260,015
 + 1 + 19
2
4

ln|5 + 2| −

1
+ 260,015 22,098 ·

668
323

·

1
2

4822
4879

9438
80,155

·

1
3

ln|3 − 7|


 2

−1 √ 1
4
ln  +  + 5 + 37,886 · 19 tan

3146 ln|2 + 1| − 80,155 ln|3 − 7| +


75,772
+ 260,015√19 tan−1 √1 (2 + 1) + 
19

ln|5 + 2| −

334
323

1
2

ln|2 + 1| −

194

11,049
260,015




1
+ 2
+



ln 2 +  + 5

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 7.5

STRATEGY FOR INTEGRATION

¤

343

Using a CAS, we get
4822 ln(5 + 2)
334 ln(2 + 1)
3146 ln(3 − 7)
−
−
4879
323
80,155
√

√
2
19
3988 19
11,049 ln( +  + 5)
−1
+ tan (2 + 1)
+
260,015
260,015
19
The main difference in this answer is that the absolute value signs and the constant of integration have been omitted. Also, the fractions have been reduced and the denominators rationalized.
71. There are only finitely many values of  where () = 0 (assuming that  is not the zero polynomial). At all other values of

,  ()() = ()(), so  () = (). In other words, the values of  and  agree at all except perhaps finitely many values of . By continuity of  and , the polynomials  and  must agree at those values of  too.
More explicitly: if  is a value of  such that () = 0, then () 6= 0 for all  sufficiently close to . Thus,
[by continuity of  ]

e

 () = lim  ()
= lim ()
→

[by continuity of ]

rS

= ()

[whenever () 6= 0]

al

→

73. If  6= 0 and  is a positive integer, then  () =

1
2
1


=
+ 2 + ··· +  +
. Multiply both sides by
 ( − )



−

 ( − ) to get 1 = 1 −1 ( − ) + 2 −2 ( − ) + · · · +  ( − ) +  . Let  =  in the last equation to get
⇒  = 1 . So

Fo

1 = 

 () −


1
1
 − 
 − 
= 
− 
=  
=−  
−
 ( − )
 ( − )
  ( − )
  ( − )
( − )(−1 + −2  + −3 2 + · · · + −2 + −1 )
  ( − )
 −1


−2  −3 2
−2
−1
=−
+   +   + ··· +   +  
 
 
 
 
 

ot

=−

N

=−

Thus, () =

1
1
1
1
1
− −1 2 − −2 3 − · · · − 2 −1 − 
 




 


1
1
1
1
1
= −  − −1 2 − · · · −  + 
.
 ( − )
  


 ( − )

7.5 Strategy for Integration
1. Let  = sin , so that  = cos  . Then
3.



sin  + sec 
 = tan 

 

sec  sin 
+
tan  tan 

5. Let  = 2 . Then  = 2 




 =
4 + 2



1
2 + 2







 cos (1 + sin2 )  = (1 + 2 )  =  + 1 3 +  = sin  +
3
 =



(cos  + csc )  = sin  + ln |csc  − cot | + 

⇒



 2 

1


1 1
1
 = √ tan−1 √
+  [by Formula 17] = √ tan−1 √
+
2
2 2
2
2 2
2

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

1
3

sin3  + .

¤

CHAPTER 7

TECHNIQUES OF INTEGRATION

7. Let  = arctan . Then  =

9.

3
1

4 ln  



 = ln ,

 =



1 + 2

 = 4 ,
 = 1 5
5



=
=





1
2

−1

1
5

arctan 
 =
1 + 2

5 ln 

243
5

( − 2) + 1
 =
( − 2)2 + 1

3

ln 3 −

 

−

1

3



−4

1 4

1 5

 243
25

−

1
25

4

 =



1
2

 4
  =  −4 = 4 − −4 .
243
5

243
5

=


1
+ 2
2 + 1
 +1

ln(2 + 1) + tan−1  +  =



ln 3 − 0 −

ln 3 −



242
25




1 5 3
 1
25

[ =  − 2,  = ]

ln(2 − 4 + 5) + tan−1 ( − 2) + 




0




=
(1 − 2 )32

 cos2   =
=

cos  
=
(cos )3

 sec2   = tan  +  = √
+ .
1 − 2

  1

 2 (1 + cos 2)  =
0
1
2





1
4


0

  +



1
2



 cos 2 

sin 2 
0

1
2

1
0 2

 = ,  = cos 2 
 = ,  = 1 sin 2
2


 1
− 2 cos 2 0 = 1 2 + 1 (1 − 1) = 1 2
4
8
4

+  =





   =







  =  +  =  + .

ot

19. Let  =  . Then

1
2


 1 2  1  1
 0 + 2 2  sin 2 0 −
2

= 1 2 + 0 −
4

√
√


, so that 2 =  and 2  = . Then arctan   = arctan  (2 ) = . Now use parts with
1
,  = 2 . Thus,
1 + 2

N

21. Let  =



Then  = cos   and (1 − 2 )12 = cos ,

al

so



2.

e

 sin4  cos4  sin   = (sin2 )2 cos4  sin  


= (1 − cos2 )2 cos4  sin   = (1 − 2 )2 4 (−) [ = cos ,  = − sin  ]

= (−4 + 26 − 8 )  = − 1 5 + 2 7 − 1 9 +  = − 1 cos5  + 2 cos7  − 1 cos9  + 
5
7
9
5
7
9

sin5  cos4   =

15. Let  = sin , where −  ≤  ≤
2

17.

1

Fo

13.

−1
 =
2 − 4 + 5
=

11.



⇒

rS

344

 = arctan ,  = 2  ⇒  =
 = 2 arctan  −
=  arctan
23. Let  = 1 +



2
 = 2 arctan  −
1 + 2

 
1−

1
1 + 2



 = 2 arctan  −  + arctan  + 



√
√
√
√
√
 −  + arctan  +  or ( + 1) arctan  −  + 
√
. Then  = ( − 1)2 ,  = 2( − 1)  ⇒

√ 8
2

2
2
1
 = 1 8 · 2( − 1)  = 2 1 (9 − 8 )  = 1 10 − 2 · 1 9 1 =
1+ 
5
9
0

1024
5

−

1024
9

−

1
5

+

2
9

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

=

4097
.
45

SECTION 7.5

25.

32 − 2
6 + 22


=3+
=3+
+
− 2 − 8
( − 4)( + 2)
−4
+2

2

 = 4 gives 46 = 6, so  =


32 − 2
 =
2 − 2 − 8


23
.
3

STRATEGY FOR INTEGRATION

¤

345

⇒ 6 + 22 = ( + 2) + ( − 4). Setting

Setting  = −2 gives 10 = −6, so  = − 5 . Now
3


 
53
233
−
 = 3 +
3+
−4
+2

27. Let  = 1 +  , so that  =   = ( − 1) . Then



23
3

ln | − 4| −

1
 =
1 + 



5
3

ln | + 2| + .


1
·
=
 −1



1
 = . Now
( − 1)

1


=
+
⇒ 1 = ( − 1) + . Set  = 1 to get 1 = . Set  = 0 to get 1 = −, so  = −1.
( − 1)

−1

 
1
−1
+
 = − ln || + ln | − 1| +  = − ln(1 +  ) + ln  +  =  − ln(1 +  ) + .
Thus,  =

−1

form − ln(− + 1) + .
√

2 − 1 ,  =  ⇒

al



29. Use integration by parts with  = ln  +

e

Another method: Multiply numerator and denominator by − and let  = − + 1. This gives the answer in the

 

1+
 =
1−

Fo

31. As in Example 5,

rS



√ 2

 −1+
1

1
1
√
√
√
 =
1+ √
 =
 = √
,  = . Then
2 −1
2 −1
2 −1
2 −1
2 −1
+ 

+ 


 



 
 




√
 =  ln  + 2 − 1 − 2 − 1 + . ln  + 2 − 1  =  ln  + 2 − 1 −
2 − 1
√



 √

1+
1+
1+

 
√
√
√
√
·√
 =
 =
+
= sin−1  − 1 − 2 + .
2
2
2
1−
1+
1−
1−
1−

(1 + )(1 − ).

ot

Another method: Substitute  =

33. 3 − 2 − 2 = −(2 + 2 + 1) + 4 = 4 − ( + 1)2 . Let  + 1 = 2 sin ,

N

where −  ≤  ≤  . Then  = 2 cos   and
2
2


√
2  =
3 − 2 − 
4 − ( + 1)2  =
4 − 4 sin2  2 cos  


= 4 cos2   = 2 (1 + cos 2) 

= 2 + sin 2 +  = 2 + 2 sin  cos  + 
√


3 − 2 − 2
+1
−1  + 1
= 2 sin
+2·
·
+
2
2
2


+1
 + 1√
+
3 − 2 − 2 + 
= 2 sin−1
2
2

35. Using product formula 2(c) in Section 7.2,

cos 2 cos 6 = 1 [cos(2 − 6) + cos(2 + 6)] = 1 [cos(−4) + cos 8] = 1 (cos 4 + cos 8). Thus,
2
2
2




1
1 1
1
1 cos 2 cos 6  = 2 (cos 4 + cos 8)  = 2 4 sin 4 + 8 sin 8 +  = 1 sin 4 + 16 sin 8 + .
8
c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

346

CHAPTER 7

TECHNIQUES OF INTEGRATION

37. Let  = tan . Then  = sec2  

 4

⇒

0

39. Let  = sec , so that  = sec  tan  . Then

tan3  sec2   =


1
0

3  =

sec  tan 
 = sec2  − sec 





1 4
4

1

= 1.
4

0

1
 =
2 − 



1
 = . Now
( − 1)



1
=
+
⇒ 1 = ( − 1) + . Set  = 1 to get 1 = . Set  = 0 to get 1 = −, so  = −1.
( − 1)

−1

 
1
−1
+
 = − ln || + ln | − 1| +  = ln |sec  − 1| − ln |sec | +  [or ln |1 − cos | + ].
Thus,  =

−1




41. Let  = ,  = tan2   = sec2  − 1 

⇒  =  and  = tan  − . So

 tan2   = (tan  − ) − (tan  − )  =  tan  − 2 − ln |sec | + 1 2 + 
2



=  tan  − 1 2 − ln |sec | + 
2

√
1
 so that  = √ . Then
2 





 √

1

2
1

 =
(2 ) = 2
 = 2
1 + 3
1 + 6
1 + (3 )2
1 + 2 3

e

43. Let  =

tan−1 (32 ) + 
√
so that 2 = 3 and  = 3 12  ⇒
  =
2
2
3

tan−1 3 +  =

al

tan−1  +  =

2
3

Another method: Let  = 32

 √
2

2
2
3
 =
 = tan−1  +  = tan−1 (32 ) + .
1 + 3
1 + 2
3
3

 = − 1 − +
3

1
3



⇒ =

5 −  =

1
3



− . Now integrate by parts with  = ,  = − :

3

ot




3 ( − 1)−4  = ( + 1)3 −4  = (3 + 32 + 3 + 1)−4  = (−1 + 3−2 + 3−3 + −4 ) 

= ln || − 3−1 − 3 −2 − 1 −3 +  = ln | − 1| − 3( − 1)−1 − 3 ( − 1)−2 − 1 ( − 1)−3 + 
2
3
2
3

√
4 + 1 ⇒ 2 = 4 + 1 ⇒ 2  = 4  ⇒  = 1  . So
2


51. Let 2 = tan 

N

49. Let  =



3

. Then

−  = − 1 − − 1 − +  = − 1 − (3 + 1) + .
3
3
3

47. Let  =  − 1, so that  = . Then





2
3



Fo

45. Let  = 3 . Then  = 32 

 = 3
 = 32 

rS

2
3

=



1
√
 =
 4 + 1

⇒ =


√
=
 42 + 1



1
 
2
1
(2 − 1) 
4

=2




√
 4 + 1 − 1 

= ln √
 4 + 1 + 1  + 

tan ,  =

sec2  
=
tan  sec 

1
2
1
2

1
2





1
2

sec2  ,

sec 
 = tan 



1  − 1


+
= 2 2 ln
2 − 1
 + 1

[by Formula 19]

√
42 + 1 = sec , so


csc  

= − ln |csc  + cot | + 
[or ln |csc  − cot | + ]

√

√


 42 + 1
 42 + 1
1 
1 
+
+
+
or ln
−
= − ln


2
2 
2
2  c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 7.5

53.



1 2
2
 cosh() −



2 sinh() =





 = 2 ,
 = 2 

 cosh() 

STRATEGY FOR INTEGRATION
 = sinh() ,
1
 =  cosh()

¤

347





 
2 1
1 
 = ,
 = cosh() , sinh() 
 sinh() −
1
 = 
 =  sinh()
 

2
2
 sinh() + 3 cosh() + 
2




√
2

2 
√ =
=
 = .
55. Let  = , so that  = 2 and  = 2 . Then
2 + 2 · 
(1 + )
+ 
1 2
 cosh() −

1
= 2 cosh() −

=



2
= +
⇒ 2 = (1 + ) + . Set  = −1 to get 2 = −, so  = −2. Set  = 0 to get 2 = .
(1 + )

1+

 

√
√ 
2
2
−
 = 2 ln || − 2 ln |1 + | +  = 2 ln  − 2 ln 1 +  + .
Thus,  =

1+

Now

√
3
 + . Then  = 3 −  ⇒


 √
 3  +   = (3 − ) · 32  = 3 (6 − 3 )  = 3 7 − 3 4 +  = 3 ( + )73 − 3 ( + )43 + 
7
4
7
4

e

57. Let  =




=
1 + cos 

 



cos3   =



 cos2  cos   = (1 − sin2 ) cos  


= (cos  − sin2  cos )  = sin  −

1
1 − cos 
·
1 + cos  1 − cos 



 =



sin3  +  = sin(sin ) −

1
3

1 − cos 
 =
1 − cos2 


= (csc2  − cot  csc )  = − cot  + csc  + 



1 − cos 
 = sin2 

1
3

 

sin3 (sin ) + 
1
cos 
−
sin2  sin2 

Fo

61.

cos  cos3 (sin )  =

rS



al

59. Let  = sin , so that  = cos  . Then

Another method: Use the substitutions in Exercise 7.4.59.
 




2(1 + 2 ) 


2 
=
=
=  =  +  = tan
+
1 + cos 
1 + (1 − 2 )(1 + 2 )
(1 + 2 ) + (1 − 2 )
2
√
 so that  =

ot

√
1
√  ⇒  = 2   = 2 . Then
2 
 √ √


    =  (2 ) = 2 2  

N

63. Let  =

= 2 2  −



4 





 = 22 ,
 = 4 

 = 4,
 = 4 

 =  ,
 = 

 =  ,
 = 








= 2 2  − 4 − 4  = 2 2  − 4 + 4 + 

 √
√
= 2( 2 − 2 + 2) +  = 2  − 2  + 2   + 
65. Let  = cos2 , so that  = 2 cos  (− sin ) . Then


67.



sin 2
 =
1 + cos4 



2 sin  cos 
 =
1 + (cos2 )2

1
(−) = − tan−1  +  = − tan−1 (cos2 ) + .
1 + 2

√
 √ 
 
√ 
+1−  
√ =
√
 =
 + 1 −  
√
√
+1+ 
+1− 


= 2 ( + 1)32 − 32 + 
3


 



1
√
√ ·
+1+ 

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.





¤

348

CHAPTER 7

TECHNIQUES OF INTEGRATION

69. Let  = tan , so that  = sec2  ,  =



1

√

3


3,

and  = 1 ⇒  =


4.

Then

√

 3 
 3
 3
1 + 2 sec  sec  (tan2  + 1) sec  tan2  sec  sec2   =
 =
+

 =
2
2 tan2  tan2  tan2 
4 tan 
4
4
=



3

4


3
(sec  + csc  cot )  = ln |sec  + tan | − csc 
4

 
√ 
= ln 2 + 3 −

2
√
3

71. Let  =  . Then  = ln ,  = 



√
3 ⇒ =

2
 =
1 + 



2 
=
1+ 





 √  √
 √
− ln  2 + 1 − 2 = 2 −

⇒


 =
1+

 
1−

1
1+



2
√
3

√ 
√ 


+ ln 2 + 3 − ln 1 + 2

 =  − ln|1 + | +  =  − ln(1 +  ) + .

1
 and  = sin . Then
1 − 2

75.



(sin  + )  = − cos  + 1 2 + 
2

al

 + arcsin 
√
 =
1 − 2

√
= − 1 − 2 + 1 (arcsin )2 + 
2

rS



e

73. Let  = arcsin , so that  = √

1

 + 
=
+ 2
( − 2)(2 + 4)
−2
 +4

⇒ 1 = (2 + 4) + ( + )( − 2) = ( + )2 + ( − 2) + (4 − 2).

=



ln| − 2| −

1
16

ln(2 + 4) −

1
8

tan−1 (2) + 

√
1 +  , so that  2 = 1 +  , 2  =  ,  =  2 − 1, and  = ln( 2 − 1). Then


√
 =
1 + 





ot

77. Let  =

1
8

Fo

So 0 =  +  =  − 2, 1 = 4 − 2. Setting  = 2 gives  = 1 ⇒  = − 1 and  = − 1 . So
8
8
4
  1




1
1
− − 4
1

1
2 
1

1
8
 =
+ 82
 =
−
−
( − 2)(2 + 4)
−2
 +4
8
−2
16
2 + 4
4
2 + 4

ln( 2 − 1)
(2 ) = 2


[ln( + 1) + ln( − 1)] 

N

= 2[( + 1) ln( + 1) − ( + 1) + ( − 1) ln( − 1) − ( − 1)] + 

[by Example 7.1.2]

= 2[ ln( + 1) + ln( + 1) −  − 1 +  ln( − 1) − ln( − 1) −  + 1] + 
= 2[(ln( + 1) + ln( − 1)) + ln( + 1) − ln( − 1) − 2] + 
√




√
√
+1
1 +  + 1
− 2 1 +  + 
= 2  ln(2 − 1) + ln
− 2 +  = 2 1 +  ln( ) + ln √
−1
1 +  − 1
√
√
√
√
√
1 +  + 1
1 +  + 1
− 4 1 +  +  = 2( − 2) 1 +  + 2 ln √
+
= 2 1 +  + 2 ln √
 −1
1+
1 +  − 1
79. Let  = ,  = sin2  cos  

⇒  = ,  = 1 sin3 . Then
3
 1


2
3
3
1
1
 sin  cos   = 3  sin  − 3 sin   = 3  sin3  − 1 (1 − cos2 ) sin  
3



1
1
 = cos ,
2
3
=  sin  +
(1 −  ) 
 = − sin  
3
3
= 1  sin3  + 1  − 1  3 +  = 1  sin3  +
3
3
9
3

1
3

cos  −

1
9

cos3  + 

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS

81.



√
1 − sin   =

 





1 − sin  1 + sin 
·
 =
1
1 + sin 



 cos2  cos  
√
 =
1 + sin 
1 + sin 




 = 1 + sin 
√
=
 = cos  

√
√
= 2  +  = 2 1 + sin  + 
=



1 − sin2 

1 + sin 
[assume cos   0]


√
Another method: Let  = sin  so that  = cos   = 1 − sin2   = 1 − 2 . Then

 


√
√
√
√

1
√
 = 2 1 +  +  = 2 1 + sin  + .
1 − sin   =
1− √
=
2
1+
1−
2

83. The function  = 2 does have an elementary antiderivative, so we’ll use this fact to help evaluate the integral.
2

22   +


2

=  −

2



2

  =

  +



2


 
 2
2
 2  +  


e



 = ,
 = 

 

2

= 2 ,



al

2

(22 + 1)  =

2

= 

2

=  + 

rS



7.6 Integration Using Tables and Computer Algebra Systems

Keep in mind that there are several ways to approach many of these exercises, and different methods can lead to different forms of the answer.

3.

0

2

 sin(5 − 2) sin(5 + 2)
 = 5
+
=2
2(5 − 2)
2(5 + 2) 0
2 

 sin 7
1
−7 − 3
5
sin 3
1
+
−0=
=−
=
= − −
6
14 0
6
14
42
21
80

cos 5 cos 2  =

2√
42 − 3  =
1



Fo

 2

√ 2
4
2 −
3 
2



ot

1.

1
2

 = 2  = 2 



 
√ 2 
4


√ 2
√ 2 
3
1 
2 −
 + 2 −
=

3 −
3  ln 

2 2
2
2
√  1 
√ 
 √

 √

1
3
3
= 2 2 13 − 2 ln 4 + 13 − 2 1 − 2 ln 3 = 13 − 3 ln 4 + 13 −
4

N

39

5.

7.

 8
0



arctan 2  =

1
2

 4
0

arctan  



1
2

+

3
4

ln 3



 = 2  = 2 

4




 1 
1
1
1

2
=
 arctan  − ln(1 + 2 ) tan−1 − ln 1 +
−0
2
2
2
4
4
2
16
0


2

1

= − ln 1 +
8
4
16
89

=

cos 
 = sin2  − 9



1

2 − 9





 = sin 
 = cos  

20

=





 − 3


1
 +  = 1 ln sin  − 3  +  ln  + 3
 sin  + 3 
2(3)
6

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

349

¤

350

CHAPTER 7

TECHNIQUES OF INTEGRATION

9. Let  = 2 and  = 3. Then  = 2  and



11.



0





1
2



=2




28
√
= −2
2 2 + 2

2 √ 2
 + 2
4
√
√
42 + 9
42 + 9
= −2
+ =−
+
9 · 2
9

√
2 + 2
+
2 

0
0

 0
 0
1 2 −
2
1
96
 
−
−  =  + 2
−  =  + 2
(− − 1) −
−1
−1 −1
(−1)2
−1
−1
−1
 0

=  + 2 − + 0 =  − 2
97

2 −  =

−1

13.


√
=
2 42 + 9



tan3 (1)

2



 = 1,
 = − 2



=−



69

tan3   = − 1 tan2  − ln |cos | +  = − 1 tan2
2
2

2 arctan( )  =



2 arctan 








=



  1 
− ln cos   + 

92

 arctan   =


2 + 1 arctan  − + 
2
2

al





e

15. Let  =  , so that  =   and 2 = 2 . Then

1

= 1 (2 + 1) arctan( ) − 1  + 
2
2





√
7. Then  = 2 − 2 ,  = 2 ,


√
 √

 √
√
6 + 4 − 4 2  =    = 1 ( + 1) 2 − 2 1  = 1  2 − 2  + 1
2 − 2 
2
2
4
4
√
√

= 1
2 − 2  − 1 (−2) 2 − 2 
4
8


 1  √
2
 = 2 − 2 ,
30  √ 2
=
 − 2 +
  sin−1 −
 = −2 
8
8

8

Fo

and

rS

17. Let  = 6 + 4 − 4 2 = 6 − (4 2 − 4 + 1) + 1 = 7 − (2 − 1)2 ,  = 2 − 1, and  =

2 − 1
2 − 1 
7
1 2
6 + 4 − 4 2 + sin−1 √
− · 32 + 
8
8
8 3
7

2 − 1
7
2 − 1
1
=
6 + 4 − 4 2 + sin−1 √
− (6 + 4 − 4 2 )32 + 
8
8
12
7

ot

=

N

This can be rewritten as



1
2 − 1
1
7
(2 − 1) −
(6 + 4 − 4 2 ) + sin−1 √
6 + 4 − 4 2
+
8
12
8
7




5 
1
7
1 2
−1 2 − 1
2 +
√
+
 − − sin 6 + 4 − 4
=
3
12
8
8
7



1
7
2 − 1
√
=
+
(8 2 − 2 − 15) 6 + 4 − 4 2 + sin−1
24
8
7

19. Let  = sin . Then  = cos  , so

sin2  cos  ln(sin )  =



=



1
9

101

2 ln   =

2+1
[(2 + 1) ln  − 1] +  = 1 3 (3 ln  − 1) + 
9
(2 + 1)2

sin3  [3 ln(sin ) − 1] + 

√
3. Then  =   and
  √ 





 + 
1

19 1
 +  = √ ln   + √3  + .
 =
=
ln 
  − 3 
2 − 2
2   −  
2 3

21. Let  =  and  =




3 − 2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 7.6 INTEGRATION USING TABLES AND COMPUTER ALGEBRA SYSTEMS

23.



77 1
4

tan  sec3  +

3
4

14 1
4

tan  sec3  +

3
8

sec5   =
=



77 1
4

sec3   =

tan  sec  +

tan  sec3  +

3
8

3
4

1

tan  sec  +

2

ln|sec  + tan | + 

1
2



 sec  

25. Let  = ln  and  = 2. Then  =  and

27.



 
 



4 + (ln )2
2 
21 
 = ln  + 2 + 2 + 
2 + 2  =
2 + 2 +

2
2




= 1 (ln ) 4 + (ln )2 + 2 ln ln  + 4 + (ln )2 + 
2
1 cos−1 (−2 )
 = −
3
2





 = −2 
 = −2−3 

cos−1  



√
√


= − 1  cos−1  − 1 − 2 +  = − 1 −2 cos−1 (−2 ) + 1 1 − −4 + 
2
2
2

88

e

29. Let  =  . Then  = ln ,  = , so






1

4 
= 5 ,

√
=
= 54 
5
2 − 2
(5 )2 − 2




√
√
43
= 1 ln + 2 − 2  +  = 1 ln5 + 10 − 2  + 
5
5

4 
√
=
10 − 2



rS

31.

al

 
 √ 2


 −1
41
2 − 1  =

 = 2 − 1 − cos−1 (1) +  = 2 − 1 − cos−1 (− ) + .


33. Use disks about the -axis:


0

Fo



73 
(sin2 )2  =  0 sin4   =  − 1 sin3  cos  0 +
4

 

 

63
=  0 + 3 1  − 1 sin 2 0 =  3 1  − 0 = 3 2
4 2
4
4 2
8

 =

3
4


0

 sin2  

N

ot


 



2
2
1
 1
2
 +  −
− 2 ln | + | +  = 3  +
35. (a)
−
 3
 + 

( + )2
( + )


2
2
1 ( + ) +  − ( + )2
= 3
2

( + )


3 2
 
1
2
= 3
=
2
 ( + )
( + )2
−
1 and  = .




 2
 

1
1
1
2
( − )2
 − 2 + 2
2
2 
= 3
 = 3
 = 3
+ 2 
1−
( + )2

2

2







2
2


1
1
+  = 3  +  −
− 2 ln | + | + 
= 3  − 2 ln || −



 + 

(b) Let  =  +  ⇒  =  . Note that  =

37. Maple and Mathematica both give



sec4   =

2
3

tan  +

1
3

tan  sec2 , while Derive gives the second

sin 
1 sin  1
1
=
= tan  sec2 . Using Formula 77, we get
3 cos3 
3 cos  cos2 
3


4
2
1
2
sec   = 3 tan  sec  + 3 sec2   = 1 tan  sec2  + 2 tan  + .
3
3

term as

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

351

¤

352

CHAPTER 7

TECHNIQUES OF INTEGRATION

√
√
√

2 + 4  = 1 (2 + 2) 2 + 4 − 2 ln 2 + 4 +  . Maple gives
4
√
 
1
(2 + 4)32 − 1  2 + 4 − 2 arcsinh 1  . Applying the command convert(%,ln); yields
4
2
2
√
√
√





 
2
32
1
1
2 + 4 − 2 ln 1  + 1
( + 4)
− 2 
2 + 4 = 1 (2 + 4)12 (2 + 4) − 2 − 2 ln  + 2 + 4 2
4
2
2
4
√
√

= 1 (2 + 2) 2 + 4 − 2 ln 2 + 4 +  + 2 ln 2
4

39. Derive gives



2

√
Mathematica gives 1 (2 + 2 ) 3 + 2 − 2 arcsinh(2). Applying the TrigToExp and Simplify commands gives
4
√
√
√
√



 

1
(2 + 2 ) 4 + 2 − 8 log 1  + 4 + 2
= 1 (2 + 2) 2 + 4 − 2 ln  + 4 + 2 + 2 ln 2, so all are
4
2
4
equivalent (without constant).



cos4   =

sin  cos3 
3 sin  cos 
3
+
+
, while Mathematica gives
4
8
8

rS

41. Derive and Maple give

al

e

Now use Formula 22 to get


√
√


24 
2 22 + 2  = (22 + 22 ) 22 + 2 − ln  + 22 + 2 + 
8
8
√
√



= (2)(2 + 2 ) 4 + 2 − 2 ln  + 4 + 2 + 
8
√
√

= 1 (2 + 2) 2 + 4 − 2 ln 2 + 4 +  + 
4

Fo

3
1
1
3
1
1
+ sin(2) + sin(4) =
+ (2 sin  cos ) + (2 sin 2 cos 2)
8
4
32
8
4
32
1
1
3
+ sin  cos  + [2 sin  cos  (2 cos2  − 1)]
=
8
2
16
1
3
1
1
=
+ sin  cos  + sin  cos3  − sin  cos ,
8
2
4
8 so all are equivalent.
Using tables,

=

43. Maple gives





1
4

cos3  sin  +

3
4



64 1
4

cos2   =

cos3  sin  +

ot

74 1
4

cos4   =

cos3  sin  + 3  +
8

tan5   =

1
4

3
(2 sin 
16

tan4  −

N



1
2

cos ) +  =

tan2  +

1
2

1
4

3
4

1

2

+

1
4

 sin 2 + 

cos3  sin  + 3  +
8

3
8

sin  cos  + 

ln(1 + tan2 ), Mathematica gives

tan5   = 1 [−1 − 2 cos(2)] sec4  − ln(cos ), and Derive gives
4



tan5   =

1
4

tan4  −

1
2

tan2  − ln(cos ).

These expressions are equivalent, and none includes absolute value bars or a constant of integration. Note that Mathematica’s and Derive’s expressions suggest that the integral is undefined where cos   0, which is not the case. Using Formula 75,





tan5   =

1
5−1

tan3   =

1
2

45. (a)  () =



tan5−1  −



tan5−2   =

tan2  + ln |cos | + , so

 ()  =





1
4

tan4  −

tan5   =

1
4



tan3  . Using Formula 69,

tan4  −

1
2

tan2  − ln |cos | + .

√
√




2

1
1  1 + 1 − 2 
35
 +  = − ln  1 + 1 −   + .
√
 = − ln 



1 


 1 − 2



 has domain  |  6= 0, 1 − 2  0 = { |  6= 0, ||  1} = (−1 0) ∪ (0 1).  has the same domain. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 7.7

APPROXIMATE INTEGRATION

¤

√

√


(b) Derive gives  () = ln 1 − 2 − 1 − ln  and Mathematica gives  () = ln  − ln 1 + 1 − 2 .
Both are correct if you take absolute values of the logarithm arguments, and both would then have the

 √ same domain. Maple gives  () = − arctanh 1 1 − 2 . This function has domain

√
√
√
  
  

 
  ||  1 −1  1 1 − 2  1 =   ||  1, 1 1 − 2  1 =   ||  1 1 − 2  1 = ∅,

the empty set! If we apply the command convert(%,ln); to Maple’s answer, we get




1
1
1
1
+ 1 + ln 1 − √
, which has the same domain, ∅.
− ln √
2
2
1 − 2
1 − 2

7.7 Approximate Integration

2


2 =

 (−1 ) ∆ =  (0 ) · 2 +  (1 ) · 2 = 2 [ (0) +  (2)] = 2(05 + 25) = 6

=1

2 =

2


 ( ) ∆ =  (1 ) · 2 +  (2 ) · 2 = 2 [ (2) + (4)] = 2(25 + 35) = 12

=1

( )∆ =  (1 ) · 2 +  (2 ) · 2 = 2 [ (1) +  (3)] ≈ 2(16 + 32) = 96
2 is an underestimate, since the area under the small rectangles is less than

Fo

(b)

al

=1

rS

2


2 =

e

1. (a) ∆ = ( − ) = (4 − 0)2 = 2

the area under the curve, and 2 is an overestimate, since the area under the large rectangles is greater than the area under the curve. It appears that 2

is an overestimate, though it is fairly close to . See the solution to

2

ot

1


∆ [(0 ) + 2(1 ) +  (2 )] = 2 [ (0) + 2 (2) + (4)] = 05 + 2(25) + 35 = 9.
2

N

(c) 2 =

Exercise 47 for a proof of the fact that if  is concave down on [ ], then

the Midpoint Rule is an overestimate of   () .

This approximation is an underestimate, since the graph is concave down. Thus, 2 = 9  . See the solution to
Exercise 47 for a general proof of this conclusion.
(d) For any , we will have          .
 
3.  () = cos 2 , ∆ = 1 − 0 = 1
4
4

1
 
 

1
(a) 4 = 4 · 2 (0) + 2 4 + 2 2 + 2 3 +  (1) ≈ 0895759
4
4
  
 
 
 
(b) 4 = 1  1 +  3 +  5 +  7 ≈ 0908907
4
8
8
8
8
The graph shows that  is concave down on [0 1]. So 4 is an

underestimate and 4 is an overestimate. We can conclude that
 
1
0895759  0 cos 2   0908907. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

353

354

¤

CHAPTER 7

TECHNIQUES OF INTEGRATION

2−0
1

−
=
=
, ∆ =
1 + 2

10
5
 1
3
5
 
10 = 1  10 +  10 +  10 + · · · +  19 ≈ 0806598
5
10

5. (a)  () =

(b) 10 =

1
5·3


 
 
 
 
 

 (0) + 4 1 + 2 2 + 4 3 + 2 4 + · · · + 4 9 + (2) ≈ 0804779
5
5
5
5
5

Actual:  =



2

0

=

1
2

2



 = 1 ln 1 + 2  0
2
1 + 2

ln 5 −

1
2

ln 1 =

1
2

[ = 1 + 2 ,  = 2 ]

ln 5 ≈ 0804719

Errors:  = actual − 10 =  − 10 ≈ −0001879
 = actual − 10 =  − 10 ≈ −0000060
√
−
2−1
1
3 − 1, ∆ =
=
=

10
10
(a) 10 = 101· 2 [ (1) + 2 (11) + 2 (12) + 2 (13) + 2 (14) + 2 (15)
+ 2 (16) + 2 (17) + 2 (18) + 2 (19) +  (2)]

e

7.  () =

al

≈ 1506361
1
(b) 10 = 10 [ (105) +  (115) +  (125) +  (135) +  (145) +  (155) +  (165) +  (175) +  (185) +  (195)]

(c) 10 =

1
10 · 3 [ (1)

rS

≈ 1518362
+ 4 (11) + 2 (12) + 4 (13) + 2 (14)

+ 4 (15) + 2 (16) + 4 (17) + 2 (18) + 4 (19) + (2)]

Fo

≈ 1511519

2−0
1

−
=
=
, ∆ =
1 + 2

10
5
1
(a) 10 = 5 · 2 [ (0) + 2 (02) + 2 (04) + 2 (06) + 2 (08) + 2 (1)

9.  () =

+ 2 (12) + 2 (14) + 2 (16) + 2 (18) + (2)]

(c) 10 =

N

≈ 2664377

ot

≈ 2660833
(b) 10 = 1 [ (01) +  (03) +  (05) +  (07) +  (09) +  (11) +  (13) + (15) +  (17) +  (19)]
5

1
[(0)
5·3

+ 4 (02) + 2 (04) + 4 (06) + 2 (08)

+ 4 (1) + 2 (12) + 4 (14) + 2 (16) + 4 (18) +  (2)]

≈ 2663244
√
11.  () = ln , ∆ =
(a) 6 =

1
[ (1)
2·2

4−1
6

=

1
2

+ 2 (15) + 2 (2) + 2(25) + 2(3) + 2 (35) +  (4)] ≈ 2591334

(b) 6 = 1 [ (125) +  (175) +  (225) +  (275) +  (325) +  (375)] ≈ 2681046
2
(c) 6 =

1
[(1)
2·3

√

13.  () = 

+ 4 (15) + 2 (2) + 4 (25) + 2 (3) + 4 (35) + (4)] ≈ 2631976



sin , ∆ = 4 − 0 = 1
8
2

1
 
 
 

(a) 8 = 2 1 2 (0) + 2 2 + 2 (1) + 2 3 + 2(2) + 2 5 + 2 (3) + 2 7 +  (4) ≈ 4513618
·
2
2
2
  
 
 
 
 
 
 
 
(b) 8 = 1  1 +  3 +  5 +  7 +  9 +  11 +  13 +  15 ≈ 4748256
2
4
4
4
4
4
4
4
4
c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 7.7

(c) 8 =

1
2·3

APPROXIMATE INTEGRATION

¤

355


 
 
 
 

 (0) + 4 1 + 2 (1) + 4 3 + 2 (2) + 4 5 + 2 (3) + 4 7 +  (4) ≈ 4675111
2
2
2
2

cos 
, ∆ = 5 − 1 = 1
8
2


3
 

1
(a) 8 = 2 · 2 (1) + 2 2 + 2 (2) + · · · + 2(4) + 2 9 +  (5) ≈ −0495333
2
  
 
 
 
 
 
 
 
(b) 8 = 1  5 +  7 +  9 +  11 +  13 +  15 +  17 +  19 ≈ −0543321
2
4
4
4
4
4
4
4
4

15.  () =

(c) 8 =

1
2·3



 
 
 
 

 (1) + 4 3 + 2 (2) + 4 5 + 2 (3) + 4 7 + 2 (4) + 4 9 +  (5) ≈ −0526123
2
2
2
2

17.  () =  , ∆ =

(a) 10 =

1 − (−1)
10

1
[ (−1)
5·2

=

1
5

+ 2 (−08) + 2 (−06) + 2 (−04) + 2 (−02) + 2 (0)
+ 2 (02) + 2 (04) + 2(06) + 2(08) +  (1)]

≈ 8363853
(b) 10 = 1 [ (−09) +  (−07) +  (−05) +  (−03) +  (−01) +  (01) +  (03) +  (05) +  (07) +  (09)]
5

(c) 10 =

1
[(−1)
5·3

e

≈ 8163298

+ 4(−08) + 2 (−06) + 4(−04) + 2 (−02)

al

+ 4 (0) + 2 (02) + 4 (04) + 2 (06) + 4(08) +  (1)]
≈ 8235114
1−0
8

=

1
8

rS

19.  () = cos(2 ), ∆ =


  
 
 

 (0) + 2  1 +  2 + · · · +  7 +  (1) ≈ 0902333
8
8
8
 1
3
5
 
8 = 1  16 +  16 +  16 + · · · +  15 = 0905620
8
16

(a) 8 =

1
8·2

Fo

(b)  () = cos(2 ),  0 () = −2 sin(2 ),  00 () = −2 sin(2 ) − 42 cos(2 ). For 0 ≤  ≤ 1, sin and cos are positive,
  so | 00 ()| = 2 sin(2 ) + 42 cos(2 ) ≤ 2 · 1 + 4 · 1 · 1 = 6 since sin(2 ) ≤ 1 and cos 2 ≤ 1 for all , and 2 ≤ 1 for 0 ≤  ≤ 1. So for  = 8, we take  = 6,  = 0, and  = 1 in Theorem 3, to get
1
128

= 00078125 and | | ≤

ot

| | ≤ 6 · 13 (12 · 82 ) =

1
256

= 000390625. [A better estimate is obtained by noting

from a graph of  that | ()| ≤ 4 for 0 ≤  ≤ 1.]
00

00

N

(c) Take  = 6 [as in part (b)] in Theorem 3. | | ≤
1
1
≤ 4
22
10

⇔ 22 ≥ 104

Theorem 3 to get | | ≤ 10−4
21.  () = sin , ∆ =

−0
10

( − )3
≤ 00001 ⇔
122

6(1 − 0)3
≤ 10−4
122

⇔

⇔ 2 ≥ 5000 ⇔  ≥ 71. Take  = 71 for  . For  , again take  = 6 in
⇔ 42 ≥ 104

⇔ 2 ≥ 2500 ⇔  ≥ 50. Take  = 50 for  .


= 10



 
 

 (0) + 2 10 + 2 2 + · · · + 2 9 +  () ≈ 1983524
10
10
 
 
 



10 = 10  20 +  3 +  5 + · · · +  19 ≈ 2008248
20
20
20


 
 
 

10 = 10· 3 (0) + 4 10 + 2 2 + 4 3 + · · · + 4 9 + () ≈ 2000110
10
10
10

(a) 10 =


10 · 2

Since  =


0



sin   = − cos  0 = 1 − (−1) = 2,  =  − 10 ≈ 0016476,  =  − 10 ≈ −0008248,

and  =  − 10 ≈ −0000110.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

356

¤

CHAPTER 7

TECHNIQUES OF INTEGRATION





(b)  () = sin  ⇒  () () ≤ 1, so take  = 1 for all error estimates.
| | ≤

( − )3
1( − 0)3
3
≈ 0025839.
=
=
122
12(10)2
1200

| | ≤

( − )5
1( − 0)5
5
=
=
≈ 0000170.
4
4
180
180(10)
1,800,000

| | ≤

3
| |
=
≈ 0012919.
2
2400

The actual error is about 64% of the error estimate in all three cases.
3
1
≤ 5
122
10

⇔

| | ≤ 000001 ⇔

3
1
≤ 5
242
10

⇔ 2 ≥

| | ≤ 000001

5
1
≤ 5
1804
10

⇔

105  3
12

2 ≥

105 3
24

⇔ 4 ≥

⇒  ≥ 5083. Take  = 509 for .
⇒  ≥ 3594. Take  = 360 for .

105 5
180

⇒  ≥ 203.

e

(c) | | ≤ 000001 ⇔

23. (a) Using a CAS, we differentiate  () = cos  twice, and find that

al

Take  = 22 for  (since  must be even).

rS

 00 () = cos  (sin2  − cos ). From the graph, we see that the maximum value of | 00 ()| occurs at the endpoints of the interval [0 2].
Since  00 (0) = −, we can use  =  or  = 28.

Fo

(b) A CAS gives 10 ≈ 7954926518. (In Maple, use Student[Calculus1][RiemannSum] or
Student[Calculus1][ApproximateInt].)

(c) Using Theorem 3 for the Midpoint Rule, with  = , we get | | ≤
28(2 − 0)3
= 0 289391916.
24 · 102

ot

With  = 28, we get | | ≤

(2 − 0)3
≈ 0280945995.
24 · 102

N

(d) A CAS gives  ≈ 7954926521.

(e) The actual error is only about 3 × 10−9 , much less than the estimate in part (c).
(f) We use the CAS to differentiate twice more, and then graph
 (4) () = cos  (sin4  − 6 sin2  cos  + 3 − 7 sin2  + cos ).




From the graph, we see that the maximum value of  (4) () occurs at the endpoints of the interval [0 2]. Since  (4) (0) = 4, we can use  = 4 or  = 109.
(g) A CAS gives 10 ≈ 7953789422. (In Maple, use Student[Calculus1][ApproximateInt].)
(h) Using Theorem 4 with  = 4, we get | | ≤
With  = 109, we get | | ≤

4(2 − 0)5
≈ 0059153618.
180 · 104

109(2 − 0)5
≈ 0059299814.
180 · 104

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 7.7

APPROXIMATE INTEGRATION

¤

357

(i) The actual error is about 7954926521 − 7953789422 ≈ 000114. This is quite a bit smaller than the estimate in part (h), though the difference is not nearly as great as it was in the case of the Midpoint Rule.

( j) To ensure that | | ≤ 00001, we use Theorem 4: | | ≤

4(2)5
≤ 00001 ⇒
180 · 4

4(2)5
≤ 4
180 · 00001

⇒

4 ≥ 5,915,362 ⇔  ≥ 493. So we must take  ≥ 50 to ensure that | −  | ≤ 00001.

( = 109 leads to the same value of .)
25.  =

1
0

  = [( − 1) ]1
0

[parts or Formula 96] = 0 − (−1) = 1,  () =  , ∆ = 1

5 = 1 [ (0) + (02) +  (04) +  (06) +  (08)] ≈ 0742943
5

 = 5:

5 = 1 [ (02) +  (04) +  (06) + (08) +  (1)] ≈ 1286599
5

5 =

1
5 · 2 [ (0) + 2(02) + 2(04) + 2(06) + 2(08) +  (1)]
1
[ (01) +  (03) +  (05) + (07) +  (09)] ≈ 0992621
5

 =  − 5 ≈ 1 − 0742943 = 0257057

al

 ≈ 1 − 1286599 = −0286599
 ≈ 1 − 1014771 = −0014771

10 =
10 =

1
10 [(0) +  (01) + (02) + · · · +  (09)] ≈ 0867782
1
[(01) +  (02) + · · · +  (09) +  (1)] ≈ 1139610
10
1
{ (0) + 2[ (01) +  (02) + · · · + (09)] +  (1)} ≈ 1003696
10 · 2
1
[(005) + (015) + · · · + (085) + (095)] ≈ 0998152
10

Fo

10 =

rS

 ≈ 1 − 0992621 = 0007379
 = 10: 10 =

≈ 1014771

e

5 =

 =  − 10 ≈ 1 − 0867782 = 0132218
 ≈ 1 − 1139610 = −0139610

ot

 ≈ 1 − 1003696 = −0003696
 ≈ 1 − 0998152 = 0001848
 = 20: 20 =

N

20 =

1
[(0) +  (005) +  (010) + · · · +  (095)] ≈ 0932967
20
1
[(005) + (010) + · · · + (095) + (1)] ≈ 1068881
20
1
20 · 2 { (0) + 2[ (005) +  (010) + · · · +  (095)] +  (1)} ≈ 1000924
1
[(0025) +  (0075) +  (0125) + · · · +  (0975)] ≈ 0999538
20

20 =
20 =

 =  − 20 ≈ 1 − 0932967 = 0067033
 ≈ 1 − 1068881 = −0068881
 ≈ 1 − 1000924 = −0000924
 ≈ 1 − 0999538 = 0000462










5 0742943 1286599 1014771 0992621
10 0867782 1139610 1003696 0998152
20 0932967 1068881 1000924 0999538











5 0257057 −0286599 −0014771 0007379

10 0132218 −0139610 −0003696 0001848
20 0067033 −0068881 −0000924 0000462
[continued]

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

358

¤

CHAPTER 7

TECHNIQUES OF INTEGRATION

Observations:
1.  and  are always opposite in sign, as are  and  .
2. As  is doubled,  and  are decreased by about a factor of 2, and  and  are decreased by a factor of about 4.
3. The Midpoint approximation is about twice as accurate as the Trapezoidal approximation.
4. All the approximations become more accurate as the value of  increases.
5. The Midpoint and Trapezoidal approximations are much more accurate than the endpoint approximations.
27.  =

2
0

1

2
2
5 0 = 32 − 0 = 64,  () = 4 , ∆ = 2 − 0 = 
5


  
 
 
 
 

6 = 6 2 2  (0) + 2  1 +  2 +  3 +  4 +  5 +  (2) ≈ 6695473
·
3
3
3
3
3
  
 
 
 
 
 
6 = 2  1 +  3 +  5 +  7 +  9 +  11 ≈ 6252572
6
6
6
6
6
6
6

 
 
 
 
 

6 = 6 2 3  (0) + 4 1 + 2 2 + 4 3 + 2 4 + 4 5 +  (2) ≈ 6403292
·
3
3
3
3
3

4  =

 = 6:

5

e

 =  − 6 ≈ 64 − 6695473 = −0295473
 ≈ 64 − 6252572 = 0147428
 = 12:

rS

al

 ≈ 64 − 6403292 = −0003292

  
 
 
 

12 = 122· 2  (0) + 2  1 +  2 +  3 + · · · +  11 +  (2) ≈ 6474023
6
6
6
6
 1
3
5
 
2
6 = 12  12 +  12 +  12 + · · · +  23 ≈ 6363008
12

1
2
3
 
 

2
6 = 12 · 3 (0) + 4 6 + 2 6 + 4 6 + 2 4 + · · · + 4 11 +  (2) ≈ 6400206
6
6
 =  − 12 ≈ 64 − 6474023 = −0074023

Fo

 ≈ 64 − 6363008 = 0036992

 ≈ 64 − 6400206 = −0000206




6

6695473

6252572

12

6474023











6403292

6

−0295473

0147428

−0003292

ot



Observations:

6363008

6400206

12

−0074023

0036992

−0000206

N

1.  and  are opposite in sign and decrease by a factor of about 4 as  is doubled.
2. The Simpson’s approximation is much more accurate than the Midpoint and Trapezoidal approximations, and  seems to decrease by a factor of about 16 as  is doubled.
29. ∆ = ( − ) = (6 − 0)6 = 1

(a) 6 =

∆
[(0)
2

+ 2 (1) + 2 (2) + 2 (3) + 2 (4) + 2 (5) +  (6)]

≈ 1 [3 + 2(5) + 2(4) + 2(2) + 2(28) + 2(4) + 1]
2
= 1 (396) = 198
2
(b) 6 = ∆[ (05) + (15) +  (25) +  (35) +  (45) + (55)]
≈ 1[45 + 47 + 26 + 22 + 34 + 32]
= 206

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 7.7

(c) 6 =

∆
3 [ (0)

APPROXIMATE INTEGRATION

¤

+ 4 (1) + 2 (2) + 4(3) + 2 (4) + 4 (5) +  (6)]

≈ 1 [3 + 4(5) + 2(4) + 4(2) + 2(28) + 4(4) + 1]
3
= 1 (616) = 2053
3
31. (a)

5
1

 ()  ≈ 4 =

5−1
4 [ (15)

+  (25) +  (35) +  (45)] = 1(29 + 36 + 40 + 39) = 144

(b) −2 ≤  00 () ≤ 3 ⇒ | 00 ()| ≤ 3 ⇒  = 3, since | 00 ()| ≤ . The error estimate for the Midpoint Rule is
| | ≤
33. ave =

( − )3
3(5 − 1)3
1
=
= .
2
24
24(4)2
2
 24

1
24 − 0

0

 ()  ≈

1
24 12

=

1 24 − 0
24 3(12) [ (0)

+ 4 (2) + 2 (4) + 4 (6) + 2 (8) + 4 (10) + 2 (12)

+ 4 (14) + 2 (16) + 4 (18) + 2 (20) + 4 (22) +  (24)]
≈

1
[67
36

=

1
(2317)
36

+ 4(65) + 2(62) + 4(58) + 2(56) + 4(61) + 2(63) + 4(68)

e

+ 2(71) + 4(69) + 2(67) + 4(66) + 64]

al

= 64361◦ F.

The average temperature was about 644◦ F.

6

() . We use Simpson’s Rule with  = 6 and

rS

35. By the Net Change Theorem, the increase in velocity is equal to

0

∆ = (6 − 0)6 = 1 to estimate this integral:
6
()  ≈ 6 = 1 [(0) + 4(1) + 2(2) + 4(3) + 2(4) + 4(5) + (6)]
3
0

Fo

≈ 1 [0 + 4(05) + 2(41) + 4(98) + 2(129) + 4(95) + 0] = 1 (1132) = 3773 fts
3
3

37. By the Net Change Theorem, the energy used is equal to

∆ =

6−0
12

=

1
2

to estimate this integral:

12
3 [ (0)

0

 () . We use Simpson’s Rule with  = 12 and

ot

6

6

 ()  ≈ 12 =

+ 4 (05) + 2 (1) + 4 (15) + 2 (2) + 4 (25) + 2 (3)
+ 4 (35) + 2 (4) + 4 (45) + 2 (5) + 4 (55) +  (6)]

N

0

= 1 [1814 + 4(1735) + 2(1686) + 4(1646) + 2(1637) + 4(1609) + 2(1604)
6
+ 4(1611) + 2(1621) + 4(1666) + 2(1745) + 4(1886) + 2052]

=

1
6 (61,064)

= 10,1773 megawatt-hours

39. (a) Let  = () denote the curve. Using disks,  =

Now use Simpson’s Rule to approximate 1 :
1 ≈ 8 =

10 − 2
[(2)
3(8)

 10
2

[ ()]2  = 

 10
2

()  = 1 .

+ 4(3) + 2(4) + 4(5) + 2(6) + 4(7) + (8)]

≈ 1 [02 + 4(15)2 + 2(19)2 + 4(22)2 + 2(30)2 + 4(38)2 + 2(40)2 + 4(31)2 + 02 ]
3

= 1 (18178)
3

Thus,  ≈  · 1 (18178) ≈ 1904 or 190 cubic units.
3

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

359

360

¤

CHAPTER 7

TECHNIQUES OF INTEGRATION

(b) Using cylindrical shells,  =

 10
2

 10

2()  = 2

2

Now use Simpson’s Rule to approximate 1 :
1 ≈ 8 =

10 − 2
3(8) [2 (2)

 ()  = 21 .

+ 4 · 3 (3) + 2 · 4(4) + 4 · 5 (5) + 2 · 6 (6)
+ 4 · 7 (7) + 2 · 8 (8) + 4 · 9(9) + 10 (10)]

≈ 1 [2(0) + 12(15) + 8(19) + 20(22) + 12(30) + 28(38) + 16(40) + 36(31) + 10(0)]
3

= 1 (3952)
3

Thus,  ≈ 2 · 1 (3952) ≈ 8277 or 828 cubic units.
3
41. Using disks,  =

1 . 1 ≈ 8 =

5
1

5−1
3(8)

(−1 )2  = 

5
1

−2  = 1 . Now use Simpson’s Rule with () = −2 to approximate

[ (1) + 4 (15) + 2 (2) + 4 (25) + 2 (3) + 4 (35) + 2 (4) + 4(45) +  (5)] ≈ 1 (114566)
6

Thus,  ≈  · 1 (114566) ≈ 60 cubic units.
6

e

 2 sin2 
 sin 
(104 )2 sin2 
,  = 10,000,  = 10−4 , and  = 6328 × 10−9 . So () =
, where  =
,
2

2

where  =

(104 )(10−4 ) sin 
10−6 − (−10−6 )
. Now  = 10 and ∆ =
= 2 × 10−7 , so
6328 × 10−9
10

al

43. () =

rS

10 = 2 × 10−7 [(−00000009) + (−00000007) + · · · + (00000009)] ≈ 594.
45. Consider the function  whose graph is shown. The area

is close to 2. The Trapezoidal Rule gives
2−0
2·2

[ (0) + 2 (1) +  (2)] =

The Midpoint Rule gives 2 =

2−0
2

1
2

0

() 

[1 + 2 · 1 + 1] = 2.

Fo

2 =

2

[(05) +  (15)] = 1[0 + 0] = 0,

so the Trapezoidal Rule is more accurate.

ot

47. Since the Trapezoidal and Midpoint approximations on the interval [ ] are the sums of the Trapezoidal and Midpoint

approximations on the subintervals [−1   ],  = 1 2     , we can focus our attention on one such interval. The condition

N

 00 ()  0 for  ≤  ≤  means that the graph of  is concave down as in Figure 5. In that figure,  is the area of the

trapezoid ,   ()  is the area of the region  , and  is the area of the trapezoid , so

    ()    . In general, the condition  00  0 implies that the graph of  on [ ] lies above the chord joining the

points (  ()) and (  ()). Thus,   ()    . Since  is the area under a tangent to the graph, and since  00  0

 implies that the tangent lies above the graph, we also have     () . Thus,     ()    .

49.  =

1
2

∆ [ (0 ) + 2 (1 ) + · · · + 2 (−1 ) +  ( )] and

 = ∆ [(1 ) +  (2 ) + · · · +  (−1 ) +  ( )], where  = 1 (−1 +  ). Now
2


1 1
2 = 2 2 ∆ [ (0 ) + 2 (1 ) + 2 (1 ) + 2 (2 ) + 2 (2 ) + · · · + 2 (−1 ) + 2 (−1 ) + 2( ) + ( )] so
1
(
2

+  ) = 1  + 1 
2
2

= 1 ∆[ (0 ) + 2 (1 ) + · · · + 2 (−1 ) +  ( )] + 1 ∆[2 (1 ) + 2 (2 ) + · · · + 2 (−1 ) + 2( )]
4
4
= 2 c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 7.8

IMPROPER INTEGRALS

¤

7.8 Improper Integrals
1. (a) Since  =



(b) Since

∞

0



(c) Since

∞

 has an infinite discontinuity at  = 1,
−1



2


 is a Type 2 improper integral.
−1

1

1
 has an infinite interval of integration, it is an improper integral of Type 1.
1 + 3
2

2 −  has an infinite interval of integration, it is an improper integral of Type 1.

−∞

(d) Since  = cot  has an infinite discontinuity at  = 0,

 4
0

cot   is a Type 2 improper integral.

3. The area under the graph of  = 13 = −3 between  = 1 and  =  is

() =


1



 
−3  = − 1 −2 1 = − 1 −2 − − 1 =
2
2
2

1
2

 
− 1 22 . So the area for 1 ≤  ≤ 10 is

e

(10) = 05 − 0005 = 0495, the area for 1 ≤  ≤ 100 is (100) = 05 − 000005 = 049995, and the area for



∞

3

1
 = lim
→∞
( − 2)32
= lim

→∞

7.



0

−∞

1
 = lim
→−∞
3 − 4



3





0



Divergent

11.

2




→∞ 2

−5  = lim

∞

0

13.

∞
0

−  =
2

−∞

→∞

−2
2
√
+√
−2
1



= 0 + 2 = 2.

0

−  = lim

[ =  − 2,  = ]

Convergent

0


1
 = lim − 1 ln |3 − 4| = lim − 1 ln 3 +
4
4
→−∞
→−∞
3 − 4


→∞





1
4

 ln |3 − 4| = ∞

0

→∞

 

 

2
2
3
√
 = lim
1+
= lim 2 1 + 3 − 2 = ∞. Divergent
3
3
→∞ 3
→∞
1 + 3
0

2

−∞

3





−5  = lim − 1 −5 2 = lim − 1 −5 + 1 −10 = 0 + 1 −10 = 1 −10 . Convergent
5
5
5
5
5

2
√
 = lim
→∞
1 + 3
2

−∞



( − 2)−32  = lim −2 ( − 2)−12

ot

∞

N

9.



rS

5.

→∞

Fo

→∞

al

1 ≤  ≤ 1000 is (1000) = 05 − 00000005 = 04999995. The total area under the curve for  ≥ 1 is

 lim () = lim 1 − 1(22 ) = 1 .
2
2

−  +

∞
0

2

− .


 1  −2 0
 
2
−2 
= lim − 1 1 − − = − 1 · 1 = − 1 , and
2
2
2

→−∞



→−∞



 
  2
 ∞ −2
2 

 = lim − 1 −
= lim − 1 − − 1 = − 1 · (−1) = 1 .
2
2
2
2
0
→∞

Therefore,
15.

∞
0

∞

−∞

2

−  = − 1 +
2
 1
(1
→∞ 0 2

sin2   = lim

Divergent

0

1
2

→∞

= 0.

Convergent

− cos 2)  = lim

→∞

1
−
2

1
2

 

sin 2 0 = lim 1  −
2
→∞

1
2



sin 2 − 0 = ∞.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

361

CHAPTER 7

∞

 


1
1
[partial fractions]
−

 +1
1
1

 




   

 = lim ln  − ln 1 = 0 − ln 1 = ln 2.
= lim ln || − ln | + 1| = lim ln 
→∞
→∞
 + 1  1 →∞
+1
2
2
1

1
 = lim
→∞
2 + 

1

Convergent
19.

TECHNIQUES OF INTEGRATION

0



0

2  = lim
−∞



1
 = lim
→∞
( + 1)

2  = lim

→−∞ 

→−∞

1

2 − 1 2
2
4



0

integration by parts with
 = ,  = 2 




 

0 − 1 − 1 2 − 1 2 = − 1 − 0 + 0
= lim
4
2
4
4
→−∞





(ln )2 ln 
 = lim
→∞

2
1

∞

1

23.



∞

−∞

2
 =
9 + 6


Now

so 2



∞

0



0



2
 +
9 + 6

−∞



2 
9 + 6





 = 3
 = 32 

2
 = 2 lim
→∞
9 + 6





0

27.



1

0

29.



−2



3

−2



1

0



2
 [since the integrand is even].
9 + 6

 = 3
 = 3 

=



1
(3 )
3
9 + 9 2

=

1
9




 3 
 3
2
1
1


2 

tan−1
= · = .
 = 2 lim
= 2 lim tan−1
→∞ 9
→∞ 9
9 + 6
3
3
9 2
9
0






1


3
3
1
− 4 = − lim 1 − 4 = ∞.
4 
4 →0+

→0+

3−5  = lim




√
= lim
4
 + 2 →−2+
=

31.


9 + 2

∞



ln 

 ln 
1
1
 = ln ,
= lim − 2
 = lim
−3 
3
 = 
→∞ 1
→∞
2 1
 (ln )


1
1
1
1
=0+ = .
Convergent
= lim −
+
→∞
2
2
2
2 (ln )2

1
 = lim
→∞
(ln )3

3
 = lim
5
→0+

14

1
3





Fo



0



2
 = 2
9 + 6

ot

∞

∞

N





Divergent

= lim


1 + 2
 3


1
1
1
+  = tan−1
+ ,
= tan−1  +  = tan−1
9
9
3
9
3

=

Convergent
25.

(ln )2
= ∞.
→∞
2

by substitution with
 = ln ,  = 

= − 1 . Convergent
4

[by l’Hospital’s Rule]

rS

21.



e

17.



al

¤

362


=
4



0

−2

4
(8
3


+
4



14

→−2+



− 0) =


0

( + 2)−14  = lim

3

32
.
3


, but
4



4
( + 2)34
3

14

=



Convergent


0

−2

Divergent



4 lim 1634 − ( + 2)34
3 →−2+

 −3 


1


1
= lim −
= lim − 3 −
= ∞.
4
3 −2 →0−
3
24
→0−

Divergent

 1
 9
1
√
 =
( − 1)−13  +
( − 1)−13 .
3
−1
0
0
1





1
Here 0 ( − 1)−13  = lim 0 ( − 1)−13  = lim 3 ( − 1)23 = lim 3 ( − 1)23 − 3 = − 3
2
2
2
2

33. There is an infinite discontinuity at  = 1.

→1−



9

→1−

0

→1−

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 7.8

and


9

9

0

1

( − 1)−13  = lim

→1+

9


( − 1)−13  = lim

→1+



3
2 (

9

− 1)23



IMPROPER INTEGRALS



= lim 6 − 3 ( − 1)23 = 6. Thus,
2
→1+

1
9
3
√
 = − + 6 = . Convergent
3
2
2
−1

35.  =



3

0

Now


=
2 − 6 + 5



3




= 1 + 2 =
( − 1)( − 5)

0

1


=
+
( − 1)( − 5)
−1
−5

1

0


+
( − 1)( − 5)



3

1


.
( − 1)( − 5)

⇒ 1 = ( − 5) + ( − 1).

Set  = 5 to get 1 = 4, so  = 1 . Set  = 1 to get 1 = −4, so  = − 1 . Thus
4
4
 

= ∞, since lim

→1−

Since 1 is divergent,  is divergent.
0

−1

1
 = lim
3
→0−





1 1 1

· 2  = lim


→0−
−1


−1
use parts
= lim ( − 1) 1



 = 1,
 = −2

 (−)







1
−2−1 −
− 1 1

→0−



= lim

−1 H 2
1
2
= − − lim
= − − lim
 →−∞ −
 →−∞ −−

Fo

2
= − − lim ( − 1)
 →−∞





−1

or Formula 96

→0−

1

rS

37.



 1

− 4 ln | − 1| = ∞.

al

→1−

e


1


−1
4
+ 4
 = lim − 1 ln | − 1| + 1 ln | − 5| 0
4
4
−1
−5
→1− 0
→1−

 

= lim − 1 ln | − 1| + 1 ln | − 5| − − 1 ln |−1| + 1 ln |−5|
4
4
4
4

1 = lim

2
2
=− −0=− .


2
0

 2 ln   = lim

= lim

9

→0+

(3 ln 2 − 1) −



 2 ln   = lim

1 3
 (3 ln 
9

N

→0+

8

2

Convergent

ot

39.  =

[ = 1]

→0+


− 1) =



8
3

2
3
(3 ln  − 1)
32


ln 2 −

8
9

−

1
9



integrate by parts or use Formula 101



lim 3 (3 ln  − 1) =

→0+



8
3

ln 2 −

8
9

− 1 .
9





3 ln  − 1 H
3
= lim
= lim −3 = 0.
Now  = lim 3 (3 ln  − 1) = lim
4
−3
→0+
→0+
→0+ −3
→0+
Thus,  = 0 and  =

41.

8
3

ln 2 − 8 .
9

Convergent

Area =

∞
1

−  = lim

→∞ 1

−

= lim (−
→∞



−1

+



−  = lim −−
→∞

−1

) =0+

1

= 1

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

363

364

¤

CHAPTER 7

TECHNIQUES OF INTEGRATION

 
1
1
 = lim

→∞ 1 (2 + 1)
3 + 
1

 
1

− 2

[partial fractions]
= lim
→∞ 1
  +1




 
1 

= lim ln || − ln 2 + 1 = lim ln √
→∞
→∞
2
2 + 1 1
1


1
1

− ln √
= lim ln √
= ln 1 − ln 2−12 = ln 2
→∞
2
2 + 1
2

43.

45.

Area =



Area =

=

∞

 2
0

sec2   =

lim

→(2)−

lim

→(2)−

(tan  − 0) = ∞


0

sec2   =

lim

→(2)−



tan 

0

e

Infinite area

sin2 
.
2
It appears that the integral is convergent.

() =

2



5

0577101

10

0621306

100

0668479

1000

0672957

10,000

0673407

1

() 

rS

0447453

Fo



al

47. (a)

(b) −1 ≤ sin  ≤ 1 ⇒ 0 ≤ sin2  ≤ 1 ⇒ 0 ≤
∞



1

∞

1
 is convergent
2

2

sin 
 is convergent by the Comparison Theorem.
2

ot



sin2 
1
≤ 2 . Since
2


[Equation 2 with  = 2  1],

1

N

(c)


1

 3 = 2.
3 + 1


 1 by the Comparison Theorem.

49. For   0,

0



Since

∞

 ()  is finite and the area under () is less than the area under  ()
∞
on any interval [1 ], 1 ()  must be finite; that is, the integral is convergent.
1

 ∞
1

 is convergent by Equation 2 with  = 2  1, so
 is convergent
2
3 + 1
1
1
 ∞
 1
 ∞




 is a constant, so
 =
 +
 is also
3 + 1
3 + 1
3 + 1
3 + 1
0
0
1
∞

convergent.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 7.8



∞

IMPROPER INTEGRALS

+1

1
+1
 2 = , so
 ()  diverges by comparison with
 √
4 −
4




2
2
∞
∞
by Equation 2 with  = 1 ≤ 1. Thus, 1 ()  = 1  ()  + 2  ()  also diverges.

51. For   1,  () = √



2

∞

¤

1
, which diverges


sec2 
1
√  32 . Now

 


 1
 1

1
2
=
−32  = lim
−32  = lim − 2−12 = lim −2 + √
= ∞, so  is divergent, and by

→0+ 
→0+
→0+

0
 1 sec2 
√ is divergent. comparison, 
0 

53. For 0   ≤ 1,

 1
 
 1
 ∞





√
√
√
√
√
55.
=
+
= lim
+ lim
. Now
 (1 + )
 (1 + )
 (1 + ) →0+ 
 (1 + ) →∞ 1
 (1 + )
0
0
1





√
√


2 
 = ,  = 2 ,
√
= 2 tan−1  +  = 2 tan−1  + , so
=2
=
2)
2
 = 2 
(1 + 
1+
 (1 + )
 ∞


√ 1
√ 

√
= lim 2 tan−1   + lim 2 tan−1  1
→∞
 (1 + )
→0+
0
√ 
√
  

 
 
= lim 2  − 2 tan−1  + lim 2 tan−1  − 2  =  − 0 + 2  −  = .
4
4
2
2
2
∞

→∞

57. If  = 1, then



1

0



1

0


= lim

→0+



1





1


= lim [ln ]1 = ∞.


→0+



Divergent

[note that the integral is not improper if   0]

Fo

If  6= 1, then


= lim

→0+

rS

→0+

al

e



= lim

→0+





1

−+1
− + 1

= lim

→0+





1
1
1 − −1
1−


1
→ ∞ as  → 0+ , and the integral diverges.
−1


 1


1
1
1
1− 
+
= lim 1 − 
.
=
If   1, then  − 1  0, so −1 → 0 as  → 0 and


1 −  →0+
1−
0

N

ot

If   1, then  − 1  0, so

Thus, the integral converges if and only if   1, and in that case its value is
59. First suppose  = −1. Then



0

1

 ln   =



0

1

ln 
 = lim

→0+





1

365

1
.
1−


1
ln 
 = lim 1 (ln )2  = − 1 lim (ln )2 = −∞, so the
2
+ 2

→0
→0+

integral diverges. Now suppose  6= −1. Then integration by parts gives


+1
+1
+1

 ln   =
+ . If   −1, then  + 1  0, so ln  −
 = ln  −
+1
+1
+1
( + 1)2



 +1
1


 1
1

+1
−1
1
ln  − lim +1 ln  −
= ∞.
 ln   = lim
=
−
( + 1)2 
 + 1 →0+
+1
→0+  + 1
( + 1)2
0
If   −1, then  + 1  0 and

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366

¤

CHAPTER 7

1

TECHNIQUES OF INTEGRATION

0

Thus, the integral converges to −
  =

so  is divergent.
(b)



−

  =

63. Volume =



∞

1
2



1

65. Work =



∞

2

0

−∞



−

→0+

ln  − 1( + 1) H
−1
=
−
( + 1)2
−(+1)



1
+1



1
−( + 1)−(+2)

lim

→0+

1 if   −1 and diverges otherwise.
( + 1)2

  +

∞
0

 , and

∞
0


→∞ 0

  = lim


→∞ −

= 1 2 − 1 2 = 0, so lim
2
2

  = lim

→∞

  = 0. Therefore,

∞

1

−∞

2

2



0

= lim

→∞

1
2


→∞ −

  6= lim


2 − 0 = ∞,

 .





 2
 

1
1
1
=   ∞.
 =  lim
=  lim −
=  lim 1 −
→∞ 1 2
→∞
→∞

 1




1

1

−
=
. The initial kinetic energy provides the work,
 = lim 
→∞ 
→∞
2




2
⇒ 0 =
.


  = lim










rS

2 so 1 0 =
2

lim

e

−∞



al

∞

1
+1

−1
1
−1
+
lim +1 =
( + 1)2
( + 1)2 →0+
( + 1)2

=

61. (a)  =



−1
−
( + 1)2

 ln   =

67. We would expect a small percentage of bulbs to burn out in the first few hundred hours, most of the bulbs to burn out after

close to 700 hours, and a few overachievers to burn on and on.
(a)

Fo

(b) () =  0 () is the rate at which the fraction  () of burnt-out bulbs increases as  increases. This could be interpreted as a fractional burnout rate.
∞
(c) 0 ()  = lim  () = 1, since all of the bulbs will eventually burn out.





∞

  0001 ⇒
71. (a)  () =



∞







1
 = lim tan−1   = lim tan−1  − tan−1  =  − tan−1 .
2
→∞
→∞
+1



− tan−1   0001 ⇒ tan−1    − 0001 ⇒   tan  − 0001 ≈ 1000.
2
2

1
 = lim
→∞
2 + 1



2

N

69.  =

ot

→∞


2

 ()−  =

0



∞

 − 
 −



1
1
−
+
. This converges to only if   0.
= lim
→∞
→∞
 0
−



−  = lim

0

1 with domain { |   0}.


 ∞
 ∞
−
 −
 ()
 =

 = lim
(b)  () =
Therefore  () =

0

= lim

→∞



0

(1−)


1
−
1−
1−

→∞





(1−)



 = lim

0

This converges only if 1 −   0 ⇒   1, in which case  () =

→∞



1 (1−)

1−


0

1 with domain { |   1}.
−1

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SECTION 7.8

(c)  () =

∞
0


→∞ 0

 ()−  = lim

¤

Therefore,  () =

− . Use integration by parts: let  = ,  = −  ⇒  = ,

∞
0

1 and the domain of  is { |   0}.
2

 0 ()− . Integrate by parts with  = − ,  =  0 ()  ⇒  = −− ,  =  ():


∞
() = lim  ()− 0 +  0  ()−  = lim  ()− −  (0) +  ()
→∞

→∞

⇒ 0 ≤  ()− ≤  − and lim (−) = 0 for   . So by the Squeeze Theorem,

But 0 ≤  () ≤ 

→∞

lim  ()− = 0 for    ⇒ () = 0 −  (0) +  () =  () −  (0) for   .

→∞








2
2
1

1 ∞ −2
1 ∞ −2
1 ∞ −2
2 −  = lim − −
+

 = lim − 2 +

 =


→∞
→∞
2
2 0
2 0
2 0
2
0

al

0

(The limit is 0 by l’Hospital’s Rule.)

77. For the first part of the integral, let  = 2 tan 

1
√
 =
2 + 4



2 sec2 
 =
2 sec 



⇒  = 2 sec2  .

sec   = ln |sec  + tan |.

Fo



2

 = ,  = − 1 − . So
2

rS

∞

⇒

e

2

75. We use integration by parts: let  = ,  = − 



√
2 + 4

, and sec  =
. So
2
2

 √ 2


 ∞
  +4
1


√
 = lim ln 
+  −  ln| + 2|
=
−

→∞
+2
2
2
2 + 4
0
0

 √2
 +4+
−  ln( + 2) − (ln 1 −  ln 2)
= lim ln
→∞
2
√
 √ 2




 +4+
 + 2 + 4

= lim ln
+ ln 2 = ln lim
+ ln 2−1
→∞
→∞
2 ( + 2)
( + 2)
√
√
 + 2 + 4 H
1 +  2 + 4
2
Now  = lim
= lim
=
.
→∞
→∞  ( + 2)−1
( + 2)
 lim ( + 2)−1

N

ot

From the figure, tan  =

→∞

If   1,  = ∞ and  diverges.
If  = 1,  = 2 and  converges to ln 2 + ln 20 = ln 2.
If   1,  = 0 and  diverges to −∞.
79. No,  =

367





−
1
1
1

−
1
− 2  + 0 + 2 = 2 only if   0.
. Then  () = lim − − − 2 − = lim
→∞
→∞




 


0

=−

73. () =

IMPROPER INTEGRALS

∞
0

 ()  must be divergent. Since lim  () = 1, there must exist an  such that if  ≥ , then  () ≥ 1 .
2
→∞



∞

() , where 1 is an ordinary definite integral that has a finite value, and 2 is
∞
improper and diverges by comparison with the divergent integral  1 .
2

Thus,  = 1 + 2 =

0

 ()  +



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¤

368

CHAPTER 7 TECHNIQUES OF INTEGRATION

7 Review

1. See Formula 7.1.1 or 7.1.2. We try to choose  =  () to be a function that becomes simpler when differentiated (or at least

not more complicated) as long as  = 0 ()  can be readily integrated to give .
2. See the Strategy for Evaluating
3. If



sin  cos   on page 497.

√
√
√
2 − 2 occurs, try  =  sin ; if 2 + 2 occurs, try  =  tan , and if 2 − 2 occurs, try  =  sec . See the

Table of Trigonometric Substitutions on page 502.
4. See Equation 2 and Expressions 7, 9, and 11 in Section 7.4.
5. See the Midpoint Rule, the Trapezoidal Rule, and Simpson’s Rule, as well as their associated error bounds, all in Section 7.7.

e

We would expect the best estimate to be given by Simpson’s Rule.

7. See Definitions 3(b), (a), and (c) in Section 7.8.

Fo

rS

8. See the Comparison Theorem after Example 8 in Section 7.8.

al

6. See Definitions 1(a), (b), and (c) in Section 7.8.



 2 + 4
8


=+ 2
=+
+
.
2 − 4
 −4
+2
−2

1. False.

Since the numerator has a higher degree than the denominator,

3. False.

It can be put in the form

5. False.

This is an improper integral, since the denominator vanishes at  = 1.
 1
 4
 4



 =
 +
 and
2 − 1
2 − 1
2 − 1
0
0
1
 1
 


 


 = lim
 = lim 1 ln2 − 1 = lim
2
2 −1
2 −1
0
→1− 0 
→1−
→1−
0 

N

ot




+ 2 +
.


−4

So the integral diverges.

7. False.

See the end of Section 7.5.

(b) False.

13. False.



ln2 − 1 = ∞

See Exercise 61 in Section 7.8.

9. (a) True.

11. False.

1
2

2

Examples include the functions  () =  , () = sin(2 ), and () =

sin 
.


If  () = 1, then  is continuous and decreasing on [1 ∞) with lim  () = 0, but
→∞

∞
Take  () = 1 for all  and () = −1 for all . Then   ()  = ∞ [divergent]
∞
∞ and  ()  = −∞ [divergent], but  [ () + ()]  = 0 [convergent].

∞
1

 ()  is divergent.

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CHAPTER 7 REVIEW

2

1

3.

2


 2
2 + 2 + 1
1
1 2
+2+
 =
 =
 + 2 + ln ||


2
1
1
1
1
 7
= (2 + 4 + ln 2) − 2 + 2 + 0 = 2 + ln 2

( + 1)2
 =


 2
0



sin  cos   =
=

5.

7.




=
22 + 3 + 1

 2
0



2

0
1

1
0



 = cos 
 = − sin  

 (−)

 1
  =  = 1 − 0 =  − 1
0

=

 2
0

1
0

(2 − 4 )  =

1

1
3

3 − 1 5
5

sin(ln )
 =




1
0

0

=





[partial fractions]

(1 − 2 )2 (−)
1
1
3

−

= ln |2 + 1| − ln | + 1| + 


 = cos ,
 = − sin  

1
5



−0=



2
15

sin   = − cos  +  = − cos(ln ) + .

rS



11. Let  = sec . Then
2

2
1
−
2 + 1
+1

(1 − cos2 ) cos2  sin   =

9. Let  = ln ,  = . Then



 

1
 =
(2 + 1)( + 1)

sin3  cos2   =



e



√
 3
 3
 3

3 √
2 − 1 tan  tan2   =
(sec2  − 1)  = tan  −  0 = 3 −  .
 = sec  tan   =
3
 sec 
0
0
0

 √

√
3
3
. Then 3 =  and 32  = , so    =  · 32  = 3. To evaluate , let  = 2 ,


 =   ⇒  = 2 ,  =  , so  = 2   = 2  − 2 . Now let  = ,  =  



 = ,  =  . Thus,  = 2  − 2  −   = 2  − 2 + 2 + 1 , and hence

Fo

13. Let  =

√
3

N

−1
−1


=
=
+
⇒  − 1 = ( + 2) + . Set  = −2 to get −3 = −2, so  = 3 . Set  = 0
2
2 + 2
( + 2)

+2

  1

3
−2
−1
 =
+ 2
 = − 1 ln || + 3 ln | + 2| + . to get −1 = 2, so  = − 1 . Thus,
2
2
2
2 + 2

+2

17. Integrate by parts with  = ,  = sec  tan  


19.

⇒

(23 − 213 + 2) + .

ot

3 = 3 (2 − 2 + 2) +  = 3

15.

369

al

1.

¤



 sec  tan   =  sec  −



⇒

 = ,  = sec :

14

sec   =  sec  − ln|sec  + tan | + .




+1
+1
= 3 + 1,
 =

= 3 
(92 + 6 + 1) + 4
(3 + 1)2 + 4

1




( − 1) + 1 1
1 1
( − 1) + 3
3
 = ·

=
2 + 4
3
3 3
2 + 4
 


1
1

1
2
1 1
2 1
=
 +
 +
 = · ln(2 + 4) + · tan−1
9
2 + 4
9
2 + 22
9 2
9 2
2


1
= 18 ln(92 + 6 + 5) + 1 tan−1 1 (3 + 1) + 
9
2

+1
 =
92 + 6 + 5



c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

370



CHAPTER 7 TECHNIQUES OF INTEGRATION


√
=
2 − 4




=

21.





=
2 − 4 + 4) − 4
(



( − 2)2 − 22




2 sec  tan  
2 tan 

 − 2 = 2 sec ,
 = 2 sec  tan  



=



sec   = ln |sec  + tan | + 1
√


 − 2
2 − 4 
 + 1
+
= ln 

 2
2


√
= ln  − 2 + 2 − 4  + , where  = 1 − ln 2

23. Let  = tan , so that  = sec2  . Then

25.

sec2  
=
tan  sec 





sec 

tan 

csc   = ln |csc  − cot | + 
√
√


 2 + 1 − 1 
 2 + 1
1
+
= ln 
−  +  = ln 







33 − 2 + 6 − 4
 + 
 + 
= 2
+ 2
(2 + 1)(2 + 2)
 +1
 +2





⇒ 33 − 2 + 6 − 4 = ( + ) 2 + 2 + ( + ) 2 + 1 .

rS

=



e


√
=
 2 + 1

al



Equating the coefficients gives  +  = 3,  +  = −1, 2 +  = 6, and 2 +  = −4 ⇒

 2
0

cos3  sin 2  =

 2
0

cos3  (2 sin  cos )  =


3

ot

27.

Fo

 = 3,  = 0,  = −3, and  = 2. Now





√


33 − 2 + 6 − 4
3 
−1

+ .
 = 3
 + 2
= ln 2 + 1 − 3 tan−1  + 2 tan−1 √
(2 + 1)(2 + 2)
2 + 1
2 + 2
2
2

29. The integrand is an odd function, so

−3

0


2
2 cos4  sin   = − 2 cos5  0 =
5

2
5


 = 0 [by 4.5.6(b)].
1 + ||



ln 10

0

33. Let  = 2 sin 

2
(4 −

2 )32

N

√ 
 − 1. Then 2 =  − 1 and 2  =  . Also,  + 8 = 2 + 9. Thus,

31. Let  =



 2

√

 3
 3
 3
  − 1
 · 2 
2
9
1− 2
 =
=2
 = 2

2
 + 8
2 + 9
 +9
0
0  +9
0


 3



3

9
−1 
=6−
= 2 (3 − 3 tan−1 1) − 0 = 2 3 − 3 ·
= 2  − tan
3
3 0
4
2

⇒
 =

32

4 − 2
= (2 cos )3 ,  = 2 cos  , so



4 sin2 
2 cos   =
8 cos3 



tan2   =




 2 sec  − 1 



+
− sin−1
= tan  −  +  = √
2
4 − 2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 7 REVIEW

35.

37.









=
√
√ 
√
 (1 +  )
 1+ 

√
√
=4 + =4 1+ +

1
√
 =
 + 32






√ 
 = 1 + ,

 =





2−12 


(cos  + sin )2 cos 2  = (cos  + sin )2 (cos2  − sin2 ) 

= (cos  + sin )3 (cos  − sin )  = 1 (cos  + sin )4 + 1
4



2
 by parts with  = 2 and  =
. Then  = ( · 22 + 2 · 1) 
(1 + 2)2
(1 + 2)2



12

0

∞

1





 
1
1
−3
1
 = lim
(2 + 1) 2  = lim −
3
→∞ 1 2
→∞
4(2 + 1)2 1
1 (2 + 1)




1
1
1
1
1
1
= − lim
−
=−
0−
=
4 →∞ (2 + 1)2
9
4
9
36

1
 = lim
→∞
(2 + 1)3


 ln 
∞

2




= lim
 ln  →∞





=




= ln || +  = ln |ln | +  so





= lim ln |ln | = lim [ln(ln ) − ln(ln 2)] = ∞, so the integral is divergent.
→∞
→∞
 ln 
2

2



 √
√ 4 ln 
∗
√  = lim 2  ln  − 4 

→0+


√  ∗∗

 √
= lim (2 · 2 ln 4 − 4 · 2) − 2  ln  − 4  = (4 ln 4 − 8) − (0 − 0) = 4 ln 4 − 8

ln 
√  = lim
→0+


4

N

4



 = ln ,
 = 



ot







Fo



12



 

2
1

1
1
1
1
1
−
−
−1
−0 = − .
 = 2
=
(1 + 2)2
4
4 + 2 0
4
8
4
8
4

rS

Thus,




 
1
2
1 1

1 2 (2 + 1)
 = −
+ · 2 +  = 2
−
+
− ·
2
1 + 2
4 + 2
2 2
4
4 + 2

al

1 2
−
 =− ·
2 1 + 2

e

1
1
and  = − ·
, so
2 1 + 2

45.

2 
√ =


 2

 cos  + 2 sin  cos  + sin2  cos 2  = (1 + sin 2) cos 2 


= cos 2  + 1 sin 4  = 1 sin 2 − 1 cos 4 + 
2
2
8

39. We’ll integrate  =

43.



(cos  + sin )2 cos 2  =

Or:

41.


√
2 

 =

0

→0+

(∗)

(∗∗)

47.

¤



0

1

√
1
1
Let  = ln ,  = √  ⇒  = ,  = 2 . Then




√
√
√
 ln 
√  = 2  ln  − 2 √ = 2  ln  − 4  + 



 √
 √ 

2 ln  H
2
= lim −4  = 0 lim 2  ln  = lim −12 = lim
+
+ 
+ − 1 −32
+
→0
→0
→0
→0
2
 1


 1

1

1
√ − √  = lim
(12 − −12 )  = lim 2 32 − 212
3

→0+ 
→0+





  2 32
= lim 2 − 2 − 3 
− 212 = − 4 − 0 = − 4
3
3
3

−1
√  = lim
→0+

→0+

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

371

¤

372

CHAPTER 7 TECHNIQUES OF INTEGRATION

49. Let  = 2 + 1. Then



∞

−∞


=
42 + 4 + 5
=



∞

−∞

1
2
2


=
+4

1 lim 2 →−∞

1
2

1



0

−∞


+
2 + 4

 0 tan−1 1   +
2
2

1
2

1 lim 2 →∞



∞

0

1
2


2 + 4

tan−1


 1 





 0 = 1 0 − − + 1  − 0 =
2
4
2
4 2


.
4



51. We first make the substitution  =  + 1, so ln(2 + 2 + 2) = ln ( + 1)2 + 1 = ln(2 + 1). Then we use parts

with  = ln(2 + 1),  = :


ln(2 + 1)  =  ln(2 + 1) −



(2) 
=  ln(2 + 1) − 2
2 + 1



2 
=  ln(2 + 1) − 2
2 + 1

 
1−

1
2 + 1





=  ln(2 + 1) − 2 + 2 arctan  + 

e

= ( + 1) ln(2 + 2 + 2) − 2 + 2 arctan( + 1) + , where  =  − 2

al

[Alternatively, we could have integrated by parts immediately with
 = ln(2 + 2 + 2).] Notice from the graph that  = 0 where  has a

 2
0

cos2  sin3   is equal to 0.

Fo

53. From the graph, it seems as though

rS

horizontal tangent. Also,  is always increasing, and  ≥ 0.

To evaluate the integral, we write the integral as
 2
 = 0 cos2  (1 − cos2 ) sin   and let  = cos  ⇒
1

2 (1 − 2 )(−) = 0.


√
42 − 4 − 3  =
(2 − 1)2 − 4 

N

55.

1

ot

 = − sin  . Thus,  =





 = 2 − 1,
 = 2 

=



√
2 − 22 1 
2





√
√
√
22 
√ 2
 − 22 − ln  + 2 − 22  +  = 1  2 − 4 − ln  + 2 − 4 + 
4
2
2


√
√
1
= 4 (2 − 1) 42 − 4 − 3 − ln 2 − 1 + 42 − 4 − 3 + 
39

=

1
2



57. Let  = sin , so that  = cos  . Then



59. (a)

cos 


√

 √
22 
21  √ 2 ln  + 22 + 2 + 
4 + sin2   =
22 + 2  =
2 + 2 +
2
2




= 1 sin  4 + sin2  + 2 ln sin  + 4 + sin2  + 
2





1
1
1
1√ 2
1 √
 − 2 − sin−1
−
·
−
+  = 2 2 − 2 + √




2 − 2
1 − 22 
 √ 2

−12 1  2


 − 2
 − 2 + 1 − 1 =
= 2 − 2
2

2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 7 REVIEW

¤

373



(b) Let  =  sin  ⇒  =  cos  , 2 − 2 = 2 1 − sin2  = 2 cos2 .
 √ 2
 2


 − 2
 cos2 
1 − sin2 
 =
 =
 = (csc2  − 1)  = − cot  −  + 
2
2 sin2  sin2 
√

2 − 2
− sin−1
+
=−


61. For  ≥ 0,

∞
0



∞
1
∞
  = lim +1( + 1) 0 = ∞. For   0, 0   = 0   + 1  . Both integrals are
→∞

improper. By (7.8.2), the second integral diverges if −1 ≤   0. By Exercise 7.8.57, the first integral diverges if  ≤ −1.
∞
Thus, 0   is divergent for all values of .
1
−
4−2
1
, ∆ =
=
= ln 

10
5

(a) 10 =

1
5 · 2 { (2)

+ 2[(22) +  (24) + · · · +  (38)] +  (4)} ≈ 1925444

(b) 10 = 1 [ (21) +  (23) +  (25) + · · · +  (39)] ≈ 1920915
5

1 ln 

+ 4 (22) + 2 (24) + · · · + 2 (36) + 4 (38) +  (4)] ≈ 1922470

⇒  0 () = −

1
(ln )2

⇒  00 () =

al

65.  () =

1
5 · 3 [ (2)

2 + ln 
2
1
= 2
+ 2
. Note that each term of
2 (ln )3
 (ln )3
 (ln )2

rS

(c) 10 =

e

63.  () =

 00 () decreases on [2 4], so we’ll take  =  00 (2) ≈ 2022. | | ≤
( − )3
= 000674. | | ≤ 000001 ⇔
242

Take  = 368 for  . | | ≤ 000001 ⇔ 2 ≥

105 (2022)(8)
24

⇔ 2 ≥

105 (2022)(8)
12

⇒  ≥ 3672.

⇒  ≥ 2596. Take  = 260 for  .


1
− 0 10 = 60 .
 10
Distance traveled = 0   ≈ 10
60

ot

67. ∆ =

 10

2022(8)
1
≤ 5
122
10

Fo

| | ≤

( − )3
2022(4 − 2)3
≈
= 001348 and
122
12(10)2

1
[40
60 · 3

+ 4(42) + 2(45) + 4(49) + 2(52) + 4(54) + 2(56) + 4(57) + 2(57) + 4(55) + 56]

N

=

=

1
(1544)
180

= 857 mi

69. (a)  () = sin(sin ). A CAS gives

 (4) () = sin(sin )[cos4  + 7 cos2  − 3]


+ cos(sin ) 6 cos2  sin  + sin 




From the graph, we see that  (4) ()  38 for  ∈ [0 ].


(b) We use Simpson’s Rule with () = sin(sin ) and ∆ = 10 :


 
 


 ()  ≈ 10· 3  (0) + 4 10 + 2 2 + · · · + 4 9 + () ≈ 1786721
10
10
0




From part (a), we know that  (4) ()  38 on [0 ], so we use Theorem 7.7.4 with  = 38, and estimate the error

as | | ≤

38( − 0)5
≈ 0000646.
180(10)4

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374

¤

CHAPTER 7 TECHNIQUES OF INTEGRATION

(c) If we want the error to be less than 000001, we must have | | ≤

385
≤ 000001,
1804

385
≈ 646,0416 ⇒  ≥ 2835. Since  must be even for Simpson’s Rule, we must have  ≥ 30
180(000001)

so 4 ≥

to ensure the desired accuracy.


2 + sin 
1
√
≥ √ for  in [1 ∞).



71. (a)

∞

1

1
1
√  is divergent by (7.8.2) with  = ≤ 1. Therefore,
2




∞

1

2 + sin 
√
 is


divergent by the Comparison Theorem.
 ∞
1
1
1
1
 √ = 2 for  in [1 ∞).
 is convergent by (7.8.2) with  = 2  1. Therefore,
(b) √
4
4

2
1+

1
 ∞
1
√
 is convergent by the Comparison Theorem.
1 + 4
1

0

(cos  − cos2 )  +


= sin  − 1  −
2

1
4



e

 2




, 0 ≤ cos2  ≤ cos . For  in    , cos  ≤ 0 ≤ cos2 . Thus,
2
2

(cos2  − cos ) 

2  sin 2 0 + 1  +
2

1
4

75. Using the formula for disks, the volume is

=


4


4

3

 [ ()]2  = 

2

1
2

+

1
4

 2
0

Fo

=

0






  sin 2 − sin  2 = 1 −  − 0 +  −  − 1 = 2
4
2
4

 2  1
2
(cos2 )2  =  0
(1 + cos 2) 
2

 2
 2 
(1 + cos2 2 + 2 cos 2)  =  0
1 + 1 (1 + cos 4) + 2 cos 2 
4
2
0

 2

 =



2 sin 4 + 2 1 sin 2 0 =
2

77. By the Fundamental Theorem of Calculus,


→∞ 0

 0 ()  = lim

79. Let  = 1



0

Therefore,

0

 3
4

+

1
8



·0+0 −0 =

3 2
16 

→∞

→∞

⇒  = 1 ⇒  = −(12 ) .

∞




4

 0 ()  = lim [ () −  (0)] = lim  () −  (0) = 0 −  (0) = − (0).

ot

0

N

∞

al

area =


2

rS



73. For  in 0

ln 
 =
1 + 2

∞



0

∞

ln 
 = −
1 + 2

ln (1)
1 + 12



0

∞


  0
 ∞
 0
− ln  ln  ln 

 = −

− 2 =
(−) =

2 + 1
1 + 2
1 + 2
∞
∞
0

ln 
 = 0.
1 + 2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

PROBLEMS PLUS
1.

By symmetry, the problem can be reduced to finding the line  =  such that the shaded area is one-third of the area of the
√
√ quarter-circle. An equation of the semicircle is  = 49 − 2 , so we require that 0 49 − 2  = 1 · 1 (7)2 ⇔
3
4
49
2

 sin−1 (7) 0 =

49

12

[by Formula 30] ⇔

1

2

√
49 − 2 +

49
2

sin−1 (7) =

e

1 √
 49 − 2 +
2

49
.
12

al

This equation would be difficult to solve exactly, so we plot the left-hand side as a function of , and find that the equation

rS

holds for  ≈ 185. So the cuts should be made at distances of about 185 inches from the center of the pizza.
3. The given integral represents the difference of the shaded areas, which appears to

be 0. It can be calculated by integrating with respect to either  or , so we find 

=

√
7
1 − 3

⇒ =

√
3
1 − 7

⇒ =


7
1 −  3 and

Fo

in terms of  for each curve:  =


3
1 −  7 , so



√
 1 
 1 √

3
1 −  7 − 7 1 −  3  = 0 7 1 − 3 − 3 1 − 7  But this
0
√

 1 √
3
1 − 7 − 7 1 − 3  = 0.
0

ot

equation is of the form  = −. So

5. The area  of the remaining part of the circle is given by




  

 2
2 − 2 −
 − 2  = 4 1 −
2 − 2 


0
0


2

√ 2
30 4
= ( − )
 − 2 + sin−1 
2
2
 0




N

 = 4 = 4

=

4
( − )





 2 
2 
4
 
0+
− 0 = ( − )
= ( − ),
2 2

4

which is the area of an ellipse with semiaxes  and  − .
Alternate solution: Subtracting the area of the ellipse from the area of the circle gives us 2 −  =  ( − ), as calculated above. (The formula for the area of an ellipse was derived in Example 2 in Section 7.3.)

c
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375

¤

376

CHAPTER 7 PROBLEMS PLUS

7. Recall that cos  cos  =

 () =


0

1
[cos(
2


2

1
2

cos  cos( − )  =


= 1  cos  +
2

=

+ ) + cos( − )]. So

cos  +

1
4

1
2


0

[cos( +  − ) + cos( −  + )]  =

 sin(2 − ) 0 =
1
4

sin(−) −


2

cos  +

sin(−) =


2

1
4

sin(2 − ) −

1
4

sin(−)

1
2


0

[cos  + cos(2 − )] 

cos 

The minimum of cos  on this domain is −1, so the minimum value of  () is  () = −  .
2
 1(1−2 )

9. In accordance with the hint, we let  =

by parts with  = (1 − 2 )+1

0

, and we find an expression for +1 in terms of  . We integrate +1

⇒  = ( + 1)(1 − 2 ) (−2),  =  ⇒  = , and then split the remaining

integral into identifiable quantities:

e

1
1
1
+1 = (1 − 2 )+1 0 + 2( + 1) 0 2 (1 − 2 )  = (2 + 2) 0 (1 − 2 ) [1 − (1 − 2 )] 
2 + 2
 . Now to complete the proof, we use induction:
2 + 3

20 (0!)2
, so the formula holds for  = 0. Now suppose it holds for  = . Then
1!
+1 =
=



2( + 1)22 (!)2
2( + 1) 2( + 1)22 (!)2
2 + 2
2 + 2 22 (!)2
 =
=
=
·
2 + 3
2 + 3 (2 + 1)!
(2 + 3)(2 + 1)!
2 + 2 (2 + 3)(2 + 1)!

Fo

0 = 1 =

⇒ +1 =

rS

So +1 [1 + (2 + 2)] = (2 + 2)

al

= (2 + 2)( − +1 )

22(+1) [( + 1)!]2
[2( + 1)]2 22 (!)2
=
(2 + 3)(2 + 2)(2 + 1)!
[2( + 1) + 1]!

ot

So by induction, the formula holds for all integers  ≥ 0.
11. 0    . Now

0

1

[ + (1 − )]  =








+1
+1 − +1
 [ =  + (1 − )] =
.
=
( − )
( + 1)( − ) 
( + 1)( − )

N



Now let  = lim

→0





+1 − +1
( + 1)( − )

1

. Then ln  = lim

→0




1
+1 − +1 ln . This limit is of the form 00,

( + 1)( − )

so we can apply l’Hospital’s Rule to get ln  = lim

→0




1
+1 ln  − +1 ln 
 ln  −  ln 
 ln 
 ln 
(−)
−
=
−1=
−
− ln  = ln (−) .
+1 − +1

+1
−
−
−


Therefore,  = −1






1(−)

.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 7 PROBLEMS PLUS



13. Write  =

8
 =
(1 + 6 )2

 = 32 ,  = −


1
2
 =
1 + 6
3





3 ·

1
6(1 + 6 )

¤

377

5
5
. Integrate by parts with  = 3 ,  =
. Then
6 )2
(1 + 
(1 + 6 )2

⇒  =−

3
1
+
6(1 + 6 )
2



2
. Substitute  = 3 in this latter integral.
1 + 6


1
1
1
3
+ tan−1 (3 ) + .
= tan−1  +  = tan−1 (3 ) + . Therefore  = −
2
1+
3
3
6(1 + 6 )
6

Returning to the improper integral,


∞

4
1 + 6

2



8
1
3
+ tan−1 (3 )
 = lim −
→∞ −1 (1 + 6 )2
→∞
6(1 + 6 )
6
−1


3
−1

1
1
−1 3
−1
= lim −
+ tan ( ) +
− tan (−1)
→∞
6(1 + 6 )
6
6(1 + 1)
6
 
 
1
1
1

1


1
=0+
−
−
−
=
−
+
= −
6 2
12
6
4
12
12
24
8
12

 = lim





e

−1



An equation of the circle with center (0 ) and radius 1 is 2 + ( − )2 = 12 , so
√
an equation of the lower semicircle is  =  − 1 − 2 . At the points of tangency,

al

15.

Fo

rS

the slopes of the line and semicircle must be equal. For  ≥ 0, we must have
√

= 2 ⇒  = 2 1 − 2 ⇒ 2 = 4(1 − 2 ) ⇒
0 = 2 ⇒ √
1 − 2
√
√
 √ 
52 = 4 ⇒ 2 = 4 ⇒  = 2 5 and so  = 2 2 5 = 4 5.
5
5
5
5

The slope of the perpendicular line segment is − 1 , so an equation of the line segment is  −
2
 = −1 +
2

1
5

√
(25) 5


(25)√5
√



√
√
1
30
−1
2
2
2 −
5 − 1 −  − 2  = 2 5  −
1−
sin  − 
2
2
0
√




5 1
2
1
4
=2 2−
−
· √ − sin−1 √
− 2(0)
5
2
5
5
5





2
2
1
= 2 − sin−1 √
= 2 1 − sin−1 √
2
5
5

N

0

ot

2

√
√ 

5 = −1  − 2 5
⇔
2
5

√
√
√
√
√
√
5 + 4 5 ⇔  = − 1  + 5, so  = 5 and an equation of the lower semicircle is  = 5 − 1 − 2 .
5
2

Thus, the shaded area is


4
5

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

N ot e

al

rS

Fo

8

FURTHER APPLICATIONS OF INTEGRATION

8.1 Arc Length
1.  = 2 − 5

⇒ =

√
√
3 
3 
1 + ()2  = −1 1 + (2)2  = 5 [3 − (−1)] = 4 5.
−1

The arc length can be calculated using the distance formula, since the curve is a line segment, so
 = [distance from (−1 −7) to (3 1)] =

So  =



4



7.  = 1 + 632



2
1
√ − 1  ≈ 36095.
2 

⇒  = 912

1
1√
 82
1 + 81  = 1 12 81
0

9.



√
1 + cos2   ≈ 38202.
0

2
1
√ −1 .
2 

⇒ 1 + ()2 = 1 + 81. So



82

 = 1 + 81,
1
= 81 · 2 32

=
3
 = 81 

1
3
+
3
4

Fo

=

1 + ()2 = 1 + cos2 . So  =

 √ 
⇒  = 1 2  − 1 ⇒ 1 + ()2 =

1+

1

⇒

e

√
−

 = cos 

al

5.  =

⇒

rS

3.  = sin 


√
√
[3 − (−1)]2 + [1 − (−7)]2 = 80 = 4 5

=

⇒

 0 = 2 −

1

1
42


2
243

 √

82 82 − 1

⇒

N

ot

2


1
1
1
1
1
= 4 + +
= 2 + 2 . So
1 + ( 0 )2 = 1 + 4 − +
2
164
2 164
4



 2
 2
 2
 2

1
 + 1   =
1 + ( 0 )2  =
=
2 + 2 

42 
4
1
1
1
2 
 


1 1
8 1
1 3
7
1
59
1
 −
−
−
−
= + =
=
=
3
4 1
3
8
3
4
3
8
24

1√
3  (

11.  =

− 3) =

1 32
3

1 + ()2 = 1 + 1  −
4
=

 9 1
1

2

12

−  12

1
2

13.  = ln(sec )

 =

+ 1  −1 = 1  +
4
4

1
2

1 12
2

− 1  −12
2

+ 1  −1 =
4



1 12

2

⇒

2
+ 1  −12 . So
2



9

 

+ 1  −12  = 1 2  32 + 2 12 = 1 2 · 27 + 2 · 3 − 2 · 1 + 2 · 1
2
2 3
2
3
3
1



 
= 24 − 8 = 1 64 =
3
2 3
1
2

⇒  =

⇒

32
.
3

sec  tan 

=
= tan  ⇒ 1 +

sec 






2

= 1 + tan2  = sec2 , so


4
 4
 4
 4 √ sec2   = 0 |sec |  = 0 sec   = ln(sec  + tan )
0
0

√

√

= ln 2 + 1 − ln(1 + 0) = ln 2 + 1

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

379

380

¤

CHAPTER 8

FURTHER APPLICATIONS OF INTEGRATION



2
1 2 1
1
1
1
1
1
1
.
 − + 2 = 2 + + 2 =
+
4
2
4
4
2 4
2
2


 2
 2
 2

1
1
1
  + 1   =
+

1 + ( 0 )2  =
=
2
2 
2
2
1
1
1
2 
 


1
3
1
1
1 2 1
 + ln || = 1 + ln 2 −
+ 0 = + ln 2
=
4
2
2
4
4
2
1

1 2 1
1
1
 − ln  ⇒ 0 =  −
4
2
2
2

15.  =

So

17.  = ln(1 − 2 )



2

⇒ 0 =

⇒ 1 + ( 0 )2 = 1 +



1
· (−2) ⇒
1 − 2

42
1 − 22 + 4 + 42
1 + 22 + 4
(1 + 2 )2
=
=
=
⇒
2 )2
2 )2
2 )2
(1 − 
(1 − 
(1 − 
(1 − 2 )2

 2 
2

1 + 2
1
1 + 2
2
1
+
[partial fractions].
1+
=
=
= −1 +
[by division] = −1 +
2

1−
1 − 2
1 − 2
1+ 1−

 12 


12 
1
1
So  =
−1 +
+
 = − + ln |1 + | − ln |1 − | 0 = − 1 + ln 3 − ln 1 − 0 = ln 3 − 1 .
2
2
2
2
1+
1−
0

=

⇒  =  ⇒ 1 + ()2 = 1 + 2 . So
1 √
1√
1 + 2  = 2 0 1 + 2  [by symmetry]
−1

 √
=2 1 2+
2

21.

e

1 2

2

19.  =

=1+

1
2

√  

ln 1 + 2 − 0 +

1
2

al




 √
= 2  1 + 2 +
2

21

1
2

√
1
 ln  + 1 + 2 0

rS

1+

√ 
 √

ln 1 = 2 + ln 1 + 2





or substitute
 = tan 

From the figure, the length of the curve is slightly larger than the hypotenuse

Fo

of the triangle formed by the points (1 2), (1 12), and (2 12). This length
√
is about 102 + 12 ≈ 10, so we might estimate the length to be 10.

ot

 = 2 + 3 ⇒  0 = 2 + 32 ⇒ 1 + (0 )2 = 1 + (2 + 32 )2 .
2
So  = 1 1 + (2 + 32 )2  ≈ 100556.

⇒  =  cos  + (sin )(1) ⇒ 1 + ()2 = 1 + ( cos  + sin )2 . Let


 2
−
 () = 1 + ()2 = 1 + ( cos  + sin )2 . Then  = 0  () . Since  = 10, ∆ = 210 0 =
 ≈ 10
=

5
3


5.

Now

N

23.  =  sin 


 
 
 
 
 
 
 
 
 

 (0) + 4  + 2 2 + 4 3 + 2 4 + 4 5 + 2 6 + 4 7 + 2 8 + 4 9 +  (2)
5
5
5
5
5
5
5
5
5

≈ 15498085

The value of the integral produced by a calculator is 15374568 (to six decimal places).
25.  = ln(1 + 3 )

∆ =

5−0
10

⇒  =

= 1 . Now
2

1
· 32
1 + 3

⇒ =

5
0

 () , where  () =


1 + 94 (1 + 3 )2 . Since  = 10,

 ≈ 10
=

12
[ (0)
3

+ 4 (05) + 2 (1) + 4(15) + 2(2) + 4 (25) + 2 (3) + 4 (35) + 2 (4) + 4 (45) + (5)]

≈ 7094570
The value of the integral produced by a calculator is 7118819 (to six decimal places).

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 8.1

ARC LENGTH

¤

381

27. (a)

(b)

Let  () =  = 

√
3
4 − . The polygon with one side is just

the line segment joining the points (0  (0)) = (0 0) and
(4 (4)) = (4 0), and its length 1 = 4.
The polygon with two sides joins the points (0 0),

al

e

√ 

(2 (2)) = 2 2 3 2 and (4 0). Its length

√
√ 2
 √
2 

2 = (2 − 0)2 + 2 3 2 − 0 + (4 − 2)2 + 0 − 2 3 2 = 2 4 + 283 ≈ 643

rS

√ 
 √  
Similarly, the inscribed polygon with four sides joins the points (0 0), 1 3 3 , 2 2 3 2 , (3 3), and (4 0),

so its length

4 =


√ 2 
√ 2 √
 √ 2 
 √

1 + 3 3 + 1 + 2 3 2 − 3 3 + 1 + 3 − 2 3 2 + 1 + 9 ≈ 750

Fo

 √

12 − 4

, the length of the curve is
=  1 (4 − )−23 (−1) + 3 4 −  =
3

3(4 − )23


2
 2

 4
 4

12 − 4
1+
 =
1+
.
=

3(4 − )23
0
0

(c) Using the arc length formula with

ot

(d) According to a calculator, the length of the curve is  ≈ 77988. The actual value is larger than any of the approximations in part (b). This is always true, since any approximating straight line between two points on the curve is shorter than the

N

length of the curve between the two points.
29.  = ln 

⇒  = 1 ⇒ 1 + ()2 = 1 + 12 = (2 + 1)2 ⇒
√



 2 2
 2√
 1 + 1 + 2  2
 +1
1 + 2
23
2 − ln 

=
 =
 =
1+


2


1
1
1
√ 

√
√
√ 

1+ 5
− 2 + ln 1 + 2
= 5 − ln
2

31.  23 = 1 − 23

⇒  = (1 − 23 )32 ⇒



= 3 (1 − 23 )12 − 2 −13 = −−13 (1 − 23 )12
2
3

 2

= −23 (1 − 23 ) = −23 − 1. Thus


=4

⇒


1
1
1
1 + (−23 − 1)  = 4 0 −13  = 4 lim 3 23 = 6.
2
0
→0+



c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 8

⇒  0 = 312

33.  = 232

() =

FURTHER APPLICATIONS OF INTEGRATION

⇒ 1 + (0 )2 = 1 + 9. The arc length function with starting point 0 (1 2) is



√
2
1 + 9  = 27 (1 + 9)32 =
1
1

√
1 − 2

35.  = sin−1  +

1 + ( 0 )2 = 1 +

1 + ( 0 )2 =
() =





0

2
27


√ 
(1 + 9)32 − 10 10 .

1

1−
⇒ 0 = √
−√
= √
1 − 2
1 − 2
1 − 2

⇒

(1 − )2
2(1 − )
1 − 2 + 1 − 2 + 2
2 − 2
2
=
=
=
=
2
1−
1 − 2
1 − 2
(1 + )(1 − )
1+



2
. Thus, the arc length function with starting point (0 1) is given by
1+



1 + [ 0 ()]2  =



0



37. The prey hits the ground when  = 0

√  √
√ √


2
 = 2 2 1 +  0 = 2 2 1 +  − 1 .
1+

⇔ 180 −

1 2

45

= 0 ⇔ 2 = 45 · 180 ⇒  =

2

2 since  must be positive.  0 = − 45  ⇒ 1 + ( 0 ) = 1 +



4
2 ,
452

90

rS

1+

√
8100 = 90,

so the distance traveled by the prey is



 4
2


 = 45 ,
4 2
  =
1 + 2 45 
2
2
 = 45 
452
0
0
√ 
√
√
 √
4
 √


21
= 45 1  1 + 2 + 1 ln  + 1 + 2 0 = 45 2 17 + 1 ln 4 + 17 = 45 17 +
2 2
2
2
2

=



⇒

e

¤

al

382

45
4

√ 

ln 4 + 17 ≈ 2091 m

Fo

39. The sine wave has amplitude 1 and period 14, since it goes through two periods in a distance of 28 in., so its equation is


 

 = 1 sin 2  = sin   . The width  of the flat metal sheet needed to make the panel is the arc length of the sine curve
14
7 from  = 0 to  = 28. We set up the integral to evaluate  using the arc length formula with

=


7

  cos   :
7


 2
 14 

 2
 28 
1 +  cos  
 = 2 0
1 +  cos  
. This integral would be very difficult to evaluate exactly,
7
7
7
7
0

ot

=




so we use a CAS, and find that  ≈ 2936 inches.

=

√
√
3 − 1  ⇒  = 3 − 1 [by FTC1] ⇒
1

N

41.  =


4
4
4√
3  = 1 32  = 2 52 = 2 (32 − 1) =
5
5
1
1

62
5

1 + ()2 = 1 +
= 124

√
2
3 − 1 = 3

⇒

8.2 Area of a Surface of Revolution
1. (a) (i)  = tan 

⇒  = sec2  ⇒  =


√
1 + ()2  = 1 + sec4  . By (7), an integral for the

area of the surface obtained by rotating the curve about the -axis is  =



2  =

 3
0

√
2 tan  1 + sec4  .

(ii) By (8), an integral for the area of the surface obtained by rotating the curve about the -axis is
=



2  =

(b) (i) 105017

 3
0

√
2 1 + sec4  .

(ii) 79353

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 8.2

3. (a) (i)  = −

2

2

⇒  = − · (−2) ⇒  =

1
2
 = 2  = −1 2− 1 + 42 −22 .

(ii) By (8),  =
(b) (i) 110753



2  =

(ii) 39603

1
0

AREA OF A SURFACE OF REVOLUTION

¤

383



1 + ()2  = 1 + 42 −22 . By (7),


2 1 + 42 −22 

[symmetric about the y-axis]

⇒  0 = 32 . So

5.  = 3

=
=

2
0

2
36


2 √
1 + ( 0 )2  = 2 0 3 1 + 94 
[ = 1 + 94 ,  = 363 ]

145
√
 145 √




  = 18 2 32
= 27 145 145 − 1
3
1

2

1

 = 4 

al

e




√
5 + 4
2
4
0
−12
1
0 )2 =
=
. So
⇒
7.  = 1 + 4 ⇒  = 2 (1 + 4)
(4) = √
1 + (
1+
1 + 4
1 + 4
1 + 4


5√
5
5√
5 + 4
 = 1 2 1 + ( 0 )2  = 2 1 1 + 4
 = 2 1 4 + 5 
1 + 4



25
 25 √  1 
 = 4 + 5,
 4 
=  (2532 − 932 ) =  (125 − 27) = 98 
= 2 2 32
= 2 9
4 3
3
3
3
9

⇒  0 =  cos  ⇒ 1 + (0 )2 = 1 +  2 cos2 (). So



1
1
 = 0 2 1 + ( 0 )2  = 2 0 sin  1 + 2 cos2 () 

rS

9.  = sin 

 =  cos 
 = − 2 sin  





1
2  
1 + 2 − 2  =
1 + 2 

 −



 
 

4 
21 4 
1 + 2  =
1 + 2 + 1 ln  + 1 + 2
=
2
 0
 2
0

  √
√
√
√



4
2 
=
1 + 2 + 1 ln  + 1 +  2
− 0 = 2 1 + 2 + ln  + 1 + 2
2
 2



−

Fo

= 2



ot


+ 2)32 ⇒  = 1 ( 2 + 2)12 (2) =   2 + 2 ⇒ 1 + ()2 = 1 +  2 ( 2 + 2) = ( 2 + 1)2 .
2
2



2
So  = 2 1 ( 2 + 1)  = 2 1  4 + 1  2 1 = 2 4 + 2 − 1 − 1 = 21 .
4
2
4
2
2
2
1
3 (

13.  =

√
3
 ⇒  = 3

N

11.  =

 = 2
=


27

⇒ 1 + ()2 = 1 + 9 4 . So

2 
2 
 1 + ()2  = 2 1  3 1 + 9 4  =
1

√ 
√

145 145 − 10 10

2
36

2
1 + 9 4 36 3  =
1


18

 
32 2
2
1 + 9 4
3



2 −  2 ⇒  = 1 (2 −  2 )−12 (−2) = − 2 −  2 ⇒
2
 2

2
2 −  2
2
2
=1+ 2
= 2
+ 2
= 2
⇒
1+

 − 2
 − 2
 − 2
 − 2
 2
 2



 2

2 − 2 
− 0 = 2 .
2 
 = 2
  = 2  0 = 2
=
2
2 −  2
0
0

15.  =

Note that this is

1
4

the surface area of a sphere of radius , and the length of the interval  = 0 to  = 2 is

1

1
4

the length of the

interval  = − to  = . c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

384

¤

CHAPTER 8

FURTHER APPLICATIONS OF INTEGRATION

⇒  = 4 ⇒ 1 + ()2 = 1 + 8 ⇒  =
√
−
Let () = 2 5 1 + 8 . Since  = 10, ∆ = 5 10 0 = 1 . Then
5
2

17.  =

1 5
5

5
0

2

1

5
5

√
1 + 8 .

 ≈ 10
=

12
3 [ (0)

+ 4 (05) + 2 (1) + 4 (15) + 2 (2) + 4 (25) + 2 (3) + 4(35) + 2(4) + 4 (45) + (5)]

≈ 1,230,507
The value of the integral produced by a calculator is approximately 1,227,192.
19.  = 

⇒  =  +  ⇒ 1 + ()2 = 1 + ( +  )2 ⇒  =

−
1
Let () = 2 1 + ( +  )2 . Since  = 10, ∆ = 1 10 0 = 10 . Then

1
0

2


1 + ( +  )2 .

 ≈ 10
=

110
[ (0)
3

+ 4 (01) + 2 (02) + 4 (03) + 2 (04) + 4 (05) + 2 (06) + 4 (07) + 2 (08) + 4 (09) +  (1)]

e

≈ 24145807

al

The value of the integral produced by a calculator is 24144251 (to six decimal places).




⇒  = 1 + ()2  = 1 + (−12 )2  = 1 + 14  ⇒

 2√ 4
 4√ 2
 2
 +1
 + 1 1 
1
1
2 ·
1 + 4  = 2
 = 2
[ = 2 ,  = 2 ]
=
2 
3



2
1
1
1
 √
 4√
4


1 + 2
1 + 2
24
=
 =  −
+ ln  + 1 + 2
2

1
1
 √
√  √2
√   
√ 
√



√
 √
17
4 ln 17 + 4 − 4 ln 2 + 1 − 17 + 4 2
=  − 4 + ln 4 + 17 + 1 − ln 1 + 2 =
4

23.  = 3 and 0 ≤  ≤ 1

21

1
0

2


3√
1 + (32 )2  = 2 0 1 + 2

= [or use CAS]

1

∞




3



1 √
 1 + 2 +
2


2

1
2

1+

1
6






 = 32 ,
 = 6 



√
3
 ln  + 1 + 2 0 =


N

25.  = 2



⇒ 0 = 32 and 0 ≤  ≤ 1.

ot

=

Fo

rS

21.  = 1

 = 2

1

∞

1




1+


3



=

3 √
10 +
2

1
 = 2
4



∞

1

1
2


3

3√
1 + 2 
0

√ 

ln 3 + 10 =


6

√ 
 √

3 10 + ln 3 + 10

√
4 + 1
. Rather than trying to evaluate this
3

√
√
integral, note that 4 + 1  4 = 2 for   0. Thus, if the area is finite,
 ∞ 2
 ∞
 ∞√ 4
 +1

1
. But we know that this integral diverges, so the area 
  2
 = 2
 = 2
3
3

1
1
1 is infinite.

27. Since   0, the curve 3 2 = ( − )2 only has points with  ≥ 0.

[32 ≥ 0 ⇒ ( − )2 ≥ 0 ⇒  ≥ 0.]
The curve is symmetric about the x-axis (since the equation is unchanged when  is replaced by −).  = 0 when  = 0 or , so the curve’s loop extends from  = 0 to  = .

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 8.2

AREA OF A SURFACE OF REVOLUTION

¤





( − )[−2 +  − ]
(3 2 ) =
[( − )2 ] ⇒ 6
=  · 2( − )(−1) + ( − )2 ⇒
=




6


 2
( − )2 ( − 3)2
( − )2 ( − 3)2
3
( − 3)2

the last fraction
=
=
·
=
⇒
2
2 2
2
2 is 1

36
36
( − )
12
 2
12
2 − 6 + 92
2 + 6 + 92
( + 3)2

2 − 6 + 92
=
+
=
= for  6= 0.
=1+
1+

12
12
12
12
12
 
 √
 
 ( − )  + 3
( − )( + 3)
√
(a)  =

2  = 2
·√
 = 2
6
3
12
=0
0
0
 


 2
 3

2
  + 2 − 3 0 =
=
(2 + 2 − 32 )  =
( + 3 − 3 ) =
· 3 =
.
3 0
3
3
3
3

385

⇒

Note that we have rotated the top half of the loop about the x-axis. This generates the full surface.

2 
2
2
 ()


=− 2
⇒
+ 2 =1 ⇒
=− 2 ⇒
2
2





 


 2
4 2 + 4 2 1 − 22
4 2 + 4 2 − 2 2 2
4 2
4 2 + 4  2

=
= 1+ 4 2 =
=
1+
4 2
4 2 (1 − 22 )

 


4 2 − 2 2 2
 2
 2
4 −  − 2 
4 + 2 2 − 2 2
=
=
4 − 2 2
2 (2 − 2 )

Fo

29. (a)

rS

al

e

(b) We must rotate the full loop about the -axis, so we get double the area obtained by rotating the top half of the loop:
 
 
 
 
 + 3
2
4
  = 4
√
12 ( + 3)  = √
(12 + 332 ) 
 = √
 = 2 · 2
12
2 3 0
3 0
=0
0
√ 
√ 
√  




2 3 2
6 2
2 3 2 52 6 52
2 3 28 2
2 2 32 6 52
=


+
 =

+ 
= √
+ 
= √
5
5
3
3
5
3
15
3  3
3 3
0
√
56 3 2
=
45

Thus,

ot

The ellipsoid’s surface area is twice the area generated by rotating the first-quadrant portion of the ellipse about the -axis.




4 − (2 − 2 )2
4   4
√
=2
2 1 +
 = 4
 = 2
 − (2 − 2 )2 

 2 − 2
0
0
0
√
 
 √ 2 2
  2 −2

 30
√


4   −  4
4
4
−1 
√
 = 2 − 2  =
= 2 sin  − 2 √
4 − 2 +
 0
2
2 0
2 − 2
2 2 − 2 2


√
2 − 2
√
 √ 2

2  sin−1
4
2
2 − 2



  −
4



√
sin−1
= 2 2 +
= √
4 − 2 (2 − 2 ) +


2
2

2 2 − 2
2 − 2







N



(b)

2





 2
 − 2


2
2 
2
 ()


=− 2
⇒
+ 2 =1 ⇒
=− 2 ⇒
2
2





 
 2

2 4 − 2 2  2 + 4  2
4  2
4 2 + 4  2
4 2 (1 −  22 ) + 4  2
=
= 1+ 4 2 =
=
1+
4 2
4 2 (1 −  22 )

 


2 4 − 2 2  2
=

4 − (2 − 2 ) 2
4 − 2  2 + 2  2
=
4 − 2  2
2 (2 −  2 )

The oblate spheroid’s surface area is twice the area generated by rotating the first-quadrant portion of the ellipse about the

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

386

¤

CHAPTER 8

FURTHER APPLICATIONS OF INTEGRATION

-axis. Thus,


 2
  
 
4 − (2 − 2 ) 2



2  1 +
 = 4
2 −  2

=2

 2 −  2
0
0 

 √2 2
4   4

4   −  4
2 − 2 ) 2  =
= 2
 − (
 − 2 √
2 − 2

2 0

0
√

 2 −2

√ 4
4
4
30
sin−1 2
= √
 − 2 +
2
2 − 2
2
2

 
0



=


√
2 − 2 


√
2 − 2
2
−1
√
 √ 2

4
2 
2 − 2
 2  sin

  −
4

−1

√
sin
= 2 +
= √
4 − 2 (2 − 2 ) +

2
2

2 2 − 2
2 − 2

Notice that this result can be obtained from the answer in part (a) by interchanging  and .




→∞ =1

2[ −  (∗ )]





1 + [ 0 (∗ )]2 ∆ =  2[ −  ()] 1 + [ 0 ()]2 .


al

 = lim

e

31. The analogue of  (∗ ) in the derivation of (4) is now  −  (∗ ), so



√
√
2 − 2 ,  0 () = − 2 − 2 . The surface area generated is

 
 






2
 − 2 − 2 √
1 =
2  − 2 − 2
1+ 2
 = 4

2
 −
2 − 2
−
0

 
2
√
= 4
−  
2 − 2
0

 
√
2

√
, so 2 = 4
+  .
For the lower semicircle, () = − 2 − 2 and  0 () = √
2 − 2
2 − 2
0

 
  
 

2
√
= 42 2 .
= 82
 = 8 2 sin−1
Thus, the total area is  = 1 + 2 = 8
2 − 2
 0
2

0

ot

Fo

rS

33. For the upper semicircle,  () =

35. In the derivation of (4), we computed a typical contribution to the surface area to be 2

−1 + 
|−1  |,
2

N

the area of a frustum of a cone. When () is not necessarily positive, the approximations  =  ( ) ≈  (∗ ) and

−1 =  (−1 ) ≈  (∗ ) must be replaced by  = | ( )| ≈ |(∗ )| and −1 = |(−1 )| ≈ | (∗ )|. Thus,




−1 + 
|−1  | ≈ 2 | (∗ )| 1 + [ 0 (∗ )]2 ∆. Continuing with the rest of the derivation as before,


2


we obtain  =  2 | ()| 1 + [ 0 ()]2 .

2

8.3 Applications to Physics and Engineering
1. The weight density of water is  = 625 lbft3 .

(a)  =  ≈ (625 lbft3 )(3 ft) = 1875 lbft2
(b)  =  ≈ (1875 lbft2 )(5 ft)(2 ft) = 1875 lb. ( is the area of the bottom of the tank.)
(c) As in Example 1, the area of the th strip is 2 (∆) and the pressure is  =  . Thus,
3

 
3
3
 = 0  · 2  ≈ (625)(2) 0   = 125 1 2 0 = 125 9 = 5625 lb.
2
2

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




SECTION 8.3 APPLICATIONS TO PHYSICS AND ENGINEERING

¤

In Exercises 3 – 9,  is the number of subintervals of length ∆ and ∗ is a sample point in the th subinterval [−1   ].

3. Set up a vertical x-axis as shown, with  = 0 at the water’s surface and  increasing in the

downward direction. Then the area of the th rectangular strip is 6 ∆ and the pressure on the strip is ∗ (where  ≈ 625 lbft3 ). Thus, the hydrostatic force on the strip is




∗ · 6 ∆ and the total hydrostatic force ≈

 = lim




→∞ =1

∗ · 6 ∆ =


6
2

=1

∗ · 6 ∆. The total force


 · 6  = 6

6
2

  = 6


2 6
2 2

1

= 6(18 − 2) = 96 ≈ 6000 lb

5. Set up a vertical x-axis as shown. The base of the triangle shown in the figure



32 − (∗ )2 , so  = 2 9 − (∗ )2 , and the area of the th



rectangular strip is 2


9 − (∗ )2 ∆. The th rectangular strip is (∗ − 1) m



e

has length

al

below the surface level of the water, so the pressure on the strip is (∗ − 1).



The hydrostatic force on the strip is (∗ − 1) · 2 9 − (∗ )2 ∆ and the total



 = lim




=1




=1

(∗ − 1) · 2


(∗ − 1) · 2



9 − (∗ )2 ∆. The total force


rS

force on the plate ≈


√
3
9 − (∗ )2 ∆ = 2 1 ( − 1) 9 − 2 


32
3

√
2  −

9

2

+2

√

 
 √
2  + 9 sin−1 1  = 38 2 −
3
3

9
2

+ 9 sin−1

ot

=

Fo

3

 √
3√
3 √
30
 9 − 2  − 2 1 9 − 2  = 2 − 1 (9 − 2 )32 − 2
9 − 2 +
3
1
2
1
 √ 
  √
 


= 2 0 + 1 8 8 − 2 0 + 9 ·  − 1 8 + 9 sin−1 1
3
2
2
2
2
3

= 2

≈ 6835 · 1000 · 98 ≈ 67 × 104 N

 1 

N

Note: If you set up a typical coordinate system with the water level at  = −1, then  =
7. Set up a vertical x-axis as shown. Then the area of the th rectangular strip is

3

9
2

sin−1

  3
1



 −1
−3

(−1 − )2


9 −  2 .

√




2
3 − ∗
2

√  , so  = 2 − √ ∗ .
2 − √ ∗ ∆. By similar triangles,
=


2
3
3
3
The pressure on the strip is ∗ , so the hydrostatic force on the strip is







2
2
∗ 2 − √ ∗ ∆.
∗ 2 − √ ∗ ∆ and the hydrostatic force on the plate ≈




3
3
=1
The total force






 √3
 √3 
2 ∗
2
2
2 − √  ∆ =
2 − √ 2 
 = lim
 2 − √   = 
→∞ =1
3
3
3
0
0
√3

2
=  2 − √ 3
=  [(3 − 2) − 0] =  ≈ 1000 · 98 = 98 × 103 N
3 3
0



3

∗


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387

¤

388

CHAPTER 8

FURTHER APPLICATIONS OF INTEGRATION

9. Set up coordinate axes as shown in the figure. The length of the th strip is

2



∗
∗
25 − ( )2 and its area is 2 25 − ( )2 ∆. The pressure on this strip is

∗ approximately  = 625(7 −  ) and so the force on the strip is approximately

∗
∗
625(7 −  )2 25 − ( )2 ∆. The total force



 = lim

→∞ =1

= 125

∗
625(7 −  )2



5
∗
25 − ( )2 ∆ = 125 0 (7 − ) 25 −  2 


5 


 
5 
5
5
7 25 −  2  − 0  25 −  2  = 125 7 0 25 −  2  − − 1 (25 −  2 )32
3
0
0




 
= 125 7 1  · 52 + 1 (0 − 125) = 125 175 −
4
3
4

125
3



4

≈ 11,972 ≈ 12 × 10 lb

11. Set up a vertical x-axis as shown. Then the area of the th rectangular strip is




∗


=1


(2 − ∗ ) ∆. The total force




 
   

(2 − )  =
2 − 2 
(2 − ∗ ) ∆ = 

0
0
→∞ =1



 3
  3 1 3   2
  2 1 3 
 − 3  0 =
 − 3 =
= 2 2
=
3



3

 = lim




∗


Fo

≈

rS

The pressure on the strip is ∗ , so the hydrostatic force on the plate


al

e





2

(2 − ∗ ) ∆. By similar triangles,
, so  = (2 − ∗ )
=



2 − ∗
2



2∗
⇒  = √  . The area of the th
3

 √
2∗
rectangular strip is √  ∆ and the pressure on it is  4 3 − ∗ .

3


8

√ = ∗

4 3

ot

13. By similar triangles,

 4√3
 √
 √
 2
2 4 3 2
 =
 4 3 −  √  = 8
  − √
 
3
3 0
0
0
√
 4√3
2  4√3
2
= 4 2 0 − √ 3 0 = 192 − √ 64 · 3 3 = 192 − 128 = 64
3 3
3 3

N

√
4 3

≈ 64(840)(98) ≈ 527 × 105 N

15. (a) The top of the cube has depth  = 1 m − 20 cm = 80 cm = 08 m.

 =  ≈ (1000)(98)(08)(02)2 = 3136 ≈ 314 N
(b) The area of a strip is 02 ∆ and the pressure on it is ∗ .

 =

1

08

(02)  = 02


2 1
2  08

1

= (02)(018) = 0036 = 0036(1000)(98) = 3528 ≈ 353 N

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SECTION 8.3 APPLICATIONS TO PHYSICS AND ENGINEERING

¤

389

17. (a) The area of a strip is 20 ∆ and the pressure on it is  .

=

3
0

20  = 20

1
2

2

3
0

9
2

= 20 ·

= 90

3

= 90(625) = 5625 lb ≈ 563 × 10 lb
(b)  =

9
0

20  = 20


2 9
2 0

1

= 20 ·

81
2

= 810 = 810(625) = 50,625 lb ≈ 506 × 104 lb.

(c) For the first 3 ft, the length of the side is constant at 40 ft. For 3   ≤ 9, we can use similar triangles to find the length :

9−
9−
=
⇒  = 40 ·
.
40
6
6
9
9

3
9

3
9−
 = 40 1 2 0 + 20  3 (9 − 2 )  = 180 + 20  9 2 − 1 3 3
 = 0 40  + 3 (40)
2
3
3
2
3
6

 

= 180 + 20  729 − 243 − 81 − 9 = 180 + 600 = 780 = 780(625) = 48,750 lb ≈ 488 × 104 lb
3
2
2
∆ hypotenuse ⇒

al

sin  =

e

(d) For any right triangle with hypotenuse on the bottom,




()  =

Simpson’s Rule to estimate  , we get
 ≈ 64

04
[70(70)
3

=

256
3 [7(12)

=

256
(48604)
3

 94
70

64() . From the table, we see that ∆ = 04, so using

Fo

19. From Exercise 18, we have  =

rS

√
√
402 + 62
409
=
∆.
hypotenuse = ∆ csc  = ∆
6
3
√
 √
9
9
 
 = 3 20 409  = 1 20 409  1 2 3
3
3
2
√
= 1 · 10 409 (81 − 9) ≈ 303,356 lb ≈ 303 × 105 lb
3

+ 4(74)(74) + 2(78)(78) + 4(82)(82) + 2(86)(86) + 4(90)(90) + 94(94)]

ot

+ 296(18) + 156(29) + 328(38) + 172(36) + 36(42) + 94(44)]
≈ 4148 lb

N

21. The moment  of the system about the origin is  =

The mass  of the system is  =

2


=1
2


  = 1 1 + 2 2 = 6 · 10 + 9 · 30 = 330.

 = 1 + 2 = 6 + 9 = 15.

=1

The center of mass of the system is  =  =
23. The mass is  =

3


330
15

= 22.

 = 4 + 2 + 4 = 10. The moment about the -axis is  =

=1

3


  = 4(−3) + 2(1) + 4(5) = 10.

=1

The moment about the -axis is  =

3


  = 4(2) + 2(−3) + 4(3) = 14. The center of mass is

=1

( ) =



 

 



=



14 10

10 10



= (14 1).

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390

¤

CHAPTER 8

FURTHER APPLICATIONS OF INTEGRATION

25. The region in the figure is “right-heavy” and “bottom-heavy,” so we know that

  05 and   1, and we might guess that  = 07 and  = 07.
=

1
0

=

1


=

1


 1
2  = 2 0 = 1 − 0 = 1.

1
0

1
1

(2)  =

1

1
(2)2
0 2

 =

2
3

1
1

3

1
0

Thus, the centroid is ( ) =

1

= 2
3

0

2

22  =

2
3



2
3

3


.

3

1
0

= 2.
3

27. The region in the figure is “right-heavy” and “bottom-heavy,” so we know

  05 and   1, and we might guess that  = 06 and  = 09.
1
 = 0   = [ ]1 =  − 1.
0

=

  =

1
[0
−1
1


1

1
[
−1

− (−1)] =

1
( )2
0 2

 =

1
.
−1
1
−1

Thus, the centroid is ( ) =
29.  =

1
0

(12 − 2 )  =

1



−  ]1
0
1
4

·



[by parts]

 2 1
 0=

1
 +1
−1
4

2 32
3

− 1 3
3

1
4( − 1)



1

 2

 −1 =

1


=

0

2
3

−

0

=

1

3
2

− 0 = 1.
3


 1
 1 1  12 2
( ) − (2 )2  = 3 1 0 ( − 4 ) 
2
0 2

1
2

2 − 1 5
5

1
0

=

3
2

1
2

 4
0

 = −1

−

1
5





=

3
2



3
10



9
 9
20 20


.

=

9
.
20

N

Thus, the centroid is ( ) =
31.  =



ot

=

1
3

Fo

0

+1
.
4

≈ (058 093).

1
(12 − 2 )  = 3 0 (32 − 3 ) 

1


3
9
= 3 2 52 − 1 4 = 3 2 − 1 = 3 20 = 20 .
5
4
5
4

=

e

0

al

=

1

1


rS

=


4 √
(cos  − sin )  = sin  + cos  0 = 2 − 1.

 4
0

(cos  − sin ) 


4
= −1 (sin  + cos ) + cos  − sin  0
√
1

 √
 2−1
−1 
4
√
=
.
2−1 =
4
2−1
 = −1

 4
0

1
(cos2
2

 − sin2 )  =

Thus, the centroid is ( ) =



1
2

 4
0

[integration by parts]

cos 2  =

1
4


4
1
1
sin 2 0 =
= √
.
4
4 2−1


√
1
 2−4
√
  √
 ≈ (027 060).
4 2−1 4 2−1

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SECTION 8.3 APPLICATIONS TO PHYSICS AND ENGINEERING

33. The curves intersect when 2 −  =  2

¤

391

⇔ 0 = 2 +  − 2 ⇔

0 = ( + 2)( − 1) ⇔  = −2 or  = 1.
=
=

1

−2

1


=

1
9

=

1


=

2
9

1

(2 −  − 2 )  = 2 − 1  2 − 1  3 −2 =
2
3

1

1
[(2
−2 2

− )2 − ( 2 )2 ]  =

2
9


1
4 − 22 + 1  3 − 1  5 −2 =
3
5
1

−2

(2 −  −  2 )  =

2
9

1

−2

1
9

·

1
2

1

 32
15

−2

7
6



− − 10 = 9 .
3
2

(4 − 4 + 2 −  4 ) 



− − 184 = 8 .
15
5

(2 − 2 −  3 ) 

 2 1 3 1 4 1
5

 − 3  − 4  −2 = 2 12 − 8 = − 1 .
9
3
2

Thus, the centroid is ( ) = ( 8  − 1 ).
5
2

 = 
 = 

15
2

4
0

 =

45
16

1

2  =

15
2

1


3 4
3 0


3 4
3 0

=

45
16

 64 
3

= 60

=

15
2

 64 
3

= 160





160
8
60
=
= and  =
=
= 1. Thus, the centroid is ( ) = 8  1 .
3

60
3

60
1

[(3 − ) − (2 − 1)]  =
−1

=2

1
0

1

rS

37.  =

 4 3 
 4   =
0

9 2
32 

(1 − 2 ) 
−1

1

 
(1 − 2 )  = 2  − 1 3 0 = 2 2 = 4 .
3
3
3



odd-degree terms drop out

Fo

=

 4 1  3 2
4
  = 10 0
0 2 4

e

 = 1 (4)(3) = 6, so  =  = 10(6) = 60.
2

al

35. The line has equation  = 3 .
4

1
(3 −  − 2 + 1)  = 3 −1 (4 − 2 − 3 + ) 
4
1
1
= 3 −1 (4 − 2 )  = 3 · 2 0 (4 − 2 ) 
4
4

1
 2
= 3 1 5 − 1 3 0 = 3 − 15 = − 1 .
2 5
3
2
5

=

1


=

3
8

−1

1

ot

1


1
− )2 − (2 − 1)2 ]  = 3 · 1 −1 (6 − 24 + 2 − 4 + 22 − 1) 
4 2



1
1
· 2 0 (6 − 34 + 32 − 1)  = 3 1 7 − 3 5 + 3 −  0 = 3 − 16 = − 12 .
4 7
5
4
35
35
1
[(3
−1 2

N

=

1





Thus, the centroid is ( ) = − 1  − 12 .
5
35

39. Choose - and -axes so that the base (one side of the triangle) lies along

the -axis with the other vertex along the positive -axis as shown. From geometry, we know the medians intersect at a point

2
3

of the way from each

vertex (along the median) to the opposite side. The median from  goes to

 the midpoint 1 ( + ) 0 of side , so the point of intersection of the
2
 

 medians is 2 · 1 ( + ) 1  = 1 ( + ) 1  .
3
2
3
3
3

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[continued]

392

¤

CHAPTER 8

FURTHER APPLICATIONS OF INTEGRATION

This can also be verified by finding the equations of two medians, and solving them simultaneously to find their point of intersection. Now let us compute the location of the centroid of the triangle. The area is  = 1 ( − ).
2
=


 0

 

 

1


 
1  0
 · ( − )  +
 · ( − )  =
( − 2 )  +
 − 2 
 


  
 0
0

=

=

and

0







 1 2 1 3

1
 1 2 1 3
 1 3 1 3
1
+
=
 − 
 − 
− 3 + 3 +
 − 
 2
3
 2
3
 2
3
 2
3

0
2
−3
2
3
1
+
·
+
·
=
(2 − 2 ) =
 ( − )
6
 ( − ) 6
3 ( − )
3

=


2
2 
  
0 
1 
1 
1
( − )  +
( − ) 
  2 
0 2 



 0
 
1 2
2
(2 − 2 + 2 )  + 2
(2 − 2 + 2 ) 
 22 
2 0

 2
0

1   2
2 
=
  − 2 + 1 3  + 2 2  − 2 + 1 3 0
3
3
 22
2






1 2  3
2 
2
( − )2

1 2
=
− + 3 − 1 3 + 2 3 − 3 + 1 3 =
(− + ) =
·
=
3
3
2
 2
2
 6
( − ) 
6
3
Thus, the centroid is ( ) =



rS

al

e

=


+ 

, as claimed.
3
3

we see that the centroid is

1
3

Fo

Remarks: Actually the computation of  is all that is needed. By considering each side of the triangle in turn to be the base, of the way from each side to the opposite vertex and must therefore be the intersection of the

medians.

The computation of  in this problem (and many others) can be

ot

simplified by using horizontal rather than vertical approximating rectangles.
If the length of a thin rectangle at coordinate  is (), then its area is

N

() ∆, its mass is () ∆, and its moment about the -axis is
∆ = () ∆. Thus,

 =



() 

and

=



() 
1 
=
() 



−
( − ) by similar triangles, so


2 

1  −
2 
2 3
( − )  = 2 0 ( −  2 )  = 2 1  2 − 1  3 0 = 2 ·
=
=
2
3
0





6
3

In this problem, () =

Notice that only one integral is needed when this method is used.

41. Divide the lamina into two triangles and one rectangle with respective masses of 2, 2 and 4, so that the total mass is 8. Using







the result of Exercise 39, the triangles have centroids −1 2 and 1 2 . The centroid of the rectangle (its center) is 0 − 1 .
3
3
2

3
  
 
 
 
1 

=
  = 1 2 2 + 2 2 + 4 − 1 = 1 2 =
8
3
3
2
8 3

 =1
 1 since the lamina is symmetric about the line  = 0. Thus, the centroid is ( ) = 0 12 .

So, using Formulas 5 and 7, we have  =

1
,
12

and  = 0,

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SECTION 8.4

43.




APPLICATIONS TO ECONOMICS AND BIOLOGY

¤

393





  ()  +   ()  =     ()  +    ()  =  +   ()  [by (8)]



=    ()  +    ()  = ( + )  () 

( + )  ()  =




45. A cone of height  and radius  can be generated by rotating a right triangle

about one of its legs as shown. By Exercise 39,  = 1 , so by the Theorem of
3
Pappus, the volume of the cone is

 

 =  = 1 · base · height · (2) = 1  · 2 1  = 1 2 .
2
2
3
3
47. Suppose the region lies between two curves  =  () and  = () where  () ≥ (), as illustrated in Figure 13.

e

Choose points  with  = 0  1  · · ·   =  and choose ∗ to be the midpoint of the th subinterval; that is,



∗
1
 =  = 2 (−1 +  ). Then the centroid of the th approximating rectangle  is its center  =   1 [ ( ) + ( )] .
2

Its area is [ ( ) − ( )] ∆, so its mass is

rS

al

[ ( ) − ( )] ∆. Thus,  ( ) = [ ( ) − ( )] ∆ ·  =  [ ( ) − ( )] ∆ and


 ( ) = [ ( ) − ( )] ∆ · 1 [ ( ) + ( )] =  · 1  ( )2 − ( )2 ∆. Summing over  and taking the limit
2
2

 as  → ∞, we get  = lim
  [ ( ) − ( )] ∆ =   [ () − ()]  and
→∞

→∞

Thus,  =

1
2


1

=
=








[ () − ()]  and  =

Fo

 = lim



 

 ( )2 − ( )2 ∆ =   1 ()2 − ()2 .
·
2






1

=
=









1
2



 ()2 − ()2 .

8.4 Applications to Economics and Biology

ot

1. By the Net Change Theorem, (4000) − (0) =

 4000

 4000
0

 0 ()  ⇒

(082 − 0000 03 + 0000 000 0032 ) 

4000
= 18,000 + 3104 = $21,104
= 18,000 + 082 − 0000 0152 + 0000 000 0013 0

N

(4000) = 18,000 +

0

3. By the Net Change Theorem, (50) − (0) =

 50
0

(06 + 0008)  ⇒


50
(50) = 100 + 06 + 00042 0 = 100 + (40 − 0) = $140,000. Similarly,
100

(100) − (50) = 06 + 00042 50 = 100 − 40 = $60,000.
450
= 10 ⇒  + 8 = 45 ⇒  = 37.
+8

 37 
 37
450
− 10 
[() − 10]  =
Consumer surplus =
+8
0
0

37
= 450 ln ( + 8) − 10 0 = (450 ln 45 − 370) − 450 ln 8
 
= 450 ln 45 − 370 ≈ $40725
8

5. () = 10

⇒

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394

¤

CHAPTER 8

FURTHER APPLICATIONS OF INTEGRATION

⇒ 400 = 200 + 0232

⇒  = 100023 = 100.

 100
 100
 100 
200 − 1 32 
Producer surplus = 0 [ −  ()]  = 0 [400 − (200 + 0232 )]  = 0
5

100
2
= 200 − 25 52
= 20,000 − 8,000 = $12,000

7.  =  ()

⇒ 200 = 0232

⇒ 1000 = 32

0

800,000−5000
= 16 ⇒  = 1 ≈ 372704.
 + 20,000

Consumer surplus = 0 1 [() − 16]  ≈ $37,753

9. () =

4

 0 ()  =

8

 √
8√

  = 2 32 = 2 16 2 − 8 ≈ $975 million
3
3
4
4

 −+1 
 


−
13.  =
  = 
=
(1− − 1− ).
− + 1 
1−


e

8

al

11.  (8) −  (4) =

 2− 


(2− − 2− ).
1−  = 
=
2− 
2−

 
1
[(2 − )](2− − 2− )
(1 − )(2− − 2− )
1−  =
Thus,  =
=
.
1− − 1− )
 
[(1 − )](
(2 − )(1− − 1− )


(4000)(0008)4
 4
=
≈ 119 × 10−4 cm3s
8
8(0027)(2)

Fo

15.  =



rS

Similarly,


6
, where
=
20
 () 
0
10 


 10
1
integrating
=
=
−06  =
(−06 − 1) −06 by parts
(−06)2
0
0

−6
1
036 (−7

+ 1)

0108
6(036)
=
≈ 01099 Ls or 6594 Lmin.
20(1 − 7−6 )
1 − 7−6

N

Thus,  =

ot

17. From (3),  =  

19. As in Example 2, we will estimate the cardiac output using Simpson’s Rule with ∆ = (16 − 0)8 = 2.

 16
0

()  ≈ 2 [(0) + 4(2) + 2(4) + 4(6) + 2(8) + 4(10) + 2(12) + 4(14) + (16)]
3
≈ 2 [0 + 4(61) + 2(74) + 4(67) + 2(54) + 4(41) + 2(30) + 4(21) + 15]
3
= 2 (1091) = 7273 mg· sL
3

Therefore,  ≈


7
=
≈ 00962 Ls or 577 Lmin.
7273
7273

8.5 Probability
1. (a)

(b)

 40,000
30,000

∞

25,000

 ()  is the probability that a randomly chosen tire will have a lifetime between 30,000 and 40,000 miles.
 ()  is the probability that a randomly chosen tire will have a lifetime of at least 25,000 miles.

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SECTION 8.5

PROBABILITY

¤

395

3. (a) In general, we must satisfy the two conditions that are mentioned before Example 1 — namely, (1)  () ≥ 0 for all , and

(2)

∞

−∞

 ()  = 1. For 0 ≤  ≤ 1,  () = 302 (1 − )2 ≥ 0 and  () = 0 for all other values of , so () ≥ 0 for

all . Also,
∞

 ()  =

−∞

1
0

302 (1 − )2  =

1
0

302 (1 − 2 + 2 )  =

1

= 103 − 154 + 65 0 = 10 − 15 + 6 = 1

1
0

(302 − 603 + 304 ) 

Therefore,  is a probability density function.

  13
13
 13


(b)   ≤ 1 = −∞  ()  = 0 302 (1 − )2  = 103 − 154 + 65 0 =
3

10
27

−

15
81

+

6
243

=

17
81

5. (a) In general, we must satisfy the two conditions that are mentioned before Example 1—namely, (1)  () ≥ 0 for all ,

 ()  = 1. If  ≥ 0, then  () ≥ 0, so condition (1) is satisfied. For condition (2), we see that

 ()  =

−∞





∞

Similarly,


 and
1 + 2

−∞







 =  lim tan−1  0 =  lim tan−1  = 
2
→∞
→∞
2
0 1+
 ∞




 = 
 = 2
, so
= .
2
2
1+
2
2
−∞ 1 + 

∞


 = lim
→∞
1 + 2

0



e

∞

−∞

al



∞

0

−∞



rS

and (2)

Since  must equal 1, we must have  = 1 so that  is a probability density function.


1

−1



 1
1
2
2 
1
tan−1  0 =
 =
−0 =
1 + 2

 4
2

Fo

(b)  (−1    1) =

1
2
 =
1 + 2


1

0

7. (a) In general, we must satisfy the two conditions that are mentioned before Example 1—namely, (1)  () ≥ 0 for all ,

∞

−∞

∞

−∞

 ()  = 1. Since  () = 0 or  () = 01, condition (1) is satisfied. For condition (2), we see that

ot

and (2)

 ()  =

 10
0

01  =



10
1
 0
10

= 1. Thus, () is a probability density function for the spinner’s values.

N

(b) Since all the numbers between 0 and 10 are equally likely to be selected, we expect the mean to be halfway between the endpoints of the interval; that is,  = 5.
=
9. We need to find  so that

(−1)(0 − −5 ) =

1
2

∞


∞

−∞

 ()  =

 ()  =

⇒ −5 =

 10
0

(01)  =

  1 −5 lim 
→∞  5

1
2

⇒

1
2

⇒ −5 = ln 1
2




1 2 10
20  0

 =

1
2

=
⇒

= 5, as expected.

 lim 1 (−5)−5
=
5

100
20

→∞



1
2

⇒  = −5 ln 1 = 5 ln 2 ≈ 347 min.
2

11. We use an exponential density function with  = 25 min.

(a)  (  4) =

∞
4



1 −25

→∞ 4 25

 ()  = lim



 = lim −−25 = 0 + −425 ≈ 0202
→∞

4


2
2
(b)  (0 ≤  ≤ 2) = 0  ()  = −−25 = −−225 + 1 ≈ 0551
0

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

⇒

¤

396

CHAPTER 8

FURTHER APPLICATIONS OF INTEGRATION

(c) We need to find a value  so that  ( ≥ ) = 002, or, equivalently,  (0 ≤  ≤ ) = 098 ⇔



 ()  = 098 ⇔ −−25 = 098 ⇔ −−25 + 1 = 098 ⇔ −25 = 002 ⇔
0
0

1
−25 = ln 002 ⇔  = −25 ln 50 = 25 ln 50 ≈ 978 min ≈ 10 min. The ad should say that if you aren’t served

within 10 minutes, you get a free hamburger.


1
( − 94)2
√ exp −
. To avoid the improper integral we approximate it by the integral from
13.  ( ≥ 10) =
2 · 422
2
10 42


 100
1
( − 94)2
√ exp −
10 to 100. Thus,  ( ≥ 10) ≈
 ≈ 0443 (using a calculator or computer to estimate
2 · 422
42 2
10


∞

the integral), so about 44 percent of the households throw out at least 10 lb of paper a week.

Note: We can’t evaluate 1 −  (0 ≤  ≤ 10) for this problem since a significant amount of area lies to the left of  = 0.


100



1
( − 112)2
√ exp −
 ≈ 00668 (using a calculator or computer to estimate the
2 · 82
8 2

e

15. (a)  (0 ≤  ≤ 100) =

0

al

integral), so there is about a 668% chance that a randomly chosen vehicle is traveling at a legal speed.


 ∞
( − 112)2
1
√ exp −
 () . In this case, we could use a calculator or computer
(b)  ( ≥ 125) =
 =
2
2·8
2
125 8
125
 125
 300 to estimate either 125  ()  or 1 − 0  () . Both are approximately 00521, so about 521% of the motorists are
∞

rS



targeted.



2

−2



+2

−2

2
1
1
√ −  /2 ( ) = √
 2
2

2

−2

2

−  /2  ≈ 09545.

4 2 −20
 
≥ 0 for  ≥ 0. Next,
3
0
 ∞
 ∞
 
4 2 −20
4
()  =
 
 = 3 lim
2 −20 
3
0 →∞ 0
−∞
0
0

N

ot

19. (a) First () =





1
−
( − )2
1
√ exp −
. Substituting  = and  =  gives us
22


 2

Fo

17.  ( − 2 ≤  ≤  + 2) =



2   = (3 )(2 2 − 2 + 2). ()


4 3
0
(−2) = 1. This satisfies the second condition for
Next, we use () (with  = −20 ) and l’Hospital’s Rule to get 3
0 −8
By using parts, tables, or a CAS , we find that

a function to be a probability density function.
(b) Using l’Hospital’s Rule,

2
2
2
4
4
2
lim lim = 2 lim
= 0.
3 →∞ 20 = 3 →∞
0
0
0 →∞ (20 )20

(20 )20

To find the maximum of , we differentiate:






4
4

2
+ −20 (2) = 3 −20 (2) − + 1
0 () = 3 2 −20 −
0
0
0
0
0 () = 0 ⇔  = 0 or 1 =


0

⇔  = 0 [0 ≈ 559 × 10−11 m].

0 () changes from positive to negative at  = 0 , so () has its maximum value at  = 0 . c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 8 REVIEW

¤

397

(c) It is fairly difficult to find a viewing rectangle, but knowing the maximum value from part (b) helps.
(0 ) =

4 2 −20 0
4 −2
 
=
 ≈ 9,684,098,979
3 0
0
0

With a maximum of nearly 10 billion and a total area under the curve of 1, we know that the “hump” in the graph must be extremely narrow.
(d)  () =





0

4 2 −20
 
 ⇒  (40 ) =
3
0

 (40 ) =

4
3
0





0

40

4 2 −20
 
. Using () from part (a) [with  = −20 ],
3
0




40
−20 4 2
4
4 3
0
[−8 (64 + 16 + 2) − 1(2)] = − 1 (82−8 − 2)
 + +2
= 3
2
−83 2
0
0 −8
0
0
0





∞

()  =

−∞

4 lim 3 →∞
0





3 −20 . Integrating by parts three times or using a CAS, we find that

0

al

(e)  =

e

= 1 − 41−8 ≈ 0986


  3 3
2
  − 32 2 + 6 − 6 . So with  = − , we use l’Hospital’s Rule, and get
4
0
 4


4
 = 3 − 0 (−6) = 3 0 .
2
0
16

rS

3   =

Fo

8 Review

1. (a) The length of a curve is defined to be the limit of the lengths of the inscribed polygons, as described near Figure 3 in

Section 8.1.

ot

(b) See Equation 8.1.2.

N

(c) See Equation 8.1.4.
2. (a)  =





2 () 1 + [ 0 ()]2 

(b) If  = (),  ≤  ≤ , then  =
(c)  =




2




2


1 + [0 ()]2 .




1 + [ 0 ()]2  or  =  2() 1 + [ 0 ()]2 

3. Let () be the cross-sectional length of the wall (measured parallel to the surface of the fluid) at depth . Then the hydrostatic

force against the wall is given by  = and  is the weight density of the fluid.




() , where  and  are the lower and upper limits for  at points of the wall

4. (a) The center of mass is the point at which the plate balances horizontally.

(b) See Equations 8.3.8.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

398

¤

CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION

5. If a plane region R that lies entirely on one side of a line  in its plane is rotated about , then the volume of the resulting solid

is the product of the area of R and the distance traveled by the centroid of R.
6. See Figure 3 in Section 8.4, and the discussion which precedes it.
7. (a) See the definition in the first paragraph of the subsection Cardiac Output in Section 8.4.

(b) See the discussion in the second paragraph of the subsection Cardiac Output in Section 8.4.
8. A probability density function  is a function on the domain of a continuous random variable  such that




 () 

measures the probability that  lies between  and . Such a function  has nonnegative values and satisfies the relation

 ()  = 1, where  is the domain of the corresponding random variable . If  = R, or if we define () = 0 for real

∞ numbers  ∈ , then −∞  ()  = 1. (Of course, to work with  in this way, we must assume that the integrals of  exist.)

 ()  represents the probability that the weight of a randomly chosen female college student is less than

130 pounds.
∞
∞
(b)  = −∞  ()  = 0  () 

(c) The median of  is the number  such that

e

0

∞


⇒  = 1 (2 + 4)12 (2) ⇒
4

2

2
1 + ()2 = 1 + 1 (2 + 4)12 = 1 + 1 2 (2 + 4) = 1 4 + 2 + 1 = 1 2 + 1 .
2
4
4
2

+ 4)32

Thus,  =

 3  1
2

0

2


3
3
2 + 1  = 0 1 2 + 1  = 1 3 +  0 =
2
6

ot

1.  =

2
1
6 (

Fo

10. See the discussion near Equation 3 in Section 8.5.

 ()  = 1 .
2

al

 130

rS

9. (a)

15
.
2

1
1
4
1 4 1 −2

+ 2 =
 + 
= 3 − −3 ⇒
⇒
16
2
16
2

4
2

1
1
1 + ()2 = 1 + 1 3 − −3 = 1 + 16 6 − 1 + −6 = 16 6 +
4
2

N

3. (a)  =

Thus,  =
2

 2 1
4

1

1
2

+ −6 =


1
2 
 1

3 + −3  = 16 4 − 1 −2 1 = 1 − 1 − 16 − 1 =
2
8
2

21
.
16

1
4

2
3 + −3 .



2

1

1 2
2 1 3 + −3  = 2 1 1 4 + −2  = 2 20 5 −  1
4
4
 1




 
1
= 2 32 − 1 − 20 − 1 = 2 8 − 1 − 20 + 1 = 2 41 = 41 
20
2
5
2
20
10

(b)  =

1

5.  = sin 

=


0

⇒  0 = cos  ⇒ 1 + (0 )2 = 1 + cos2 . Let  () =

 ()  ≈ 10
=

( − 0)10
3

≈ 3820188

√
1 + cos2 . Then



 
 
 
 (0) + 4 10 + 2 2 + 4 3 + 2 4
10
10
10
 6 
 7 
 8 
 

 5 
+ 4 10 + 2 10 + 4 10 + 2 10 + 4 9 +  ()
10

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 8 REVIEW

7.  =

  √
 − 1 
1

Thus,  =

√
 =
−1

⇒

⇒


16
 16
 16 √
  = 1 14  = 4 54
= 4 (32 − 1) =
5
5
1
1

1

=
2−
2

Thus,  =
11.  =

2
0

(5 − )  = 

5

2
2

⇒ 2 = 2 −  and  = 2(15 + ) = 3 + 2 = 3 + 2 −  = 5 − .

4

 4 √

 − 1   = 2 32 − 1 2 =
2
3
4
0



3
4

1


2 52

5

4

1
0 2

− 1 3
6

4

=

0

3
4

64
6

−

5





√ 2  1 2
 =
 − 2
8
5

−4 =

=

3
4


1 .

3
4

 64 
30

=

 3 2 
 3   =
0

2
27

· 625 ≈ 458 lb.

8
5

 4 1


 − 1 2  = 3 1 2 −
4
8 2
0 2

 3 3
 0 = 2 and  =

1
3

2
3 .

 3 1  2 2
  =
0 2 3

Fo

1
3

22
3

4
3

13. An equation of the line passing through (0, 0) and (3, 2) is  =

=

[ = ] ≈

 4  32 1 2 
− 2  

0

3
4

 64

Thus, the centroid is ( ) =

16
3

22
3 


1 3 4
12  0

=

3
8


8−

al

=


 4 √
  − 1   =
2
0



=  10 − 8 =
3

0

rS

=

1


2

− 1 3
3

0

=

399

124
.
5

e

9. As in Example 1 of Section 8.3,

¤

√
 √
1 + ()2 = 1 +
 − 1 = .

=

2
81

1
2

16
3



=

3
8

8
3

=1

· 3 · 2 = 3. Therefore, using Equations 8.3.8,

 3 3


 0 = 2 . Thus, the centroid is ( ) = 2 2 .
3
3

15. The centroid of this circle, (1 0), travels a distance 2(1) when the lamina is rotated about the -axis. The area of the circle

is (1)2 . So by the Theorem of Pappus,  = (2) = (1)2 2(1) = 22 .
⇒  = 2000 − 01(100) − 001(100)2 = 1890

ot

17.  = 100

 100 

2000 − 01 − 0012 − 1890 
0

100
= 110 − 0052 − 001 3 0 = 11,000 − 500 − 10,000 ≈ $716667
3
3

N

Consumer surplus =

19.  () =



20

0

  sin 10 

 100
0

[() −  ]  =

if 0 ≤  ≤ 10

if   0 or   10

(a)  () ≥ 0 for all real numbers  and
∞

−∞

 ()  =

 10
0


20

  sin 10   =

Therefore,  is a probability density function.
(b)  (  4) =

4

−∞

 ()  =

4


0 20


20

·

10



  10
− cos 10  0 = 1 (− cos  + cos 0) = 1 (1 + 1) = 1
2
2


  4

 

sin 10   = 1 − cos 10  0 = 1 − cos 2 + cos 0
2
2
5

≈ 1 (−0309017 + 1) ≈ 03455
2

c
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400

¤

CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION

(c)  =
=
=

∞

−∞



 ()  =


0 20

5



0

 10
0

 
· 10 (sin ) 10 


82 5


 sin   =







20  sin 10 

[ =



10 ,


10

 =

[sin  −  cos ] =
0

5
[0


]

− (−1)] = 5

This answer is expected because the graph of  is symmetric about the line  = 5.

21. (a) The probability density function is  () =

10

3

 = −−8 = −−38 + 1 ≈ 03127
0



1 −8

 = lim −−8
= lim (−−8 + −108 ) = 0 + −54 ≈ 02865
8
→∞

(c) We need to find  such that  ( ≥ ) =
1
2

1
2

⇒ −8 =

∞

1 −8

 8

⇒
1
2

 =

1
2

⇒



lim −−8
=

⇒ −8 = ln 1
2

→∞



1
2

⇒

⇒  = −8 ln 1 = 8 ln 2 ≈ 555 minutes.
2

N

ot

Fo

lim (−−8 + −8 ) =

→∞

→∞

10

e

∞

1 −8

0 8

if  ≥ 0

al

(b)  (  10) =

3

if   0

1 −8
8

rS

 (0 ≤  ≤ 3) =


0

c
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PROBLEMS PLUS
1. 2 +  2 ≤ 4

⇔ 2 + ( − 2)2 ≤ 4, so  is part of a circle, as shown

in the diagram. The area of  is


1

1
113
4 −  2  = −2 4 − 2 + 2 cos−1 2−
2
2
0

[ = 2]

0

Another method (without calculus): Note that  = ∠ =


,
3

so the area is

(area of sector ) − (area of 4) =

1
2

√
 2 
2 3 − 1 (1) 3 =
2

2
3

−

√

3
2

e

√
 
= − 1 3 + 2 cos−1 1 − 2 cos−1 1
2
2
√
√

3
= − 2 + 2 3 − 2(0) = 2 − 23
3

al

3. (a) The two spherical zones, whose surface areas we will call 1 and 2 , are

generated by rotation about the -axis of circular arcs, as indicated in the figure.

rS

The arcs are the upper and lower portions of the circle 2 +  2 = 2 that are

obtained when the circle is cut with the line  = . The portion of the upper arc

ot

Fo

in the first quadrant is sufficient to generate the upper spherical zone. That

portion of the arc can be described by the relation  = 2 −  2 for

 ≤  ≤ . Thus,  = − 2 −  2 and



 2
2

2
 
 = 1 +
 = 1 + 2
 =
 = 

 − 2
2 − 2
2 − 2

N

From Formula 9.2.8 we have

 2
 
 
 


 
2 1 +
 =
2 2 −  2 
=
2  = 2( − )
1 =

2 −  2






Similarly, we can compute 2 = − 2 1 + ()2  = − 2  = 2( + ). Note that 1 + 2 = 42 , the surface area of the entire sphere.

(b)  = 3960 mi and  =  (sin 75◦ ) ≈ 3825 mi, so the surface area of the Arctic Ocean is about
2(−) ≈ 2(3960)(135) ≈ 336×106 mi2 .

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

401

402

¤

CHAPTER 8 PROBLEMS PLUS

(c) The area on the sphere lies between planes  = 1 and  = 2 , where 2 − 1 = . Thus, we compute the surface area on

 2
 2
 2

the sphere to be  =
2 1 +
 =
2  = 2(2 − 1 ) = 2.

1
1
This equals the lateral area of a cylinder of radius  and height , since such a cylinder is obtained by rotating the line  =  about the -axis, so the surface area of the cylinder between the planes  = 1 and  = 2 is

 2
 2
 2


2 1 +
 =
2 1 + 02 
=

1
1
2

= 2(2 − 1 ) = 2
= 2
=1

e

(d)  = 2 sin 2345◦ ≈ 3152 mi, so the surface area of the

rS

al

Torrid Zone is 2 ≈ 2(3960)(3152) ≈ 784 × 107 mi2 .

5. (a) Choose a vertical -axis pointing downward with its origin at the surface. In order to calculate the pressure at depth ,

Fo

consider  subintervals of the interval [0 ] by points  and choose a point ∗ ∈ [−1   ] for each . The thin layer of

water lying between depth −1 and depth  has a density of approximately (∗ ), so the weight of a piece of that layer

with unit cross-sectional area is (∗ ) ∆. The total weight of a column of water extending from the surface to depth 


ot

(with unit cross-sectional area) would be approximately




(∗ ) ∆. The estimate becomes exact if we take the limit


=1

N

as  → ∞; weight (or force) per unit area at depth  is  = lim




→∞ =1

(∗ ) ∆. In other words,  () =


More generally, if we make no assumptions about the location of the origin, then  () = 0 + the pressure at  = 0. Differentiating, we get  = ().
(b)


0


0

() .

() , where 0 is



√
 ( + ) · 2 2 − 2 
 √
 +
 
0    · 2 2 − 2 
= − 0 + 0
 √
 
 √
= 0 − 2 2 − 2  + 0  − (+) − 1 · 2 2 − 2 

 =

−

√
 √

= (0 − 0 ) − 2 2 − 2  + 0  − (+) · 2 2 − 2 
√



= (0 − 0 ) 2 + 0  −  · 2 2 − 2 

c
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CHAPTER 8 PROBLEMS PLUS

¤

403

7. To find the height of the pyramid, we use similar triangles. The first figure shows a cross-section of the pyramid passing

through the top and through two opposite corners of the square base. Now || = , since it is a radius of the sphere, which has diameter 2 since it is tangent to the opposite sides of the square base. Also, || =  since 4 is isosceles. So the
√
√ height is || = 2 + 2 = 2 .

e

We first observe that the shared volume is equal to half the volume of the sphere, minus the sum of the four equal volumes
(caps of the sphere) cut off by the triangular faces of the pyramid. See Exercise 5.2.49 for a derivation of the formula for the

al

volume of a cap of a sphere. To use the formula, we need to find the perpendicular distance  of each triangular face from the surface of the sphere. We first find the distance  from the center of the sphere to one of the triangular faces. The third figure

rS

shows a cross-section of the pyramid through the top and through the midpoints of opposite sides of the square base. From similar triangles we find that

√

6

3

=

Fo

So  =  −  =  −

√
2
||

=
= 
√ 2

||
2 +
2

√
3− 6
.
3



√ 2
√
2
6
= √
=

2
3
3

So, using the formula  = 2 ( − 3) from Exercise 5.2.49 with  = , we find that

√ 2 
3− 6


3

4

observation, the shared volume is  =

1
2

−

√ 
3− 6

3·3



3 − 4 2 −
3

ot

the volume of each of the caps is 

⇒

3

7
27

=

√
15 − 6 6
9

·

√
6+ 6
3
9

=

2
3

√  3  28 √

6  = 27 6 − 2 3 .

−

7
27

√  3
6  . So, using our first

N

9. We can assume that the cut is made along a vertical line  =   0, that the

disk’s boundary is the circle 2 +  2 = 1, and that the center of mass of the smaller piece (to the right of  = ) is decimal places. We have

numerator gives us −

1


1
==
2

1

1


. We wish to find  to two

20

√
 · 2 1 − 2 
. Evaluating the
1 √
2 1 − 2 




(1 − 2 )12 (−2)  = − 2
3




32 1
32  2
= −2 0 − 1 − 2
1 − 2
= 3 (1 −  2 )32 .
3


Using Formula 30 in the table of integrals, we find that the denominator is

 √
  √
1 

1
 1 − 2 + sin−1   = 0 +  −  1 −  2 + sin−1  . Thus, we have =  =
2
2

−  2 )32
√
, or,
−  1 −  2 − sin−1 
2
(1
3


2

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404

¤

CHAPTER 8 PROBLEMS PLUS

equivalently, 2 (1 −  2 )32 =
3


4

− 1
2

√
1 − 2 −

1
2

sin−1 . Solving this equation numerically with a calculator or CAS, we

obtain  ≈ 0138173, or  = 014 m to two decimal places.
11. If  = , then  =

area under  =  sin 
=
area of rectangle



 sin  
[− cos ]
− (−1) + 1
2
0
=
=
= .





0

area under  = 1  sin 
2
If  = 2, then  =
=
area of rectangle
13. Solve for : 2 + ( +  + 1)2 = 1



1
 sin  
0 2



⇒ ( +  + 1)2 = 1 − 2

√
 = − − 1 ± 1 − 2 .
 1 
 



− − 1 + 1 − 2 − − − 1 − 1 − 2 
=
2

−1

=

1





 
1 − 2  = 2 
2

1

−1

·2



area of semicircle 

=

⇒

e

1

√
⇒  +  + 1 = ± 1 − 2


1 − 2  = 0 [odd integrand]

rS

=

[− cos ]
2
1
0
=
= .
2
2


al

−1



=



2 
2 





1 
1 1 1
2
2
 =
− − 1 + 1 − 
−4 1 − 2 − 4 1 − 2 
− − − 1 − 1 − 
 −1 2
−1 2
 1  
 1 



2
2 1 
2
 1 − 2 + 1 − 2  = −
 1 − 2  −
1 − 2 
=−
 −1
 −1
 −1


2
2   area of
= − (0) [odd integrand] −
= −1 semicircle 
 2


1

Fo

1
=


N

located at its center.

ot

Thus, as expected, the centroid is ( ) = (0 −1). We might expect this result since the centroid of an ellipse is

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

9

DIFFERENTIAL EQUATIONS

9.1 Modeling with Differential Equations
1.  =

2 

3

+ −2

⇒  0 = 2  − 2−2 . To show that  is a solution of the differential equation, we will substitute the
3

expressions for  and  0 in the left-hand side of the equation and show that the left-hand side is equal to the right-hand side.


LHS = 0 + 2 = 2  − 2−2 + 2 2  + −2 = 2  − 2−2 + 4  + 2−2
3
3
3
3
= 6  = 2 = RHS
3

3. (a)  = 

⇒  0 = 

⇒  00 = 2  . Substituting these expressions into the differential equation

(2 − 1)( + 1) = 0 [since  is never zero] ⇒  =
1
2

or −1.

al

(b) Let 1 =

1
2

e

2 00 +  0 −  = 0, we get 22  +  −  = 0 ⇒ (22 +  − 1) = 0 ⇒

and 2 = −1, so we need to show that every member of the family of functions  = 2 + − is a

rS

solution of the differential equation 2 00 +  0 −  = 0.

⇒  0 = 1 2 − − ⇒  00 = 1 2 + − .
2
4
 


LHS = 2 00 +  0 −  = 2 1 2 + − + 1 2 − − − (2 + − )
4
2

 = 2 + −

Fo

= 1 2 + 2− + 1 2 − − − 2 − −
2
2
1
 2
1
= 2 + 2 −  
+ (2 −  − )−
= 0 = RHS

⇒  0 = cos  ⇒  00 = − sin .

ot

5. (a)  = sin 

LHS =  00 +  = − sin  + sin  = 0 6= sin , so  = sin  is not a solution of the differential equation.

N

(b)  = cos  ⇒  0 = − sin  ⇒ 00 = − cos .
LHS =  00 +  = − cos  + cos  = 0 6= sin , so  = cos  is not a solution of the differential equation.
(c)  = 1  sin  ⇒  0 = 1 ( cos  + sin ) ⇒  00 = 1 (− sin  + cos  + cos ).
2
2
2
LHS = 00 +  = 1 (− sin  + 2 cos ) + 1  sin  = cos  6= sin , so  = 1  sin  is not a solution of the
2
2
2
differential equation.
(d)  = − 1  cos  ⇒  0 = − 1 (− sin  + cos ) ⇒ 00 = − 1 (− cos  − sin  − sin ).
2
2
2


LHS =  00 +  = − 1 (− cos  − 2 sin ) + − 1  cos  = sin  = RHS, so  = − 1  cos  is a solution of the
2
2
2
differential equation.

7. (a) Since the derivative  0 = − 2 is always negative (or 0 if  = 0), the function  must be decreasing (or equal to 0) on any

interval on which it is defined.

(b)  =

1
+

⇒ 0 = −

1
1
0
=−
2 . LHS =  = −
( + )
( + )2



1
+

2

= −2 = RHS

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

405

406

¤

CHAPTER 9

DIFFERENTIAL EQUATIONS

(c)  = 0 is a solution of  0 = −2 that is not a member of the family in part (b).
1
1
1
1
1
1
, then (0) =
= . Since (0) = 05,
=
⇒  = 2, so  =
.
+
0+


2
+2







9. (a)
= 12 1 −
. Now
0 ⇒ 1−
 0 [assuming that   0] ⇒
1 ⇒

4200

4200
4200
(d) If () =

  4200 ⇒ the population is increasing for 0    4200.

(b)


 0 ⇒   4200


(c)


= 0 ⇒  = 4200 or  = 0


11. (a) This function is increasing and also decreasing. But  =  ( − 1)2 ≥ 0 for all , implying that the graph of the

solution of the differential equation cannot be decreasing on any interval.

e

(b) When  = 1,  = 0, but the graph does not have a horizontal tangent line.

2 − 2

(b)  0 = −

al

13. (a)  0 = 1 + 2 +  2 ≥ 1 and  0 → ∞ as  → ∞. The only curve satisfying these conditions is labeled III.

 0 if   0 and  0  0 if   0. The only curve with negative tangent slopes when   0 and positive

(c)  0 =

rS

tangent slopes when   0 is labeled I.

1
 0 and  0 → 0 as  → ∞. The only curve satisfying these conditions is labeled IV.
1 + 2 +2

Fo

(d)  0 = sin() cos() = 0 if  = 0, which is the solution graph labeled II.

15. (a)  increases most rapidly at the beginning, since there are usually many simple, easily-learned sub-skills associated with

learning a skill. As  increases, we would expect  to remain positive, but decrease. This is because as time

(b)

ot

progresses, the only points left to learn are the more difficult ones.


= ( −  ) is always positive, so the level of performance 


(c)

N

is increasing. As  gets close to  ,  gets close to 0; that is, the performance levels off, as explained in part (a).

9.2 Direction Fields and Euler's Method
1. (a)

(b) It appears that the constant functions  = 05 and  = 15 are equilibrium solutions. Note that these two values of  satisfy the given differential equation 0 =  cos .

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 9.2

DIRECTION FIELDS AND EULER’S METHOD

¤

407

3.  0 = 2 − . The slopes at each point are independent of , so the slopes are the same along each line parallel to the -axis.

Thus, III is the direction field for this equation. Note that for  = 2, 0 = 0.
5.  0 =  +  − 1 = 0 on the line  = − + 1. Direction field IV satisfies this condition. Notice also that on the line  = − we

have 0 = −1, which is true in IV.
7. (a) (0) = 1

(b) (0) = 2

e

(c) (0) = −1

Note that for  = 0,  0 = 0. The three solution curves sketched go

9.

0 = 1 
2


0

0

1

05

0

2

1

0

−3

through (0 0), (0 1), and (0 −1).

0

−15

−2

−1

11.
0

Fo

0

rS

0

al



Note that  0 = 0 for any point on the line  = 2. The slopes are



−2

−2

2

2

6

2

2

−2

positive to the left of the line and negative to the right of the line. The solution curve in the graph passes through (1 0).

N

−2

 =  − 2

ot



2

−2

−6

c
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¤

CHAPTER 9

DIFFERENTIAL EQUATIONS

Note that  0 = ( + 1) = 0 for any point on  = 0 or on  = −1.

13.



1

±2

±2

0

 0 =  + 

±2



∓4

±2

−3

The slopes are positive when the factors  and  + 1 have the same sign and negative when they have opposite signs. The solution curve

±4

in the graph passes through (0 1).

al

15. In Maple, we can use either directionfield (in Maple’s share library) or

e

408

DEtools[DEplot] to plot the direction field. To plot the solution, we can

rS

either use the initial-value option in directionfield, or actually solve the equation. In Mathematica, we use PlotVectorField for the direction field, and the

Fo

Plot[Evaluate[  ]] construction to plot the solution, which is

 3
 = 2 arctan  3 · tan 1 .
2

In Derive, use Direction_Field (in utility file ODE_APPR) to plot the direction field. Then use

The direction field is for the differential equation  0 =  3 − 4.

N

17.

ot

DSOLVE1(-xˆ2*SIN(y),1,x,y,0,1) (in utility file ODE1) to solve the equation. Simplify each result.

19. (a)  0 =  ( ) =  and (0) = 1

 = lim () exists for −2 ≤  ≤ 2;
→∞

 = ±2 for  = ±2 and  = 0 for −2    2.
For other values of ,  does not exist.

⇒ 0 = 0, 0 = 1.

(i)  = 04 and 1 = 0 +  (0  0 ) ⇒ 1 = 1 + 04 · 1 = 14. 1 = 0 +  = 0 + 04 = 04, so 1 =  (04) = 14.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 9.2

DIRECTION FIELDS AND EULER’S METHOD

(ii)  = 02 ⇒ 1 = 02 and 2 = 04, so we need to find 2 .
1 = 0 +  (0  0 ) = 1 + 020 = 1 + 02 · 1 = 12,
2 = 1 +  (1  1 ) = 12 + 021 = 12 + 02 · 12 = 144.
(iii)  = 01 ⇒ 4 = 04, so we need to find 4 . 1 = 0 +  (0  0 ) = 1 + 010 = 1 + 01 · 1 = 11,
2 = 1 +  (1  1 ) = 11 + 011 = 11 + 01 · 11 = 121,
3 = 2 +  (2  2 ) = 121 + 012 = 121 + 01 · 121 = 1331,
4 = 3 +  (3  3 ) = 1331 + 013 = 1331 + 01 · 1331 = 14641.
We see that the estimates are underestimates since

(b)

al

e

they are all below the graph of  =  .

rS

(c) (i) For  = 04: (exact value) − (approximate value) = 04 − 14 ≈ 00918

(ii) For  = 02: (exact value) − (approximate value) = 04 − 144 ≈ 00518
(iii) For  = 01: (exact value) − (approximate value) = 04 − 14641 ≈ 00277

Fo

Each time the step size is halved, the error estimate also appears to be halved (approximately).
21.  = 05, 0 = 1, 0 = 0, and  ( ) =  − 2.

Note that 1 = 0 +  = 1 + 05 = 15, 2 = 2, and 3 = 25.
1 = 0 +  (0  0 ) = 0 + 05 (1 0) = 05[0 − 2(1)] = −1.

ot

2 = 1 +  (1  1 ) = −1 + 05 (15 −1) = −1 + 05[−1 − 2(15)] = −3.
3 = 2 +  (2  2 ) = −3 + 05 (2 −3) = −3 + 05[−3 − 2(2)] = −65.

N

4 = 3 +  (3  3 ) = −65 + 05 (25 −65) = −65 + 05[−65 − 2(25)] = −1225.
23.  = 01, 0 = 0, 0 = 1, and  ( ) =  + .

Note that 1 = 0 +  = 0 + 01 = 01, 2 = 02, 3 = 03, and 4 = 04.
1 = 0 +  (0  0 ) = 1 + 01 (0 1) = 1 + 01[1 + (0)(1)] = 11.
2 = 1 +  (1  1 ) = 11 + 01 (01 11) = 11 + 01[11 + (01)(11)] = 1221
3 = 2 +  (2  2 ) = 1221 + 01 (02 1221) = 1221 + 01[1221 + (02)(1221)] = 136752.
4 = 3 +  (3  3 ) = 136752 + 01 (03 136752) = 136752 + 01[136752 + (03)(136752)]
= 15452976.
5 = 4 +  (4  4 ) = 15452976 + 01 (04 15452976)
= 15452976 + 01[15452976 + (04)(15452976)] = 1761639264.
Thus, (05) ≈ 17616.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

409

410

¤

CHAPTER 9

DIFFERENTIAL EQUATIONS

25. (a)  + 32  = 62

⇒  0 = 62 − 32 . Store this expression in Y1 and use the following simple program to

evaluate (1) for each part, using H =  = 1 and N = 1 for part (i), H = 01 and N = 10 for part (ii), and so forth.
 → H: 0 → X: 3 → Y:

For(I, 1, N): Y + H × Y1 → Y: X + H → X:

End(loop):

Display Y. [To see all iterations, include this statement in the loop.]
(i) H = 1, N = 1 ⇒ (1) = 3
(ii) H = 01, N = 10 ⇒ (1) ≈ 23928
(iii) H = 001, N = 100 ⇒ (1) ≈ 23701
(iv) H = 0001 N = 1000 ⇒ (1) ≈ 23681
⇒  0 = −32 −

3

e

3

(b)  = 2 + −

al



3
3
3
3
LHS =  0 + 32  = −32 − + 32 2 + − = −32 − + 62 + 32 − = 62 = RHS
3

(c) The exact value of (1) is 2 + −1 = 2 + −1 .

rS

(0) = 2 + −0 = 2 + 1 = 3

(i) For  = 1: (exact value) − (approximate value) = 2 + −1 − 3 ≈ −06321

Fo

(ii) For  = 01: (exact value) − (approximate value) = 2 + −1 − 23928 ≈ −00249
(iii) For  = 001: (exact value) − (approximate value) = 2 + −1 − 23701 ≈ −00022
(iv) For  = 0001: (exact value) − (approximate value) = 2 + −1 − 23681 ≈ −00002

27. (a) 

ot

In (ii)–(iv), it seems that when the step size is divided by 10, the error estimate is also divided by 10 (approximately).
1

1
+  = () becomes 50 +
 = 60


005

N

or 0 + 4 = 12.

(b) From the graph, it appears that the limiting value of the charge  is about 3.
(c) If 0 = 0, then 4 = 12 ⇒  = 3 is an equilibrium solution.

(d)

c
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SECTION 9.3

SEPARABLE EQUATIONS

¤

411

(e) 0 + 4 = 12 ⇒ 0 = 12 − 4. Now (0) = 0, so 0 = 0 and 0 = 0.
1 = 0 +  (0  0 ) = 0 + 01(12 − 4 · 0) = 12
2 = 1 +  (1  1 ) = 12 + 01(12 − 4 · 12) = 192
3 = 2 +  (2  2 ) = 192 + 01(12 − 4 · 192) = 2352
4 = 3 +  (3  3 ) = 2352 + 01(12 − 4 · 2352) = 26112
5 = 4 +  (4  4 ) = 26112 + 01(12 − 4 · 26112) = 276672
Thus, 5 = (05) ≈ 277 C.

9.3 Separable Equations

3.  2  0 =  + 1
1 3
3

⇒ =

⇒ 2

=  + ln || + 

5. ( + sin )  0 =  + 3
1 2
2



  ⇒ − −1 = 1 2 + 
2

⇒

2
1
=
, where  = −2.  = 0 is also a solution.
 − 2
− 1 2 − 
2


+1
=



⇒ 2  =





 
1
1
 ⇒
 2  =
 ⇒
1+
1+



⇒  3 = 3 + 3 ln || + 3
⇒ ( + sin )

⇒ =


=  + 3


⇒




3
3 + 3 ln || + , where  = 3

( + sin )  =



( + 3 )  ⇒

− cos  = 1 2 + 1 4 + . We cannot solve explicitly for .
2
4




= +2 =  2


 

ot

7.

 −2  =

e

1
= − 1 2 − 
2





=   [ 6= 0] ⇒
2

al

⇒

rS


= 2


Fo

1.

2

⇒   = −  ⇒



  =



2

−  ⇒

2

2

[by parts] = − 1 − + . The solution is given implicitly by the equation  ( − 1) =  − 1 − .
2
2

N

 − 

We cannot solve explicitly for .
9.


= 2  −  + 2 − 1 = (2 − 1) + 1(2 − 1) = ( + 1)(2 − 1) ⇒



1
 =
+1



(2 − 1)  ⇒ ln | + 1| = 1 3 −  + 
3

1
 = (2 − 1)  ⇒
+1
3

⇒ | + 1| =  3−+

3

⇒  + 1 = ±  3−

⇒

3

 =  3− − 1, where  = ± . Since  = −1 is also a solution,  can equal 0, and hence,  can be any real number.
11.



=


1
(−3)2
2

⇒   =   ⇒
= 1 (0)2 + 
2



  =



  ⇒

⇒  = 9 , so 1  2 = 1 2 +
2
2
2

9
2

1 2
2

= 1 2 + . (0) = −3 ⇒
2

√
⇒  2 = 2 + 9 ⇒  = − 2 + 9 since (0) = −3  0.

c
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¤

412

13.

CHAPTER 9

DIFFERENTIAL EQUATIONS





2 + sec2 
=
, (0) = −5.
2  = 2 + sec2   ⇒ 2 = 2 + tan  + ,

2

where [(0)]2 = 02 + tan 0 + 

√
⇒  = (−5)2 = 25. Therefore, 2 = 2 + tan  + 25, so  = ± 2 + tan  + 25.

√
Since (0) = −5, we must have  = − 2 + tan  + 25.







3 +  2  0 , (1) = 1.  ln   =
 +  3 +  2 



15.  ln  =  1 +

⇒

1 2
2

ln  −



1
2  

[use parts with  = ln ,  = ] = 1  2 + 1 (3 +  2 )32
2
3

⇒

1 2

2

ln  − 1 2 +  = 1  2 + 1 (3 +  2 )32 .
4
2
3

Now (1) = 1 ⇒ 0 −

1
2

8
3

1
4

41
12


=
+



+ 1 (4)32
3

⇒ =

+

+

=

41
,
12

so

= 1  2 + 1 (3 +  2 )32 . We do not solve explicitly for .
2
3

17.  0 tan  =  + , 0    2



1
2

⇒

+

=

tan 

⇒

cos 
 ⇒ ln | + | = ln |sin | +  sin 


= cot   [ +  6= 0] ⇒
+

e

ln  − 1 2 +
4

+ =

⇒ | + | = ln|sin |+ = ln|sin | ·  =  |sin | ⇒

al

1 2

2

1
4

 = − by allowing  to be zero.)

rS

 +  =  sin , where  = ± . (In our derivation,  was nonzero, but we can restore the excluded case
√
 
3
(3) =  ⇒  +  =  sin
⇒ 2 = 
3
2

Fo

4
4
Thus,  +  = √ sin  and so  = √ sin  − .
3
3
19. If the slope at the point ( ) is , then we have


= 



=   [ 6= 0] ⇒

2

⇒  = 0. Thus, || =  2

ot

ln || = 1 2 + . (0) = 1 ⇒ ln 1 = 0 + 
2

⇒



4
⇒ = √ .
3


=




  ⇒

2

2

⇒  = ± 2 , so  =  2

N

since (0) = 1  0. Note that  = 0 is not a solution because it doesn’t satisfy the initial condition (0) = 1.






() =
( + ) ⇒
=1+
, but
=  +  = , so
= 1+ ⇒










=  [ 6= −1] ⇒
=  ⇒ ln |1 + | =  +  ⇒ |1 + | = + ⇒
1+
1+

21.  =  + 

⇒

1 +  = ± 

⇒  = ±  − 1 ⇒  +  = ±  − 1 ⇒  =  −  − 1, where  = ± 6= 0.

If  = −1, then −1 =  + 

⇒  = − − 1, which is just  =  −  − 1 with  = 0. Thus, the general solution

is  =  −  − 1, where  ∈ R.
23. (a)  0 = 2


1 − 2

⇒



= 2 1 −  2


sin−1  = 2 +  for −  ≤ 2 +  ≤
2


2.

⇒



= 2  ⇒
1 − 2





=
1 − 2



2  ⇒

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SECTION 9.3

SEPARABLE EQUATIONS

¤

413

(b) (0) = 0 ⇒ sin−1 0 = 02 + 

⇒  = 0,

 so sin−1  = 2 and  = sin(2 ) for − 2 ≤  ≤ 2.


1 −  2 to be a real number, we must have −1 ≤  ≤ 1; that is, −1 ≤ (0) ≤ 1. Thus, the initial-value problem

 0 = 2 1 −  2 , (0) = 2 does not have a solution.

(c) For

25.

 sin 

=
, (0) = .

sin 
2
− cos  = − cos  + 



sin   =



sin   ⇔

⇔ cos  = cos  − . From the initial condition,

= cos 0 − 

⇒  = 1, so the solution is

⇒ 0=1−

e

we need cos


2

So

cos  = cos  − 1. Note that we cannot take cos−1 of both sides, since that would

al

unnecessarily restrict the solution to the case where −1 ≤ cos  − 1 ⇔ 0 ≤ cos , as cos−1 is defined only on [−1 1]. Instead we plot the graph using Maple’s

rS

plots[implicitplot] or Mathematica’s Plot[Evaluate[· · · ]].
(b)  0 =  2

27. (a) , (c)

⇒


= 2


Fo

−−1 =  + 



 −2  =

1
= − − 




 ⇒

⇒

1
, where  = −.  = 0 is also a solution.
−

ot

=

⇒

⇒

29. The curves 2 + 2 2 = 2 form a family of ellipses with major axis on the -axis. Differentiating gives

N


−

(2 + 2 2 ) =
(2 ) ⇒ 2 + 4 0 = 0 ⇒ 40 = −2 ⇒ 0 =
. Thus, the slope of the tangent line


2
at any point ( ) on one of the ellipses is  0 = must satisfy 0 =



=2


|| = ln 






2 +
1

2


⇔


2
=



⇔

−
, so the orthogonal trajectories
2



=2=



⇔ ln || = 2 ln || + 1

⇔

⇔ ln || = ln ||2 + 1

⇔

⇔  = ± 2 · 1 = 2 . This is a family of parabolas.

31. The curves  =  form a family of hyperbolas with asymptotes  = 0 and  = 0. Differentiating gives



() =



 


⇒ 0 = − 2



⇒ 0 = −


2


[since  =  ⇒  = ] ⇒ 0 = − . Thus, the slope


c
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¤

414

CHAPTER 9

DIFFERENTIAL EQUATIONS

of the tangent line at any point ( ) on one of the hyperbolas is  0 = −, so the orthogonal trajectories must satisfy 0 = 
  =   ⇔
 2 = 2 + 2

⇔



asymptotes  = ±.


=
1−

= 1 2 + 1
2

2 −  2 = . This is a family of hyperbolas with

[ − ()]  ⇒ 0 () =  − () [by FTC 1]

Thus, − ln |1 − | = 1 2 − 2 ⇒ ln |1 − | = 2 − 1 2
2
2


= (1 − ) ⇒




0

2 2

⇒  = 1 + 2 − 

2 2

⇒ 0= 2+

⇒ |1 − | = 2 − 

[(2) = 2].



2 ()  ⇒  0 () = 2 () ⇒

⇒  = −2
2 2


√
= 2 


⇒



⇒


√ =


rS

1 −  = ±2 − 


⇒

  ⇒ − ln |1 − | = 1 2 + . Letting  = 2 in the original integral equation
2

gives us (2) = 2 + 0 = 2. Thus, − ln |1 − 2| = 1 (2)2 + 
2

35. () = 4 +

⇔



2



1 2
2

  ⇔

⇔

e









=



al

33. () = 2 +

  =

⇔



2  ⇒

√
2  = 2 + . Letting  = 0 in the original integral equation gives us (0) = 4 + 0 = 4.
√

2
√
√
 = 1 2 + 2 ⇒  = 1 2 + 2 .
Thus, 2 4 = 02 +  ⇒  = 4. 2  = 2 + 4 ⇒
2
2

= 12 − 4 ⇔


ln|12 − 4| = −4 − 4
4 = 12 − −4




=
12 − 4



Fo

37. From Exercise 9.2.27,

⇔ |12 − 4| = −4−4

 ⇔ − 1 ln|12 − 4| =  + 
4

⇔

⇔ 12 − 4 = −4 [ = ±−4 ] ⇔

⇔  = 3 − −4 [ = 4]. (0) = 0 ⇔ 0 = 3 −  ⇔  = 3 ⇔


= ( −  ) ⇔





=
 −

N

39.

ot

() = 3 − 3−4 . As  → ∞, () → 3 − 0 = 3 (the limiting value).


(−)  ⇔ ln| −  | = − + 

⇔ | − | = −+

⇔

 −  = − [ = ± ] ⇔  =  + − . If we assume that performance is at level 0 when  = 0, then
 (0) = 0 ⇔ 0 =  +  ⇔  = −

⇔  () =  − − .

lim  () =  −  · 0 =  .

→∞



= ( − )( − )12 becomes
= ( − )32 ⇒ ( − )−32  =   ⇒




√
2
= − ⇒
( − )−32  =   ⇒ 2( − )−12 =  +  [by substitution] ⇒
 + 

2
4
2
=  −  ⇒ () =  −
. The initial concentration of HBr is 0, so (0) = 0 ⇒
 + 
( + )2

41. (a) If  = , then

0=−

4
2

⇒

Thus, () =  −

4
4
=  ⇒ 2 =
2


√
⇒  = 2  [ is positive since  +  = 2( − )−12  0].

4
√ 2.
( + 2  )

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 9.3

(b)



SEPARABLE EQUATIONS

¤

415






√
√
= ( − )( − )12 ⇒
=   ⇒
=   ().

( − )  − 
( − )  − 
√
From the hint,  =  −  ⇒ 2 =  −  ⇒ 2  = −, so






−2 

√
= −2 √
=
= −2
2
[ − ( − 2 )]
 −  + 2
( − )  − 
 −  + 2


1

17
= −2 √ tan−1 √
−
−

⇒


= −( − ) ⇒





=
 − 

⇒ | − | = −+2

ln| − | = − + 2

() = 4 − + . (0) = 0
() = (0 − )− + .



− ⇒ (1) ln| − | = − + 1

al


=  − 


⇒  −  = 3 −

rS

43. (a)

e

√
√
−

−2
−2
So () becomes √ tan−1 √
=  + . Now (0) = 0 ⇒  = √ tan−1 √ and we have
−
−
−
−
√


√


−


−
2
2
−2
√ tan−1 √
=  − √ tan−1 √
⇒ √ tan−1 − tan−1
=  ⇒
−
−
−
−
−
−
−




2

−
() = √ tan−1 − tan−1
.
−
−
 −

⇒ 0 = 4 + 

⇒  = 3 − + 

⇒ 4 = 0 − 

⇒
⇒

⇒

Fo

(b) If 0  , then 0 −   0 and the formula for () shows that () increases and lim () = .
→∞

As  increases, the formula for () shows how the role of 0 steadily diminishes as that of  increases.
45. (a) Let () be the amount of salt (in kg) after  minutes. Then (0) = 15. The amount of liquid in the tank is 1000 L at all


1
=−

100



 ⇒ ln  = −

N



ot

times, so the concentration at time  (in minutes) is ()1000 kgL and




() kg
L
() kg

=−
10
=−
.

1000 L min 100 min



+ , and (0) = 15 ⇒ ln 15 = , so ln  = ln 15 −
.
100
100

 


It follows that ln
=−
and
= −100 , so  = 15−100 kg.
15
100
15

(b) After 20 minutes,  = 15−20100 = 15−02 ≈ 123 kg.
47. Let () be the amount of alcohol in the vat after  minutes. Then (0) = 004(500) = 20 gal. The amount of beer in the vat

is 500 gallons at all times, so the percentage at time  (in minutes) is ()500 × 100, and the change in the amount of alcohol





gal
()
gal

30 −  gal with respect to time  is
= rate in − rate out = 006 5
−
5
= 03 −
=
.

min
500
min
100
100 min




1
= and − ln |30 − | = 100  + . Because (0) = 20, we have − ln 10 = , so
Hence,
30 − 
100
− ln |30 − | =

1
100 

− ln 10 ⇒ ln |30 − | = −100 + ln 10

⇒

ln |30 − | = ln −100 + ln 10 ⇒

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416

¤

CHAPTER 9

DIFFERENTIAL EQUATIONS

ln |30 − | = ln(10−100 )

⇒

|30 − | = 10−100 . Since  is continuous, (0) = 20, and the right-hand side is

never zero, we deduce that 30 −  is always positive. Thus, 30 −  = 10−100

⇒  = 30 − 10−100 . The

percentage of alcohol is () = ()500 × 100 = ()5 = 6 − 2−100 . The percentage of alcohol after one hour is
(60) = 6 − 2−60100 ≈ 49.
49. Assume that the raindrop begins at rest, so that (0) = 0.  =  and ()0 = 

 0 + () =  ⇒  0 +  = 
− (1) ln| − | =  + 

1 2
1 1
=
1 
2  ln 1 = ln  + 
2

⇒ ln | − | = − − 



⇒


=
 − 



 ⇒

 −  = − . (0) = 0 ⇒  = .

⇒

 = ()(1 − − ). Since   0, as  → ∞, − → 0 and therefore, lim () = .
→∞

⇒
⇒



(ln 1 ) = ( ln 2 )






1 = ln 2 +  = ln 2 

⇒



⇒

1 =  , where  =  .
2


(ln 1 )  =





(ln  ) 
2


e

51. (a)

⇒


=  − 


⇒

al

So  =  − −

⇒

⇒ 0 + 0 =  ⇒

rS

(b) From part (a) with 1 = , 2 =  , and  = 00794, we have  =  00794 .


() and  − (); that is, the rate is proportional to the
√
product of those two quantities. So for some constant ,  =   ( − ). We are interested in the maximum of

53. (a) The rate of growth of the area is jointly proportional to

Fo

the function  (when the tissue grows the fastest), so we differentiate, using the Chain Rule and then substituting for

ot

 from the differential equation:




√
 


1 −12 
=   (−1)
+ ( − ) · 2 
= 1 −12
[−2 + ( − )]
2
 




 √
= 1 −12  ( − ) [ − 3] = 1 2 ( − )( − 3)
2
2

N

This is 0 when  −  = 0 [this situation never actually occurs, since the graph of () is asymptotic to the line  = , as in the logistic model] and when  − 3 = 0 ⇔ () = 3. This represents a maximum by the First Derivative


  goes from positive to negative when () = 3.
Test, since
 
(b) From the CAS, we get () = 



√

   − 1
√
   + 1

2

we substitute  = 0 and  = 0 = (0): 0 = 


√
√
√
√
0 + 0 =   − 

√
√
√
√ 
 + 0 = 
 − 0

. To get  in terms of the initial area 0 and the maximum area  ,



−1
+1

2

⇔ ( + 1)

√
√
0 = ( − 1) 

⇔

√
√
√
√
 + 0 =   −  0 ⇔
√
√
 + 0
⇔ = √
√ . [Notice that if 0 = 0, then  = 1.]
 − 0

⇔

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SECTION 9.4

MODELS FOR POPULATION GROWTH

¤

417

9.4 Models for Population Growth
1. (a)  = 005 − 00005 2 = 005 (1 − 001 ) = 005 (1 − 100). Comparing to Equation 4,

 =  (1 − ), we see that the carrying capacity is  = 100 and the value of  is 005.
(b) The slopes close to 0 occur where  is near 0 or 100. The largest slopes appear to be on the line  = 50. The solutions are increasing for 0  0  100 and decreasing for 0  100.
(c)

All of the solutions approach  = 100 as  increases. As in part (b), the solutions differ since for 0  0  100 they are increasing, and for 0  100 they are decreasing. Also, some have an IP and some don’t. It appears that the solutions which

e

have 0 = 20 and 0 = 40 have inflection points at  = 50.

all nonzero solutions approach  = 100 as  → ∞.


 


 − (0)
=  1 −
⇒ () =
. With  = 8 × 107 ,  = 071, and with  =


1 + −
(0)

rS

3. (a)

al

(d) The equilibrium solutions are  = 0 (trivial solution) and  = 100. The increasing solutions move away from  = 0 and

(b) () = 4 × 107

5. Using (7),  =

⇒ =

and  (4) =

⇒ 2 = 1 + 3−071

⇒ −071 =

⇒

1
3

⇒

ln 3
≈ 155 years
071

 − 0
10,000 − 1000
10,000
=
.  (1) = 2500
= 9, so  () =
0
1000
1 + 9−
9− = 3

N

1 + 9− = 4

8 × 107
= 4 × 107
1 + 3−071

ot

−071 = ln 1
3

⇒

8 × 107
8 × 107
, so (1) =
≈ 323 × 107 kg.
−071
1 + 3
1 + 3−071

Fo

(0) = 2 × 107 , we get the model () =

⇒

− =

1
3

⇒

− = ln 1
3

⇒

⇒

2500 =

10,000
1 + 9−(1)

⇒

 = ln 3. After another three years,  = 4,

10,000
10,000
10,000
10,000
10,000
=
=
=
= 10 = 9000.
1 + 9(3)−4
1 + 9−(ln 3)4
1+ 1
1 + 9 (ln 3 )−4
9
9

7. (a) We will assume that the difference in the birth and death rates is 20 millionyear. Let  = 0 correspond to the year 1990

1
1
1 
=
(002) =
, so
 
53
265





1


=  1 −
=
 1−
,
 in billions


265
100

and use a unit of 1 billion for all calculations.  ≈

(b)  =

 − 0
100 − 53
947

=
=
=
≈ 178679.  () =
0
53
53
1 + −
1+

100

947 −(1265) ,
53 

so  (10) ≈ 549 billion.

c
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418

¤

CHAPTER 9

DIFFERENTIAL EQUATIONS

(c)  (110) ≈ 781, and  (510) ≈ 2772. The predictions are 781 billion in the year 2100 and 2772 billion in 2500.
(d) If  = 50, then  () =

50

1+

447 −(1265) .

53

So  (10) ≈ 548,  (110) ≈ 761, and  (510) ≈ 2241. The predictions

become 548 billion in the year 2000, 761 billion in 2100, and 2241 billion in the year 2500.

= (1 − ), where  is the fraction of the population that has heard the rumor.








(b) Using the logistic equation (4),
=  1 −
, we substitute  =
,  = , and
=
,






9. (a) Our assumption is that


= ()(1 − ) ⇔


Now the solution to (4) is  () =


= (1 − ), our equation in part (a).



 − 0
, where  =
.
1 + −
0

 − 0 −
1+

0

⇒ =

0
.
0 + (1 − 0 )−

al

We use the same substitution to obtain  =

e

to obtain 

rS

Alternatively, we could use the same steps as outlined in the solution of Equation 4.
(c) Let  be the number of hours since 8 AM. Then 0 = (0) =

80
1000

= 008 and (4) = 1 , so
2

1
008
= (4) =
. Thus, 008 + 092−4 = 016, −4 =
2
008 + 092−4
008
4

008 + 092(223)


=


2
23 ,

and − =




2 14
,
23

2 + 23(223)4

. Solving this equation for , we get

 4
 4−1
2 − 2
2 1−
1−
2
2
=2 ⇒
=
=
=
⇒
·
⇒
.
23
23
23

23



ln[(1 − )] ln((1 − ))

.
, so  = 4 1 +
It follows that − 1 =
2
2
4
ln 23 ln 23


1−
ln 9
1
When  = 09,
≈ 76 h or 7 h 36 min. Thus, 90% of the population will have heard
= 9 , so  = 4 1 −
2
 ln 23
2
23

4

N

ot

2
2 + 23
23

4

2

=

Fo

so  =

008
092

the rumor by 3:36 PM.
11. (a)





=  1 −
⇒



 




 
2
 



1 
+ 1−
=
−
+1−
=  −
2
 
 








 
2
2


1−
= 2  1 −
1−
=   1 −





(b)  grows fastest when  0 has a maximum, that is, when  00 = 0. From part (a),  00 = 0 ⇔  = 0,  =  , or  = 2. Since 0    , we see that  00 = 0 ⇔  = 2.

c
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SECTION 9.4

MODELS FOR POPULATION GROWTH

¤

419

13. Following the hint, we choose  = 0 to correspond to 1960 and subtract

94,000 from each of the population figures. We then use a calculator to obtain the models and add 94,000 to get the exponential function
 () = 15783(10933) + 94,000 and the logistic function
 () =

32,6585
+ 94,000.  is a reasonably accurate
1 + 1275−01706

model, while  is not, since an exponential model would only be used for the first few data points.







=  −  =   −
. Let  =  − , so
=
and the differential equation becomes
= .






 
  
 
  
= 0 −

+ 0 −
 .
⇒  () =
The solution is  = 0  ⇒  −





(c) The population will be constant if 0 −


 0 ⇔   0 .


al

(b) Since   0, there will be an exponential expansion ⇔ 0 −

e

15. (a)



= 0 ⇔  = 0 . It will decline if 0 −
 0 ⇔   0 .



rS

(d) 0 = 8,000,000,  =  −  = 0016,  = 210,000 ⇒   0 (= 128,000), so by part (c), the population was declining. are caught.

Fo

17. (a) The term −15 represents a harvesting of fish at a constant rate—in this case, 15 fishweek. This is the rate at which fish

(c) From the graph in part (b), it appears that  () = 250 and  () = 750

(b)

(d)

N

ot

are the equilibrium solutions. We confirm this analytically by solving the equation  = 0 as follows: 008 (1 − 1000) − 15 = 0 ⇒
008 − 000008 2 − 15 = 0 ⇒
−000008( 2 − 1000 + 187,500) = 0 ⇒
( − 250)( − 750) = 0 ⇒  = 250 or 750.

For 0  0  250,  () decreases to 0. For 0 = 250,  () remains constant. For 250  0  750,  () increases and approaches 750.
For 0 = 750,  () remains constant. For 0  750,  () decreases and approaches 750.

(e)






100,000 

100,000
= 008 1 −
− 15 ⇔ −
·
= (008 − 000008 2 − 15) · −
⇔

1000
8

8
−12,500


=  2 − 1000 + 187,500 ⇔


1

=−
 ⇔
( − 250)( − 750)
12,500

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420

¤

CHAPTER 9

DIFFERENTIAL EQUATIONS


 
1
1
1
1
 ⇔
−
 = 25  ⇔
12,500
 − 250
 − 750





  − 250 

1
 = 1  +  ⇔   − 250  = 25+ = 25 ln| − 250| − ln| − 750| = 25  +  ⇔ ln
  − 750  25
  − 750 
 

−1500
1500
+
 − 250
 − 750

 − 250
= 25
 − 750

 () =



 = −

⇔  − 250 =  25 − 75025

⇔  −  25 = 250 − 75025

250 − 75025
250 − 750
. If  = 0 and  = 200, then 200 =
1−
1 − 25

550 = 50 ⇔  =

1
.
11

⇔

⇔

⇔ 200 − 200 = 250 − 750

⇔

Similarly, if  = 0 and  = 300, then

 = − 1 . Simplifying  with these two values of  gives us
9




 

= ( ) 1 −
1−
. If     , then  = (+)(+)(+) = + ⇒  is increasing.




al

19. (a)

250(325 − 11)
750(25 + 3) and  () =
.
25 − 11
25 + 9

e

 () =

rS

If 0    , then  = (+)(+)(−) = − ⇒  is decreasing.

 = 008,  = 1000, and  = 200 ⇒




200

= 008 1 −
1−

1000


(b)

Fo

For 0  0  200, the population dies out. For 0 = 200, the population is steady. For 200  0  1000, the population increases and approaches

1000. For 0  1000, the population decreases and approaches 1000.
The equilibrium solutions are  () = 200 and  () = 1000.


= 






ot



 

=  1 −
1−





=
( −  )( − )

 −


 −




=


( −  )( − ) ⇔



1


. By partial fractions,
=
+
, so

( −  )( − )
 −
 −

N

(c)

( − ) + ( −  ) = 1.
If  = ,  =

1
(− ln | −  | + ln | − |) =
 −


 −

 = ( − )   + 1 ⇔ ln 
 −

Let  = 0:



1
1

+
 =
 ⇒
 −
 −



 −
1
 =  + ⇒ ln 
 −   −  


1
1
1
; if  = ,  =
, so
 −
 −
 −

+


⇒

 

 −
= (−)( )
 −

[ = ±1 ].

0 −  (−)( )
0 − 
 −
=
= . So

.
 − 0
 −
 − 0

Solving for  , we get  () =

( − 0 ) + (0 − )(−)()
.
 − 0 + (0 − )(−)( )

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SECTION 9.5

LINEAR EQUATIONS

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421

(d) If 0  , then 0 −   0. Let () be the numerator of the expression for  () in part (c). Then
(0) = 0 ( − )  0, and 0 −   0 ⇔

lim (0 − )(−)( ) = −∞ ⇒

→∞

lim () = −∞.

→∞

Since  is continuous, there is a number  such that () = 0 and thus  () = 0. So the species will become extinct.
21. (a)  =  cos( − )

⇒ ( ) =  cos( − )  ⇒



( ) = 



cos( − )  ⇒

ln  = () sin( − ) + . (Since this is a growth model,   0 and we can write ln  instead of ln| |.) Since
 (0) = 0 , we obtain ln 0 = () sin(−) +  = − () sin  + 

⇒  = ln 0 + () sin . Thus,

ln  = () sin( − ) + ln 0 + () sin , which we can rewrite as ln(0 ) = ()[sin( − ) + sin ] or, after exponentiation,  () = 0 ()[sin(−)+sin ] .
(b) As  increases, the amplitude

As  increases, the amplitude and the period decrease.

increases, but the minimum value stays the same.

A change in  produces slight adjustments in the phase shift and

Fo

rS

al

e

amplitude.

 () oscillates between 0 ()(1+sin ) and 0 ()(−1+sin ) (the extreme values are attained when  −  is an odd multiple of


),
2

so lim  () does not exist.
→∞


. By comparison, if  = (ln ) and  = 1 ( − ), then
2
1 + −

ot

23. By Equation 7,  () =

 − −
 + −
 − −
2
−
2
= 
+ 
= 
· − =
 + −
 + −
 + −
 + − 
1 + −2

N

1 + tanh  = 1 +

and −2 = −(−) =  − = ln  − = − , so
1

2



 



2
1 + tanh 1 ( − ) = [1 + tanh ] =
=
=
=  ().
·
2
2
2 1 + −2
1 + −2
1 + −

9.5 Linear Equations
1.  −  0 = 
3.  0 =

⇔

0 +  =  is linear since it can be put into the standard linear form (1),  0 +  ()  = ().

1
1
+ is not linear since it cannot be put into the standard linear form (1),  0 +  ()  = ().
 

5. Comparing the given equation,  0 +  = 1, with the general form,  0 +  ()  = (), we see that  () = 1 and the

integrating factor is () = 



 () 

=



1 

=  . Multiplying the differential equation by () gives

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¤

422

CHAPTER 9

DIFFERENTIAL EQUATIONS

  0 +   = 

⇒ ( )0 = 

⇒

  =

 = 1 + − .
7.  0 =  − 

⇒  0 +  =  (). () = 

  0 +   = 
 =  − 1 + −

⇒ ( )0 = 
[divide by  ].





 

⇒

  =  + 

⇒

 


=  + 




⇒



=  1  =  . Multiplying the differential equation () by () gives

⇒   =   ⇒   =  −  +  [by parts] ⇒
 () 

9. Since  () is the derivative of the coefficient of  0 [ () = 1 and the coefficient is ], we can write the differential equation

 0 +  =

√
√
 in the easily integrable form ()0 =  ⇒  = 2 32 + 
3

⇒ =

2√

3

+ .



+ (cos )  = sin(2 ) ⇒ [(sin ) ]0 = sin(2 ) ⇒ (sin )  = sin(2 )  ⇒  =
11. sin 


1
+
 = 1 (), which has the

1+

form 0 +  ()  = (). The integrating factor is () = 



 () 



e


+  = 1 + ,   0 [divide by 1 + ] ⇒


sin(2 )  + 
.
sin 

al

13. (1 + )



=

[1(1+)] 

= ln(1+) = 1 + .


(1 + ) = (1 + )  =  + 1 2 + 
2

 + 1 2 + 
2 + 2 + 2
2
or  =
.
1+
2( + 1)

⇒ (2 )0 = ln  ⇒ 2  =

12 (2) = 1 ln 1 − 1 + 



ln   ⇒ 2  =  ln  −  +  [by parts]. Since (1) = 2,

Fo

15. 2  0 + 2 = ln 

⇒ =

rS

Multiplying () by () gives us our original equation back. We rewrite it as [(1 + )]0 = 1 + . Thus,

⇒ 2 = −1 + 

⇒  = 3 so 2  =  ln  −  + 3, or  =

1
1
3 ln  − + 2 .






3
1
= 2 + 3 ⇒ 0 −  =  (). () =  −3  = −3 ln|| = (ln|| )−3 = −3 [  0] = 3 . Multiplying ()




0

1
3
1
1
1
1
1
1
1
by () gives 3 0 − 4  = 2 ⇒
 = 2 ⇒ 3=
 ⇒ 3  = − + . Since (2) = 4,



3


2



⇒  = 1, so

1
1
 = − + 1, or  = −2 + 3 .
3


N

1
1
(4) = − + 
23
2

ot

17. 


−1
1
1
 =  sin . () =  (−1)  = − ln  = ln  = .



0
1
1
1
1
1 0
Multiplying by gives  − 2  = sin  ⇒
 = sin  ⇒
 = − cos  + 






19.  0 =  + 2 sin 

⇒ 0 −

⇒  = − cos  + .

() = 0 ⇒ − · (−1) +  = 0 ⇒  = −1, so  = − cos  − .

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SECTION 9.5

LINEAR EQUATIONS

¤



21.  0 + 2 = 

() = 




2
=
.



2
= 2 ln|| = ln|| = ||2 = 2 .

⇒ 0 +

(2) 

Multiplying by () gives 2  0 + 2 =  ⇒ (2 )0 = 

2  =   = ( − 1) +  [by parts] ⇒

⇒

 = [( − 1) + ]2 . The graphs for  = −5, −3, −1, 1, 3, 5, and 7 are shown.  = 1 is a transitional value. For   1, there is an inflection point and for   1, there is a local minimum. As || gets larger, the “branches” get further from the origin.


(1−) 
+  ()1(1−) = ()(1−) or
+ (1 − ) () = ()(1 − ).
1 −  


al

becomes




  
(1−) 
= (1 − )  − or =
=
. Then the Bernoulli differential equation



1 −  
1 −  

e

23. Setting  =  1− ,

3
1
2
2
2
4
 = 2 , so  = 3,  () = and () = 2 . Setting  =  −2 ,  satisfies 0 −
= − 2.











2
2
2
(−4) 
−4
4
4
− 6  +  = 
=  and  = 
+  = 4 +
Then () = 
.

55
5

rS

25. Here  0 +

27. (a) 2

Fo

−12

2
.
Thus,  = ± 4 +
5




+ 10 = 40 or
+ 5 = 20. Then the integrating factor is  5  = 5 . Multiplying the differential equation



by the integrating factor gives 5

⇒ (5 )0 = 205

⇒


205  +  = 4 + −5 . But 0 = (0) = 4 + , so () = 4 − 4−5 .

ot




+ 55 = 205


() = −5

29. 5

N

(b) (01) = 4 − 4−05 ≈ 157 A



+ 20 = 60 with (0) = 0 C. Then the integrating factor is  4  = 4 , and multiplying the differential


equation by the integrating factor gives 4
() = −4




+ 44  = 124


⇒ (4 )0 = 124

⇒


124  +  = 3 + −4 . But 0 = (0) = 3 +  so () = 3(1 − −4 ) is the charge at time 

and  =  = 12−4 is the current at time .
31.



+  = , so () =    =  . Multiplying the differential equation



+   =  ⇒ (  )0 =  ⇒



 () = −   +  =  + − ,   0. Furthermore, it is by () gives 

reasonable to assume that 0 ≤  (0) ≤ , so − ≤  ≤ 0.

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423

424

¤

CHAPTER 9

DIFFERENTIAL EQUATIONS



 kg L kg Salt is added at a rate of 04
5
=2
 Since solution is drained from the tank at a rate of
L
min min 33. (0) = 0 kg.

3 Lmin, but salt solution is added at a rate of 5 Lmin, the tank, which starts out with 100 L of water, contains (100 + 2) L of liquid after  min. Thus, the salt concentration at time  is


() kg
. Salt therefore leaves the tank at a rate of
100 + 2 L



L
3
kg
3
=
. Combining the rates at which salt enters and leaves the tank, we get min 100 + 2 min



3

3
=2−
. Rewriting this equation as
+
 = 2, we see that it is linear.

100 + 2

100 + 2




3 
= exp 3 ln(100 + 2) = (100 + 2)32
() = exp
2
100 + 2
() kg
100 + 2 L

⇒

⇒ (100 + 2)32  = 2 (100 + 2)52 + 
5

⇒

al

[(100 + 2)32 ]0 = 2(100 + 2)32


+ 3(100 + 2)12  = 2(100 + 2)32


e

Multiplying the differential equation by () gives (100 + 2)32

at time  to be () =

()
=
100 + 2



−40,000

(100 + 2)52

+

2
5



2 kg −40,000 kg + ≈ 02275
. In particular, (20) =
L
5
L
14052

Fo

and (20) = 2 (140) − 40,000(140)−32 ≈ 3185 kg.
5
35. (a)

rS

1
 = 2 (100 + 2) + (100 + 2)−32 . Now 0 = (0) = 2 (100) +  · 100−32 = 40 + 1000  ⇒  = −40,000, so
5
5


 = 2 (100 + 2) − 40,000(100 + 2)−32 kg. From this solution (no pun intended), we calculate the salt concentration
5




+  =  and () =  ()  = () , and multiplying the differential equation by




0

()
+
= () ⇒ ()  = () . Hence,




() = −() ()  +  =  + −() . But the object is dropped from rest, so (0) = 0 and

ot

() gives ()

N

 = −. Thus, the velocity at time  is () = ()[1 − −() ].

(b) lim () = 
→∞

(c) () =



()  = ()[ + ()−() ] + 1 where 1 = (0) − 2 2 .

(0) is the initial position, so (0) = 0 and () = ()[ + ()−() ] − 2 2 .
1
0
0
1
⇒  0 = − 2 . Substituting into  0 =  (1 − ) gives us − 2 = 






1


0
 = − 1 −
⇒  0 = − +
⇒  0 +  =
().




37. (a)  =

1


⇒  =



1
1−
⇒


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SECTION 9.6


(b) The integrating factor is 
  =
 =



 

=  . Multiplying () by  gives   0 +   =

1 
 
  ⇒   =
 +



1
1
+ −


⇒  =

⇒ =

PREDATOR-PREY SYSTEMS




⇒ ( )0 =

¤

 



425

⇒

1
1
+ − . Since  = , we have





, which agrees with Equation 9.4.7,  =
, when  = .
1 + −
1 + −

9.6 Predator-Prey Systems
1. (a)  = −005 + 00001. If  = 0, we have  = −005, which indicates that in the absence of ,  declines at

a rate proportional to itself. So  represents the predator population and  represents the prey population. The growth of the prey population, 01 (from  = 01 − 0005), is restricted only by encounters with predators (the term

e

−0005). The predator population increases only through the term 00001; that is, by encounters with the prey and

al

not through additional food sources.

(b)  = −0015 + 000008. If  = 0, we have  = −0015, which indicates that in the absence of ,  would

rS

decline at a rate proportional to itself. So  represents the predator population and  represents the prey population. The growth of the prey population, 02 (from  = 02 − 000022 − 0006 = 02(1 − 0001) − 0006), is restricted by a carrying capacity of 1000 [from the term 1 − 0001 = 1 − 1000] and by encounters with predators (the

Fo

term −0006). The predator population increases only through the term 000008; that is, by encounters with the prey and not through additional food sources.

3. (a)  = 05 − 00042 − 0001 = 05(1 − 125) − 0001.

 = 04 − 00012 − 0002 = 04(1 − 400) − 0002.

ot

The system shows that  and  have carrying capacities of 125 and 400. An increase in  reduces the growth rate of  due to the negative term −0002. An increase in  reduces the growth rate of  due to the negative term −0001. Hence

N

the system describes a competition model.

(b)  = 0 ⇒ (05 − 0004 − 0001) = 0 ⇒ (500 − 4 − ) = 0 (1) and  = 0 ⇒
(04 − 0001 − 0002) = 0 ⇒ (400 −  − 2) = 0 (2).
From (1) and (2), we get four equilibrium solutions.
(i)  = 0 and  = 0: If the populations are zero, there is no change.
(ii)  = 0 and 400 −  − 2 = 0 ⇒  = 0 and  = 400: In the absence of an -population, the -population stabilizes at 400.

(iii) 500 − 4 −  = 0 and  = 0 ⇒  = 125 and  = 0: In the absence of -population, the -population stabilizes at 125.

(iv) 500 − 4 −  = 0 and 400 −  − 2 = 0 ⇒  = 500 − 4 and  = 400 − 2 ⇒

500 − 4 = 400 − 2 ⇒

100 = 2 ⇒  = 50 and  = 300: A -population of 300 is just enough to support a constant -population of 50.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

426

¤

CHAPTER 9

DIFFERENTIAL EQUATIONS

5. (a) At  = 0, there are about 300 rabbits and 100 foxes. At  = 1 , the number

of foxes reaches a minimum of about 20 while the number of rabbits is about 1000. At  = 2 , the number of rabbits reaches a maximum of about
2400, while the number of foxes rebounds to 100. At  = 3 , the number of rabbits decreases to about 1000 and the number of foxes reaches a maximum of about 315. As  increases, the number of foxes decreases greatly to 100, and the number of rabbits decreases to 300 (the initial populations), and the cycle starts again.

al

e

(b)


−002 + 000002
=

008 − 0001

⇔ (008 − 0001 )  = (−002 + 000002)  ⇔
 

ot

9.

Fo

rS

7.

008 − 0001
−002 + 000002
 =
 ⇔





 
008
002
− 0001  =
−
+ 000002  ⇔



N

008 ln| | − 0001 = −002 ln|| + 000002 +  ⇔ 008 ln  + 002 ln  = 0001 + 000002 + 

 ln  008 002 = 000002 + 0001 +  ⇔  008 002 = 000002+0001 + ⇔
002  008 = 000002 0001

⇔

− + 
  
002  008

=
, then  =   .
= . In general, if
000002 0001

 − 
 

11. (a) Letting  = 0 gives us  = 008(1 − 00002).  = 0

⇔  = 0 or 5000. Since   0 for

0    5000, we would expect the rabbit population to increase to 5000 for these values of . Since   0 for
  5000, we would expect the rabbit population to decrease to 5000 for these values of . Hence, in the absence of wolves, we would expect the rabbit population to stabilize at 5000.
(b)  and  are constant ⇒  0 = 0 and  0 = 0 ⇒


0 = 008(1 − 00002) − 0001
0 = −002 + 000002

⇒


0 = [008(1 − 00002) − 0001 ]
0 =  (−002 + 000002)

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

⇔

CHAPTER 9 REVIEW

The second equation is true if  = 0 or  =
1
00002

=

002
000002

¤

427

= 1000. If  = 0 in the first equation, then either  = 0 or

= 5000 [as in part (a)]. If  = 1000, then 0 = 1000[008(1 − 00002 · 1000) − 0001 ] ⇔

0 = 80(1 − 02) − 

⇔  = 64.

Case (i):

 = 0,  = 0: both populations are zero

Case (ii):

 = 0,  = 5000: see part (a)

Case (iii):  = 1000,  = 64: the predator/prey interaction balances and the populations are stable.
(c) The populations of wolves and rabbits fluctuate around

(d)

64 and 1000, respectively, and eventually stabilize at

al

e

those values.

rS

9 Review

1. (a) A differential equation is an equation that contains an unknown function and one or more of its derivatives.

(b) The order of a differential equation is the order of the highest derivative that occurs in the equation.

Fo

(c) An initial condition is a condition of the form (0 ) = 0 .

2.  0 = 2 +  2 ≥ 0 for all  and .  0 = 0 only at the origin, so there is a horizontal tangent at (0 0), but nowhere else. The

graph of the solution is increasing on every interval.

ot

3. See the paragraph preceding Example 1 in Section 9.2.

N

4. See the paragraph following Figure 14 in Section 9.2.
5. A separable equation is a first-order differential equation in which the expression for  can be factored as a function of 

times a function of , that is,  = () (). We can solve the equation by integrating both sides of the equation
 () = () and solving for .
6. A first-order linear differential equation is a differential equation that can be put in the form


+  ()  = (), where 


and  are continuous functions on a given interval. To solve such an equation, multiply it by the integrating factor


() =   () to put it in the form [() ]0 = () () and then integrate both sides to get ()  = () () ,

that is, 
7. (a)



 () 

=







 () 

() . Solving for  gives us  = −

1 

=  ; the relative growth rate,
, is constant.

 



 () 







 () 

() .

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

428

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CHAPTER 9

DIFFERENTIAL EQUATIONS

(b) The equation in part (a) is an appropriate model for population growth, assuming that there is enough room and nutrition to support the growth.
(c) If (0) = 0 , then the solution is () = 0  .
8. (a)  =  (1 − ), where  is the carrying capacity.

(b) The equation in part (a) is an appropriate model for population growth, assuming that the population grows at a rate proportional to the size of the population in the beginning, but eventually levels off and approaches its carrying capacity because of limited resources.
9. (a)  =  −   and  = − +  .

(b) In the absence of sharks, an ample food supply would support exponential growth of the fish population, that is,

e

 =  , where  is a positive constant. In the absence of fish, we assume that the shark population would decline at a

rS

al

rate proportional to itself, that is,  = −, where  is a positive constant.

Since  4 ≥ 0, 0 = −1 −  4  0 and the solutions are decreasing functions.

3. False.

 +  cannot be written in the form () ().

5. True.

  0 = 

⇒  0 = − 

equation is linear.
By comparing

⇒  0 + (−− ) = 0, which is of the form  0 +  ()  = (), so the




= 2 1 − with the logistic differential equation (9.4.4), we see that the carrying

5

ot

7. True.

Fo

1. True.

capacity is 5; that is, lim  = 5.

N

→∞

1. (a)

(b) lim () appears to be finite for 0 ≤  ≤ 4. In fact
→∞

lim () = 4 for  = 4, lim () = 2 for 0    4, and

→∞

→∞

lim () = 0 for  = 0. The equilibrium solutions are

→∞

() = 0, () = 2, and () = 4.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 9 REVIEW

3. (a)

¤

429

We estimate that when  = 03,  = 08, so (03) ≈ 08.

e



2
(b)  = 01, 0 = 0, 0 = 1 and  ( ) = 2 −  2 . So  = −1 + 01 2 − −1 . Thus,
−1






1 = 1 + 01 02 − 12 = 09, 2 = 09 + 01 012 − 092 = 082, 3 = 082 + 01 022 − 0822 = 075676.
This is close to our graphical estimate of (03) ≈ 08.

al

(c) The centers of the horizontal line segments of the direction field are located on the lines  =  and  = −.

5.  0 = − sin  −  cos 


rS

When a solution curve crosses one of these lines, it has a local maximum or minimum.

⇒ 0 + (cos )  = − sin  (). This is a linear equation and the integrating factor is

= sin  . Multiplying () by sin  gives sin   0 + sin  (cos )  =  ⇒ (sin  )0 =  ⇒


sin   = 1 2 +  ⇒  = 1 2 +  − sin  .
2
2 cos  

Fo

() = 


√
√ 
√
2 
2
 ⇒ 2
= 2 + 3  ⇒ 2  = 2 + 3   ⇒

√ 


2
2
2 + 3   ⇒  = 2 + 232 +  ⇒  2 = ln(2 + 232 + ) ⇒
2  =
2

7. 2  0 = 2 + 3


+ 2 = 


⇒


=  − 2 = (1 − 2) ⇒


N

9.

ot


 = ± ln(2 + 232 + )

2 +

|| = −

2




=




(1 − 2)  ⇒ ln || =  − 2 + 

⇒

2

= − . Since (0) = 5, 5 = 0 = . Thus, () = 5− .

11.  0 −  =  ln 

⇒ 0 −

−1


1
 = ln . () =  (−1)  = − ln|| = ln||
= ||−1 = 1 since the condition


(1) = 2 implies that we want a solution with   0. Multiplying the last differential equation by () gives

0

1
1
1
1 0
1
1
1
ln 
 − 2  = ln  ⇒
 = ln  ⇒
=
 ⇒
 = 1 (ln )2 +  ⇒
2








 = 1 (ln )2 + . Now (1) = 2 ⇒ 2 = 0 + 
2
13.

⇒  = 2, so  = 1 (ln )2 + 2.
2

1



1
() =
( ) ⇒ 0 =  = , so the orthogonal trajectories must have  0 = −
⇒
=−
⇒







  = − ⇒
  = −  ⇒ 1  2 = − +  ⇒  =  − 1  2 , which are parabolas with a horizontal axis.
2
2 c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

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DIFFERENTIAL EQUATIONS





= 01 1 − with  (0) = 100, we have  = 01,

2000

15. (a) Using (4) and (7) in Section 9.4, we see that for

 = 2000, 0 = 100, and  =

2000
2000
and  (20) =
≈ 560.
1 + 19−01
1 + 19−2

 () =

(b)  = 1200 ⇔ 1200 =
−01 =

2000
1 + 19−01

2
2
19 ⇔ −01 = ln 57
3


∝ ∞ −  ⇒


17. (a)

2000 − 100
= 19. Thus, the solution of the initial-value problem is
100

⇔ 1 + 19−01 =

⇔ 19−01 =

5
−1 ⇔
3

2
⇔  = −10 ln 57 ≈ 335.


= (∞ − ) ⇒





=
∞ − 

⇒ |∞ − | = −−

ln |∞ − | = − − 

2000
1200


  ⇒ − ln |∞ − | =  + 

⇒

⇒  = ∞ − − .

⇒ ∞ −  = −

e

At  = 0,  = (0) = ∞ −  ⇒  = ∞ − (0) ⇒ () = ∞ − [∞ − (0)]− .

al

(b) ∞ = 53 cm, (0) = 10 cm, and  = 02 ⇒ () = 53 − (53 − 10)−02 = 53 − 43−02 .

19. Let  represent the population and  the number of infected people. The rate of spread  is jointly proportional to  and


= ( − )


growth in Section 9.4].

⇒

() =

0 
[from the discussion of logistic
0 + ( − 0 )− 

rS

to  − , so for some constant ,

Now, measuring  in days, we substitute  = 7,  = 5000, 0 = 160 and (7) = 1200 to find :
160 · 5000
160 + (5000 − 160)−5000·7·

−35,000 =

2000 − 480
14,520

⇔

2000
160 + 4840−35,000

Fo

1200 =

⇔ 3=

−35,000 = ln

=

N

160 + 4840−5000 = 200 ⇔ −5000 =
1
−1 ln =
5000 121

1
7

⇔

=

200 − 160
4840

480 + 14,520−35,000 = 2000

⇔

−1
38
ln
≈ 000006448. Next, let
35,000 363

160 · 5000
160 + (5000 − 160)−·5000·

ot

 = 5000 × 80% = 4000, and solve for : 4000 =

38
363

⇔

⇔ −5000 = ln

⇔ 1=
1
121

200
160 + 4840−5000

⇔

⇔

1 ln 121
1
=7·
≈ 14875. So it takes about 15 days for 80% of the population
38 · ln
121
ln 363 ln 363
38

to be infected.
21.



=−






+

 +  ln  = −



⇒



+
 =





 
 



1  ⇒
 ⇒
 = −
−
1+





 + . This equation gives a relationship between  and , but it is not possible to isolate  and express it in


terms of .
23. (a)  = 04(1 − 0000005) − 0002,  = −02 + 0000008. If  = 0, then

 = 04(1 − 0000005), so  = 0 ⇔  = 0 or  = 200,000, which shows that the insect population

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 9 REVIEW

¤

431

increases logistically with a carrying capacity of 200,000. Since   0 for 0    200,000 and   0 for
  200,000, we expect the insect population to stabilize at 200,000.
(b)  and  are constant ⇒ 0 = 0 and  0 = 0 ⇒



0 = 04(1 − 0000005) − 0002
0 = −02 + 0000008

The second equation is true if  = 0 or  = or  =

1
0000005

02
0000008

⇒


0 = 04[(1 − 0000005) − 0005]
0 = (−02 + 0000008)

= 25,000. If  = 0 in the first equation, then either  = 0

= 200,000. If  = 25,000, then 0 = 04(25,000)[(1 − 0000005 · 25,000) − 0005] ⇒

0 = 10,000[(1 − 0125) − 0005] ⇒ 0 = 8750 − 50

⇒  = 175.

Case (i):

 = 0,  = 0: Zero populations

Case (ii):

 = 0,  = 200,000: In the absence of birds, the insect population is always 200,000.

around 175 and 25,000, respectively, and



1+






2

Fo

 2
25. (a)
=
2

rS

eventually stabilize at those values.

(d)

al

(c) The populations of the birds and insects fluctuate

e

Case (iii):  = 25,000,  = 175: The predator/prey interaction balances and the populations are stable.

. Setting  =

√


, we get
=  1 + 2




√
=  . Using Formula 25 gives
1 + 2

[where  =  ] ⇒

ot

√
√


ln  + 1 +  2 =  +  ⇒  + 1 +  2 = 

⇒

√
1 +  2 =  − 

⇒

 
1 −
 −

. Now
2
2

⇒ 2  =  2 2 − 1 ⇒  =



1 −
=  −


2
2

 
1 −
 +

+  0 . From the diagram in the text, we see that (0) = 
2
2

N

1 +  2 =  2 2 − 2  +  2

⇒ =

and (±) = .  = (0) =
=

1

+
+0
2
2

⇒ 0 = −

1

−
2
2

⇒

1
 
1
 
( − 1) +
(− − 1) + . From  = (±), we find  =
( − 1) +
(− − 1) + 
2
2
2
2

 −
1
− 1) +
(
( − 1) + . Subtracting the second equation from the first, we get
2
2


1  − −
1
1
  − −
−
=
−
sinh .
0=

2

2


and  =

Now   0 and   0, so sinh   0 and  = ±1. If  = 1, then
=

1 
1
1
1 −
1  + −
( − 1) +
(
− +  =  + (cosh  − 1). If  = −1,
− 1) +  =
2
2

2



c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

CHAPTER 9

then  = −

DIFFERENTIAL EQUATIONS

1
1
1 
1 −
−1  + −
( − 1) −
(
+ +  =  − (cosh  − 1).
− 1) +  =
2
2

2



Since   0, cosh  ≥ 1, and  ≥ , we conclude that  = 1 and  =  +
 = () =  +

1
(cosh  − 1), where


1
(cosh  − 1). Since cosh() = cosh(−), there is no further information to extract from the


condition that () = (−). However, we could replace  with the expression  −
 =+

1
(cosh  − 1), obtaining


1
(cosh  − cosh ). It would be better still to keep  in the expression for , and use the expression for  to


solve for  in terms of , , and . That would enable us to express  in terms of  and the given parameters , , and .
Sadly, it is not possible to solve for  in closed form. That would have to be done by numerical methods when specific

e

parameter values are given.

ot

Fo

0

rS



= 2 (1) sinh  = (2) sinh 

al

(b) The length of the cable is
 


 
 = − 1 + ()2  = − 1 + sinh2   = − cosh   = 2 0 cosh  

N

432

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

PROBLEMS PLUS
1. We use the Fundamental Theorem of Calculus to differentiate the given equation:

[ ()]2 = 100 +



2
2
[ ()]2 + [ 0 ()]  ⇒ 2 () 0 () = [ ()]2 + [ 0 ()]
0

2

⇒

2

[ ()]2 + [ 0 ()] − 2 () 0 () = 0 ⇒ [() −  0 ()] = 0 ⇔  () =  0 (). We can solve this as a separable equation, or else use Theorem 9.4.2 with  = 1, which says that the solutions are () =  . Now [ (0)]2 = 100, so
 (0) =  = ±10, and hence  () = ±10 are the only functions satisfying the given equation.
( + ) −  ()
 () [ () − 1]
= lim
→0



=  () lim

→0

[since  ( + ) = () ()]

e

→0

 () − 1
 () −  (0)
=  () lim
=  () 0 (0) =  ()
→0

−0

al

3.  0 () = lim

Therefore,  0 () =  () for all  and from Theorem 9.4.2 we get  () =  .

rS

Now  (0) = 1 ⇒  = 1 ⇒ () =  .

5. “The area under the graph of  from 0 to  is proportional to the ( + 1)st power of  ()” translates to

0

 ()  = [ ()]+1 for some constant . By FTC1,








 ()  =

0

Fo




 
[()]+1
⇒


 () = ( + 1)[ ()]  0 () ⇒ 1 = ( + 1)[ ()]−1  0 () ⇒ 1 = ( + 1)−1


( + 1)−1  =

ot

( + 1) −1  =  ⇒

Now  (0) = 0 ⇒ 0 = 0 + 






⇒

1
 ⇒ ( + 1)   =  + .


⇒  = 0 and then  (1) = 1 ⇒ ( + 1)

1

=1 ⇒ =
,

+1

N

so   =  and  =  () = 1 .

7. Let () denote the temperature of the peach pie  minutes after 5:00 PM and  the temperature of the room. Newton’s Law of

Cooling gives us  = ( − ) Solving for  we get
| − | = +

⇒  −  = ± · 


=   ⇒ ln| − | =  + 
−

⇒

⇒  =  + , where  is a nonzero constant. We are given

temperatures at three times.
(0) = 100

⇒

100 =  + 

⇒

 = 100 − 

(10) = 80

⇒

80 = 10 + 

(1)

(20) = 65

⇒

65 = 20 + 

(2)

Substituting 100 −  for  in (1) and (2) gives us
−20 = 10 −  (3) and −35 = 20 −  (4) c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

433

434

¤

CHAPTER 9 PROBLEMS PLUS



 10 − 1
−20
=
−35
 (20 − 1)

4
10 − 1
= 20
⇒ 420 − 4 = 710 − 7 ⇒
7

−1



420 − 710 + 3 = 0. This is a quadratic equation in 10 . 410 − 3 10 − 1 = 0 ⇒ 10 = 3 or 1 ⇒
4
Dividing (3) by (4) gives us

10 = ln 3 or ln 1 ⇒  =
4

−20 =  ·

3
4

−

1
10

⇒

ln 3 since  is a nonzero constant of proportionality. Substituting
4

⇒ −20 = − 1 
4

3
4

for 10 in (3) gives us

⇒  = 80. Now  = 100 −  so  = 20◦ C.

9. (a) While running from ( 0) to ( ), the dog travels a distance

=

 
 
1 + ()2  = −  1 + ()2 , so




= − 1 + ()2 . The dog and rabbit run at the same speed, so the


rabbit’s position when the dog has traveled a distance  is (0 ). Since the


−
=
(see the figure).

0−
 2



 
 2


=
−  2 +1
= − 2 . Equating the two expressions for







al

rS


⇒


 2

 2 gives us  2 = 1 +
, as claimed.



Thus,  =  − 

e

dog runs straight for the rabbit,

√




, we obtain the differential equation 
= 1 +  2 , or √
. Integrating:
=



1 + 2





25
√
ln  =
= ln  + 1 +  2 + . When  = ,  =  = 0, so ln  = ln 1 + . Therefore,
1 + 2

 √

√

√
⇒  =  1 + 2 +  ⇒
 = ln , so ln  = ln 1 +  2 +  + ln  = ln  1 +  2 + 
  2 2
  2

√

−  ⇒ 1 + 2 =
+ 2 ⇒
−1 = 0 ⇒
1 + 2 =
−
− 2






Fo

(b) Letting  =

()2 − 1
2 − 2

1

2

=
=
−
[for   0]. Since  =
,=
− ln  + 1 .
2()
2
2
2

4
2

ot

=

=



− ln  + 1
4
2

N

Since  = 0 when  = , 0 =

⇒ 1 =



ln  − . Thus,
2
4




2 − 2
 
2
− ln  + ln  − =
− ln

4
2
2
4
4
2


(c) As  → 0+ ,  → ∞, so the dog never catches the rabbit.
11. (a) We are given that  =

1
2 ,
3

  = 60,000 ft3h, and  = 15 = 3 . So  = 1 
2
3

 3 2
  = 3 3
2
4





4( )
240,000
80,000
= 3  · 32
= 9 2
. Therefore,
=
=
=
() ⇒
4
4




92
92
32

 2
3  = 80,000  ⇒ 3 = 80,000 + . When  = 0,  = 60. Thus,  = 603 = 216,000, so

3 = 80,000 + 216,000. Let  = 100. Then 1003 = 1,000,000 = 80,000 + 216,000 ⇒
80,000 = 784,000 ⇒  = 98, so the time required is 98 hours.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

⇒

CHAPTER 9 PROBLEMS PLUS

¤

435

(b) The floor area of the silo is  =  · 2002 = 40,000 ft2 , and the area of the base of the pile is
 = 2 = 

 3 2
 =
2

9 2
 .
4

So the area of the floor which is not covered when  = 60 is

 −  = 40,000 − 8100 = 31,900 ≈ 100,217 ft2 . Now  = and from () in part (a) we know that when  = 60,  =
 =



9
(2)(60) 200
4
27



9 2

4

80,000
3(60)2

=

⇒  =

200
27

9
4

· 2 (),

fth. Therefore,

= 2000 ≈ 6283 ft2 h.

(c) At  = 90 ft,   = 60,000 − 20,000 = 40,000 ft3h. From () in part (a),
4( )

4(40,000)
160,000
=
=
=

92
92
92

⇒



92  =



160,000  ⇒ 33 = 160,000 + . When  = 0,

 = 90; therefore,  = 3 · 729,000 = 2,187,000. So 33 = 160,000 + 2,187,000. At the top,  = 100 ⇒
813,000
160,000

≈ 51. The pile reaches the top after about 51 h.

e

3(100)3 = 160,000 + 2,187,000 ⇒  =

normal line at  is  −  = −

al

13. Let  ( ) be any point on the curve. If  is the slope of the tangent line at  , then  =  0 (), and an equation of the

1
1

( − ), or equivalently,  = −  +  + . The -intercept is always 6, so








=6− ⇒ =
. We will solve the equivalent differential equation
=

6−

6−


(6 − )  =   ⇒
(6 − )  =   ⇒ 6 − 1  2 = 1 2 +  ⇒ 12 −  2 = 2 + .
2
2

=6 ⇒


rS

+

⇒  = 11. So the curve is given by 12 −  2 = 2 + 11 ⇒

Fo

Since (3 2) is on the curve, 12(2) − 22 = 32 + 

2 +  2 − 12 + 36 = −11 + 36 ⇒ 2 + ( − 6)2 = 25, a circle with center (0 6) and radius 5.
15. From the figure, slope  =


. If triangle  is isosceles, then slope


N

ot


 must be − , the negative of slope . This slope is also equal to  0 (),







so we have
=−
⇒
=−
⇒



 ln || = − ln || + 
|| = (ln|| )−1 

⇒ || = − ln||+

⇒ || =

1 

||

⇒

⇒ =

⇒


,  6= 0.


c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

N ot e

al

rS

Fo

10

PARAMETRIC EQUATIONS AND POLAR COORDINATES

10.1 Curves Defined by Parametric Equations
1.  = 2 + ,

 = 2 − , −2 ≤  ≤ 2

−2

−1

0

1

2

2

0

0

2

6



6

2

0

0

2

0

6



1

34



1

3
14

12

1−

2
0

√

3
2

≈ 013

rS



al

 = 1 − sin , 0 ≤  ≤ 2

0

Fo

3.  = cos2 ,

e




5.  = 3 − 4,  = 2 − 3

(a)
−1

0

1

2



7

3

−1

−5



5

2

−1

−4

ot



N

(b)  = 3 − 4 ⇒ 4 = − + 3 ⇒  = − 1  + 3 , so
4
4

 1
3
3
9
 = 2 − 3 = 2 − 3 − 4  + 4 = 2 + 4  − 4 ⇒  = 3  −
4

1
4

7.  = 1 − 2 ,  =  − 2, −2 ≤  ≤ 2

(a)


−2

−1

0

1

2



−3

0

1

0

−3



−4

−3

−2

−1

0

(b)  =  − 2 ⇒  =  + 2, so  = 1 − 2 = 1 − ( + 2)2

⇒

 = −( + 2) + 1, or  = − − 4 − 3, with −4 ≤  ≤ 0
2

2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

437

438

¤

CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

9.  =

(a)

√
,  = 1 − 


1

2

3

4



0

1

1414

1732

2


(b)  =

0
1

0

−1

√
 ⇒  = 2

−2

−3

⇒  = 1 −  = 1 − 2 . Since  ≥ 0,  ≥ 0.

So the curve is the right half of the parabola  = 1 − 2 .
(b)

11. (a)  = sin 1 ,  = cos 1 , − ≤  ≤ .
2
2
1
2 +  2 = sin2 2  + cos2 1  = 1. For − ≤  ≤ 0, we have
2

−1 ≤  ≤ 0 and 0 ≤  ≤ 1. For 0   ≤ , we have 0   ≤ 1


2,

 = csc  =

1
1
= . sin 


we have 0    1 and   1. Thus, the curve is the

portion of the hyperbola  = 1 with   1.

⇒ 2 = ln  ⇒  =
1
2

ln  + 1.

ln .

ot

 =+1 =

1
2

17. (a)  = sinh ,  = cosh 

(b)

Fo

15. (a)  = 2

(b)

rS

For 0   


.
2

al

13. (a)  = sin   = csc , 0   

e

and 1   ≥ 0. The graph is a semicircle.

(b)

⇒ 2 − 2 = cosh2  − sinh2  = 1. Since

N

 = cosh  ≥ 1, we have the upper branch of the hyperbola  2 − 2 = 1.

19.  = 3 + 2 cos ,  = 1 + 2 sin , 2 ≤  ≤ 32.

By Example 4 with  = 2,  = 3, and  = 1, the motion of the particle

takes place on a circle centered at (3 1) with a radius of 2. As  goes from


2

to

3
2 ,

the particle starts at the point (3 3) and

moves counterclockwise along the circle ( − 3) + ( − 1) = 4 to (3 −1) [one-half of a circle].
2

21.  = 5 sin ,  = 2 cos 

⇒ sin  =

2

  2   2


+
= 1. The motion of the
, cos  = . sin2  + cos2  = 1 ⇒
5
2
5
2

particle takes place on an ellipse centered at (0 0). As  goes from − to 5, the particle starts at the point (0 −2) and moves clockwise around the ellipse 3 times.

23. We must have 1 ≤  ≤ 4 and 2 ≤  ≤ 3. So the graph of the curve must be contained in the rectangle [1 4] by [2 3]. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 10.1

CURVES DEFINED BY PARAMETRIC EQUATIONS

¤

439

25. When  = −1, ( ) = (0 −1). As  increases to 0,  decreases to −1 and 

increases to 0. As  increases from 0 to 1,  increases to 0 and  increases to 1.
As  increases beyond 1, both  and  increase. For   −1,  is positive and decreasing and  is negative and increasing. We could achieve greater accuracy by estimating - and -values for selected values of  from the given graphs and plotting the corresponding points.
27. When  = 0 we see that  = 0 and  = 0, so the curve starts at the origin. As 

increases from 0 to 1 , the graphs show that  increases from 0 to 1 while 
2
increases from 0 to 1, decreases to 0 and to −1, then increases back to 0, so we arrive at the point (0 1). Similarly, as  increases from

1
2

to 1,  decreases from 1

e

to 0 while  repeats its pattern, and we arrive back at the origin. We could achieve greater accuracy by estimating - and

Fo

rS

29. Use  =  and  =  − 2 sin  with a -interval of [− ].

al

-values for selected values of  from the given graphs and plotting the corresponding points.

31. (a)  = 1 + (2 − 1 ),  = 1 + (2 − 1 ), 0 ≤  ≤ 1. Clearly the curve passes through 1 (1  1 ) when  = 0 and

ot

through 2 (2  2 ) when  = 1. For 0    1,  is strictly between 1 and 2 and  is strictly between 1 and 2 . For every value of ,  and  satisfy the relation  − 1 =

N

1 (1  1 ) and 2 (2  2 ).

Finally, any point ( ) on that line satisfies

2 − 1
( − 1 ), which is the equation of the line through
2 − 1

 − 1
 − 1
=
; if we call that common value , then the given
2 − 1
2 − 1

parametric equations yield the point ( ); and any ( ) on the line between 1 (1  1 ) and 2 (2  2 ) yields a value of
 in [0 1]. So the given parametric equations exactly specify the line segment from 1 (1  1 ) to 2 (2  2 ).
(b)  = −2 + [3 − (−2)] = −2 + 5 and  = 7 + (−1 − 7) = 7 − 8 for 0 ≤  ≤ 1.
33. The circle 2 + ( − 1)2 = 4 has center (0 1) and radius 2, so by Example 4 it can be represented by  = 2 cos ,

 = 1 + 2 sin , 0 ≤  ≤ 2. This representation gives us the circle with a counterclockwise orientation starting at (2 1).
(a) To get a clockwise orientation, we could change the equations to  = 2 cos ,  = 1 − 2 sin , 0 ≤  ≤ 2.
(b) To get three times around in the counterclockwise direction, we use the original equations  = 2 cos ,  = 1 + 2 sin  with the domain expanded to 0 ≤  ≤ 6.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

440

¤

CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

(c) To start at (0 3) using the original equations, we must have 1 = 0; that is, 2 cos  = 0. Hence,  =
 = 2 cos ,  = 1 + 2 sin ,


2

≤≤


2.

So we use

3
.
2

Alternatively, if we want  to start at 0, we could change the equations of the curve. For example, we could use
 = −2 sin ,  = 1 + 2 cos , 0 ≤  ≤ .
35. Big circle: It’s centered at (2 2) with a radius of 2, so by Example 4, parametric equations are

 = 2 + 2 cos 

 = 2 + 2 sin 

0 ≤  ≤ 2

Small circles: They are centered at (1 3) and (3 3) with a radius of 01. By Example 4, parametric equations are
(left)

 = 3 + 01 sin 

(right)

and

 = 1 + 01 cos 

0 ≤  ≤ 2

 = 3 + 01 cos 

 = 3 + 01 sin 

0 ≤  ≤ 2

Semicircle: It’s the lower half of a circle centered at (2 2) with radius 1. By Example 4, parametric equations are
 = 2 + 1 sin 

 ≤  ≤ 2

e

 = 2 + 1 cos 

al

To get all four graphs on the same screen with a typical graphing calculator, we need to change the last -interval to[0 2] in order to match the others. We can do this by changing  to 05. This change gives us the upper half. There are several ways to

 = 2 + 1 cos(05)
37. (a)  = 3

⇒  = 13 , so  = 2 = 23 .

rS

get the lower half—one is to change the “+” to a “−” in the -assignment, giving us
 = 2 − 1 sin(05)
(b)  = 6

Since  = 6 ≥ 0, we only get the right half of the curve  = 23 .

N

ot

right direction.

⇒  = 16 , so  = 4 = 46 = 23 .

Fo

We get the entire curve  = 23 traversed in a left to

0 ≤  ≤ 2

(c)  = −3 = (− )3

[so − = 13 ],

 = −2 = (− )2 = (13 )2 = 23 .
If   0, then  and  are both larger than 1. If   0, then  and  are between 0 and 1. Since   0 and   0, the curve never quite reaches the origin.
39. The case


2

    is illustrated.  has coordinates ( ) as in Example 7,

and  has coordinates (  +  cos( − )) = ( (1 − cos ))

[since cos( − ) = cos  cos  + sin  sin  = − cos ], so  has

coordinates ( −  sin( − ) (1 − cos )) = (( − sin ) (1 − cos ))

[since sin( − ) = sin  cos  − cos  sin  = sin ]. Again we have the

parametric equations  = ( − sin ),  = (1 − cos ).

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 10.1

CURVES DEFINED BY PARAMETRIC EQUATIONS

¤

441

41. It is apparent that  = || and  = | | = | |. From the diagram,

 = || =  cos  and  = | | =  sin . Thus, the parametric equations are
 =  cos  and  =  sin . To eliminate  we rearrange: sin  =  ⇒ sin2  = ()2 and cos  =  ⇒ cos2  = ()2 . Adding the two equations: sin2  + cos2  = 1 = 2 2 +  2 2 . Thus, we have an ellipse.

43.  = (2 cot  2), so the -coordinate of  is  = 2 cot . Let  = (0 2).

Then ∠ is a right angle and ∠ = , so || = 2 sin  and
 = ((2 sin ) cos  (2 sin ) sin ). Thus, the -coordinate of  is  = 2 sin2 .
There are 2 points of intersection:

e

45. (a)

rS

al

(−3 0) and approximately (−21 14).

(b) A collision point occurs when 1 = 2 and 1 = 2 for the same . So solve the equations:

Fo

3 sin  = −3 + cos  (1)

2 cos  = 1 + sin 

(2)

From (2), sin  = 2 cos  − 1. Substituting into (1), we get 3(2 cos  − 1) = −3 + cos  ⇒ 5 cos  = 0 () ⇒

2

or

3
2 .

We check that  =

3
2

satisfies (1) and (2) but  =


2

does not. So the only collision point

ot

cos  = 0 ⇒  = occurs when  =

3
2 ,

and this gives the point (−3 0). [We could check our work by graphing 1 and 2 together as

functions of  and, on another plot, 1 and 2 as functions of . If we do so, we see that the only value of  for which both

N

pairs of graphs intersect is  =

3
.]
2

(c) The circle is centered at (3 1) instead of (−3 1). There are still 2 intersection points: (3 0) and (21 14), but there are no collision points, since () in part (b) becomes 5 cos  = 6 ⇒ cos  =

6
5

 1.

47.  = 2   = 3 − . We use a graphing device to produce the graphs for various values of  with − ≤  ≤ . Note that all

the members of the family are symmetric about the -axis. For   0, the graph does not cross itself, but for  = 0 it has a cusp at (0 0) and for   0 the graph crosses itself at  = , so the loop grows larger as  increases.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

442

¤

CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

49.  =  +  cos   =  +  sin    0. From the first figure, we see that

curves roughly follow the line  = , and they start having loops when  is between 14 and 16. The loops increase in size as  increases.

While not required, the following is a solution to determine the exact values for which the curve has a loop, that is, we seek the values of  for which there exist parameter values  and  such that    and
( +  cos   +  sin ) = ( +  cos   +  sin ).
In the diagram at the left,  denotes the point ( ),  the point ( ),

e

and  the point ( +  cos   +  sin ) = ( +  cos   +  sin ).

al

Since   =   = , the triangle    is isosceles. Therefore its base angles,  = ∠   and  = ∠   are equal. Since  =  −

+=

3
4

−=

5
4

and

− , the relation  =  implies that

rS

 = 2 −


4

3
2

(1).

ot

Fo


√
Since   = distance(( ) ( )) = 2( − )2 = 2 ( − ), we see that
√
1
√

( − ) 2 cos  = 2
, so  −  = 2  cos , that is,
=


√










 −  = 2  cos  − 4 (2). Now cos  −  = sin  −  −  = sin 3 −  ,
4
2
4
4
√


so we can rewrite (2) as  −  = 2  sin 3 −  (20 ). Subtracting (20 ) from (1) and
4
dividing by 2, we obtain  =

3
4

−

√



2
 sin 3
2
4


−  , or

3
4

−=


√
2



sin 3 −  (3).
4

N



Since   0 and   , it follows from (20 ) that sin 3 −   0. Thus from (3) we see that  
4

3
.
4

[We have

implicitly assumed that 0     by the way we drew our diagram, but we lost no generality by doing so since replacing 

by  + 2 merely increases  and  by 2. The curve’s basic shape repeats every time we change  by 2.] Solving for  in
√  3

√
√
2 4 −
2
 3
 . Write  = 3 − . Then  =
, where   0. Now sin    for   0, so   2.
(3), we get  =
4
sin  sin 4 − 


√
 −
As  → 0+ , that is, as  → 3 ,  → 2 .
4

51. Note that all the Lissajous figures are symmetric about the -axis. The parameters  and  simply stretch the graph in the

- and -directions respectively. For  =  =  = 1 the graph is simply a circle with radius 1. For  = 2 the graph crosses

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 10.2

CALCULUS WITH PARAMETRIC CURVES

¤

itself at the origin and there are loops above and below the -axis. In general, the figures have  − 1 points of intersection, all of which are on the -axis, and a total of  closed loops.

==1

=2

=3

3.  = 1 + 4 − 2 ,  = 2 − 3 ;  = 1.

al




2 + 1

= 2 + 1,
=  cos  + sin , and
=
=
.




 cos  + sin 

⇒



−32


= −32 ,
= 4 − 2, and
=
=
. When  = 1,




4 − 2

rS

1.  =  sin ,  = 2 + 

e

10.2 Calculus with Parametric Curves

( ) = (4 1) and  = − 3 , so an equation of the tangent to the curve at the point corresponding to  = 1 is
2

Fo

 − 1 = − 3 ( − 4), or  = − 3  + 7.
2
2
5.  =  cos ,  =  sin ;  = .




 cos  + sin 

=  cos  + sin ,
= (− sin ) + cos , and
=
=
.




− sin  + cos 

When  = , ( ) = (− 0) and  = −(−1) = , so an equation of the tangent to the curve at the point

ot

corresponding to  =  is  − 0 = [ − (−)], or  =  + 2 .

N

7. (a)  = 1 + ln ,  = 2 + 2; (1 3).


1


2

= 2
=  and
=
=
= 22 . At (1 3),





1

 = 1 + ln  = 1 ⇒ ln  = 0 ⇒  = 1 and


= 2, so an equation of the tangent is  − 3 = 2( − 1),


or  = 2 + 1.

(b)  = 1 + ln  ⇒ ln  =  − 1 ⇒  = −1 , so  = 2 + 2 = (−1 )2 + 2 = 2−2 + 2, and  0 = 2−2 · 2.
At (1 3),  0 = 2(1)−2 · 2 = 2, so an equation of the tangent is  − 3 = 2( − 1), or  = 2 + 1.
9.  = 6 sin ,  = 2 + ; (0 0).



2 + 1
=
=
. The point (0 0) corresponds to  = 0, so the


6 cos  slope of the tangent at that point is 1 . An equation of the tangent is therefore
6
 − 0 = 1 ( − 0), or  = 1 .
6
6

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

443

444

¤

CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

11.  = 2 + 1,  = 2 + 

The curve is CU when

13.  =  ,  = −

⇒


2 + 1
1

=
=
=1+


2
2

⇒

 
 
 
2 
−1(22 )
1
=
= − 3.
=
2


2
4

2 
 0, that is, when   0.
2

⇒


−− + −

− (1 − )
=
=
=
= −2 (1 − ) ⇒






 
 
 
2 
−2 (−1) + (1 − )(−2−2 )
−2 (−1 − 2 + 2)
=
=
=
= −3 (2 − 3). The curve is CU when
2



2 
 0, that is, when   3 .
2
2

e

15.  = 2 sin ,  = 3 cos , 0    2.

rS

al

 
 
− 3 sec2 
 


−3 sin 
3
2 
3
=
=
= − tan , so 2 =
= 2
= − sec3 .


2 cos 
2


2 cos 
4
The curve is CU when sec3   0 ⇒ sec   0 ⇒ cos   0 ⇒
17.  = 3 − 3,  = 2 − 3.



3
.
2



= 2, so
=0 ⇔ =0 ⇔





= 32 − 3 = 3( + 1)( − 1), so
=0 ⇔



Fo

( ) = (0 −3).


2

 = −1 or 1 ⇔ ( ) = (2 −2) or (−2 −2). The curve has a horizontal

N

ot

tangent at (0 −3) and vertical tangents at (2 −2) and (−2 −2).

19.  = cos ,  = cos 3. The whole curve is traced out for 0 ≤  ≤ .



= −3 sin 3, so
= 0 ⇔ sin 3 = 0 ⇔ 3 = 0, , 2, or 3



 

 = 0,  , 2 , or  ⇔ ( ) = (1 1), 1  −1 , − 1  1 , or (−1 −1).
3
3
2
2


= − sin , so
= 0 ⇔ sin  = 0 ⇔  = 0 or 



( ) = (1 1) or (−1 −1). Both

⇔

⇔



and equal 0 when  = 0 and .




−3 sin 3 H
−9 cos 3
= lim
= 9, which is the same slope when  = .
= lim
→0 − cos 
 →0 − sin 




Thus, the curve has horizontal tangents at 1  −1 and − 1  1 , and there are no vertical tangents.
2
2

To find the slope when  = 0, we find lim

→0

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 10.2

CALCULUS WITH PARAMETRIC CURVES

¤

445

21. From the graph, it appears that the rightmost point on the curve  =  − 6 ,  = 

is about (06 2). To find the exact coordinates, we find the value of  for which the
√
graph has a vertical tangent, that is, 0 =  = 1 − 65 ⇔  = 1 5 6.
Hence, the rightmost point is
 √


√ 
 √ 
5
−15
1 5 6 − 1 6 5 6  1 6 = 5 · 6−65  6
≈ (058 201).
23. We graph the curve  = 4 − 23 − 22 ,  = 3 −  in the viewing rectangle [−2 11] by [−05 05]. This rectangle

al

e

corresponds approximately to  ∈ [−1 08].

rS

We estimate that the curve has horizontal tangents at about (−1 −04) and (−017 039) and vertical tangents at about (0 0) and (−019 037). We calculate



32 − 1
=
= 3
. The horizontal tangents occur when


4 − 62 − 4

1
 = 32 − 1 = 0 ⇔  = ± √3 , so both horizontal tangents are shown in our graph. The vertical tangents occur when

Fo

 = 2(22 − 3 − 2) = 0 ⇔ 2(2 + 1)( − 2) = 0 ⇔  = 0, − 1 or 2. It seems that we have missed one vertical
2
tangent, and indeed if we plot the curve on the -interval [−12 22] we see that there is another vertical tangent at (−8 6).
25.  = cos ,  = sin  cos .

 = − sin ,  = − sin2  + cos2  = cos 2.

ot

( ) = (0 0) ⇔ cos  = 0 ⇔  is an odd multiple of
 = −1 and  = −1, so  = 1. When  =


.
2

3
2 ,

When  =


,
2

 = 1 and

N

 = −1. So  = −1. Thus,  =  and  = − are both tangent to the curve at (0 0).

27.  =  −  sin ,  =  −  cos .

(a)



 sin 

=  −  cos ,
=  sin , so
=
.



 −  cos 

(b) If 0    , then | cos | ≤   , so  −  cos  ≥  −   0. This shows that  never vanishes, so the trochoid can have no vertical tangent if   .
29.  = 23 ,  = 1 + 4 − 2

⇒




4 − 2
. Now solve
=
=
=1 ⇔


62


62 + 2 − 4 = 0 ⇔ 2(3 − 2)( + 1) = 0 ⇔  = the point is (−2 −4).

2
3

4 − 2
=1 ⇔
62


or  = −1. If  = 2 , the point is 16  29 , and if  = −1,
3
27 9

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

446

¤

CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

31. By symmetry of the ellipse about the - and -axes,



=4

0

  = 4


= 2  −

1
2

0

2

sin 2

 sin  (− sin )  = 4

2
0

 
= 2  = 
2

 2
0

sin2   = 4

 2
0

1
(1
2

− cos 2) 

33. The curve  = 1 +  ,  =  − 2 = (1 − ) intersects the -axis when  = 0,

that is, when  = 0 and  = 1. The corresponding values of  are 2 and 1 + .

=2

=1

=0

=

1
0

=3

[() − 0] 0 ()  =

  −

1
0

1
0

2   =

1
0

1
0

( − 2 ) 

1

1
  − 2  0 + 2 0  

1

  − ( − 0) = 3 ( − 1) 0 − 

[Formula 96 or parts]

al

= 3[0 − (−1)] −  = 3 − 
35.  =  −  sin ,  =  −  cos .

 2

[Formula 97 or parts]

e

The shaded area is given by
 =1+

( −  )  =

 2
( −  cos )( −  cos )  = 0 (2 − 2 cos  + 2 cos2 ) 
2


= 2  − 2 sin  + 1 2  + 1 sin 2 0 = 22 + 2
2
2
  =

0

 2
0

rS

=

37.  =  + − ,  =  − − , 0 ≤  ≤ 2.

 = 1 − − and  = 1 + − , so

Fo

()2 + ()2 = (1 − − )2 + (1 + − )2 = 1 − 2− + −2 + 1 + 2− + −2 = 2 + 2−2 .

2√
Thus,  =  ()2 + ()2  = 0 2 + 2−2  ≈ 31416.

39.  =  − 2 sin ,  = 1 − 2 cos , 0 ≤  ≤ 4.

 = 1 − 2 cos  and  = 2 sin , so

N

ot

()2 + ()2 = (1 − 2 cos )2 + (2 sin )2 = 1 − 4 cos  + 4 cos2  + 4 sin2  = 5 − 4 cos .

 4 √
Thus,  =  ()2 + ()2  = 0
5 − 4 cos   ≈ 267298.

41.  = 1 + 32 ,  = 4 + 23 , 0 ≤  ≤ 1.  = 6 and  = 62 , so ()2 + ()2 = 362 + 364

Thus,  =



1

0



362 + 364  =

1

6

0



1 + 2  = 6

1

2

 √

= 3 2 32 = 2(232 − 1) = 2 2 2 − 1
3

2

√ 1 
 2  [ = 1 + 2 ,  = 2 ]

1

43.  =  sin ,  =  cos , 0 ≤  ≤ 1.






2

+

Thus,  =






2



=  cos  + sin  and
= − sin  + cos , so



= 2 cos2  + 2 sin  cos  + sin2  + 2 sin2  − 2 sin  cos  + cos2 
= 2 (cos2  + sin2 ) + sin2  + cos2  = 2 + 1.

√
1√
21 
2 + 1  = 1  2 + 1 +
2
0

1
2

√
√

1
ln  + 2 + 1 0 = 1 2 +
2

1
2

√ 

ln 1 + 2 .

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 10.2

CALCULUS WITH PARAMETRIC CURVES

¤

447

 =  cos ,  =  sin , 0 ≤  ≤ .
  2   2
+  = [ (cos  − sin )]2 + [ (sin  + cos )]2


45.

= ( )2 (cos2  − 2 cos  sin  + sin2 )

+ ( )2 (sin2  + 2 sin  cos  + cos2 
= 2 (2 cos2  + 2 sin2 ) = 22
Thus,  =

√   √
√
√
22  = 0 2   = 2  0 = 2 ( − 1).
0

The figure shows the curve  = sin  + sin 15,  = cos  for 0 ≤  ≤ 4.

47.

 = cos  + 15 cos 15 and  = − sin , so

e

()2 + ()2 = cos2  + 3 cos  cos 15 + 225 cos2 15 + sin2 .
 4 √
1 + 3 cos  cos 15 + 225 cos2 15  ≈ 167102.
Thus,  = 0



+

  2


Set  () =
≈

= (1 −  )2 + (1 +  )2 = (1 − 2 + 2 ) + (1 + 2 + 2 ) = 2 + 22 , so  =

√
2 + 22 . Then by Simpson’s Rule with  = 6 and ∆ =

2
3 [ (−6)

rS

  2

al

49.  =  −  ,  =  +  , −6 ≤  ≤ 6.

6−(−6)
6

= 2, we get

6 √
2 + 22 .
−6

+ 4 (−4) + 2 (−2) + 4 (0) + 2 (2) + 4 (4) +  (6)] ≈ 6123053.

Fo

51.  = sin2 ,  = cos2 , 0 ≤  ≤ 3.

()2 + ()2 = (2 sin  cos )2 + (−2 cos  sin )2 = 8 sin2  cos2  = 2 sin2 2 ⇒
Distance =

2
√  2
√ 
√
√
 3 √
2 |sin 2|  = 6 2 0 sin 2  [by symmetry] = −3 2 cos 2
= −3 2 (−1 − 1) = 6 2.
0
0

because the curve is the segment of  +  = 1 that lies in the first quadrant
√
 2
(since ,  ≥ 0), and this segment is completely traversed as  goes from 0 to  . Thus,  = 0 sin 2  = 2, as above.
2

ot

The full curve is traversed as  goes from 0 to


2,

N

53.  =  sin ,  =  cos , 0 ≤  ≤ 2.

  2

  2

= ( cos )2 + (− sin )2 = 2 cos2  + 2 sin2  = 2 (1 − sin2 ) + 2 sin2 


2
= 2 − (2 − 2 ) sin2  = 2 − 2 sin2  = 2 1 − 2 sin2  = 2 (1 − 2 sin2 )

 2  

 2 
So  = 4 0
2 1 − 2 sin2   [by symmetry] = 4 0
1 − 2 sin2  .


+



55. (a)  = 11 cos  − 4 cos(112),  = 11 sin  − 4 sin(112).

Notice that 0 ≤  ≤ 2 does not give the complete curve because
(0) 6= (2). In fact, we must take  ∈ [0 4] in order to obtain the complete curve, since the first term in each of the parametric equations has period 2 and the second has period

2
112

=

4
11 ,

and the least common

integer multiple of these two numbers is 4. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

448

¤

CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

(b) We use the CAS to find the derivatives  and , and then use Theorem 6 to find the arc length. Recent versions
 √ 
 4  of Maple express the integral 0
()2 + ()2  as 88 2 2  , where () is the elliptic integral
 1√
√
1 − 2 2
√
 and  is the imaginary number −1.
2
1−
0
Some earlier versions of Maple (as well as Mathematica) cannot do the integral exactly, so we use the command

evalf(Int(sqrt(diff(x,t)ˆ2+diff(y,t)ˆ2),t=0..4*Pi)); to estimate the length, and find that the arc length is approximately 29403. Derive’s Para_arc_length function in the utility file Int_apps simplifies the
 4 
 
  integral to 11 0
−4 cos  cos 11 − 4 sin  sin 11 + 5 .
2
2
 =  cos  + sin  and  = − sin  + cos , so

57.  =  sin ,  =  cos , 0 ≤  ≤ 2.
2

2

2

2

() + () =  cos  + 2 sin  cos  + sin2  + 2 sin2  − 2 sin  cos  + cos2 


 2

2  =

0

√
2 cos  2 + 1  ≈ 47394.

al

=

59.  = 1 +  ,  = (2 + 1) , 0 ≤  ≤ 1.

=



1

2  =

0

+

  2


1

2

0

= 2



13

4







=

2
1215

−4
9



+


+



 2
0

 2
= 32 + (2)2 = 94 + 42 .



  2



 =

1

22

0


√ 1
 18 

− 8 32
3

13

=





94 + 42  = 2

4


81

·

2

 = 9 + 4,  = ( − 4)9,
1
 = 18 , so   = 18 

2
15


13
352 − 2032


.
2

+

2 ·  sin3  · 3 sin  cos   = 62

65.  = 32 ,  = 23 , 0 ≤  ≤ 5

  2



2


2 (92 + 4) 

=

2
9 · 18



13

4

(32 − 412 ) 

4

  2


1

0

2

√
√ 


3 · 132 13 − 20 · 13 13 − (3 · 32 − 20 · 8) =

63.  =  cos3 ,  =  sin3 , 0 ≤  ≤

=

  2

N


81

  2

 
 2

2 52
5

=


√
1
2 ( + 1)2 (2 + 2 + 2)  = 0 2(2 + 1)2 ( + 1) 2 + 2 + 2  ≈ 1035999

ot



so

= 2 ( + 1)2 + 2 ( + 1)4 = 2 ( + 1)2 [1 + ( + 1)2 ],

2(2 + 1)

61.  = 3 ,  = 2 , 0 ≤  ≤ 1.

=

= ( +  )2 + [(2 + 1) +  (2)]2 = [ ( + 1)]2 + [ (2 + 2 + 1)]2

rS



Fo

  2

e

= 2 (cos2  + sin2 ) + sin2  + cos2  = 2 + 1

  2


 2
0

  2

2
1215

√


247 13 + 64

= (−3 cos2  sin )2 + (3 sin2  cos )2 = 92 sin2  cos2 .


2
sin4  cos   = 6 2 sin5  0 = 6 2
5
5

= (6)2 + (62 )2 = 362 (1 + 2 ) ⇒

√
5
5 √
5
 = 0 2 ()2 + ()2  = 0 2(32 )6 1 + 2  = 18 0 2 1 + 2 2 


26

 26
 26
√
 = 1 + 2 
= 18 1 (32 − 12 )  = 18 2 52 − 2 32
= 18 1 ( − 1)  
5
3
⇒



+



 = 2 

= 18

 2
5

√
· 676 26 −

2
3

√  

· 26 26 − 2 − 2 =
5
3

1

24

5

√


949 26 + 1

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 10.3

POLAR COORDINATES

¤

449

67. If  0 is continuous and  0 () 6= 0 for  ≤  ≤ , then either  0 ()  0 for all  in [ ] or  0 ()  0 for all  in [ ]. Thus, 

is monotonic (in fact, strictly increasing or strictly decreasing) on [ ]. It follows that  has an inverse. Set  =  ◦  −1 , that is, define  by  () = ( −1 ()). Then  = () ⇒  −1 () = , so  = () = ( −1 ()) =  ().
 
  

 


1



˙
= tan−1 =
. But
=
=
⇒



1 + ()2  



˙


 
 
 
  − 
¨˙ ¨ ˙
¨ − 
˙ ¨ ˙
 
˙
  − 
¨˙ ¨ ˙
1

= 2
=
=
=
⇒
. Using the Chain Rule, and the
 
 
˙
2
˙

1 + ()2
˙ ˙
2
˙
 + 2
˙
˙

69. (a)  = tan−1



fact that  =






⇒

 
  2


0



=
=





+

¨ − 
˙ ¨ ˙
2 + 2
˙
˙

  2




 ⇒




=

 
 2


+

  2


12

= 2 + 2
˙
˙
, we have that


  
    ¨ −  
˙ ¨ ˙ 
1
|¨ − |
˙ ¨ ˙
¨ − 
˙ ¨ ˙
= 2
. So  =   =  2
    ( + 2 )32  = (2 + 2 )32 .
(2 +  2 )12
˙
˙
( +  2 )32
˙
˙
˙
˙
˙
˙

e


2 
, =
¨
.

2
 2



 2 
1 · (2 2 ) − 0 · ()
=
.
So  =
[1 + ()2 ]32
[1 + ()2 ]32

rS

al

(b)  =  and  =  () ⇒  = 1,  = 0 and  =
˙
¨
˙

⇒  = 1 − cos  ⇒  = sin , and  = 1 − cos  ⇒  = sin  ⇒  = cos . Therefore,
˙
¨
˙
¨




cos  − cos2  − sin2 
cos  − (cos2  + sin2 )
|cos  − 1|
=
=
=
. The top of the arch is
2 + sin2 ]32
(2 − 2 cos )32
[(1 − cos )
(1 − 2 cos  + cos2  + sin2 )32

71.  =  − sin 

Fo

characterized by a horizontal tangent, and from Example 2(b) in Section 10.2, the tangent is horizontal when  = (2 − 1), so take  = 1 and substitute  =  into the expression for :  =

|cos  − 1|
|−1 − 1|
1
=
= .
4
(2 − 2 cos )32
[2 − 2(−1)]32

ot

73. The coordinates of  are ( cos   sin ). Since   was unwound from

arc  ,   has length . Also ∠   = ∠   − ∠  = 1  − ,
2

N



so  has coordinates  =  cos  +  cos 1  −  = (cos  +  sin ),
2



 =  sin  −  sin 1  −  = (sin  −  cos ).
2

10.3 Polar Coordinates


1. (a) 2


3





we obtain the point 2 7 . The direction
3


is 4 , so −2 4 is a point that satisfies the   0
3
3

By adding 2 to opposite 
3


3,

requirement.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

450

¤

CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES



 
  0: 1 − 3 + 2 = 1 5
4
4

 

  0: −1 − 3 +  = −1 
4
4



(c) −1 
2



 
  0: −(−1)  +  = 1 3
2
2

 

  0: −1  + 2 = −1 5
2
2

3. (a)

 = 1 cos  = 1(−1) = −1 and

al

 = 1 sin  = 1(0) = 0 give us

e



(b) 1 − 3
4

rS

the Cartesian coordinates (−1 0).



 
 = 2 cos − 2 = 2 − 1 = −1 and
3
2
 √ 
√
 2 
 = 2 sin − 3 = 2 − 23 = − 3

Fo

(b)

ot

√ 

give us −1 − 3 .

 √  √
 = −2 cos 3 = −2 − 22 = 2 and
4

N

(c)

√ 
√
 = −2 sin 3 = −2 22 = − 2
4
gives us

5. (a)  = 2 and  = −2

⇒ =

√ 
√
2 − 2 .


√
 
22 + (−2)2 = 2 2 and  = tan−1 −2 = −  . Since (2 −2) is in the fourth
2
4

 √ quadrant, the polar coordinates are (i) 2 2

(b)  = −1 and  =

7
4



 √

and (ii) −2 2 3 .
4


√ 
√
√ 2
3
3 ⇒  = (−1)2 +
3 = 2 and  = tan−1 −1 =



 quadrant, the polar coordinates are (i) 2 2 and (ii) −2
3

5
3


.

2
.
3

√ 

Since −1 3 is in the second

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 10.3

The curve  = 1 represents a circle with center

7.  ≥ 1.

POLAR COORDINATES

¤

451

9.  ≥ 0, 4 ≤  ≤ 34.

 and radius 1. So  ≥ 1 represents the region on or

 =  represents a line through .

outside the circle. Note that  can take on any value.

5
3

≤≤

7
3

rS

al

e

11. 2    3,



13. Converting the polar coordinates (2 3) and (4 23) to Cartesian coordinates gives us 2 cos   2 sin
3

√ 

 
4 cos 2  4 sin 2 = −2 2 3 . Now use the distance formula.
3
3

15. 2 = 5

 √ 
= 1 3 and



√ 2 √
√
√
 √
(2 − 1 )2 + (2 − 1 )2 = (−2 − 1)2 + 2 3 − 3 = 9 + 3 = 12 = 2 3

⇔ 2 +  2 = 5, a circle of radius
⇒ 2 = 2 cos 

√
5 centered at the origin.

⇔ 2 +  2 = 2 ⇔ 2 − 2 + 1 +  2 = 1 ⇔ ( − 1)2 +  2 = 1, a circle of

ot

17.  = 2 cos 



Fo

=


3

radius 1 centered at (1 0). The first two equations are actually equivalent since 2 = 2 cos 

⇒

( − 2 cos ) = 0 ⇒

N

 = 0 or  = 2 cos . But  = 2 cos  gives the point  = 0 (the pole) when  = 0. Thus, the equation  = 2 cos  is equivalent to the compound condition ( = 0 or  = 2 cos ).
⇔ 2 (cos2  − sin2 ) = 1 ⇔ ( cos )2 − ( sin )2 = 1 ⇔ 2 −  2 = 1, a hyperbola centered at

19. 2 cos 2 = 1

the origin with foci on the -axis.
21.  = 2

⇔  sin  = 2 ⇔  =

23.  = 1 + 3

=

2 sin 

⇔  sin  = 1 + 3 cos 

⇔  = 2 csc 
⇔  sin  − 3 cos  = 1 ⇔ (sin  − 3 cos ) = 1 ⇔

1 sin  − 3 cos 

25. 2 +  2 = 2

⇔ 2 = 2 cos 

⇔ 2 − 2 cos  = 0 ⇔ ( − 2 cos ) = 0 ⇔  = 0 or  = 2 cos .

 = 0 is included in  = 2 cos  when  =


2

+ , so the curve is represented by the single equation  = 2 cos 

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452

¤

CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

27. (a) The description leads immediately to the polar equation  =

slightly more difficult to derive.


6,

  and the Cartesian equation  = tan   =
6

(b) The easier description here is the Cartesian equation  = 3.
29.  = −2 sin 

Fo
N

35.  = 4 sin 3

ot

33.  = ,  ≥ 0

rS

al

e

31.  = 2(1 + cos )

37.  = 2 cos 4

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

1
√
3

 is

SECTION 10.3

POLAR COORDINATES

39.  = 1 − 2 sin 

al

e

41. 2 = 9 sin 2

N

ot

45.  = 1 + 2 cos 2

Fo

rS

43.  = 2 + sin 3

47. For  = 0, , and 2,  has its minimum value of about 05. For  =


2

and

3
2 ,

 attains its maximum value of 2.

We see that the graph has a similar shape for 0 ≤  ≤  and  ≤  ≤ 2.

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¤

453

454

¤

CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

49.  =  cos  = (4 + 2 sec ) cos  = 4 cos  + 2. Now,  → ∞

(4 + 2 sec ) → ∞ ⇒  →

  −

or  →

2

consider 0 ≤   2], so lim  =
→∞

 3 +
2

lim  =

[since we need only

lim (4 cos  + 2) = 2. Also,

→2−

 → −∞ ⇒ (4 + 2 sec ) → −∞ ⇒  →
→−∞

⇒

  +
2

or  →

 3 −
2

, so

lim (4 cos  + 2) = 2. Therefore, lim  = 2 ⇒  = 2 is a vertical asymptote.
→±∞

→2+

51. To show that  = 1 is an asymptote we must prove lim  = 1.
→±∞

 = () cos  = (sin  tan ) cos  = sin2 . Now,  → ∞ ⇒ sin  tan  → ∞ ⇒
 −
 →  , so lim  = lim sin2  = 1. Also,  → −∞ ⇒ sin  tan  → −∞ ⇒
2
→

2

→2−

, so lim  =
→−∞

lim

→2+

sin2  = 1. Therefore, lim  = 1 ⇒  = 1 is
→±∞

e

→∞

  +

a vertical asymptote. Also notice that  = sin2  ≥ 0 for all , and  = sin2  ≤ 1 for all . And  6= 1, since the curve is not

.
2

Therefore, the curve lies entirely within the vertical strip 0 ≤   1.

al

defined at odd multiples of

rS

53. (a) We see that the curve  = 1 +  sin  crosses itself at the origin, where  = 0 (in fact the inner loop corresponds to

negative -values,) so we solve the equation of the limaçon for  = 0 ⇔  sin  = −1 ⇔ sin  = −1. Now if
||  1, then this equation has no solution and hence there is no inner loop. But if   −1, then on the interval (0 2)

Fo

the equation has the two solutions  = sin−1 (−1) and  =  − sin−1 (−1), and if   1, the solutions are
 =  + sin−1 (1) and  = 2 − sin−1 (1). In each case,   0 for  between the two solutions, indicating a loop.
(b) For 0    1, the dimple (if it exists) is characterized by the fact that  has a local maximum at  =
2  is negative at  =
2

 =  sin  = sin  +  sin2 
3
,
2

⇒

⇒

2 
= − sin  + 2 cos 2.
2

this is equal to −(−1) + 2(−1) = 1 − 2, which is negative only for   1 . A similar argument shows that
2

for −1    0,  only has a local minimum at  =
55.  = 2 sin 

So we

since by the Second Derivative Test this indicates a maximum:


= cos  + 2 sin  cos  = cos  +  sin 2


N

At  =

3
,
2

ot

determine for what -values

3
2 .

⇒


2

(indicating a dimple) for   − 1 .
2

 =  cos  = 2 sin  cos  = sin 2,  =  sin  = 2 sin2 

⇒


2 · 2 sin  cos  sin 2

=
=
=
= tan 2

 cos 2 · 2 cos 2

When  =
57.  = 1

 
√
 

,
= tan 2 ·
= tan = 3. [Another method: Use Equation 3.]
6 
6
3
⇒  =  cos  = (cos ),  =  sin  = (sin )

⇒



sin (−12 ) + (1) cos  2
− sin  +  cos 
·
=
=
=


− cos  −  sin  cos (−12 ) − (1) sin  2
When  = ,

−0 + (−1)
−

=
=
= −.

−(−1) − (0)
1

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SECTION 10.3

59.  = cos 2

⇒  =  cos  = cos 2 cos ,  =  sin  = cos 2 sin 

POLAR COORDINATES

¤

455

⇒



cos 2 cos  + sin  (−2 sin 2)
=
=

 cos 2 (− sin ) + cos  (−2 sin 2)
√
 √

√
0 22 +
22 (−2)
 
− 2
When  = ,
=  √
 √

= √ = 1.
4 
0 − 22 +
22 (−2)
− 2

61.  = 3 cos 



⇒  =  cos  = 3 cos  cos ,  =  sin  = 3 cos  sin 

⇒

= −3 sin  + 3 cos  = 3 cos 2 = 0 ⇒ 2 = or
⇔  = or



 


3
3
3
So the tangent is horizontal at √2   and − √2  3 same as √2  −  .
4
4
4



2

2


2

= −6 sin  cos  = −3 sin 2 = 0 ⇒ 2 = 0 or 

63.  = 1 + cos 

3
2


4

⇔  = 0 or


2.

3
4 .



So the tangent is vertical at (3 0) and 0  .
2

⇒  =  cos  = cos  (1 + cos ),  =  sin  = sin  (1 + cos ) ⇒




al

e

= (1 + cos ) cos  − sin2  = 2 cos2  + cos  − 1 = (2 cos  − 1)(cos  + 1) = 0 ⇒ cos  =




 =  , , or 5 ⇒ horizontal tangent at 3   , (0 ), and 3  5 .
3
3
2 3
2 3

= −(1 + cos ) sin  − cos  sin  = − sin  (1 + 2 cos ) = 0 ⇒ sin  = 0 or cos  = − 1
2
 1 4 
 1 2 
2
4
 = 0, , 3 , or 3 ⇒ vertical tangent at (2 0), 2  3 , and 2  3 .

or −1 ⇒

⇒

rS




1
2

Note that the tangent is horizontal, not vertical when  = , since lim

→


= 0.


⇒ 2 =  sin  +  cos  ⇒ 2 +  2 =  +  ⇒
 2
 2  2  2

2 
2
2 −  + 1  +  2 −  + 1  = 1  + 1 
⇒
 − 1  +  − 1  = 1 (2 + 2 ), and this is a circle
2
2
2
2
2
2
4
√
1 1 
1
2 + 2 . with center 2  2  and radius 2 

Fo

65.  =  sin  +  cos 

69.  = sin  − 2 cos(4).

The parameter interval is [0 2].

N

ot

67.  = 1 + 2 sin(2). The parameter interval is [0 4].

71.  = 1 + cos999 . The parameter interval is [0 2].

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456

¤

CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES



73. It appears that the graph of  = 1 + sin  −


6



is the same shape as

the graph of  = 1 + sin , but rotated counterclockwise about the

 origin by  . Similarly, the graph of  = 1 + sin  −  is rotated by
6
3

.
3

In general, the graph of  =  ( − ) is the same shape as that of

 =  (), but rotated counterclockwise through  about the origin.
That is, for any point (0  0 ) on the curve  =  (), the point
(0  0 + ) is on the curve  =  ( − ), since 0 = (0 ) =  ((0 + ) − ).
75. Consider curves with polar equation  = 1 +  cos , where  is a real number. If  = 0, we get a circle of radius 1 centered at

the pole. For 0   ≤ 05, the curve gets slightly larger, moves right, and flattens out a bit on the left side. For 05    1, the left side has a dimple shape. For  = 1, the dimple becomes a cusp. For   1, there is an internal loop. For  ≥ 0, the

e

rightmost point on the curve is (1 +  0). For   0, the curves are reflections through the vertical axis of the curves

Fo

rS

al

with   0.

 = 025

 = 075

=1

=2

N

ot



− tan 
− tan 

tan  − tan 
= 
77. tan  = tan( − ) =
=


1 + tan  tan  tan 
1+
1+ tan 









 sin2  sin  +  cos  − tan  cos  −  sin 
−
tan 
 cos  +  ·



cos 


 =
= 
= 





 sin2 
+
tan  cos  −  sin  + tan  sin  +  cos  cos  +
·





 cos 
=


 cos2  +  sin2 
=


 cos2  + sin2 



10.4 Areas and Lengths in Polar Coordinates
1.  = −4 , 2 ≤  ≤ .

=





2

1 2

2

 =





2

1
(−4 )2
2

 =





2

1 −2

2

 =

1
2



−2−2

2

= −1(−2 − −4 ) = −4 − −2

c
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SECTION 10.4

AREAS AND LENGTHS IN POLAR COORDINATES

3. 2 = 9 sin 2,  ≥ 0, 0 ≤  ≤ 2.
2

0

5.  =

1 2
2

 =



2
1
2 (9 sin 2) 

0



√
, 0 ≤  ≤ 2.  =



0

2

1
2



=



2
1
2

0


√ 2
  =

2

·2

0

1 2

2

0

=2



 =


1
2 (1

0

1
2



1
4

2

2
0

= 2

1
2

(16 + 9 sin2 ) 



2

(16 + 24 sin  + 9 sin2 ) 

−2

[by Theorem 4.5.6(a)]



0

2 

2 

41
2


16 + 9 · 1 (1 − cos 2) 
2

−

9
2



cos 2  = 41  −
2



(2 sin )2  =

0


− cos 2) =  −

1
2

1
2

9
4

[by Theorem 4.5.6(a)]

sin 2

2
0

=

 41
4


− 0 − (0 − 0) =

41
4

rS



=



sin 2



4 sin2  

0



=

0

Fo



1
 
2

0

9. The area is bounded by  = 2 sin  for  = 0 to  = .

=

2

9
2

−2

1
2

=



 =

 1
2
− 2 cos 2 0 = − 9 (−1 − 1) =
4

+ 3 sin )2  =

1
2 ((4

−2

=

1 2

2

9
2


.
2

7.  = 4 + 3 sin , −  ≤  ≤
2

=

2

=

e



al

=

Also, note that this is a circle with radius 1, so its area is (1)2 = .


2

0

=

1
2



2

0



2

 =



2

1
2 (3

0

+ 2 cos )2  =

1
2

(11 + 12 cos  + 2 cos 2)  =

0

= 1 (22) = 11
2

13.  =



2

0

=
=

1
2

1
2



1 2

2

2

0



2

0

1
2

 =



0

2
1
(2
2

1
2

2

+ 4 sin 4 −

1
2

2

(9 + 12 cos  + 4 cos2 ) 


2
11 + 12 sin  + sin 2 0

2

+ sin 4)  =



4 + 4 sin 4 + 1 (1 − cos 8) 
2
9



0



9 + 12 cos  + 4 · 1 (1 + cos 2) 
2

N

=

1 2
2

ot

11.  =

1
2



2

(4 + 4 sin 4 + sin2 4) 

0



cos 8  = 1 9  − cos 4 −
2 2

= 1 [(9 − 1) − (−1)] = 9 
2
2

1
16

sin 8

2
0

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¤

457

458

¤

15.  =

CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES



2

0

=

1
2

=

1
2



1 2

2

2

 =



2
1
2

0

2

1 + cos2 5 

(1 + cos2 5)  =

0

3

1
20

+

2

sin 10

2
0

=



1
2

2

0

1
(3)
2



1 + 1 (1 + cos 10) 
2

= 3
2

17. The curve passes through the pole when  = 0

=


6

⇒ 4 cos 3 = 0 ⇒ cos 3 = 0

⇒

3 =


2

+  ⇒

+  . The part of the shaded loop above the polar axis is traced out for
3

 = 0 to  = 6, so we’ll use −6 and 6 as our limits of integration.
 6
 6
1
1
=
(4 cos 3)2  = 2
(16 cos2 3) 
2
2
−6

0

6
1
(1
2

1
6

sin 6

6
0

=8


6

= 4
3

al

0


+ cos 6)  = 8  +

e

= 16



19.  = 0

0

rS

⇒ sin 4 = 0 ⇒ 4 =  ⇒  =  .
4
 4
 4
 4
2
2
1
1
1
1
(sin 4)  = 2 sin 4  = 2
(1 − cos 8) 
=
2
2
0


= −

1
8

sin 8

4
0

=

1
4

0


4

1

16

=

Fo

1
4

This is a limaçon, with inner loop traced out between  =



32

76

N

= 2

1
(1
2

+ 2 sin )2  =




1 + 4 sin  + 4 sin2   =

32 

76

√

32   
=  − 4 cos  + 2 − sin 2 76 = 9 − 7 + 2 3 −
2
2
23. 2 cos  = 1

⇒ cos  =

= 2
=

 3
0

1
2

⇒ =

1
[(2 cos )2
2


3

or

− 12 ]  =

7
6

and

11
6

[found by

solving  = 0].

ot

21.

√

3
2



=




1 + 4 sin  + 4 · 1 (1 − cos 2) 
2

32 

76
√
 − 323

5
.
3

 3
0

(4 cos2  − 1) 

 3   1


 3
4 2 (1 + cos 2) − 1  = 0 (1 + 2 cos 2) 
0


3
=  + sin 2 0 =


3

+

√
3
2

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 10.4

AREAS AND LENGTHS IN POLAR COORDINATES

25. To find the area inside the leminiscate 2 = 8 cos 2 and outside the circle  = 2,

we first note that the two curves intersect when 2 = 8 cos 2 and  = 2, that is, when cos 2 = 1 . For −   ≤ , cos 2 =
2

1
2

⇔ 2 = ±3

or ±53 ⇔  = ±6 or ±56. The figure shows that the desired area is
4 times the area between the curves from 0 to 6. Thus,

 6  1
 6
= 4 0
(8 cos 2) − 1 (2)2  = 8 0 (2 cos 2 − 1) 
2
2

6
√
√

= 8 sin 2 − 
= 8 32 − 6 = 4 3 − 43
0

27. 3 cos  = 1 + cos 

1
2

⇔ cos  =

⇒ =


3

or −  .
3

 3


3
(3 + 4 cos 2 − 2 cos )  = 3 + 2 sin 2 − 2 sin  0
√
√
=+ 3− 3=

=
=

 3
0

 3
0

1
4

√
√
sin 
⇒ tan  = 3 ⇒  =
3=
cos 
 2 1 √
2
2
1
3 cos  
2 (sin )  + 3 2
1
2


−


=

1
4


3

=


12

−

· 1 (1 − cos 2)  +
2

√

3
16

+


8

−

√
3 3
16

=

⇒

= 8·2

 8
0


=4 −

33. sin 2 = cos 2

= 4

1
3 2

 8
0

1
4

1
2

5
24

−

√
3
4

sin 2 2  = 8

sin 4

8
0

 8
0


=4  −
8

1
4

1
(1
2

√ 
3
4

sin 2 


4

⇒

− cos 4) 


·1 =

⇒ tan 2 = 1 ⇒ 2 =
1
2


.
3

· 3 · 1 (1 + cos 2) 
2

sin 2
= 1 ⇒ tan 2 = 1 ⇒ 2 = cos 2

⇒

N

=

 2


3
2 sin 2 0 + 3  + 1 sin 2 3
4
2


√ 
 
− 43 − 0 + 3  + 0 −  +
4
2
3
1
2

31. sin 2 = cos 2

8

⇒

rS

=

al

√
3 cos  = sin 

ot

29.

0

Fo

=

e

 3 1
 = 2 0 2 [(3 cos )2 − (1 + cos )2 ] 
 3
 3
= 0 (8 cos2  − 2 cos  − 1)  = 0 [4(1 + cos 2) − 2 cos  − 1] 


2


4

−1

⇒ =


8

[since 2 = sin 2]

 8


8
2 sin 2  = − cos 2 0
√
√
= − 1 2 − (−1) = 1 − 1 2
2
2

=

0

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¤

459

460

¤

CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

35. The darker shaded region (from  = 0 to  = 23) represents

From this area, we’ll subtract

1
2

1
2

of the desired area plus

1
2

of the area of the inner loop.

of the area of the inner loop (the lighter shaded region from  = 23 to  = ), and then

double that difference to obtain the desired area.

2

2 

23 1  1
=2 0
+ cos   − 23 1 1 + cos  
2 2
2 2
=

 23  1
4

0

 23  1
4

0


 

+ cos  + cos2   − 23 1 + cos  + cos2  
4


+ cos  + 1 (1 + cos 2) 
2

 
− 23 1 + cos  + 1 (1 + cos 2) 
4
2

23 



 sin 2

sin 2
−
+ sin  + +
+ sin  + +
4
2
4
4
2
4
0
23


√
√ 
√
√ 


=  + 23 +  − 83 −  +  +  + 23 +  − 83
6
3
4
2
6
3


=


4

+

3
4

√ 
√

3= 1 +3 3
4

al

=

=


6

or

⇒ 1 = 2 sin 

1
2

⇒ sin  =

5
.
6

39. 2 sin 2 = 1

1
2

2




6



and

3
2


 5 .
6

⇒ 2 =

 5 13
6, 6 , 6 ,

or

17
6 .

ot

⇒ sin 2 =

3

⇒

Fo

The other two points of intersection are

rS

37. The pole is a point of intersection.

1 + sin  = 3 sin 

e

=

By symmetry, the eight points of intersection are given by
 5 13
, , 12 ,
12 12

and

17
,
12

and

N

(1 ), where  =

(−1 ), where  =

7 11 19
12 , 12 , 12 ,

and

23
12 .

[There are many ways to describe these points.]

41. The pole is a point of intersection. sin  = sin 2 = 2 sin  cos 

sin  (1 − 2 cos ) = 0 ⇔ sin  = 0 or cos  =
 = 0, , and √


,
3

3 2
 3
2

or − 
3



1
2

⇒

⇒ the other intersection points are

[by symmetry].

√

3 
3
2

⇔


c
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SECTION 10.4

AREAS AND LENGTHS IN POLAR COORDINATES

¤

461

43.

From the first graph, we see that the pole is one point of intersection. By zooming in or using the cursor, we find the -values of the intersection points to be  ≈ 088786 ≈ 089 and  −  ≈ 225. (The first of these values may be more easily

estimated by plotting  = 1 + sin  and  = 2 in rectangular coordinates; see the second graph.) By symmetry, the total area contained is twice the area contained in the first quadrant, that is,
 2
 

 
1
1
(2)2  + 2
(1 + sin )2  =
42  +
= 2
2
2

45.  =







=





2 + ()2  =

0



0



1
4

sin 2

2


=

4 3

3



 

+

2

+


(2 cos )2 + (−2 sin )2 



4(cos2  + sin2 )  =



0

e

0



+  − 2 cos  + 1  −
2


4




−  − 2 cos  + 1  −
2

al

=




3 0
3



1 + 2 sin  + 1 (1 − cos 2) 
2
1
4

 sin 2 ≈ 34645

rS

0

4

2

√
 
4  = 2 0 = 2

As a check, note that the curve is a circle of radius 1, so its circumference is 2(1) = 2.






=





2 + ()2  =

0

2

0

2



(2 )2 + (2)2  =

Fo

47.  =



2 (2 + 4)  =

0

2


 2 + 4 

0

2





2 + 4  =

4

4 2 +4

1
2

√
  =

1
2

N




  =

ot

Now let  =  + 4, so that  = 2 
2

·

2
3

1
2

2

0


4 + 42 


 and


4(2 +1)
32
= 1 [432 ( 2 + 1)32 − 432 ] = 8 [(2 + 1)32 − 1]
3
3
4

49. The curve  = cos4 (4) is completely traced with 0 ≤  ≤ 4.

2

2 + ()2 = [cos4 (4)]2 + 4 cos3 (4) · (− sin(4)) · 1
4
= cos8 (4) + cos6 (4) sin2 (4)

= cos6 (4)[cos2 (4) + sin2 (4)] = cos6 (4)

 4 
 4  cos6 (4)  = 0 cos3 (4) 
0
 2
 2
= 2 0 cos3 (4)  [since cos3 (4) ≥ 0 for 0 ≤  ≤ 2] = 8 0 cos3  


1
 2
 = sin 
= 8 0 (1 − sin2 ) cos   = 8 0 (1 − 2 ) 

=



 = 1
4

 = cos  



1

= 8  − 1 3 0 = 8 1 − 1 =
3
3

16
3

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462

¤

CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

51. One loop of the curve  = cos 2 is traced with −4 ≤  ≤ 4.






2

= cos2 2 + (−2 sin 2)2 = cos2 2 + 4 sin2 2 = 1 + 3 sin2 2

53. The curve  = sin(6 sin ) is completely traced with 0 ≤  ≤ .

2 +






2

= sin2 (6 sin ) + 36 cos2  cos2 (6 sin ) ⇒ 

55. (a) From (10.2.6),

 = sin(6 sin ) ⇒




0

4

−4


1 + 3 sin2 2  ≈ 24221.


= cos(6 sin ) · 6 cos , so


 sin2 (6 sin ) + 36 cos2  cos2 (6 sin )  ≈ 80091.




2

(b) The curve 2 = cos 2 goes through the pole when cos 2 = 0 ⇒
⇒ =


.
4

We’ll rotate the curve from  = 0 to  =


4

and double

al


2




()2 + ()2 


[from the derivation of Equation 10.4.5]
=  2 2 + ()2 


=  2 sin  2 + ()2 

=

2 =

⇒ 

e

2 +

4

= 4

√ cos 2 sin 

0



0

4



cos2 2 + sin2 2

cos 2

 4
√


√
√ 

4
1
 = 4 cos 2 sin  √ sin   = 4 − cos  0 = −4 22 − 1 = 2 2 − 2 cos 2
0

10.5 Conic Sections

⇒ 4 = 6 ⇒  = 3 .
2
 3
The vertex is (0 0), the focus is 0 2 , and the directrix

N

1. 2 = 6 and 2 = 4

is  = − 3 .
2

4

Fo

0



√


2 cos 2 sin  cos 2 + sin2 2 cos 2  = 4

ot

=2



rS

this value to obtain the total surface area generated.
 2

 sin2 2 sin2 2
2 = cos 2 ⇒ 2
= −2 sin 2 ⇒
.
=
=


2 cos 2

⇒  2 = −2. 4 = −2 ⇒  = − 1 .
2
 1 
The vertex is (0 0), the focus is − 2  0 , and the

3. 2 = − 2

directrix is  = 1 .
2

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 10.5
2

5. ( + 2) = 8 ( − 3). 4 = 8, so  = 2. The vertex is

7.  2 + 2 + 12 + 25 = 0

(−2 3), the focus is (−2 5), and the directrix is  = 1.

CONIC SECTIONS

¤

463

⇒

2

 + 2 + 1 = −12 − 24 ⇒

( + 1)2 = −12( + 2). 4 = −12, so  = −3.

The vertex is (−2 −1), the focus is (−5 −1), and the

e

directrix is  = 1.

9. The equation has the form  2 = 4, where   0. Since the parabola passes through (−1 1), we have 12 = 4(−1), so

al



4 = −1 and an equation is  2 = − or  = − 2 . 4 = −1, so  = − 1 and the focus is − 1  0 while the directrix
4
4 is  = 1 .
4

rS

√
√
√
√
√
2
2
+
= 1 ⇒  = 4 = 2,  = 2,  = 2 − 2 = 4 − 2 = 2. The
2
4
√ 

ellipse is centered at (0 0), with vertices at (0 ±2). The foci are 0 ± 2 .

Fo

11.

ot

√
2  2
+
= 1 ⇒  = 9 = 3,
9
1
√
√
√
√
√
2 − 2 =
 = 1 = 1,  = 
9 − 1 = 8 = 2 2.
⇔

N

13. 2 + 9 2 = 9

The ellipse is centered at (0 0), with vertices (±3 0).
√
The foci are (±2 2 0).

17. The center is (0 0),  = 3, and  = 2, so an equation is

15. 92 − 18 + 4 2 = 27

⇔

9(2 − 2 + 1) + 4 2 = 27 + 9 ⇔
9( − 1)2 + 4 2 = 36 ⇔

( − 1)2
2
+
=1 ⇒
4
9

√
5 ⇒ center (1 0),
√ 

vertices (1 ±3), foci 1 ± 5

 = 3,  = 2,  =

√
√ 
√

2
2
+
= 1.  = 2 − 2 = 5, so the foci are 0 ± 5 .
4
9

c
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464

19.

¤

CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

√
√
2
2
−
= 1 ⇒  = 5,  = 3,  = 25 + 9 = 34 ⇒
25
9
√ 

center (0 0), vertices (0 ±5), foci 0 ± 34 , asymptotes  = ± 5 .
3

Note: It is helpful to draw a 2-by-2 rectangle whose center is the center of the hyperbola. The asymptotes are the extended diagonals of the rectangle.

2
2
−
= 1 ⇒  =  = 10,
100
100
√
√
 = 100 + 100 = 10 2 ⇒ center (0 0), vertices (±10 0),
√ 

foci ±10 2 0 , asymptotes  = ± 10  = ±
10

al

23. 42 −  2 − 24 − 4 + 28 = 0

e

⇔

⇔

rS

21. 2 −  2 = 100

4(2 − 6 + 9) − ( 2 + 4 + 4) = −28 + 36 − 4 ⇔

( + 2)2
( − 3)2
−
=1 ⇒
1
4
√
√
√
√
 = 1 = 1,  = 4 = 2,  = 1 + 4 = 5 ⇒
√


center (3 −2), vertices (4 −2) and (2 −2), foci 3 ± 5 −2 ,

ot

asymptotes  + 2 = ±2( − 3).

Fo

4( − 3)2 − ( + 2)2 = 4 ⇔

⇔ 2 = 1( + 1). This is an equation of a parabola with 4 = 1, so  = 1 . The vertex is (0 −1) and the
4


3 focus is 0 − 4 .

27. 2 = 4 − 2 2

N

25. 2 =  + 1

⇔ 2 + 2 2 − 4 = 0 ⇔ 2 + 2( 2 − 2 + 1) = 2 ⇔ 2 + 2( − 1)2 = 2 ⇔

 √

 √ 
2
( − 1)2
+
= 1. This is an equation of an ellipse with vertices at ± 2 1 . The foci are at ± 2 − 1 1 = (±1 1).
2
1

( + 1)2
− 2 = 1. This is an equation
4
√ 
 

√ of a hyperbola with vertices (0 −1 ± 2) = (0 1) and (0 −3). The foci are at 0 −1 ± 4 + 1 = 0 −1 ± 5 .

29.  2 + 2 = 42 + 3

⇔ 2 + 2 + 1 = 42 + 4 ⇔ ( + 1)2 − 42 = 4 ⇔

31. The parabola with vertex (0 0) and focus (1 0) opens to the right and has  = 1, so its equation is  2 = 4, or  2 = 4.
33. The distance from the focus (−4 0) to the directrix  = 2 is 2 − (−4) = 6, so the distance from the focus to the vertex is
1
2 (6)

= 3 and the vertex is (−1 0). Since the focus is to the left of the vertex,  = −3. An equation is 2 = 4( + 1) ⇒

 2 = −12( + 1). c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 10.5

CONIC SECTIONS

¤

465

35. A parabola with vertical axis and vertex (2 3) has equation  − 3 = ( − 2)2 . Since it passes through (1 5), we have

⇒  = 2, so an equation is  − 3 = 2( − 2)2 .

5 − 3 = (1 − 2)2

37. The ellipse with foci (±2 0) and vertices (±5 0) has center (0 0) and a horizontal major axis, with  = 5 and  = 2,

so 2 = 2 − 2 = 25 − 4 = 21. An equation is

2
2
+
= 1.
25
21

39. Since the vertices are (0 0) and (0 8), the ellipse has center (0 4) with a vertical axis and  = 4. The foci at (0 2) and (0 6)

are 2 units from the center, so  = 2 and  =

√
√
√
( − 0)2
( − 4)2
2 − 2 = 42 − 22 = 12. An equation is
+
=1 ⇒
2

2

2
( − 4)2
+
= 1.
12
16

⇒ 2 = 12, and the equation is

2 = 25 − 9 = 16, so the equation is

2
2
− 2 = 1. Foci (±5 0) ⇒  = 5 and 32 + 2 = 52
2
3


rS

43. An equation of a hyperbola with vertices (±3 0) is

( + 1)2
( − 4)2
+
= 1.
12
16

al

from the center, so  = 2. Thus, 2 + 22 = 42

( + 1)2
( − 4)2
+
= 1. The focus (−1 6) is 2 units
2
42

e

41. An equation of an ellipse with center (−1 4) and vertex (−1 0) is

⇒

2
2
−
= 1.
9
16

Fo

45. The center of a hyperbola with vertices (−3 −4) and (−3 6) is (−3 1), so  = 5 and an equation is

( + 3)2
( − 1)2
−
= 1. Foci (−3 −7) and (−3 9) ⇒  = 8, so 52 + 2 = 82
2
5
2
( + 3)2
( − 1)2
−
= 1.
25
39

ot

equation is

⇒ 2 = 64 − 25 = 39 and the

47. The center of a hyperbola with vertices (±3 0) is (0 0), so  = 3 and an equation is


2
2
= 2 ⇒  = 2(3) = 6 and the equation is
−
= 1.

9
36

N

Asymptotes  = ±2 ⇒

2
2
− 2 = 1.
32


49. In Figure 8, we see that the point on the ellipse closest to a focus is the closer vertex (which is a distance

 −  from it) while the farthest point is the other vertex (at a distance of  + ). So for this lunar orbit,
( − ) + ( + ) = 2 = (1728 + 110) + (1728 + 314), or  = 1940; and ( + ) − ( − ) = 2 = 314 − 110, or  = 102. Thus, 2 = 2 − 2 = 3,753,196, and the equation is

2
2
+
= 1.
3,763,600
3,753,196

51. (a) Set up the coordinate system so that  is (−200 0) and  is (200 0).

| | − | | = (1200)(980) = 1,176,000 ft =
2 = 2 − 2 =

3,339,375
121

⇒

2450
11

mi = 2 ⇒  =

1225
,
11

and  = 200 so

1212
121 2
−
= 1.
1,500,625
3,339,375

c
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466

¤

CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

(b) Due north of 

⇒  = 200 ⇒

1212
133,575
(121)(200)2
−
=1 ⇒ =
≈ 248 mi
1,500,625
3,339,375
539

53. The function whose graph is the upper branch of this hyperbola is concave upward. The function is



2
√ 2

=
 + 2 , so  0 = (2 + 2 )−12 and
2




 2
( + 2 )−12 − 2 (2 + 2 )−32 = (2 + 2 )−32  0 for all , and so  is concave upward.
 00 =

 = () = 

1+

55. (a) If   16, then  − 16  0, and

2
2
+
= 1 is an ellipse since it is the sum of two squares on the left side.

 − 16

(b) If 0    16, then  − 16  0, and left side.

2
2
+
= 1 is a hyperbola since it is the difference of two squares on the

 − 16

e

(c) If   0, then  − 16  0, and there is no curve since the left side is the sum of two negative terms, which cannot equal 1.

al

(d) In case (a), 2 = , 2 =  − 16, and 2 = 2 − 2 = 16, so the foci are at (±4 0). In case (b),  − 16  0, so 2 = ,
2 = 16 − , and 2 = 2 + 2 = 16, and so again the foci are at (±4 0).
⇒ 0 =


, so the tangent line at (0  0 ) is
2

0
2
0
=
( − 0 ). This line passes through the point ( −) on the
4
2

directrix, so − −

0
2
0
=
( − 0 ) ⇒ −42 − 2 = 20 − 22
0
0
4
2

Fo

−

⇒ 2 = 40

rS

57. 2 = 4

2 − 20 − 42 = 0 ⇔ 2 − 20 + 2 = 2 + 42
0
0

⇔

⇔

N

ot



(0 − )2 = 2 + 42 ⇔ 0 =  ± 2 + 42 . The slopes of the tangent lines at  =  ± 2 + 42

 ± 2 + 42
, so the product of the two slopes is are 2


 + 2 + 42  − 2 + 42
2 − (2 + 42 )
−42
=
= −1,
·
=
2
2
2
4
42
showing that the tangent lines are perpendicular.
59. 92 + 4 2 = 36

⇔

2
2
+
= 1. We use the parametrization  = 2 cos ,  = 3 sin , 0 ≤  ≤ 2. The circumference
4
9

is given by
 2 
 2 
 2 
()2 + ()2  = 0
(−2 sin )2 + (3 cos )2  = 0
4 sin2  + 9 cos2  
0
 2 √
4 + 5 cos2  
= 0

=

√

2 − 0
= , and  () = 4 + 5 cos2  to get
8
4


 3 
 
 
 

4 
 ≈ 8 = 3  (0) + 4 4 + 2 2 + 4 4 + 2 () + 4 5 + 2 3 + 4 7 +  (2) ≈ 159.
4
2
4
Now use Simpson’s Rule with  = 8, ∆ =

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 10.5

61.

¤

CONIC SECTIONS

467

2
2
2
2 − 2
√ 2
− 2 =1 ⇒
=
⇒ =±
 − 2 .
2
2
2





 
  
 


2 

39 2 

= 2
2 − 2  =
2 − 2 − ln  + 2 − 2 
 2
2
 




√

 √ 2
  − 2 − 2 ln  + 2 − 2  + 2 ln ||

√
Since 2 + 2 = 2  2 − 2 = 2 , and 2 − 2 = .




=  − 2 ln( + ) + 2 ln  =  + 2 (ln  − ln( + ))


=

= 2  +  ln[( + )], where 2 = 2 + 2 .

63. 92 + 4 2 = 36

⇔

2
2
+
= 1 ⇒  = 3,  = 2. By symmetry,  = 0. By Example 2 in Section 7.3, the area of the
4
9

top half of the ellipse is 1 () = 3. Solve 92 + 4 2 = 36 for  to get an equation for the top half of the ellipse:
2
√
4 − 2 . Now

e

92 + 4 2 = 36 ⇔ 4 2 = 36 − 92

⇔  2 = 9 (4 − 2 ) ⇒  =
4

3
2

2
 2  
 2
1
3
1
1 3
4 − 2
 =
(4 − 2 ) 
[ ()]2  =
2
3 −2 2 2
8 −2

2

 
 2
3
3 16
3
1
4
(4 − 2 )  =
=
·2
4 − 3 =
=
8
4
3
4 3

0
0




al

1


rS

=

so the centroid is (0 4).

line at  is −

2
2
+ 2 =1 ⇒
2


2 20
2 
+ 2 = 0 ⇒ 0 = − 2
2

 

Fo

65. Differentiating implicitly,

[ 6= 0]. Thus, the slope of the tangent

2 1
1
1 and of 2  is
. By the formula in Problem 15 on text page 202,
. The slope of 1  is
2 1
1 + 
1 − 

we have



2 using 2 2 + 2 1 = 2 2 ,
1

and 2 − 2 = 2

N

ot

1
2 1
+ 2
2  2 + 2 1 (1 + )
2 2 + 2 1
1 + 
 1
= 2 1
= 2 tan  =
2
 1 (1 + ) − 2 1 1
 1 1 + 2 1
 1 1
1− 2
 1 (1 + )


2 1 + 2
2
=
=
2)
1 (1 + 
1

and

1
2 1


−
2 1 − 2
−2  2 − 2 1 (1 − )
2
−2 2 + 2 1
2 1
1 − 
= 2 1
=
= 2
=
tan  =
2
2 
2 
2)
 1 (1 − ) −  1 1
 1 1 −  1
1 (1 − 
1
 1 1
1− 2
 1 (1 − )
−

Thus,  = .

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

468

¤

CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

10.6 Conic Sections in Polar Coordinates
1. The directrix  = 4 is to the right of the focus at the origin, so we use the form with “+  cos  ” in the denominator.

(See Theorem 6 and Figure 2.) An equation is  =

1
·4
4

2
=
=
.
1 +  cos 
2 + cos 
1 + 1 cos 
2

3. The directrix  = 2 is above the focus at the origin, so we use the form with “+  sin  ” in the denominator. An equation is

=


15(2)
6
=
=
.
1 +  sin 
1 + 15 sin 
2 + 3 sin 

5. The vertex (4 32) is 4 units below the focus at the origin, so the directrix is 8 units below the focus ( = 8), and we

use the form with “−  sin  ” in the denominator.  = 1 for a parabola, so an equation is

e

1(8)
8

=
=
.
1 −  sin 
1 − 1 sin 
1 − sin 

al

=

7. The directrix  = 4 sec  (equivalent to  cos  = 4 or  = 4) is to the right of the focus at the origin, so we will use the form

9.  =

1
(4)
2
4

2
· =
=
.
1 +  cos 
2 + cos 
1 + 1 cos  2
2

15
45
4
, where  =
·
=
5 − 4 sin  15
1 − 4 sin 
5

(a) Eccentricity =  =
(b) Since  =

4
5

4
5

and  =

4
5

⇒  = 1.

Fo

=

rS

with “+  cos ” in the denominator. The distance from the focus to the directrix is  = 4, so an equation is

4
5

 1, the conic is an ellipse.

ot

(c) Since “−  sin  ” appears in the denominator, the directrix is below the focus at the origin,  = | | = 1, so an equation of the directrix is  = −1.

N





(d) The vertices are 4  and 4  3 .
2
9 2
11.  =

2
13
23
·
=
, where  = 1 and  =
3 + 3 sin  13
1 + 1 sin 

2
3

⇒  = 2.
3

(a) Eccentricity =  = 1
(b) Since  = 1, the conic is a parabola.
(c) Since “+  sin  ” appears in the denominator, the directrix is above the focus at the origin.  = | | = 2 , so an equation of the directrix is  = 2 .
3
3
1 
(d) The vertex is at 3  2 , midway between the focus and directrix.

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SECTION 10.6

13.  =

9
16
32
, where  =
·
=
6 + 2 cos  16
1 + 1 cos 
3

(a) Eccentricity =  =
(b) Since  =

1
3

1
3

and  =

3
2

CONIC SECTIONS IN POLAR COORDINATES

⇒  = 9.
2

1
3

 1, the conic is an ellipse.

(c) Since “+ cos  ” appears in the denominator, the directrix is to the right of the focus at the origin.  = | | = 9 , so an equation of the directrix is
2
(d) The vertices are
9  that is, 16   .
15.  =

9



80

and

9

4


, so the center is midway between them,

14
34
3
·
=
, where  = 2 and  =
4 − 8 cos  14
1 − 2 cos 

(b) Since  = 2  1, the conic is a hyperbola.

⇒  = 3.
8

al

(a) Eccentricity =  = 2

3
4

e

 = 9.
2

rS

(c) Since “− cos  ” appears in the denominator, the directrix is to the left of the focus at the origin.  = | | = 3 , so an equation of the directrix is
8
 = −3.
8

17. (a)  =

Fo





(d) The vertices are − 3  0 and 1   , so the center is midway between them,
4
4

 that is, 1   .
2
1
, where  = 2 and  = 1 ⇒  = 1 . The eccentricity
2
1 − 2 sin 

ot

 = 2  1, so the conic is a hyperbola. Since “− sin  ” appears in the

N

denominator, the directrix is below the focus at the origin.  = | | = 1 ,
2


1

so an equation of the directrix is  = − 2 . The vertices are −1 2 and
 1 3 


, so the center is midway between them, that is, 2  3 .

3 2
3 2
(b) By the discussion that precedes Example 4, the equation is  =

1

1 − 2 sin  −

3
4

.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

469

470

¤

CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

19. For   1 the curve is an ellipse. It is nearly circular when  is close to 0. As 

increases, the graph is stretched out to the right, and grows larger (that is, its right-hand focus moves to the right while its left-hand focus remains at the origin.) At  = 1, the curve becomes a parabola with focus at the origin.
⇒  = [ −  cos( − )] = ( +  cos ) ⇒

23. |  | =  | |


1 −  cos 

e

(1 −  cos ) =  ⇒  =

⇒  = [ −  sin( − )] = ( +  sin ) ⇒

1 −  sin 

rS

(1 −  sin ) =  ⇒  =

al

21. |  | =  | |

=

Fo

25. We are given  = 0093 and  = 228 × 108 . By (7), we have

(1 − 2 )
228 × 108 [1 − (0093)2 ]
226 × 108
=
≈
1 +  cos 
1 + 0093 cos 
1 + 0093 cos 

27. Here 2 = length of major axis = 3618 AU

1809[1 − (097)2 ]
107
≈
. By (8), the maximum distance from the comet to the sun is
1 + 097 cos 
1 + 097 cos 

ot

=

⇒  = 1809 AU and  = 097. By (7), the equation of the orbit is

N

1809(1 + 097) ≈ 3564 AU or about 3314 billion miles.
29. The minimum distance is at perihelion, where 46 × 107 =  = (1 − ) = (1 − 0206) = (0794)

⇒

 = 46 × 1070794. So the maximum distance, which is at aphelion, is


 = (1 + ) = 46 × 1070794 (1206) ≈ 70 × 107 km.

31. From Exercise 29, we have  = 0206 and (1 − ) = 46 × 107 km. Thus,  = 46 × 1070794. From (7), we can write the

equation of Mercury’s orbit as  = 

1 − 2
. So since
1 +  cos 


(1 − 2 ) sin 
=
⇒

(1 +  cos )2
 2

2 (1 − 2 )2
2 (1 − 2 )2 2 sin2 
2 (1 − 2 )2
2 +
=
+
=
(1 + 2 cos  + 2 )
2
4

(1 +  cos )
(1 +  cos )
(1 +  cos )4 c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 10 REVIEW

the length of the orbit is

=

2

0



2 + ()2  = (1 − 2 )

0

2

¤

471

√
1 + 2 + 2 cos 
 ≈ 36 × 108 km
(1 +  cos )2

This seems reasonable, since Mercury’s orbit is nearly circular, and the circumference of a circle of radius  is 2 ≈ 36 × 108 km.

10 Review

1. (a) A parametric curve is a set of points of the form ( ) = ( () ()), where  and  are continuous functions of a

variable .

e

(b) Sketching a parametric curve, like sketching the graph of a function, is difficult to do in general. We can plot points on the curve by finding () and () for various values of , either by hand or with a calculator or computer. Sometimes, when

al

 and  are given by formulas, we can eliminate  from the equations  =  () and  = () to get a Cartesian equation relating  and . It may be easier to graph that equation than to work with the original formulas for  and  in terms of .

(b) Calculate the area as than (() ())].

(b)  =



  =




()  0 () [or




()  0 () if the leftmost point is ( () ()) rather

 

()2 + ()2  =  [ 0 ()]2 + [ 0 ()]2 





2




()2 + ()2  =  2() [ 0 ()]2 + [0 ()]2 

ot

3. (a)  =



rS



 as a function of  by calculating
=
[if  6= 0].




Fo

2. (a) You can find

4. (a) See Figure 5 in Section 10.3.

N

(b)  =  cos ,  =  sin 

(c) To find a polar representation ( ) with  ≥ 0 and 0 ≤   2, first calculate  = cos  =  and sin  = .


2 +  2 . Then  is specified by

 



 sin  +  cos 
()
( sin )


 = 

, where  = ().
5. (a) Calculate
=
=
=  





()
( cos ) cos  −  sin 




(b) Calculate  =
(c)  =



1 2

 2

 =



1
[()]2
 2






()2 + ()2  =  2 + ()2  =  [ ()]2 + [ 0 ()]2 


6. (a) A parabola is a set of points in a plane whose distances from a fixed point  (the focus) and a fixed line  (the directrix)

are equal.
(b) 2 = 4;  2 = 4 c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

472

¤

CHAPTER 10

PARAMETRIC EQUATIONS AND POLAR COORDINATES

7. (a) An ellipse is a set of points in a plane the sum of whose distances from two fixed points (the foci) is a constant.

(b)

2
2
+ 2
= 1.
2

 − 2

8. (a) A hyperbola is a set of points in a plane the difference of whose distances from two fixed points (the foci) is a constant.

This difference should be interpreted as the larger distance minus the smaller distance.
2
2
− 2
=1
2

 − 2
√
2 − 2

(c)  = ±


(b)

9. (a) If a conic section has focus  and corresponding directrix , then the eccentricity  is the fixed ratio |  |  | | for points

 of the conic section.

1. False.

al





.  = −:  =
.  = :  =
.  = −:  =
.
1 +  cos 
1 −  cos 
1 +  sin 
1 −  sin 

rS

(c)  = :  =

e

(b)   1 for an ellipse;   1 for a hyperbola;  = 1 for a parabola.

Consider the curve defined by  =  () = ( − 1)3 and  = () = ( − 1)2 . Then 0 () = 2( − 1), so 0 (1) = 0,

3. False.

Fo

but its graph has a vertical tangent when  = 1. Note: The statement is true if  0 (1) 6= 0 when 0 (1) = 0.
For example, if  () = cos  and () = sin  for 0 ≤  ≤ 4, then the curve is a circle of radius 1, hence its length
 4 
 4
 4 
[ 0 ()]2 + [ 0 ()]2  = 0
(− sin )2 + (cos )2  = 0 1  = 4, since as  increases is 2, but 0

5. True.

ot

from 0 to 4, the circle is traversed twice.

The curve  = 1 − sin 2 is unchanged if we rotate it through 180◦ about  because

1 − sin 2( + ) = 1 − sin(2 + 2) = 1 − sin 2. So it’s unchanged if we replace  by −. (See the discussion

7. False.

N

after Example 8 in Section 10.3.) In other words, it’s the same curve as  = −(1 − sin 2) = sin 2 − 1.
The first pair of equations gives the portion of the parabola  = 2 with  ≥ 0, whereas the second pair of equations traces out the whole parabola  = 2 .
9. True.

By rotating and translating the parabola, we can assume it has an equation of the form  = 2 , where   0.


The tangent at the point  2 is the line  − 2 = 2( − ); i.e.,  = 2 − 2 . This tangent meets

 the parabola at the points  2 where 2 = 2 − 2 . This equation is equivalent to 2 = 2 − 2
[since   0]. But 2 = 2 − 2 ⇔ 2 − 2 + 2 = 0 ⇔ ( − )2 = 0 ⇔  =  ⇔

 

 2 =  2 . This shows that each tangent meets the parabola at exactly one point.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 10 REVIEW

¤

473

1.  = 2 + 4,  = 2 − , −4 ≤  ≤ 1.  = 2 − , so

 = (2 − )2 + 4(2 − ) = 4 − 4 +  2 + 8 − 4 = 2 − 8 + 12 ⇔
 + 4 = 2 − 8 + 16 = ( − 4)2 . This is part of a parabola with vertex
(−4 4), opening to the right.

3.  = sec  =

1
1
= . Since 0 ≤  ≤ 2, 0   ≤ 1 and  ≥ 1. cos 


al

e

This is part of the hyperbola  = 1.

(i)  = ,  =

√


(ii)  = 4 ,  = 2
(iii)  = tan2 ,  = tan , 0 ≤   2

√
 are

rS

5. Three different sets of parametric equations for the curve  =

 
The Cartesian coordinates are  = 4 cos 2 = 4 − 1 = −2 and
3
2
√ 
√
√ 

 = 4 sin 2 = 4 23 = 2 3, that is, the point −2 2 3 .
3

N

ot

7. (a)

Fo

There are many other sets of equations that also give this curve.

(b) Given  = −3 and  = 3, we have  =
(−3 3) is in the second quadrant,  =
 √

 √ 11 
3 2 4 and −3 2 7 .
4


√
√

(−3)2 + 32 = 18 = 3 2. Also, tan  =


3
.
4

⇒ tan  =

 √
Thus, one set of polar coordinates for (−3 3) is 3 2

3
4

3
, and since
−3


, and two others are

9.  = 1 − cos . This cardioid is

symmetric about the polar axis.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

474

¤

CHAPTER 10

PARAMETRIC EQUATIONS AND POLAR COORDINATES

11.  = cos 3. This is a

three-leaved rose. The curve is traced twice.

13.  = 1 + cos 2. The curve is

symmetric about the pole and both the horizontal and vertical

3
1 + 2 sin 

⇒  = 2  1, so the conic is a hyperbola.  = 3 ⇒

al

15.  =

e

axes.

and the form “+2 sin ” imply that the directrix is above the focus at



 the origin and has equation  = 3 . The vertices are 1  and −3 3 .
2
2
2
3
2

17.  +  = 2

Fo

rS

=

⇔  cos  +  sin  = 2 ⇔ (cos  + sin ) = 2 ⇔  =

19.  = (sin ). As  → ±∞,  → 0.

As  → 0,  → 1. In the first figure,

2 cos  + sin 

ot

there are an infinite number of

-intercepts at  = ,  a nonzero

N

integer. These correspond to pole points in the second figure.

21.  = ln ,  = 1 + 2 ;  = 1.


1


2

= 2 and
= , so
=
=
= 22 .





1

When  = 1, ( ) = (0 2) and  = 2.
23.  = −

⇒  =  sin  = − sin  and  =  cos  = − cos 



=
=


When  = ,






⇒

sin  +  cos 
−− sin  + − cos  − sin  − cos 
=
·
.
=
−− cos  − − sin  − cos  + sin  cos  −  sin 

0 − (−1)
1

=
=
= −1.

−1 + 0
−1

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 10 REVIEW

25.  =  + sin ,  =  − cos 

⇒



1 + sin 
=
=


1 + cos 

¤

475

⇒

 
 
(1 + cos ) cos  − (1 + sin )(− sin )
 
(1 + cos )2
1 + cos  + sin 
  cos  + cos2  + sin  + sin2 
=
=
=
=
2

1 + cos 
(1 + cos )3
(1 + cos )3
2

27. We graph the curve  = 3 − 3,  = 2 +  + 1 for −22 ≤  ≤ 12.

By zooming in or using a cursor, we find that the lowest point is about
(14 075). To find the exact values, we find the -value at which


 = 2 + 1 = 0 ⇔  = − 1 ⇔ ( ) = 11  3 
2
8 4

1
2

⇒  = 0,

or

5
.
3

4
.
3

rS

2
,
3

, or




= 2 cos  − 2 cos 2 = 2 1 + cos  − 2 cos2  = 2(1 − cos )(1 + 2 cos ) = 0 ⇒


 = 2 sin  −  sin 2 ⇒
 = 0,


,
3

al

sin  = 0 or cos  =


= −2 sin  + 2 sin 2 = 2 sin (2 cos  − 1) = 0 ⇔


⇒

e

29.  = 2 cos  −  cos 2

Thus the graph has vertical tangents where  =


,
3

 and

5
,
3

and horizontal tangents where  =

To determine







0

√

3

2
−1
2

3

2
√
3 3

2

−1
2
3
2

√
−323
√
− 23 

ot


3
2
3

4
.
3

Fo

→0

0

and


= 0, so there is a horizontal tangent there.


what the slope is where  = 0, we use l’Hospital’s Rule to evaluate lim


2
3



N

4
3
5
3

−3

0


31. The curve 2 = 9 cos 5 has 10 “petals.” For instance, for − 10 ≤  ≤


,
10

there are two petals, one with   0 and one

with   0.
 = 10

 10

1 2

−10 2

 = 5

33. The curves intersect when 4 cos  = 2

 10

−10

9 cos 5  = 5 · 9 · 2

⇒ cos  = 1 ⇒  = ± 
2
3
 

 for − ≤  ≤ . The points of intersection are 2 3 and 2 −  .
3

 10
0


10
cos 5  = 18 sin 5 0
= 18

c
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476

¤

CHAPTER 10

PARAMETRIC EQUATIONS AND POLAR COORDINATES

35. The curves intersect where 2 sin  = sin  + cos 

sin  = cos 

⇒=


,
4

⇒

and also at the origin (at which  =

3
4

on the second curve).
 34
 4 1
 = 0 2 (2 sin )2  + 4 1 (sin  + cos )2 
2

 4
1 34
= 0 (1 − cos 2)  + 2 4 (1 + sin 2) 
4 
34

=  − 1 sin 2 0 + 1  − 1 cos 2 4 = 1 ( − 1)
2
2
4
2
37.  = 32 ,  = 23 .

√
2
2√
2√
2
()2 + ()2  = 0 (6)2 + (62 )2  = 0 362 + 364  = 0 362 1 + 2 
0
√
2
2 √
5




 = 1 + 2 ,  = 2 
= 0 6 || 1 + 2  = 6 0  1 + 2  = 6 1 12 1 
2

5
 √

= 6 · 1 · 2 32 = 2(532 − 1) = 2 5 5 − 1
2
3
1



2




2 + 1

2

al

 2 
 2 
39.  = 
2 + ()2  = 
(1)2 + (−12 )2  =

e

=



rS

 

√
√
√



 2

2 + 1
2 + 1
42 + 1
2 + 42 + 1
24
√
+ ln  + 2 + 1
−
+ ln
= −
=


2
 + 2 + 1

=

√
1
3
+ 2, 1 ≤  ≤ 4 ⇒
,  =
3
2

4
1


  √ 2
4 
2  + (2 − −3 )2 
()2 + ()2  = 1 2 1 3 + 1 −2
3
2

 4 1
1

3

3 + 1 −2
2

4

(2 + −3 )2  = 2 1 1 5 +
3

5
6


1
4
+ 1 −5  = 2 18 6 + 5  − 1 −4 1 =
2
6
8

471,295

1024

N

= 2

2

ot

41.  = 4

Fo

√
√
√


2 + 42 + 1
2  2 + 1 − 42 + 1
√
+ ln
=
2
 + 2 + 1

43. For all  except −1, the curve is asymptotic to the line  = 1. For

  −1, the curve bulges to the right near  = 0. As  increases, the bulge becomes smaller, until at  = −1 the curve is the straight line  = 1.
As  continues to increase, the curve bulges to the left, until at  = 0 there is a cusp at the origin. For   0, there is a loop to the left of the origin, whose size and roundness increase as  increases. Note that the -intercept of the curve is always −

c
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¤

CHAPTER 10 REVIEW

45.

2
2
+
= 1 is an ellipse with center (0 0).
9
8
√
 = 3,  = 2 2,  = 1 ⇒

47. 6 2 +  − 36 + 55 = 0
2

477

⇔

6( − 6 + 9) = −( + 1) ⇔

( − 3)2 = − 1 ( + 1), a parabola with vertex (−1 3),
6


1 opening to the left,  = − 24 ⇒ focus − 25  3 and
24

foci (±1 0), vertices (±3 0).

directrix  = − 23 .
24

2
2
+
= 1.
25
9

al

so 2 = 2 − 2 = 52 − 42 = 9. An equation is

e

49. The ellipse with foci (±4 0) and vertices (±5 0) has center (0 0) and a horizontal major axis, with  = 5 and  = 4,

The asymptote  = 3 has slope 3, so
8
5

3

=

1

and so 2 = 16 −

8
5

⇒  = 3 and 2 + 2 = 2

=

72
.
5

Thus, an equation is

2
2
− 2 = 1.
2


⇒ (3)2 + 2 = 42

⇒

2
5 2
52
2
−
= 1, or
−
= 1.
725
85
72
8

Fo

102 = 16 ⇒ 2 =

rS

51. The center of a hyperbola with foci (0 ±4) is (0 0), so  = 4 and an equation is

53. 2 = −( − 100) has its vertex at (0 100), so one of the vertices of the ellipse is (0 100). Another form of the equation of a

ot

parabola is 2 = 4( − 100) so 4( − 100) = −( − 100) ⇒ 4 = −1 ⇒  = − 1 . Therefore the shared focus is
4




found at 0 399 so 2 = 399 − 0 ⇒  = 399 and the center of the ellipse is 0 399 . So  = 100 − 399 = 401 and
4
4
8
8
8
8




399 2
399 2
− 8
− 8
4012 − 3992
2
2
2
2
2
= 25. So the equation of the ellipse is 2 +
=1 ⇒
 = − =
+  401 2 = 1,
82

2
25
8

N

(8 − 399)2
2
+
= 1. or 25
160,801
55. Directrix  = 4

⇒  = 4, so  =

1
3

⇒ =


4
=
.
1 +  cos 
3 + cos 

57. (a) If ( ) lies on the curve, then there is some parameter value 1 such that

31
32
1
= . If 1 = 0,
3 =  and
1 + 1
1 + 3
1

the point is (0 0), which lies on the line  = . If 1 6= 0, then the point corresponding to  =
=

1 is given by
1

3(11 )2
32
31
3(11 )
= . So ( ) also lies on the curve. [Another way to see
= 3 1 = ,  =
= 3
3
1 + (11 )
1 + 1
1 + (11 )3
1 + 1

this is to do part (e) first; the result is immediate.] The curve intersects the line  =  when
 = 2

⇒  = 0 or 1, so the points are (0 0) and

3
2

3
32
=
1 + 3
1 + 3


3 .
2

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⇒

¤

(b)

CHAPTER 10

PARAMETRIC EQUATIONS AND POLAR COORDINATES

√
(1 + 3 )(6) − 32 (32 )

6 − 34
=
=
= 0 when 6 − 34 = 3(2 − 3 ) = 0 ⇒  = 0 or  = 3 2, so there are
3 )2
3 )2

(1 + 
(1 + 
√ √ 
3
2 3 4 . Using the symmetry from part (a), we see that there are vertical tangents at horizontal tangents at (0 0) and
√ √ 
(0 0) and 3 4 3 2 .

(c) Notice that as  → −1+ , we have  → −∞ and  → ∞. As  → −1− , we have  → ∞ and  → −∞. Also
 − (− − 1) =  +  + 1 = slant asymptote.


(1 + 3 )(3) − 3(32 )
6 − 34

(2 − 3 )
3 − 63


=
=
=
=
=
and from part (b) we have
. So
.
3 )2
3 )2
3 )2

(1 + 
(1 + 

(1 + 


1 − 23
 
 
 
2(1 + 3 )4
1
2 
0 ⇔  √ .
Also 2 =
=
3


3(1 − 23 )3
2

rS

information from parts (a), (b), and (c), we sketch the curve.

al

So the curve is concave upward there and has a minimum point at (0 0)
√ √  and a maximum point at 3 2 3 4 . Using this together with the

e

(d)

( + 1)3
( + 1)2
3 + 32 + (1 + 3 )
→ 0 as  → −1. So  = − − 1 is a
=
= 2
3
3
1+
1+
 −+1

3 
3
3
32
273 + 276
273 (1 + 3 )
273
+
=
=
=
and
1 + 3
1 + 3
(1 + 3 )3
(1 + 3 )3
(1 + 3 )2



32
273
3
=
, so 3 +  3 = 3.
3 = 3
3
3
1+
1+
(1 + 3 )2


Fo

(e) 3 +  3 =

(f ) We start with the equation from part (e) and substitute  =  cos ,  =  sin . Then 3 +  3 = 3

ot

3 cos3  + 3 sin3  = 32 cos  sin . For  6= 0, this gives  =

⇒

3 cos  sin 
. Dividing numerator and denominator cos3  + sin3 



sin 
1
3 cos  cos 
3 sec  tan 
=
by cos3 , we obtain  =
.
1 + tan3  sin3 
1+
cos3 
 
(g) The loop corresponds to  ∈ 0 2 , so its area is

N

478


2



2
1 2 3 sec  tan 
9 2 sec2  tan2 
9 ∞ 2 
 =
 =
 =
2
2 0
1 + tan3 
2 0
(1 + tan3 )2
2 0 (1 + 3 )2
0


= lim 9 − 1 (1 + 3 )−1 0 = 3
2
3
2

=



2

[let  = tan ]

→∞

(h) By symmetry, the area between the folium and the line  = − − 1 is equal to the enclosed area in the third quadrant, plus twice the enclosed area in the fourth quadrant. The area in the third quadrant is 1 , and since  = − − 1 ⇒
2
1
, the area in the fourth quadrant is sin  + cos 

2 
2 

1 −4
1
3 sec  tan 
CAS 1
−
 = . Therefore, the total area is
−
2 −2 sin  + cos 
1 + tan3 
2

 sin  = − cos  − 1 ⇒  = −

1
2

 
+ 2 1 = 3.
2
2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

PROBLEMS PLUS
1.  =



1



cos 
,  =






1

sin 

cos 

sin 
, so by FTC1, we have
=
and
=
. Vertical tangent lines occur when







= 0 ⇔ cos  = 0. The parameter value corresponding to ( ) = (0, 0) is  = 1, so the nearest vertical tangent

occurs when  =

=


.
2



1

Therefore, the arc length between these points is
2






2

+






2

 =



2

1



cos2  sin2 
+ 2  =
2





1

2

2
 
= ln  1 = ln 
2


e

3. In terms of  and , we have  =  cos  = (1 +  sin ) cos  = cos  +  sin  cos  = cos  + 1  sin 2 and
2

−1 ≤  ≤ 2. Furthermore,  = 2 when  = 1 and  =

while  = −1 for  = 0 and  =

3
.
2

Therefore, we need a viewing

rS

rectangle with −1 ≤  ≤ 2.


,
2

al

 =  sin  = (1 +  sin ) sin  = sin  +  sin2 . Now −1 ≤ sin  ≤ 1 ⇒ −1 ≤ sin  +  sin2  ≤ 1 +  ≤ 2, so

To find the -values, look at the equation  = cos  + 1  sin 2 and use the fact that sin 2 ≥ 0 for 0 ≤  ≤
2


2

and

− ≤  ≤
2


2 .]

Fo

sin 2 ≤ 0 for −  ≤  ≤ 0. [Because  = 1 +  sin  is symmetric about the -axis, we only need to consider
2
So for −  ≤  ≤ 0,  has a maximum value when  = 0 and then  = cos  has a maximum value
2



of 1 at  = 0. Thus, the maximum value of  must occur on 0  with  = 1. Then  = cos  +
2
= − sin  + cos 2 = − sin  + 1 − 2 sin2 

⇒




sin 2

⇒

= −(2 sin  − 1)(sin  + 1) = 0 when sin  = −1 or

1
2

ot




1
2

N

[but sin  6= −1 for 0 ≤  ≤  ]. If sin  = 1 , then  =  and
2
2
6
√
√

1

3
 = cos 6 + 2 sin 3 = 4 3. Thus, the maximum value of  is 3 3, and,
4
√ by symmetry, the minimum value is − 3 3. Therefore, the smallest
4
viewing rectangle that contains every member of the family of polar curves
√ 
 √
 = 1 +  sin , where 0 ≤  ≤ 1, is − 3 3 3 3 × [−1 2].
4
4

5. Without loss of generality, assume the hyperbola has equation

2
2
− 2 = 1. Use implicit differentiation to get
2



2 2 0
2 
2 
− 2 = 0, so 0 = 2 . The tangent line at the point ( ) on the hyperbola has equation  −  = 2 ( − ).
2


 
 
The tangent line intersects the asymptote  =


2 

 when  −  = 2 ( − ) ⇒  − 2 2 = 2  − 2 2


 

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⇒

479

¤

CHAPTER 10 PROBLEMS PLUS

 − 2  = 2 2 − 2 2

⇒ =

 + 
  + 
 + 
2 2 − 2 2
=
and the -value is
=
.
( − )






Similarly, the tangent line intersects  = −  at





 −   − 

. The midpoint of these intersection points is



 
 
 

1  + 
 − 
1  + 
 − 
1 2 1 2
+

+
=

= ( ), the point of tangency.
2


2


2  2 

Note: If  = 0, then at (± 0), the tangent line is  = ±, and the points of intersection are clearly equidistant from the point

ot

Fo

rS

al

e

of tangency.

N

480

c
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11

INFINITE SEQUENCES AND SERIES

11.1 Sequences
1. (a) A sequence is an ordered list of numbers. It can also be defined as a function whose domain is the set of positive integers.

(b) The terms  approach 8 as  becomes large. In fact, we can make  as close to 8 as we like by taking  sufficiently large. (c) The terms  become large as  becomes large. In fact, we can make  as large as we like by taking  sufficiently large.

(−1)−1
, so the sequence is
5

7.  =

1
, so the sequence is
( + 1)!

9. 1 = 1, +1 = 5 − 3.





e

5.  =

 

2
4
6
8
10
4 3 8 5




    = 1        .
1 + 1 4 + 1 9 + 1 16 + 1 25 + 1
5 5 17 13

 
1 −1 1 −1 1
1
1 1
1
1
− 
−

 .
 2  3 4  5 =
51 5 5 5 5
5
25 125
625 3125

al

2
, so the sequence is
2 + 1

 

1 1 1 1 1
1 1 1 1
1
     =
  

 .
2! 3! 4! 5! 6!
2 6 24 120 720

rS



3.  =

Each term is defined in terms of the preceding term. 2 = 51 − 3 = 5(1) − 3 = 2.

Fo

3 = 52 − 3 = 5(2) − 3 = 7. 4 = 53 − 3 = 5(7) − 3 = 32. 5 = 54 − 3 = 5(32) − 3 = 157.
The sequence is {1 2 7 32 157   }.

13.

15.
17.



2
4
27
= . The sequence is 2 2  2  2  2     .
=
3 5 7 9
1 + 4
1 + 27
9

N

5 =

23
25

1
2
2
3
2
2
2
. 2 =
=
=
=
= . 3 =
= . 4 =
= .
1 + 
1 + 1
1+2
3
1 + 2
1 + 23
5
1 + 3
1 + 25
7

ot

11. 1 = 2, +1 =

 1 1 1 1

1 3  5  7  9     . The denominator of the nth term is the nth positive odd integer, so  =

1
.
2 − 1



 −1
−3 2 − 4  8  − 16     . The first term is −3 and each term is − 2 times the preceding one, so  = −3 − 2
.
3 9
27
3
3
1
2


 − 4  9  − 16  25     . The numerator of the nth term is 2 and its denominator is  + 1. Including the alternating signs,
3 4
5
6

we get  = (−1)+1

2
.
+1

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

481

CHAPTER 11

INFINITE SEQUENCES AND SERIES

19.



 =

3
1 + 6

1

04286

2

04615

3

04737

4

04800

5

04839

6

04865

7

04884

8

lim

→∞

04918

1

 
 = 1 + − 1
2

2

12500

3

08750

4

10625

5

09688

6

10156

7

09922

8

10039

9

09980

10

10010

05000

al



It appears that lim  = 1.
→∞


  
 
= lim 1 + lim − 1 = 1 + 0 = 1 since lim 1 + − 1
2
2

→∞

→∞

  lim − 1 = 0 by (9).
2

→∞

23.  = 1 − (02) , so lim  = 1 − 0 = 1 by (9) .

Converges

ot

→∞

3 + 52
(3 + 52 )2
5 + 32
5+0
, so  →
= 5 as  → ∞. Converges
=
=
 + 2
( + 2 )2
1 + 1
1+0

N

25.  =

→∞

Fo

21.

3
(3)
3
3
1
= lim
= lim
= =
1 + 6 →∞ (1 + 6) →∞ 1 + 6
6
2

04909

10

→∞

04898

9

It appears that lim  = 05.

e

¤

rS

482

27. Because the natural exponential function is continuous at 0, Theorem 7 enables us to write

lim  = lim 1 = lim→∞ (1) = 0 = 1 Converges

→∞

→∞

(2)
2
2

2
, then lim  = lim
= lim
=
= . Since tan is continuous at
→∞
→∞ (1 + 8)
→∞ 1 + 8
1 + 8
8
4




2
2

Theorem 7, lim tan
= tan lim
= tan = 1. Converges
→∞
→∞ 1 + 8
1 + 8
4

29. If  =


,
4

√
√
√

2
2  3
√
= 
, so  → ∞ as  → ∞ since lim  = ∞ and
= √
→∞
3 + 4
1 + 42
3 + 4 3

31.  = √

lim

→∞


1 + 42 = 1. Diverges




 (−1)  1
1
1 lim 33. lim | | = lim  √  =
= (0) = 0, so lim  = 0 by (6).
→∞
→∞  2  
→∞
2 →∞ 12
2

Converges

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

by

SECTION 11.1

SEQUENCES

¤

483

This sequence diverges since the terms don’t approach any particular real number as  → ∞.

35.  = cos( 2).

The terms take on values between −1 and 1
37.  =

(2 − 1)!
1
(2 − 1)!
=
=
→ 0 as  → ∞. Converges
(2 + 1)!
(2 + 1)(2)(2 − 1)!
(2 + 1)(2)

39.  =

 + − −
1 + −2
· − = 
→ 0 as  → ∞ because 1 + −2 → 1 and  − − → ∞. Converges
2 − 1 
 − −

43. 0 ≤

2 H
2 H
2
2
. Since lim  = lim  = lim  = 0, it follows from Theorem 3 that lim  = 0. Converges

→∞ 
→∞ 
→∞ 
→∞


cos2 
1
≤ 
2
2

45.  =  sin(1) =

1
= 0,
→∞ 2

[since 0 ≤ cos2  ≤ 1], so since lim



cos2 
2



converges to 0 by the Squeeze Theorem.

sin(1) sin  sin(1) . Since lim
= lim
[where  = 1] = 1, it follows from Theorem 3
→∞
1
1

→0+

e

41.  = 2 − =

al

that { } converges to 1.


2
1+


Fo

rS



2
⇒ ln  =  ln 1 +
, so




1
2
− 2
1 + 2

ln(1 + 2) H
2
=2 ⇒
= lim
= lim lim ln  = lim
2
→∞
→∞
→∞
→∞ 1 + 2
1
−1




2
2 lim 1 +
= lim ln  = 2 , so by Theorem 3, lim 1 +
= 2 . Converges
→∞
→∞
→∞



47.  =

49.  = ln(22 + 1) − ln(2 + 1) = ln



22 + 1
2 + 1



= ln



2 + 12
1 + 12

ot

51.  = arctan(ln ). Let  () = arctan(ln ). Then lim  () =

Thus, lim  = lim  () =
→∞


.
2


2

→ ln 2 as  → ∞. Converges

since ln  → ∞ as  → ∞ and arctan is continuous.

Converges

N

→∞

→∞



53. {0 1 0 0 1 0 0 0 1   } diverges since the sequence takes on only two values, 0 and 1, and never stays arbitrarily close to

either one (or any other value) for  sufficiently large.
55.  =
57.

( − 1) 
1 
!
1 2 3
· ≥ ·
= · · · ··· ·
2
2 2 2
2
2
2 2

[for   1] =


→ ∞ as  → ∞, so { } diverges.
4

From the graph, it appears that the sequence converges to 1.
{(−2) } converges to 0 by (7), and hence {1 + (−2) } converges to 1 + 0 = 1.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

484

¤

CHAPTER 11

INFINITE SEQUENCES AND SERIES

From the graph, it appears that the sequence converges to 1 .
2

59.

As  → ∞,


3 + 22
32 + 2
 =
=
2 +
8
8 + 1

⇒



0+2
=
8+0



1
1
= ,
4
2

so lim  = 1 .
2
→∞

From the graph, it appears that the sequence { } =

61.



2 cos 
1 + 2



is

divergent, since it oscillates between 1 and −1 (approximately). To

{ } converges to 1, and lim lim 


=
= cos , so
= . But

1


al

→∞

 does not exist. This contradiction shows that { } diverges.


rS

→∞

2
, then
1 + 2

e

prove this, suppose that { } converges to . If  =

From the graph, it appears that the sequence approaches 0.

63.

0   =

1 · 3 · 5 · · · · · (2 − 1)
1
3
5
2 − 1
=
·
·
· ··· ·
(2)
2 2 2
2

1
1
· (1) · (1) · · · · · (1) =
→ 0 as  → ∞
2
2


1 · 3 · 5 · · · · · (2 − 1) converges to 0.
So by the Squeeze Theorem,

(2)

⇒ 1 = 1060, 2 = 112360, 3 = 119102, 4 = 126248, and 5 = 133823.

ot

65. (a)  = 1000(106)

Fo

≤

(b) lim  = 1000 lim (106) , so the sequence diverges by (9) with  = 106  1.
→∞

→∞

N

67. (a) We are given that the initial population is 5000, so 0 = 5000. The number of catfish increases by 8% per month and is

decreased by 300 per month, so 1 = 0 + 8%0 − 300 = 1080 − 300, 2 = 1081 − 300, and so on. Thus,
 = 108−1 − 300.
(b) Using the recursive formula with 0 = 5000, we get 1 = 5100, 2 = 5208, 3 = 5325 (rounding any portion of a catfish), 4 = 5451, 5 = 5587, and 6 = 5734, which is the number of catfish in the pond after six months.
69. If || ≥ 1, then { } diverges by (9), so { } diverges also, since | | =  | | ≥ | |. If ||  1 then

lim  = lim

→∞

→∞

 H
1

= lim
= lim
= 0, so lim  = 0, and hence { } converges
→∞ (− ln )  −
→∞ − ln 
→∞
−

whenever ||  1.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.1

¤

SEQUENCES

485

71. Since { } is a decreasing sequence,   +1 for all  ≥ 1. Because all of its terms lie between 5 and 8, { } is a

bounded sequence. By the Monotonic Sequence Theorem, { } is convergent; that is, { } has a limit .  must be less than
8 since { } is decreasing, so 5 ≤   8.
73.  =

1
1
1
1
is decreasing since +1 =
=

=  for each  ≥ 1. The sequence is
2 + 3
2( + 1) + 3
2 + 5
2 + 3

bounded since 0   ≤

1
5

for all  ≥ 1. Note that 1 = 1 .
5

75. The terms of  = (−1) alternate in sign, so the sequence is not monotonic. The first five terms are −1, 2, −3, 4, and −5.

Since lim | | = lim  = ∞, the sequence is not bounded.
→∞


(2 + 1)(1) − (2)
1 − 2

= 2
≤0
defines a decreasing sequence since for  () = 2
,  0 () =
2 + 1
 +1
(2 + 1)2
( + 1)2

for  ≥ 1. The sequence is bounded since 0   ≤

for all  ≥ 1.

)

lim  = lim 21−(12

→∞

→∞

= 21 = 2.

al



 
√  √
√



2, 2 2, 2 2 2,    , 1 = 212 , 2 = 234 , 3 = 278 ,   , so  = 2(2 −1)2 = 21−(12 ) .

rS

79. For

1
2

e

77.  =

→∞

Alternate solution: Let  = lim  . (We could show the limit exists by showing that { } is bounded and increasing.)
→∞

81. 1 = 1, +1 = 3 −

Fo

√
Then  must satisfy  = 2 ·  ⇒ 2 = 2 ⇒ ( − 2) = 0.  6= 0 since the sequence increases, so  = 2.
1
. We show by induction that { } is increasing and bounded above by 3. Let  be the proposition


−

ot

that +1   and 0    3. Clearly 1 is true. Assume that  is true. Then +1  
1
1
1
1
 − . Now +2 = 3 −
3−
= +1
+1

+1


⇒

1
1

+1


⇒

⇔ +1 . This proves that { } is increasing and bounded

N

above by 3, so 1 = 1    3, that is, { } is bounded, and hence convergent by the Monotonic Sequence Theorem.
If  = lim  , then lim +1 =  also, so  must satisfy  = 3 − 1 ⇒ 2 − 3 + 1 = 0 ⇒  =
→∞

→∞

But   1, so  =

√
3± 5
.
2

√
3+ 5
.
2

83. (a) Let  be the number of rabbit pairs in the nth month. Clearly 1 = 1 = 2 . In the nth month, each pair that is

2 or more months old (that is, −2 pairs) will produce a new pair to add to the −1 pairs already present. Thus,
 = −1 + −2 , so that { } = { }, the Fibonacci sequence.
(b)  =

+1


⇒ −1 =


−1 + −2
−2
1
1
=
=1+
=1+
=1+
. If  = lim  ,
→∞
−1
−1
−1
−1 /−2
−2

then  = lim −1 and  = lim −2 , so  must satisfy  = 1 +
→∞

→∞

1


⇒ 2 −  − 1 = 0 ⇒  =

[since  must be positive]. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

√
1+ 5
2

486

¤

CHAPTER 11

INFINITE SEQUENCES AND SERIES

85. (a)

From the graph, it appears that the sequence



5
!



5
= 0.
→∞ !

converges to 0, that is, lim

al

e

(b)

rS

From the first graph, it seems that the smallest possible value of  corresponding to  = 01 is 9, since 5 !  01 whenever  ≥ 10, but 95 9!  01. From the second graph, it seems that for  = 0001, the smallest possible value for 

Fo

is 11 since 5 !  0001 whenever  ≥ 12.

87. Theorem 6: If lim | | = 0 then lim − | | = 0, and since − | | ≤  ≤ | |, we have that lim  = 0 by the
→∞

→∞

Squeeze Theorem.

→∞

89. To Prove: If lim  = 0 and { } is bounded, then lim (  ) = 0.
→∞

ot

→∞

N

Proof: Since { } is bounded, there is a positive number  such that | | ≤  and hence, | | | | ≤ | |  for all  ≥ 1. Let   0 be given. Since lim  = 0, there is an integer  such that | − 0| 
→∞

|  − 0| = |  | = | | | | ≤ | |  = | − 0|  

 if   . Then



·  =  for all   . Since  was arbitrary,


lim (  ) = 0.

→∞

91. (a) First we show that   1  1  .

1 − 1 =

+
2

−


√

√
√
√ 2
 = 1  − 2  +  = 1
 −   0 [since   ] ⇒ 1  1 . Also
2
2

 − 1 =  − 1 ( + ) = 1 ( − )  0 and  − 1 =  −
2
2

√
√ √
√ 
 =   −   0, so   1  1  . In the same

way we can show that 1  2  2  1 and so the given assertion is true for  = 1. Suppose it is true for  = , that is,
  +1  +1   . Then

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.2

+2 − +2 = 1 (+1 + +1 ) −
2


+1 +1

SERIES

¤

487



√
2


= 1 +1 − 2 +1 +1 + +1 = 1
+1 − +1  0,
2
2

+1 − +2 = +1 − 1 (+1 + +1 ) = 1 (+1 − +1 )  0, and
2
2
+1 − +2 = +1 −





√
+1 +1 = +1
+1 − +1  0

+1  +2  +2  +1 ,

⇒

so the assertion is true for  =  + 1. Thus, it is true for all  by mathematical induction.
(b) From part (a) we have     +1  +1    , which shows that both sequences, { } and { }, are monotonic and bounded. So they are both convergent by the Monotonic Sequence Theorem.
(c) Let lim  =  and lim  = . Then lim +1 = lim
→∞

→∞

→∞

 + 
2

⇒ =

+
2

⇒

⇒  = .

93. (a) Suppose { } converges to . Then +1 =


 + 

 lim 
⇒

lim +1 =

→∞

 + lim 

e

2 =  + 

→∞

→∞

→∞

⇒

=


+

⇒

( +  − ) = 0 ⇒  = 0 or  =  − .
 

 





(b) +1 =
=
    since 1 +   1.
 + 
1+

 
 
 2
 
 3
 






0 , 2 
1 
2 
(c) By part (b), 1 
0 , 3 
0 , etc. In general,  
0 ,






 




· 0 = 0 since   . By (7) lim  = 0 if − 1    1. Here  = ∈ (0 1) . so lim  ≤ lim
→∞
→∞
→∞



al

⇒

Fo

rS

2 +  = 

(d) Let   . We first show, by induction, that if 0   − , then    −  and +1   .
0
0 ( −  − 0 )
− 0 =
 0 since 0   − . So 1  0 .
 + 0
 + 0

ot

For  = 0, we have 1 − 0 =

Now we suppose the assertion is true for  = , that is,    −  and +1   . Then

( − ) +  −  − 
( −  −  )
=
=
 0 because    − . So
 + 
 + 
 + 

N

 −  − +1 =  −  −

+1   − . And +2 − +1 =

+1
+1 ( −  − +1 )
− +1 =
 0 since +1   − . Therefore,
 + +1
 + +1

+2  +1 . Thus, the assertion is true for  =  + 1. It is therefore true for all  by mathematical induction.
A similar proof by induction shows that if 0   − , then    −  and { } is decreasing.
In either case the sequence { } is bounded and monotonic, so it is convergent by the Monotonic Sequence Theorem.
It then follows from part (a) that lim  =  − .
→∞

11.2 Series
1. (a) A sequence is an ordered list of numbers whereas a series is the sum of a list of numbers.

(b) A series is convergent if the sequence of partial sums is a convergent sequence. A series is divergent if it is not convergent. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

488

3.

¤
∞


=1

5. For

CHAPTER 11

INFINITE SEQUENCES AND SERIES

 = lim  = lim [2 − 3(08) ] = lim 2 − 3 lim (08) = 2 − 3(0) = 2
→∞

→∞

→∞

→∞

∞
 1
1
1
1
,  = 3 . 1 = 1 = 3 = 1, 2 = 1 + 2 = 1 + 3 = 1125, 3 = 2 + 3 ≈ 11620,
3

1
2
=1 

4 = 3 + 4 ≈ 11777, 5 = 4 + 5 ≈ 11857, 6 = 5 + 6 ≈ 11903, 7 = 6 + 7 ≈ 11932, and
8 = 7 + 8 ≈ 11952. It appears that the series is convergent.
7. For

∞


=1



1
2
√ = 05, 2 = 1 + 2 = 05 +
√ ≈ 13284,
√ ,  =
√ . 1 = 1 =
1+ 
1+ 
1+ 1
1+ 2

3 = 2 + 3 ≈ 24265, 4 = 3 + 4 ≈ 37598, 5 = 4 + 5 ≈ 53049, 6 = 5 + 6 ≈ 70443,
7 = 6 + 7 ≈ 89644, 8 = 7 + 8 ≈ 110540. It appears that the series is divergent.
9.



1

−240000
−192000

3

al

2

e



−201600

4

−199680
−200064

6

rS

5

−199987
−200003

From the graph and the table, it seems that the series converges to −2. In fact, it is a geometric

−200000

8

series with  = −24 and  = − 1 , so its sum is
5

−199999

9
10

−200000

Fo

7

∞


=1

12
−24
−24
  =
= −2
=
(−5)
12
1 − −1
5

Note that the dot corresponding to  = 1 is part of both { } and { }.

TI-86 Note: To graph { } and { }, set your calculator to Param mode and DrawDot mode. (DrawDot is under

ot

GRAPH, MORE, FORMT (F3).) Now under E(t)= make the assignments: xt1=t, yt1=12/(-5)ˆt, xt2=t, yt2=sum seq(yt1,t,1,t,1). (sum and seq are under LIST, OPS (F5), MORE.) Under WIND use

11.

N

1,10,1,0,10,1,-3,1,1 to obtain a graph similar to the one above. Then use TRACE (F4) to see the values.





1

044721

2

115432

3

198637

4

288080

5

380927

6

475796

7

571948

8

668962

9

766581

10

864639

The series

∞


=1


√
diverges, since its terms do not approach 0.
2 +4


c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.2

SERIES

¤

13.



1

029289

2

042265

3

050000

4

055279

5

059175

6

062204

7

064645

8

066667

9

068377

10

069849

From the graph and the table, it seems that the series converges.
 

 





1
1
1
1
1
1
1
1
√ −√
= √ −√
+ √ −√
+··· + √ − √

+1
+1
1
2
2
3

=1
1
= 1− √
,
+1




∞

1
1
1
√ −√
= 1.
= lim 1 − √ so →∞

+1
+1
=1

→∞

→∞

(b) Since lim  =
→∞

17. 3 − 4 +

16
3

−

64
9

2
2
= , so the sequence { } is convergent by (11.1.1).
3 + 1
3
2
3

6= 0, the series

∞


 is divergent by the Test for Divergence.

=1

rS

15. (a) lim  = lim

al

e



+ · · · is a geometric series with ratio  = − 4 . Since || =
3

4
3

 1, the series diverges.

Fo

2
19. 10 − 2 + 04 − 008 + · · · is a geometric series with ratio − 10 = − 1 . Since || =
5

1
5

 1, the series converges to


10
10
50
25
=
=
=
=
.
1−
1 − (−15)
65
6
3
∞


=1

6(09)−1 is a geometric series with first term  = 6 and ratio  = 09. Since || = 09  1, the series converges to

ot

21.

23.

N


6
6
=
=
= 60.
1−
1 − 09
01


−1
∞
∞
 (−3)−1
1 
3
=
. The latter series is geometric with  = 1 and ratio  = − 3 . Since || =
−
4
4
4 =1
4
=1 converges to

25.

∞


=0

27.

3
4

 1, it

  
1
= 4 . Thus, the given series converges to 1 4 = 1 .
7
4
7
7
1 − (−34)

∞

1    

= is a geometric series with ratio  = . Since ||  1, the series diverges.
3+1
3 =0 3
3

∞
∞
 1
1
1
1
1
1
1  1
+ + +
+
+ ··· =
=
. This is a constant multiple of the divergent harmonic series, so
3
6
9
12
15
3 =1 
=1 3

it diverges.
29.

∞
 −1
−1
1 diverges by the Test for Divergence since lim  = lim
= 6= 0.
→∞
→∞ 3 − 1
3
=1 3 − 1 c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

489

490

¤

CHAPTER 11

INFINITE SEQUENCES AND SERIES

31. Converges.



    
∞
∞
∞
 1 + 2
 1

2
2
1
=
+  =
+

3
3
3
3
=1
=1 3
=1
=

33.

[sum of two convergent geometric series]

23
1
5
13
+
= +2=
1 − 13
1 − 23
2
2

∞
√
√
√
 √

2 = 2 + 2 + 3 2 + 4 2 + · · · diverges by the Test for Divergence since

=1

lim  = lim

→∞

→∞

√

2 = lim 21 = 20 = 1 6= 0.
→∞

∞
   

39.

∞


is a geometric series with ratio  =

3

=0


3

≈ 1047. It diverges because || ≥ 1.

arctan  diverges by the Test for Divergence since lim  = lim arctan  =
→∞

∞
∞
 1

=

=1
=1

→∞


2

6= 0.

rS

=1

41.

al

37.

e

 2

∞

 +1
35.
ln diverges by the Test for Divergence since
22 + 1
=1



 2
2 + 1
 +1
= ln lim
= ln 1 6= 0. lim  = lim ln
2
→∞
→∞
→∞ 22 + 1
22 + 1

 
1
1
1
1 is a geometric series with first term  = and ratio  = . Since || =  1, the series converges





∞

1
1
1

1
=
· =
. By Example 7,
= 1. Thus, by Theorem 8(ii),
1 − 1
1 − 1 
−1
=1 ( + 1)


∞
∞
∞

 1

1
1
1
1
1
−1

+
+
=
=
+1 =
+
=
.



( + 1)
−1
−1
−1
−1
=1
=1 
=1 ( + 1)

Fo

to

∞


2 are 2 − 1
=2






2
1
1
=
−
 =
−1
+1
=2 ( − 1)( + 1)
=2
 
 


 


1
1 1
1
1
1
1
1
1
+
−
+
−
+ ··· +
−
+
−
= 1−
3
2
4
3
5
−3
−1
−2


N

ot

43. Using partial fractions, the partial sums of the series

1
1
1
−
− .
2
−1



∞

2
1
1
1
3
= lim  = lim 1 + −
−
= .
Thus,
→∞
2 − 1 →∞
2
−1

2
=2
This sum is a telescoping series and  = 1 +






 1
3
3
1
,  =
=
−
[using partial fractions]. The latter sum is
+3
=1 ( + 3)
=1 ( + 3)
=1 

 
 
 

 
 
 

1
1
1
1
1
1
1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + · · · + −3 −  +  1 −  + 1 + −1 − +2 +  −
4
2
5
3
6
4
7
−2

45. For the series

∞


=1+

Thus,


3
= lim  = lim 1 +
→∞
→∞
=1 ( + 3)
∞


1
2

+

1
3

−

1
 +1

1
2

−

+

1
3

−

1
+2

1
 +1

−

1
+3

−


1
+2

−

=1+

1
2

1
+3

+

1
3

[telescoping series]

=

11
.
6

Converges

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

1
+3



SECTION 11.2

SERIES

¤

491


∞
  1

47. For the series
− 1(+1) ,
=1





  1
 =
 − 1(+1) = (1 − 12 ) + (12 − 13 ) + · · · + 1 − 1(+1) =  − 1(+1)
=1

[telescoping series]




∞
  1
Thus,

− 1(+1) = lim  = lim  − 1(+1) =  − 0 =  − 1. Converges
→∞

=1

→∞

49. (a) Many people would guess that   1, but note that  consists of an infinite number of 9s.

(b)  = 099999    =
 = 01. Its sum is

∞
 9
9
9
9
9
+
+
+
+ ··· =
, which is a geometric series with 1 = 09 and

10
100
1000
10,000
=1 10

09
09
=
= 1, that is,  = 1.
1 − 01
09

(c) The number 1 has two decimal representations, 100000    and 099999    .

8
1

810
8
8
8
+ 2 + · · · is a geometric series with  = and  =
. It converges to
=
= .
10
10
10
10
1−
1 − 110
9

53. 2516 = 2 +

rS

51. 08 =

al

example, 05 can be written as 049999    as well as 050000    .

e

(d) Except for 0, all rational numbers that have a terminating decimal representation can be written in more than one way. For

516
516
516
516
516
1
+ 6 + · · · . Now 3 + 6 + · · · is a geometric series with  = 3 and  = 3 . It converges to
103
10
10
10
10
10

55. 15342 = 153 +

42
42
42
42
42
1
+ 6 + · · · . Now 4 + 6 + · · · is a geometric series with  = 4 and  = 2 .
104
10
10
10
10
10


42104
42104
42
=
.
=
=
2
1−
1 − 110
99102
9900

Thus, 15342 = 153 +
∞


(−5)  =

=1

∞


(−5) is a geometric series with  = −5, so the series converges ⇔ ||  1 ⇔

N

57.

153
42
15,147
42
15,189
5063
42
=
+
=
+
=
or
.
9900
100
9900
9900
9900
9900
3300

ot

It converges to

Fo

516103
516103

516
516
2514
838
=
=
=
. Thus, 2516 = 2 +
=
=
.
3
1−
1 − 110
999103
999
999
999
333

=1

|−5|  1 ⇔ ||  1 , that is, − 1    1 . In that case, the sum of the series is
5
5
5
59.


−5
−5
=
=
.
1−
1 − (−5)
1 + 5



∞
∞
 ( − 2)
 −2 
−2
, so the series converges ⇔ ||  1 ⇔
=
is a geometric series with  =
3
3
3
=0
=0


 − 2
−2


 3   1 ⇔ −1  3  1 ⇔ −3   − 2  3 ⇔ −1    5. In that case, the sum of the series is

3
1
1
=
=
.
−2
3 − ( − 2)
5−
1−
3
3
 

∞
∞
 2
 2
2
61.
=
is a geometric series with  = , so the series converges ⇔ ||  1 ⇔



=0
=0

=
1−

2  ||

⇔

  2 or   −2. In that case, the sum of the series is

1


=
=
.
1−
1 − 2
−2

 
2
 1 ⇔


c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

492

63.

¤

CHAPTER 11

∞


 =

=0

∞


=0

INFINITE SEQUENCES AND SERIES

( ) is a geometric series with  =  , so the series converges ⇔ ||  1 ⇔ | |  1 ⇔

−1    1 ⇔ 0    1 ⇔   0. In that case, the sum of the series is

1

=
.
1−
1 − 

65. After defining , We use convert(f,parfrac); in Maple, Apart in Mathematica, or Expand Rational and

1
1
32 + 3 + 1
= 3 −
. So the nth partial sum is
(2 + )3

( + 1)3

 
 




 1
1
1
1
1
1
1
1
 =
−
− 3 + ··· +
−
= 1− 3 +
=1−
3
( + 1)3
2
23
3
3
( + 1)3
( + 1)3
=1

Simplify in Derive to find that the general term is

The series converges to lim  = 1. This can be confirmed by directly computing the sum using
→∞

sum(f,n=1..infinity); (in Maple), Sum[f,{n,1,Infinity}] (in Mathematica), or Calculus Sum

e

(from 1 to ∞) and Simplify (in Derive).
67. For  = 1, 1 = 0 since 1 = 0. For   1,

∞


=1

 = lim  = lim
→∞

→∞

1 − 1
= 1.
1 + 1

al

Also,

( − 1) − 1
( − 1) − ( + 1)( − 2)
2
−1
−
=
=
+1
( − 1) + 1
( + 1)
( + 1)

rS

 =  − −1 =

69. (a) The quantity of the drug in the body after the first tablet is 150 mg. After the second tablet, there is 150 mg plus 5%

Fo

of the first 150- mg tablet, that is, [150 + 150(005)] mg. After the third tablet, the quantity is
[150 + 150(005) + 150(005)2 ] = 157875 mg. After  tablets, the quantity (in mg) is
3000
150(1 − 005 )
=
(1 − 005 ).
1 − 005
19


(b) The number of milligrams remaining in the body in the long run is lim 3000 (1 − 005 ) = 3000 (1 − 0) ≈ 157895,
19
19

ot

150 + 150(005) + · · · + 150(005)−1 . We can use Formula 3 to write this as
→∞

only 002 mg more than the amount after 3 tablets.

N

71. (a) The first step in the chain occurs when the local government spends  dollars. The people who receive it spend a

fraction  of those  dollars, that is,  dollars. Those who receive the  dollars spend a fraction  of it, that is,
2 dollars. Continuing in this way, we see that the total spending after  transactions is
 =  +  + 2 + · · · + –1 =
(b) lim  = lim
→∞

→∞


=


(1 −  ) by (3).
1−


(1 −  )

= lim (1 −  ) =
1−
1 −  →∞
1−

[since  +  = 1] = 

[since  = 1]

 since 0    1

⇒

 lim  = 0

→∞

If  = 08, then  = 1 −  = 02 and the multiplier is  = 1 = 5.
73.

∞


(1 + )− is a geometric series with  = (1 + )−2 and  = (1 + )−1 , so the series converges when

=2



(1 + )−1   1 ⇔ |1 + |  1 ⇔ 1 +   1 or 1 +   −1 ⇔   0 or   −2. We calculate the sum of the c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.2

series and set it equal to 2:

(1 + )−2
=2 ⇔
1 − (1 + )−1

22 + 2 − 1 = 0 ⇔  =
So  =

√

−2 ±
4

√
12

=

√
± 3−1
.
2



1
1+

SERIES

¤

493



1
=2−2
⇔ 1 = 2(1 + )2 − 2(1 + ) ⇔
1+

However, the negative root is inadmissible because −2 

√
− 3−1
2

 0.

3−1
.
2
1

1



1

75.  = 1+ 2 + 3 +···+  = 1 12 13 · · · 1  (1 + 1) 1 +

=

2

234
+1
···
=+1
123


1
2





1
1 + 1 ··· 1 + 
3

[  1 + ]

Thus,    + 1 and lim  = ∞. Since { } is increasing, lim  = ∞, implying that the harmonic series is
→∞

→∞

divergent.
77. Let  be the diameter of  . We draw lines from the centers of the  to

al

e

the center of  (or ), and using the Pythagorean Theorem, we can write
2 
2

⇔
12 + 1 − 1 1 = 1 + 1 1
2
2

2 
2

1 = 1 + 1 1 − 1 − 1 1 = 21 [difference of squares] ⇒ 1 = 1 .
2
2
2

= (2 − 1 )(1 + 2 )

+1

⇔

2 
2

1
(1 − 1 )2
− 1 =
, 1 = 1 + 1 3 − 1 − 1 − 2 − 1 3
2
2
2 − 1
2 − 1

Fo

2 =

rS

Similarly,
2 
2

1 = 1 + 1 2 − 1 − 1 − 1 2 = 22 + 21 − 2 − 1 2
1
2
2

⇔ 3 =

[1 − (1 + 2 )]2
, and in general,
2 − (1 + 2 )

2


1 −  
1
1
=1
 and =
. If we actually calculate 2 and 3 from the formulas above, we find that they are =
2 −  
6
2·3
=1


1
1

1
1
= −
. Then
=
 = 1 −
( + 1)

+1
+1
+1
=1

N

 ≤ ,  =

ot

1
1
1
=
respectively, so we suspect that in general,  =
. To prove this, we use induction: Assume that for all
12
3·4
( + 1)
[telescoping sum]. Substituting this into our


2
1

1−
+1
( + 1)2
1
 =

= formula for +1 , we get +1 =
, and the induction is complete.
+2
( + 1)( + 2)

2−
+1
+1

Now, we observe that the partial sums   of the diameters of the circles approach 1 as  → ∞; that is,
=1
∞


 =

=1

∞


=1

1
= 1, which is what we wanted to prove.
( + 1)

79. The series 1 − 1 + 1 − 1 + 1 − 1 + · · · diverges (geometric series with  = −1) so we cannot say that

0 = 1 −1 + 1 −1 + 1 −1 + ···.
81.

∞

=1

 = lim

→∞



=1

 = lim 
→∞



=1

 =  lim

→∞



=1

 = 

∞

=1

 , which exists by hypothesis.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

494

¤

CHAPTER 11

INFINITE SEQUENCES AND SERIES




( +  ) converges. Then ( +  ) and  are convergent series. So by



Theorem 8(iii), [( +  ) −  ] would also be convergent. But [( +  ) −  ] =
 , a contradiction, since

 is given to be divergent.

83. Suppose on the contrary that

85. The partial sums { } form an increasing sequence, since  − −1 =   0 for all . Also, the sequence { } is bounded

since  ≤ 1000 for all . So by the Monotonic Sequence Theorem, the sequence of partial sums converges, that is, the series

 is convergent.

87. (a) At the first step, only the interval
9



8
9

3




 2 (length 1 ) is removed. At the second step, we remove the intervals 1  2 and
3
3
9 9

 3

 2
, which have a total length of 2 · 1 . At the third step, we remove 22 intervals, each of length 1 . In general,
3
3

at the nth step we remove 2−1 intervals, each of length
∞


=1

1
3

 2 −1
3

=

13
1 − 23

the th step, the leftmost interval that is removed is

3

, for a length of 2−1 ·

 1 
3


= 1 geometric series with  =

=

1
3

1
3

 2 −1
3

and  =

2
3

. Thus, the total


. Notice that at

 1   2  
, so we never remove 0, and 0 is in the Cantor set. Also,
 3
3

al

length of all removed intervals is

 1 

e

7

1

rS


 
   the rightmost interval removed is 1 − 2  1 − 1
, so 1 is never removed. Some other numbers in the Cantor set
3
3 are 1 , 2 , 1 , 2 , 7 , and 8 .
3 3 9 9 9
9

(b) The area removed at the first step is 1 ; at the second step, 8 ·
9
 1 

 
∞
 1 8 −1
19
= 1.
=
9 9
1 − 89
=1
∞


=1

1
9

 8 −1

 1 3
9

. In general, the area

9

, so the total area of all removed squares is

1
1
5

1
2
5
3
23
, 1 =
= , 2 = +
= , 3 = +
=
,
( + 1)!
1·2
2
2
1·2·3
6
6
1·2·3·4
24

4
119
23
( + 1)! − 1
+
=
. The denominators are ( + 1)!, so a guess would be  =
.
24
1·2·3·4·5
120
( + 1)!

N

4 =

=

; at the third step, (8)2 ·

ot

89. (a) For

9

9

Fo

removed at the th step is (8)−1

 1 2

(b) For  = 1, 1 =

+1 =
=

( + 1)! − 1
1
2! − 1
=
, so the formula holds for  = 1. Assume  =
. Then
2
2!
( + 1)!
+1
( + 1)! − 1
+1
( + 2)! − ( + 2) +  + 1
( + 1)! − 1
+
=
+
=
( + 1)!
( + 2)!
( + 1)!
( + 1)!( + 2)
( + 2)!
( + 2)! − 1
( + 2)!

Thus, the formula is true for  =  + 1. So by induction, the guess is correct.


∞

( + 1)! − 1

1
(c) lim  = lim
= lim 1 −
= 1 and so
= 1.
→∞
→∞
→∞
( + 1)!
( + 1)!
=1 ( + 1)!

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS

¤

495

11.3 The Integral Test and Estimates of Sums
1. The picture shows that 2 =

3 =

1
313





3

2

1
13

1
213





2

1

, and so on, so

1
,
13
∞


=2



1

13

∞

1

1
. The
13

integral converges by (7.8.2) with  = 13  1, so the series converges.

√
5

3. The function  () = 1  = −15 is continuous, positive, and decreasing on [1 ∞), so the Integral Test applies.

∞
1


→∞ 1

−15  = lim

−15  = lim

→∞



5 45

4



= lim

1

→∞



5 45

4

−

5
4



= ∞, so

∞


√
1 5  diverges.

=1

1 is continuous, positive, and decreasing on [1 ∞), so the Integral Test applies.
(2 + 1)3




 
1
1
1
1
1
1
1
 = lim
 = lim −
= lim −
+
=
.
→∞ 1 (2 + 1)3
→∞
(2 + 1)3
4 (2 + 1)2 1 →∞
4(2 + 1)2
36
36

∞

1

Since this improper integral is convergent, the series

∞


 is continuous, positive, and decreasing on [1 ∞), so the Integral Test applies.
2 + 1


 


1
1
 = lim
 = lim ln(2 + 1) = lim [ln(2 + 1) − ln 2] = ∞. Since this improper
→∞ 1 2 + 1
→∞ 2
2 + 1
2 →∞
1

∞

1

Fo

7. The function  () =



integral is divergent, the series

∞


=1

 is also divergent by the Integral Test.
2 + 1

√
1
√ is a -series with  =
2  1, so it converges by (1).
2
=1 

11. 1 +

13. 1 +

ot

∞


∞
 1
1
1
1
1
+
+
+
+··· =
. This is a -series with  = 3  1, so it converges by (1).
3
8
27
64
125
=1 

N

9.

1 is also convergent by the Integral Test.
(2 + 1)3

rS

=1

al



e

5. The function  () =

∞

1
1
1
1
1
1
+ + + + ··· =
. The function () = is 3
5
7
9
2 − 1
=1 2 − 1

continuous, positive, and decreasing on [1 ∞), so the Integral Test applies.
 
 ∞
∞



1
1
1
 = lim
 = lim 1 ln |2 − 1| 1 = 1 lim (ln(2 − 1) − 0) = ∞, so the series
2
2 →∞
→∞ 1 2 − 1
→∞
2 − 1
=1 2 − 1
1
diverges.

√

√
∞
∞
∞

 1
 4
+4

4
15.
=
+ 2 =
+
.
2
2
2
32



=1
=1
=1 
=1 
∞


∞


1 is a convergent -series with  =
32
=1

3
2

 1.

∞
∞
 4
 1
=4
is a constant multiple of a convergent -series with  = 2  1, so it converges. The sum of two
2
2
=1 
=1 

convergent series is convergent, so the original series is convergent.

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¤

496

CHAPTER 11

INFINITE SEQUENCES AND SERIES

1 is continuous, positive, and decreasing on [1 ∞), so we can apply the Integral Test.
2 + 4


 
 

 
1
1
1

1

1
= lim tan−1
 = lim
 = lim tan−1 − tan−1
→∞ 1 2 + 4
→∞ 2
2 + 4
2 1
2 →∞
2
2
 

1
1 
=
− tan−1
2 2
2

17. The function  () =



1

∞

Therefore, the series

∞


=1

19.

1 converges. 2 + 4

∞
∞
 ln 
 ln  ln 1 ln 
=
since
= 0. The function  () = 3 is continuous and positive on [2 ∞).
3
3
1

=1 
=2 

 0 () =

3 (1) − (ln )(32 )
2 − 32 ln 
1 − 3 ln 
=
=
 0 ⇔ 1 − 3 ln   0 ⇔ ln  
(3 )2
6
4

1
3

⇔

al

e

  13 ≈ 14, so  is decreasing on [2 ∞), and the Integral Test applies.




 
 ∞
∞
 ln  ln  ln 
()
ln 
1
1
1 () 1
 = lim
 = lim − 2 − 2 = lim − 2 (2 ln  + 1) +
= , so the series
3
→∞ 2
→∞
3
3
2
4 1 →∞
4
4
4
=2 
2

converges.

rS

():  = ln ,  = −3  ⇒  = (1) ,  = − 1 −2 , so
2


 ln 
 = − 1 −2 ln  − − 1 −2 (1)  = − 1 −2 ln  + 1 −3  = − 1 −2 ln  − 1 −2 + 
2
2
2
2
2
4
3


2
1
2 ln  + 1 H
−
= − 1 lim 2 = 0.
= − lim
4 →∞
→∞
→∞ 8
42


Fo

(): lim

1
1 + ln  is continuous and positive on [2 ∞), and also decreasing since  0 () = − 2
 0 for   2, so we can
 ln 
 (ln )2
 ∞
∞
 1
1
 = lim [ln(ln )] = lim [ln(ln ) − ln(ln 2)] = ∞, so the series diverges. use the Integral Test.
2
→∞
→∞
 ln 
=2  ln 
2

ot

21.  () =

23. The function  () = 12 is continuous, positive, and decreasing on [1 ∞), so the Integral Test applies.

N

[() = 1 is decreasing and dividing by 2 doesn’t change that fact.]
  1
 ∞


∞
 1

 ()  = lim
 = lim −1 = − lim (1 − ) = −(1 − ) =  − 1, so the series
2
→∞ 1
→∞
→∞
2
1
=1 
1
converges.

25. The function  () =

1
1
1
1
[by partial fractions] is continuous, positive and decreasing on [1 ∞),
= 2 − +
2 + 3


+1

so the Integral Test applies.



 
 ∞
1
1
1
1
 ()  = lim
− +
 = lim − − ln  + ln( + 1)
→∞ 1
→∞
2

+1

1
1


1
+1
= lim − + ln
+ 1 − ln 2 = 0 + 0 + 1 − ln 2
→∞


The integral converges, so the series

∞


1 converges. 2 + 3
=1

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SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS

27. The function  () =

satisfied for the series

497

cos 
√
is neither positive nor decreasing on [1 ∞), so the hypotheses of the Integral Test are not

∞
 cos 
√ .

=1

29. We have already shown (in Exercise 21) that when  = 1 the series

∞


=2

 () =

¤

1 diverges, so assume that  6= 1.
(ln )

1
 + ln  is continuous and positive on [2 ∞), and  0 () = − 2
 0 if   − , so that  is eventually
(ln )
 (ln )+1

decreasing and we can use the Integral Test.


2

∞



1
(ln )1−
 = lim
→∞
(ln )
1− 2

[for  6= 1] = lim

→∞



(ln )1−
(ln 2)1−
−
1−
1−



e

This limit exists whenever 1 −   0 ⇔   1, so the series converges for   1.
31. Clearly the series cannot converge if  ≥ − 1 , because then lim (1 + 2 ) 6= 0. So assume   − 1 . Then
2
2

al

→∞

 () = (1 + 2 ) is continuous, positive, and eventually decreasing on [1 ∞), and we can use the Integral Test.
∞

2 

(1 +  )  = lim

→∞

1



1 (1 + 2 )+1
·
2
+1



=

1 lim [(1 + 2 )+1 − 2+1 ].
2( + 1) →∞

rS



1

This limit exists and is finite ⇔  + 1  0 ⇔   −1, so the series converges whenever   −1.

Fo

33. Since this is a -series with  = , () is defined when   1. Unless specified otherwise, the domain of a function  is the

set of real numbers  such that the expression for () makes sense and defines a real number. So, in the case of a series, it’s the set of real numbers  such that the series is convergent.

(b)

 4
 4
∞
∞
 81
 1
3
94

=
=
= 81
= 81
4
4
90
10
=1 
=1 
=1 
∞


∞


∞
 1
1
1
1
1
4
−
= 4 + 4 + 4 + ··· =
=
4
( − 2)4
3
4
5
90
=3 

N

=5

ot

35. (a)



1
1
+ 4
14
2



[subtract 1 and 2 ] =

17
4
−
90
16

1
2
is positive and continuous and  0 () = − 3 is negative for   0, and so the Integral Test applies.
2


37. (a)  () =

∞
 1
1
1
1
1
≈ 10 = 2 + 2 + 2 + · · · + 2 ≈ 1549768.
2
1
2
3
10
=1 

10 ≤
(b) 10 +



∞

10



∞

11





−1
1
1
1
1
=
, so the error is at most 01.
 = lim
= lim − +
→∞
2
 10 →∞

10
10

1
 ≤  ≤ 10 +
2



∞

10

1
 ⇒ 10 +
2

1
11

≤  ≤ 10 +

1
10

⇒

1549768 + 0090909 = 1640677 ≤  ≤ 1549768 + 01 = 1649768, so we get  ≈ 164522 (the average of 1640677 and 1649768) with error ≤ 0005 (the maximum of 1649768 − 164522 and 164522 − 1640677, rounded up).

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498

¤

CHAPTER 11

INFINITE SEQUENCES AND SERIES

(c) The estimate in part (b) is  ≈ 164522 with error ≤ 0005. The exact value given in Exercise 34 is 2 6 ≈ 1644934.

The difference is less than 00003.
 ∞
1
1
1
1
 = . So   0001 if 
(d)  ≤
2


1000


⇔   1000.

39.  () = 1(2 + 1)6 is continuous, positive, and decreasing on [1 ∞), so the Integral Test applies. Using (3),

 ≤



∞

(2 + 1)−6  = lim

→∞



5
1
≤ 6
10(2 + 1)5
10
4 =

4


=1
∞


−1
10(2 + 1)5



=



1
. To be correct to five decimal places, we want
10(2 + 1)5

⇔ (2 + 1)5 ≥ 20,000 ⇔  ≥

1
2


√
5
20,000 − 1 ≈ 312, so use  = 4.

1
1
1
1
1
= 6 + 6 + 6 + 6 ≈ 0001 446 ≈ 000145.
(2 + 1)6
3
5
7
9
∞


=1

e

1 is a convergent -series with  = 1001  1. Using (2), we get
1001

 −0001 



 ∞
1

1
1000
−1001  = lim
= −1000 lim
= −1000 − 0001 = 0001 .
 ≤
→∞ −0001
→∞ 0001





−1001 =

=1

1000
 5 × 10−9
0001

⇔ 0001 

1000
5 × 10−9

rS

We want   0000 000 005 ⇔

al

41.



⇔

Thus,  = 1 +

Fo


1000
  2 × 1011
= 21000 × 1011,000 ≈ 107 × 10301 × 1011,000 = 107 × 1011,301 .

43. (a) From the figure, 2 + 3 + · · · +  ≤ 1  () , so with
 
1
1
1
1 1 1
 = ln .
 () = , + + + · · · + ≤
 2
3
4


1
1
1
1
1
+ + + · · · + ≤ 1 + ln .
2
3
4


ot

(b) By part (a), 106 ≤ 1 + ln 106 ≈ 1482  15 and
109 ≤ 1 + ln 109 ≈ 2172  22.
ln 

ln 

= ln 
= ln  =

N



45. ln  = ln 

ln   −1 ⇔   −1

1
. This is a -series, which converges for all  such that − ln   1 ⇔
− ln 

⇔   1 [with   0].

11.4 The Comparison Tests
1. (a) We cannot say anything about



 . If    for all  and

divergent. (See the note after Example 2.)
(b) If    for all , then
3.





 is convergent, then



 could be convergent or

 is convergent. [This is part (i) of the Comparison Test.]

∞
∞

 1



1
1
 converges by comparison with
=
 2 for all  ≥ 1, so
, which converges
3
2
3 +1
2
+1
2
2

=1 2
=1 

23

because it is a p-series with  = 2  1.

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SECTION 11.4

5.

1
2

499

9
9
  =

3 + 10
10

≤ 1.


9
10



for all  ≥ 1.

converges by the Comparison Test.
9.

¤

∞
∞
 +1
 1

1
+1
√ , which diverges because it is a
√  √ = √ for all  ≥ 1, so
√ diverges by comparison with
=1 
=1
 
 




p-series with  =
7.

THE COMPARISON TESTS

∞


=1


9 
10

 is a convergent geometric series || =

9
10

∞


 1 , so

=1

9
3 + 10

∞
∞
 ln 
 1 ln 
1
 for all k ≥ 3 [since ln   1 for  ≥ 3], so diverges by comparison with
, which diverges because it


=3 
=3 

is a -series with  = 1 ≤ 1 (the harmonic series). Thus, convergence or divergence of a series.

∞
 ln  diverges since a finite number of terms doesn’t affect the
=1 

e

√
√
√
3
3
3
∞
∞

 1



13
1
√
,
 √ = 32 = 76 for all  ≥ 1, so converges by comparison with
76
3 + 4 + 3
3 + 4 + 3




3
=1
=1 

which converges because it is a -series with  =

∞
∞
 arctan  arctan 
2
  1
 12 for all  ≥ 1, so converges by comparison with
, which converges because it is a
12
12



2 =1 12
=1

constant times a p-series with  = 12  1.
 
4+1
4 · 4
4

=4
for all  ≥ 1.
3 − 2
3
3

 
 
∞

 4 
4
4
=4
is a divergent geometric series || =
3
=1
=1 3
∞


Fo

15.

 1.

rS

13.

7
6

al

11. √

∞
 4+1 diverges by the Comparison Test.

=1 3 − 2

4
3


 1 , so

1
1
and  = :

2 + 1

∞


∞
 1


1
= 1  0. Since the harmonic series diverges, so does
= lim √
= lim 
2)
→∞
→∞

2 + 1
1 + (1
=1 

N

lim

→∞

ot

17. Use the Limit Comparison Test with  = √

1
√
.
2 +1

=1

19. Use the Limit Comparison Test with  =

1 + 4
4
and  =  :
1 + 3
3

1 + 4



3
1

1 + 4 3
1 + 4
1
=10 lim = lim 1 +3 = lim
·  = lim
·
= lim
+1 ·
1
→∞ 
→∞
4
→∞ 1 + 3
→∞
→∞
4
4
1 + 3
4
+1

3
3
∞

  4 
 1 + 4
Since the geometric series  = diverges, so does
. Alternatively, use the Comparison Test with
3

=1 1 + 3
 
1 + 4
1 4
1 + 4
4
=
 

or use the Test for Divergence.
1 + 3
3 + 3
2(3 )
2 3

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500

¤

CHAPTER 11

INFINITE SEQUENCES AND SERIES

√
+2
1 and  = 32 :
22 +  + 1


√
√
√
√
1 + 2

32  + 2
(32  + 2 )(32  )
1
1 lim = lim
=  0.
= lim
= lim
=
→∞ 
→∞ 22 +  + 1
→∞
→∞ 2 + 1 + 12
(22 +  + 1)2
2
2
√
∞
∞


 1

+2 also converges. is a convergent p-series  = 3  1 , the series
Since
2
2
32
=1 
=1 2 +  + 1

21. Use the Limit Comparison Test with  =

23. Use the Limit Comparison Test with  =

5 + 2
1
and  = 3 :
(1 + 2 )2


14

3 (5 + 2)
53 + 24
= lim
= lim
·
= lim 
→∞ 
→∞ (1 + 2 )2
→∞ (1 + 2 )2
→∞
1(2 )2 lim -series [ = 3  1], the series

∞


=1

5

1
2

∞
 1 is a convergent
2 = 2  0. Since
3
=1 
+1

+2

5 + 2 also converges.
(1 + 2 )2

al

e

√
√
√
∞
 4 + 1
4 + 1
4
2
1
= 2
=
for all  ≥ 1, so
25. 3
 2 diverges by comparison with
3
2
 + 2
 ( + 1)
 ( + 1)
+1
=1  + 
∞
 1
1
=
, which diverges because it is a -series with  = 1 ≤ 1.
=1  + 1
=2 

27. Use the Limit Comparison Test with  =

− =

=1


2
2


1
1
1+
− and  = − : lim
= lim 1 +
= 1  0. Since
→∞ 
→∞



∞

 1 is a convergent geometric series || =

=1 

1


29. Clearly ! = ( − 1)( − 2) · · · (3)(2) ≥ 2 · 2 · 2 · · · · · 2 · 2 = 2−1 , so
1
2

∞

 1
 1 , so converges by the Comparison Test.
=1 !

∞
 1
1
1
≤ −1 . is a convergent geometric
−1
!
2
=1 2

ot

 series || =


2
1
1+
− also converges.

=1

∞


 1 , the series

Fo

∞


rS

∞


 


1
1 and  = . Then  and  are series with positive terms and



lim

→∞
∞


N

31. Use the Limit Comparison Test with  = sin

∞

 sin(1) sin 
= lim
= 1  0. Since
= lim
 is the divergent harmonic series,
→∞
→0

1

=1

sin (1) also diverges. [Note that we could also use l’Hospital’s Rule to evaluate the limit:

=1



cos(1) · −12 sin(1) H
1
lim
= lim
= lim cos = cos 0 = 1.]
→∞
→∞
→∞
1
−12

33.

10


1
1
1
1
1
1
1
1
√
≈ 124856. Now √
= √ + √ + √ + ··· + √
 √ = 2 , so the error is
4 +1
4 +1
4

10,001


2
17
82

=1




 ∞
1
1
1
1
1
=
= 01.
 = lim −
= lim − +
10 ≤ 10 ≤
→∞
2
 10 →∞

10
10
10

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.5

35.

10


ALTERNATING SERIES

¤

501

cos2 2 cos2 1 cos2 3 cos2 10 cos2 
1
+
+
+ ··· +
≈ 007393. Now
≤  , so the error is
2
3
10
5
5
5
5
5
5
 − 
 −

 ∞
 
1
5
5
5−10
1
+
= 10
 64 × 10−8 .
≤
 = lim
5−  = lim −
= lim −

→∞ 10
→∞
ln 5 10 →∞ ln 5 ln 5
5 ln 5
10 5

5− cos2  =

=1

10 ≤ 10
37. Since

∞

 9

9
≤  for each , and since is a convergent geometric series || =
10
10
10
=1

1
10

will always converge by the Comparison Test.
39. Since



∞

 
 1 , 01 2 3    =

=1 10

 converges, lim  = 0, so there exists  such that | − 0|  1 for all   
→∞

all   



⇒ 0 ≤ 2 ≤  . Since


 converges, so does



⇒ 0 ≤   1 for

2 by the Comparison Test.




= ∞, there is an integer  such that
 1 whenever   . (Take  = 1 in Definition 11.1.5.)




Then    whenever    and since  is divergent,  is also divergent by the Comparison Test.

41. (a) Since lim

so by part (a),

∞


=2

(ii) If  = so ∞


1 is divergent. ln 

∞

 ln 
1
and  = , then
 is the divergent harmonic series and lim
= lim ln  = lim ln  = ∞,
→∞ 
→∞
→∞


=1

 diverges by part (a).

43. lim  = lim

→∞

Fo

=1

→∞

al



 H
1
1
1
and  = for  ≥ 2, then lim
= lim
= lim  = ∞,
= lim
= lim
→∞ 
→∞ ln 
→∞ ln 
→∞ 1
→∞
ln 


rS

(b) (i) If  =

e

→∞


1
, so we apply the Limit Comparison Test with  = . Since lim   0 we know that either both
→∞
1


divergent.


∞

 1 diverges [-series with  = 1]. Therefore,  must be

=1

 is a convergent series with positive terms, lim  = 0 by Theorem 11.2.6, and

N

45. Yes. Since

ot

series converge or both series diverge, and we also know that

→∞

series with positive terms (for large enough ). We have lim

→∞

is also convergent by the Limit Comparison Test.



 =



sin( ) is a



sin( )
= lim
= 1  0 by Theorem 2.4.2. Thus, 
→∞



11.5 Alternating Series
1. (a) An alternating series is a series whose terms are alternately positive and negative.

(b) An alternating series

∞


=1

 =

∞


(−1)−1  , where  = | |, converges if 0  +1 ≤  for all  and lim  = 0.

=1

→∞

(This is the Alternating Series Test.)
(c) The error involved in using the partial sum  as an approximation to the total sum  is the remainder  =  −  and the size of the error is smaller than +1 ; that is, | | ≤ +1 . (This is the Alternating Series Estimation Theorem.) c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

502

¤

3. −

CHAPTER 11

INFINITE SEQUENCES AND SERIES

∞

2
2
2
4
6
8
10
2
2
+ − + −
+ ··· =
. Now lim  = lim
= lim
= 6= 0. Since
(−1)
→∞
→∞  + 4
→∞ 1 + 4
5
6
7
8
9
+4
1
=1

lim  6= 0 (in fact the limit does not exist), the series diverges by the Test for Divergence.

→∞

5.

∞


 =

=1

∞


(−1)−1

=1

∞

1
1
=
 0, { } is decreasing, and lim  = 0, so the
(−1)−1  . Now  =
→∞
2 + 1 =1
2 + 1

series converges by the Alternating Series Test.
7.

∞


 =

=1

∞


(−1)

=1

∞

3 − 1
3 − 1
3
=
= 6= 0. Since lim  6= 0
(−1)  . Now lim  = lim
→∞
→∞ 2 + 1
→∞
2 + 1 =1
2

(in fact the limit does not exist), the series diverges by the Test for Divergence.
9.

∞


 =

=1

∞


(−1) − =

=1

∞


(−1)  . Now  =

=1

1
 0, { } is decreasing, and lim  = 0, so the series converges
→∞


e

by the Alternating Series Test.

2
 0 for  ≥ 1. { } is decreasing for  ≥ 2 since
+4
0

2
(3 + 4)(2) − 2 (32 )
(23 + 8 − 33 )
(8 − 3 )
=
=
= 3
 0 for   2. Also,
3 +4
3 + 4)2
3 + 4)2

(
(
( + 4)2
3

al

11.  =

rS

∞

1
2
converges by the Alternating Series Test.
= 0. Thus, the series
(−1)+1 3
→∞ 1 + 43
 +4
=1

lim  = lim

→∞

13. lim  = lim 2 = 0 = 1, so lim (−1)−1 2 does not exist. Thus, the series
→∞

→∞

Fo

→∞

Test for Divergence.

∞


(−1)−1 2 diverges by the

=1




 



.  = sin
 0 for  ≥ 2 and sin
≥ sin
, and lim sin
= sin 0 = 0, so the
(−1) sin
→∞



+1

=1
∞


N

17.

ot



sin  + 1 
(−1)
1
2
√
√ . Now  =
√  0 for  ≥ 0, { } is decreasing, and lim  = 0, so the series
15.  =
=
→∞
1+ 
1+ 
1+ 


∞
 sin  + 1 
√2
converges by the Alternating Series Test.
1+ 
=0

series converges by the Alternating Series Test.

19.


·  · ··· · 
=
≥ ⇒
!
1 · 2 · ··· · 

lim

→∞


=∞ ⇒
!

by the Test for Divergence.
21.

lim

→∞

∞

(−1) 

(−1) does not exist. So the series diverges !
!
=1

The graph gives us an estimate for the sum of the series
∞
 (−08) of −055.
!
=1

8 =

(08)
≈ 0000 004, so
8!

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.5

ALTERNATING SERIES

¤

503

∞
7
 (−08)
 (−08)
≈ 7 =
!
!
=1
=1

≈ −08 + 032 − 00853 + 001706 − 0002 731 + 0000 364 − 0000 042 ≈ −05507

Adding 8 to 7 does not change the fourth decimal place of 7 , so the sum of the series, correct to four decimal places, is −05507.
23. The series

∞
 (−1)+1
1
1
1
satisfies (i) of the Alternating Series Test because
 6 and (ii) lim 6 = 0, so the
→∞ 
6
( + 1)6

=1

series is convergent. Now 5 =

1
1
= 0000064  000005 and 6 = 6 ≈ 000002  000005, so by the Alternating Series
56
6

Estimation Theorem,  = 5. (That is, since the 6th term is less than the desired error, we need to add the first 5 terms to get the sum to the desired accuracy.)

so the series is convergent. Now 3 =

e

∞
 (−1)
1
1
1
satisfies (i) of the Alternating Series Test because +1
  and (ii) lim
= 0,

→∞ 10 !
10
( + 1)!
10 !
=0 10 !

1
1
≈ 0000 167  0000 005 and 4 = 4 = 0000 004  0000 005, so by
103 3!
10 4!

al

25. The series

the Alternating Series Estimation Theorem,  = 4 (since the series starts with  = 0, not  = 1). (That is, since the 5th term

27. 4 =

1
1
=
≈ 0000 025, so
8!
40,320

rS

is less than the desired error, we need to add the first 4 terms to get the sum to the desired accuracy.)

Fo

∞
3
 (−1)
 (−1)
1
1
1
≈ 3 =
=− +
−
≈ −0459 722
2
24
720
=1 (2)!
=1 (2)!

Adding 4 to 3 does not change the fourth decimal place of 3 , so by the Alternating Series Estimation Theorem, the sum of the series, correct to four decimal places, is −04597.
72
= 0000 004 9, so
107

ot

29. 7 =

N

∞
6
 (−1)−1 2
 (−1)−1 2
≈ 6 =
=
10
10
=1
=1

1
10

−

4
100

+

9
1000

−

16
10,000

+

25
100,000

−

36
1,000,000

= 0067 614

Adding 7 to 6 does not change the fourth decimal place of 6 , so by the Alternating Series Estimation Theorem, the sum of the series, correct to four decimal places, is 00676.
31.

∞
 (−1)−1
1
1
1
1
1
1
1
= 1 − + − + ··· +
−
+
−
+ · · · . The 50th partial sum of this series is an

2
3
4
49
50
51
52
=1
 


∞
 (−1)−1
1
1
1
1
= 50 +
−
+
−
+ · · · , and the terms in parentheses are all positive. underestimate, since

51
52
53
54
=1

The result can be seen geometrically in Figure 1.

33. Clearly  =

1 is decreasing and eventually positive and lim  = 0 for any . So the series converges (by the
→∞
+

Alternating Series Test) for any  for which every  is defined, that is,  +  6= 0 for  ≥ 1, or  is not a negative integer. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

504

35.

¤



CHAPTER 11

2 =



INFINITE SEQUENCES AND SERIES

1(2)2 clearly converges (by comparison with the -series for  = 2). So suppose that

converges. Then by Theorem 11.2.8(ii), so does




(−1)−1  +  = 2 1 +

diverges by comparison with the harmonic series, a contradiction. Therefore,
Series Test does not apply since { } is not decreasing.



1
3

+

1
5





+··· = 2

(−1)−1 

1
. But this
2 − 1

(−1)−1  must diverge. The Alternating

11.6 Absolute Convergence and the Ratio and Root Tests






+1 
= 8  1, part (b) of the Ratio Test tells us that the series
1. (a) Since lim 
→∞   



 is divergent.



 +1 
 = 08  1, part (a) of the Ratio Test tells us that the series   is absolutely convergent (and
(b) Since lim 
→∞   













al

e

therefore convergent).


 +1 

 = 1, the Ratio Test fails and the series   might converge or it might diverge.
(c) Since lim 
→∞
 


∞

 +1 


 = lim   + 1 · 5  = lim  1 ·  + 1  = 1 lim 1 + 1 = 1 (1) = 1  1, so the series   is
3. lim 
 →∞  5
 5 →∞

→∞   
→∞  5+1


1
5
5
=1 5

5.  =

rS

absolutely convergent by the Ratio Test.

∞
 (−1)
1
 0 for  ≥ 0, { } is decreasing for  ≥ 0, and lim  = 0, so converges by the Alternating
→∞
5 + 1
=0 5 + 1

1 to get

∞


1
5 + 1
1
lim
= lim
= lim
= 5  0, so diverges by the Limit Comparison Test with the
→∞ 
→∞ 1(5 + 1)
→∞

=1 5 + 1









 +1 

 1

( + 1) 2
2
+1 2
1
3
=
= lim lim 1 +
= 2 (1) =
 
3
→∞

3
3 →∞

 2
3

2
3

 1, so the series

N

+1 
= lim
7. lim 
→∞   
→∞

∞
 (−1) is conditionally convergent.
=0 5 + 1

ot

harmonic series. Thus, the series

Fo

Series Test. To determine absolute convergence, choose  =

∞
  2 
 3 is absolutely convergent by the Ratio Test. Since the terms of this series are positive, absolute convergence is the

=1

same as convergence.






+1
 +1 
4
 = lim (11)
9. lim 
·
→∞   
→∞ ( + 1)4
(11)



(11)4
1
1
= (11) lim
= (11) lim
→∞ ( + 1)4
→∞ ( + 1)4
→∞ (1 + 1)4
4

= lim

= (11)(1) = 11  1,

so the series

∞


(−1)

=1

(11) diverges by the Ratio Test.
4

 
∞
∞
 1
 1
1

1
11. Since 0 ≤
≤ 3 =  3 and is a convergent -series [ = 3  1], converges, and so
3
3
3



=1 
=1 
∞
 (−1) 1 is absolutely convergent.
3
=1

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.6

ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS

¤

505







∞
 +1 

10+1
10  + 1
10
5
( + 1) 42+1
 = lim
=  1, so the series
13. lim 
·
·
= lim

2+1
→∞   
→∞ ( + 2) 42+3
→∞ 42
10
+2
8
=1 ( + 1)4 is absolutely convergent by the Ratio Test. Since the terms of this series are positive, absolute convergence is the same as convergence. 



∞
∞
∞
 (−1) arctan   2
 2
 (−1) arctan 
  1

15. 
, so since
=
converges ( = 2  1), the given series


2
2
2
2


2 =1 
2
=1 
=1

converges absolutely by the Comparison Test.

17.

∞
 (−1)
1
converges by the Alternating Series Test since lim
= 0 and
→∞ ln 
=2 ln 



1 ln 



is decreasing. Now ln   , so

∞
∞
 1
 1
1
1
 , and since is the divergent (partial) harmonic series, diverges by the Comparison Test. Thus, ln 


=2
=2 ln 

∞
∞
 1
 cos(3)
|cos (3)|
1
≤ and converges (use the Ratio Test), so the series converges absolutely by the
!
!
!
!
=1
=1

al

19.

e

∞
 (−1) is conditionally convergent.
=2 ln 

21. lim

→∞

rS

Comparison Test.

∞


1
2 + 1
1 + 12

| | = lim
=  1, so the series
= lim
2 +1
2
→∞ 2
→∞ 2 + 1
2
=1

Root Test.

→∞

→∞

2 + 1
22 + 1



is absolutely convergent by the



2

1
1

1+
= lim 1 +
=   1 [by Equation 7.4.9 (or 7.4*.9)], so the series
→∞



Fo


23. lim  | | = lim









ot


2
1 diverges by the Root Test.
1+

=1
∞





100

+1

N

+1 
( + 1) 100
= lim 
25. lim 
→∞   
→∞ 
( + 1)! so the series

=0·1=01

·



100

100

+1
100
!
1
 = lim 100
1+
= lim
→∞  + 1
100 100  →∞  + 1



∞
 100 100 is absolutely convergent by the Ratio Test.
!
=1

27. Use the Ratio Test with the series

1−

∞

1 · 3 · 5 · · · · · (2 − 1)
1 · 3 · 5 · · · · · (2 − 1)
1·3·5
1·3·5·7
1·3
+
−
+ · · · + (−1)−1
+ ··· =
.
(−1)−1
3!
5!
7!
(2 − 1)!
(2 − 1)!
=1







 +1 

(2 − 1)!

 = lim  (−1) · 1 · 3 · 5 · · · · · (2 − 1)[2( + 1) − 1] · lim 
   →∞ 
−1 · 1 · 3 · 5 · · · · · (2 − 1) 
→∞
[2( + 1) − 1]!
(−1)


 (−1)(2 + 1)(2 − 1)! 
 = lim 1 = 0  1,

= lim 
→∞ (2 + 1)(2)(2 − 1)! 
→∞ 2

so the given series is absolutely convergent and therefore convergent.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

506

29.

¤

CHAPTER 11

INFINITE SEQUENCES AND SERIES

∞
∞
∞
∞
 2 · 4 · 6 · · · · · (2)
 (2 · 1) · (2 · 2) · (2 · 3) · · · · · (2 · )
 2 !
 
2 , which diverges by the Test for
=
=
=
!
!
!
=1
=1
=1
=1

Divergence since lim 2 = ∞.
→∞






 +1 


 = lim  5 + 1  = 5  1, so the series diverges by the Ratio Test.
31. By the recursive definition, lim 
→∞
  →∞  4 + 3  4

∞
∞
  cos 


1

=
(−1)  , where   0 for  ≥ 1 and lim  = .
→∞


2
=1
=1




+1 +1

∞
 +1 





 = lim  (−1)
 = lim   = 1 (1) = 1  1, so the series   cos  is lim 
·
 →∞ 
 →∞  + 1
→∞

+1
(−1) 
2
2

=1


33. The series

absolutely convergent by the Ratio Test.



 1( + 1)3 
3
1
 = lim
= lim
= 1. Inconclusive
→∞ (1 + 1)3
13  →∞ ( + 1)3

e

35. (a) lim 
→∞ 

al





 ( + 1) 2 
 = lim  + 1 = lim 1 + 1 = 1 . Conclusive (convergent)
(b) lim  +1 ·
→∞  2
→∞ 2
  →∞ 2
2
2







+1










Fo




rS




√

 (−3)
 

1
 = 3 lim
·
(c) lim  √
= 3 lim
= 3. Conclusive (divergent)
→∞   + 1 (−3)−1 
→∞
→∞
+1
1 + 1


 √


+1
12 + 1
1 + 2 
1
· √  = lim
1+ ·
= 1. Inconclusive
(d) lim 
→∞  1 + ( + 1)2
 12 + (1 + 1)2
  →∞

+1 

 
1
!
·   = lim 
= || lim
= || · 0 = 0  1, so by the Ratio Test the
= lim 
37. (a) lim 
→∞   
→∞  ( + 1)!
→∞  + 1
  →∞   + 1  series ∞
  converges for all .
=0 !

5


=1

 =

→∞


= 0 by Theorem 11.2.6.
!

1
1
1
1
1
1
661
= + +
+
+
=
≈ 068854. Now the ratios
2
2
8
24
64
160
960

N

39. (a) 5 =

ot

(b) Since the series of part (a) always converges, we must have lim

+1
2

=
= form an increasing sequence, since

( + 1)2+1
2( + 1)


( + 1)2 − ( + 2)
1
+1
−
=
=
 0. So by Exercise 34(b), the error
2( + 2)
2( + 1)
2( + 1)( + 2)
2( + 1)( + 2)


1 6 · 26
1
6 in using 5 is 5 ≤
=
≈ 000521.
=
1 − lim 
1 − 12
192

+1 −  =

→∞

(b) The error in using  as an approximation to the sum is  =

+1
2
=
. We want   000005 ⇔
( + 1)2+1
1− 1
2

1
 000005 ⇔ ( + 1)2  20,000. To find such an  we can use trial and error or a graph. We calculate
( + 1)2
(11 + 1)211 = 24,576, so 11 =

11


=1

1
≈ 0693109 is within 000005 of the actual sum.
2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.6

ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS

41. (i) Following the hint, we get that | |   for  ≥ , and so since the geometric series

¤

507

∞

 converges [0    1],



the series ∞ | | converges as well by the Comparison Test, and hence so does ∞ | |, so ∞  is absolutely
=
=1
=1
=1

convergent.
→∞




| | =   1, then there is an integer  such that  | |  1 for all  ≥ , so | |  1 for  ≥ . Thus,

lim  6= 0, so

→∞

(iii) Consider
43. (a) Since



∞

=1

 diverges by the Test for Divergence.

∞
∞

 1
 1
[converges]. For each sum, lim  | | = 1, so the Root Test is inconclusive.
[diverges] and
→∞

2
=1
=1

 
 
 is absolutely convergent, and since +  ≤ | | and −  ≤ | | (because + and − each equal





either  or 0), we conclude by the Comparison Test that both
Or: Use Theorem 11.2.8.



+ and




− must be absolutely convergent.


e

(ii) If lim

45. Suppose that



 is only conditionally convergent. Hence,

 is conditionally convergent.

2  is divergent: Suppose



integer   0 such that   

+ can’t converge. Similarly, neither can




⇒ 2 | |  1. For   , we have | | 

Remark: The same argument shows that


1
, so
| | converges by
2




 1
. In other words,
 converges absolutely, contradicting the
2


 is conditionally convergent. This contradiction shows that

N

− .


→∞

ot

assumption that



2  converges. Then lim 2  = 0 by Theorem 6 in Section 11.2, so there is an

comparison with the convergent -series

(b)



Fo

(a)





rS

diverges because

al




(b) We will show by contradiction that both + and − must diverge. For suppose that + converged. Then so





 + 1    1
 + 1 
1
| |, which
 − 2  =
= 1 would  − 2  by Theorem 11.2.8. But
2 ( + | |) − 2 
2



  diverges for any   1.



2  diverges.

∞
 (−1) is conditionally convergent. It converges by the Alternating Series Test, but does not converge absolutely
=2  ln 

1 by the Integral Test, since the function  () = is continuous, positive, and decreasing on [2 ∞) and
 ln 

 
 ∞




(−1)
= lim
= lim ln(ln ) = ∞ . Setting  = for  ≥ 2, we find that
 ln  →∞ 2  ln  →∞
 ln 
2
2
∞


 =

=2

∞
 (−1) converges by the Alternating Series Test.
=2 ln 

It is easy to find conditionally convergent series



 such that



 diverges. Two examples are

∞
 (−1)−1 and 
=1

∞
 (−1)−1

√
, both of which converge by the Alternating Series Test and fail to converge absolutely because | | is a

=1

-series with  ≤ 1. In both cases,  diverges by the Test for Divergence. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

508

¤

CHAPTER 11

INFINITE SEQUENCES AND SERIES

11.7 Strategy for Testing Series
1.

1
1
  =
 + 3
3

 

1 is a convergent geometric series || =
3
=1

 
1
for all  ≥ 1.
3

∞


converges by the Comparison Test.

1
3

∞


 1 , so

=1

1
 + 3

∞




(−1)
= 1, so lim  = lim (−1) does not exist. Thus, the series diverges by
→∞  + 2
→∞
→∞
+2
+2
=1

3. lim | | = lim
→∞

the Test for Divergence.





2

2 
 
2
 +1 

2
2

 = lim  ( + 1) 2 · (−5)  = lim 2( + 1) = 2 lim 1 + 1
5. lim 
= (1) =  1, so the series
→∞
  →∞  (−5)+1
2 2−1  →∞
52
5 →∞

5
5

e

∞
 2 2−1
 converges by the Ratio Test.
=1 (−5)

1
√
. Then  is positive, continuous, and decreasing on [2 ∞), so we can apply the Integral Test. ln 

 

√
1
 = ln ,
√

Since
=
−12  = 212 +  = 2 ln  + , we find
 = 
 ln 
 ∞
 
 √
 √


√


√
√
= lim
= lim 2 ln  = lim 2 ln  − 2 ln 2 = ∞. Since the integral diverges, the
→∞
2
 ln  →∞ 2  ln  →∞
2

al

1
√
diverges. ln 
=2 

∞
 2
. Using the Ratio Test, we get

=1
=1 






2
 +1 
 ( + 1)2  
+1
1
1
1
 = lim 
= 12 · =  1, so the series converges.
· 2  = lim
·
lim 
→∞   
→∞  +1
  →∞




∞


∞


=1

2 − =



1
1
+ 
3
3



=

ot

11.

∞


∞
∞
 1

+
3 =1
=1

 
1
. The first series converges since it is a -series with  = 3  1 and the second
3

N

9.

rS

given series



Fo

7. Let  () =

series converges since it is geometric with || =






1
3

 1. The sum of two convergent series is convergent.



∞
 +1 
 +1 ( + 1)2
 3 2
3( + 1)2
+1
! 
 = lim  3
13. lim 
= 3 lim
= 0  1, so the series
·  2  = lim
→∞   
→∞ 
→∞ 2
( + 1)!
3   →∞ ( + 1)2
=1 !

converges by the Ratio Test.

2−1 3+1
2 2−1 3 31
3
15.  =
=
=



2



2·3




 

6
6
. By the Root Test, lim 
= lim
= 0  1, so the series
→∞
→∞ 


 
 
∞
∞
 2−1 3+1
 3 6 
6
converges. It follows from Theorem 8(i) in Section 11.2 that the given series,
=
,



=1
=1
=1 2
∞


also converges.

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.7

STRATEGY FOR TESTING SERIES

¤

509






 +1 

 = lim  1 · 3 · 5 · · · · · (2 − 1)(2 + 1) · 2 · 5 · 8 · · · · · (3 − 1)  = lim 2 + 1
17. lim 
→∞   
→∞  2 · 5 · 8 · · · · · (3 − 1)(3 + 2) 1 · 3 · 5 · · · · · (2 − 1) 
→∞ 3 + 2
= lim

→∞

so the series

2 + 1
2
=  1
3 + 2
3

∞
 1 · 3 · 5 · · · · · (2 − 1) converges by the Ratio Test.
=1 2 · 5 · 8 · · · · · (3 − 1)

ln 

19. Let () = √ . Then  0 () =



2 − ln  ln 
 0 when ln   2 or   2 , so √ is decreasing for   2 .
232


∞

ln 
1
2 ln 
 √  = lim √ = 0, so the series
(−1) √ converges by the
By l’Hospital’s Rule, lim √ = lim
→∞
→∞
→∞
=1



1 2 

Alternating Series Test.








→∞

→∞

(−1) cos(12 ) diverges by the

=1

al

Test for Divergence.

→∞

∞


e

21. lim | | = lim (−1) cos(12 ) = lim cos(12 ) = cos 0 = 1, so the series

rS

 
1
1
23. Using the Limit Comparison Test with  = tan and  = , we have



tan(1) tan(1) H sec2 (1) · (−12 )

= lim
= lim
= lim sec2 (1) = 12 = 1  0. Since
= lim
→∞ 
→∞
→∞
→∞
→∞
1
1
−12
lim

∞


 is the divergent harmonic series,

=1

 is also divergent.

Fo

=1

∞






2
 ( + 1)! 2 
∞
 +1 
 !
( + 1)! · 
 
+1

 = lim 
= lim 2+1 = 0  1, so
25. Use the Ratio Test. lim 
·
 = lim
   →∞  (+1)2
2 +2+1 !
2
→∞
→∞ 
!  →∞ 
=1  converges. 

∞

2



ln  ln 
1
−
 = lim −
→∞
2

 1

ot

27.

H

[using integration by parts] = 1. So

∞
 ln  converges by the Integral Test, and since
2
=1 

29.

∞


N

∞
  ln 
 ln  ln 
 ln 
= 2 , the given series
3 
3 converges by the Comparison Test.
3


( + 1)
=1 ( + 1)

 =

=1

∞


(−1)

=1

∞

1
1
(−1)  . Now  =
=
 0, { } is decreasing, and lim  = 0, so the series
→∞
cosh  =1 cosh 

converges by the Alternating Series Test.
Or: Write

∞
∞
 1

1
2
2
1
=  is convergent by the
  and is a convergent geometric series, so cosh 
 + −

 cosh 
=1
=1

Comparison Test. So

∞


(−1)

=1

5
= [divide by 4 ]
→∞ 3 + 4

31. lim  = lim
→∞

Thus,

1 is absolutely convergent and therefore convergent. cosh 

∞


=1

3

 
 
(54)
3
5
= 0 and lim
= ∞.
= ∞ since lim
→∞ (34) + 1
→∞ 4
→∞ 4 lim 5 diverges by the Test for Divergence.
+ 4

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

510

¤

33. lim

→∞

CHAPTER 11 INFINITE SEQUENCES AND SERIES





| | = lim

→∞


+1

2 

= lim

→∞

→∞

converges by the Root Test.
35.  =

1
1
1
=
, so let  = and use the Limit Comparison Test.

1+1
 · 1

[see Exercise 3.4.61], so the series

37. lim

→∞

∞

1
1
1
=
=  1, so the series
[( + 1) ] lim (1 + 1)

=1

lim

→∞




+1

2


1
= lim 1 = 1  0
→∞ 


∞


1 diverges by comparison with the divergent harmonic series.
1+1
=1

∞


 √


| | = lim (21 − 1) = 1 − 1 = 0  1, so the series
2 − 1 converges by the Root Test.
→∞

=1

11.8 Power Series
∞

=0

  = 0 + 1  + 2 2 + 3 3 + · · · , where  is a variable and the  ’s are

e

1. A power series is a series of the form

al

constants called the coefficients of the series.

More generally, a series of the form ∞  ( − ) = 0 + 1 ( − ) + 2 ( − )2 + · · · is called a power series in
=0

rS

( − ) or a power series centered at  or a power series about , where  is a constant.
3. If  = (−1)  , then







 

+1

 +1 


( + 1)+1 
1
 = lim  (−1)
 = lim (−1)  + 1  = lim
|| = ||. By the Ratio Test, the lim 
1+
 →∞
 →∞ 
→∞   
→∞ 
(−1) 


∞


(−1)  converges when ||  1, so the radius of convergence  = 1. Now we’ll check the endpoints, that is,

Fo

series

=1

 = ±1. Both series

∞


(−1) (±1) =

=1

(∓1)  diverge by the Test for Divergence since lim |(∓1) | = ∞. Thus,

=1

→∞

ot

the interval of convergence is  = (−1 1).



 +1





 +1 

2 − 1 

 = lim  
 = lim 2 − 1 || = lim 2 − 1 || = ||. By
, then lim 
·
→∞   
→∞  2 + 1
→∞
2 − 1
  →∞ 2 + 1
2 + 1
∞


∞


1
converges when ||  1, so  = 1. When  = 1, the series diverges by
2 − 1
=1 2 − 1

N

5. If  =

∞


the Ratio Test, the series

=1

comparison with

∞


∞
1
1
1
1  1 since  and diverges since it is a constant multiple of the harmonic series.
2 − 1
2
2 =1 
=1 2

When  = −1, the series

∞
 (−1) converges by the Alternating Series Test. Thus, the interval of convergence is [−1 1).
=1 2 − 1




 +1



 +1 


1
! 


 = lim  
 = lim    = || lim
, then lim 
·
= || · 0 = 0  1 for all real .
7. If  =
→∞
→∞  + 1
!
  →∞  ( + 1)!   →∞   + 1 
So, by the Ratio Test,  = ∞ and  = (−∞ ∞).

2 
, then
2








2 
2 +1
2
 +1 


2 
|| 2

 = lim  ( + 1) 
 = lim  ( + 1)  = lim || 1 + 1
(1) =
· 2  lim =
→∞   
→∞ 
→∞ 
2+1
 
22  →∞ 2

2

9. If  = (−1)

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

1
2

||. By the

SECTION 11.8 POWER SERIES
∞


Ratio Test, the series

(−1)

=1

When  = ±2, both series

∞


2  converges when
2

(−1)

=1

1
2

¤

511

||  1 ⇔ ||  2, so the radius of convergence is  = 2.

∞

2 (±2)
=
(∓1) 2 diverge by the Test for Divergence since
2
=1



lim (∓1) 2  = ∞. Thus, the interval of convergence is  = (−2 2).

→∞

11. If  =

(−3) 
, then
32







32 
32


 +1 
 (−3)+1 +1

32 
1


 = lim 
 = lim −3
·
lim 
 = 3 || lim
→∞   
→∞  ( + 1)32
→∞

(−3)   →∞ 
+1
1 + 1
= 3 || (1) = 3 ||

∞
 (−3) 
√  converges when 3 ||  1 ⇔ ||  1 , so  = 1 . When  = 1 , the series
3
3
3

=1 

e

By the Ratio Test, the series

al

∞
∞
 (−1)
 1 converges by the Alternating Series Test. When  = − 1 , the series is a convergent -series
3
32
32
=1
=1 




 = 3  1 . Thus, the interval of convergence is − 1  1 .
2
3 3

[by l’Hospital’s Rule] =
 = −4,

∞


(−1)

=2

rS





 +1 

+1

ln 
4 ln   ||
||
 = lim 
=
, then lim 
·
lim
=
·1
→∞   
→∞  4+1 ln( + 1)
4 ln 
 
4 →∞ ln( + 1)
4
||
||
. By the Ratio Test, the series converges when
1
4
4

=2

∞


1 is divergent by the Comparison Test. When  = 4, ln 
=2

∞


1
=
, which converges by the Alternating Series Test. Thus,  = (−4 4].
(−1)
ln  =2 ln 

4

ot

(−1)

||  4, so  = 4. When

∞
∞
∞
 [(−1)(−4)]
 1
 1

1
1
=
=
. Since ln    for  ≥ 2,
 and is the
 ln  ln  =2
4
ln 

=2 ln 
=2 

4

divergent harmonic series (without the  = 1 term),
∞


⇔

Fo

13. If  = (−1)

N





+1
 +1 

( − 2)
2 + 1
2 + 1 

 = lim  ( − 2)
 = | − 2| lim
15. If  =
, then lim 
·
= | − 2|. By the
→∞
→∞ ( + 1)2 + 1
2 + 1
  →∞  ( + 1)2 + 1 ( − 2) 

∞
 ( − 2) converges when | − 2|  1 [ = 1] ⇔ −1   − 2  1 ⇔ 1    3. When
2
=0  + 1

Ratio Test, the series
 = 1, the series

∞


(−1)

=0

2

comparison with the p-series

∞

1
1
converges by the Alternating Series Test; when  = 3, the series converges by
2
+1
=0  + 1

∞
 1
[ = 2  1]. Thus, the interval of convergence is  = [1 3].
2
=1 



√
√


 +1 (

+1
 +1 


3 ( + 4)


 = lim  3 √ + 4)
√
17. If  =
· 
= 3 | + 4|.
, then lim 

 = 3 | + 4| lim √
→∞
→∞
  →∞ 
3 ( + 4) 
+1
+1

By the Ratio Test, the series

∞
 3 ( + 4)
√
converges when 3 | + 4|  1 ⇔ | + 4| 
=1


1
3



= 1
⇔
3

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

512

CHAPTER 11 INFINITE SEQUENCES AND SERIES

−1   + 4 
3

1
3

⇔ − 13    − 11 . When  = − 13 , the series
3
3
3

Test; when  = − 11 , the series
3
19. If  =

∞

 1
√ diverges  =
=1


1
2

∞


1
(−1) √ converges by the Alternating Series
=1





≤ 1 . Thus, the interval of convergence is  = − 13  − 11 .
3
3


| − 2|
( − 2)
, then lim  | | = lim
= 0, so the series converges for all  (by the Root Test).
→∞
→∞



 = ∞ and  = (−∞ ∞).


( − ) , where   0.





+1
 +1 
| − |

1 | − |

 = lim ( + 1) | − |
·
= lim 1 + lim =
.
→∞   
→∞
→∞
+1
 | − |




21.  =

| − |
 1 ⇔ | − |   [so  = ] ⇔ −   −    ⇔


e

By the Ratio Test, the series converges when

 −      + . When | − | = , lim | | = lim  = ∞, so the series diverges. Thus,  = ( −   + ).
→∞

al

→∞





+1 
 +1 


 = lim  ( + 1)! (2 − 1)
 = lim ( + 1) |2 − 1| → ∞ as  → ∞
23. If  = ! (2 − 1) , then lim 
 →∞
→∞
  →∞ 
!(2 − 1)
  for all  6= 1 . Since the series diverges for all  6= 1 ,  = 0 and  = 1 .
2
2
2

rS



(5 − 4)
, then
3




3
3


+1
 +1 


3

1
 = lim  (5 − 4)
 = lim |5 − 4| lim 
·
= lim |5 − 4|
→∞   
→∞  ( + 1)3
→∞
(5 − 4)  →∞
+1
1 + 1

Fo

25. If  =

= |5 − 4| · 1 = |5 − 4|

   1, so  = 1 . When  = 1, the series
5

N

3
5

∞


 (5 − 4) converges when |5 − 4|  1 ⇔  − 4  
5
3
=1

ot

By the Ratio Test,

1
5

⇔ −1   −
5

4
5



1
5

⇔

∞
 1 is a convergent -series (  = 3  1). When  = 3 , the series
5
3
=1 

∞


 (−1) converges by the Alternating Series Test. Thus, the interval of convergence is  = 3  1 .
5
3

=1


, then
1 · 3 · 5 · · · · · (2 − 1)




 +1 

+1
||
1 · 3 · 5 · · · · · (2 − 1) 
 = lim 
 = lim
·
lim 
 →∞ 2 + 1 = 0  1. Thus, by
→∞   
→∞  1 · 3 · 5 · · · · · (2 − 1)(2 + 1)


27. If  =

 converges for all real  and we have  = ∞ and  = (−∞ ∞).
=1 1 · 3 · 5 · · · · · (2 − 1)

the Ratio Test, the series

∞


29. (a) We are given that the power series

∞


=0  

is convergent for  = 4. So by Theorem 3, it must converge for at least

−4   ≤ 4. In particular, it converges when  = −2; that is,

∞

=0

 (−2) is convergent.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.8 POWER SERIES

(b) It does not follow that

∞


=0  (−4)

¤

513

is necessarily convergent. [See the comments after Theorem 3 about convergence at

the endpoint of an interval. An example is  = (−1) (4 ).]
(!) 
 , then
()!



 +1 
( + 1)
 = lim [( + 1)!] ()! || = lim lim 
||
→∞   
→∞ (!) [( + 1)]!
→∞ ( + )( +  − 1) · · · ( + 2)( + 1)


( + 1) ( + 1)
( + 1)
= lim
···
||
→∞ ( + 1) ( + 2)
( + )






+1
+1
+1
lim
· · · lim
||
= lim
→∞  + 1 →∞  + 2
→∞  + 
=

 

1
||  1


⇔

||   for convergence, and the radius of convergence is  =  

e

31. If  =

al

33. No. If a power series is centered at , its interval of convergence is symmetric about . If a power series has an infinite radius

rS

of convergence, then its interval of convergence must be (−∞ ∞), not [0 ∞).

(−1) 2+1
, then
!( + 1)! 22+1




 +1 

2+3
1
!( + 1)! 22+1    2
 = lim 
=
lim 
= 0 for all .
·
lim

→∞   
→∞  ( + 1)!( + 2)! 22+3
2+1
2 →∞ ( + 1)( + 2)

Fo

35. (a) If  =

So 1 () converges for all  and its domain is (−∞ ∞).

(b), (c) The initial terms of 1 () up to  = 5 are 0 =

3
5
7
9
, 2 =
, 3 = −
, 4 =
,
16
384
18,432
1,474,560

ot

1 = −


,
2

11
. The partial sums seem to
176,947,200

N

and 5 = −

approximate 1 () well near the origin, but as || increases, we need to take a large number of terms to get a good approximation. 37. 2−1 = 1 + 2 + 2 + 23 + 4 + 25 + · · · + 2−2 + 22−1

= 1(1 + 2) + 2 (1 + 2) + 4 (1 + 2) + · · · + 2−2 (1 + 2) = (1 + 2)(1 + 2 + 4 + · · · + 2−2 )

= (1 + 2)

1 + 2
1 − 2
[by (11.2.3) with  = 2 ] → as  → ∞ by (12.2.4)], when ||  1.
1 − 2
1 − 2

Also 2 = 2−1 + 2 → approach 1 + 2
1 + 2 since 2 → 0 for ||  1. Therefore,  → since 2 and 2−1 both
1 − 2
1 − 2

1 + 2
1 + 2 as  → ∞. Thus, the interval of convergence is (−1 1) and  () =
.
1 − 2
1 − 2 c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

514

¤

CHAPTER 11

INFINITE SEQUENCES AND SERIES

39. We use the Root Test on the series

||  1, so  = 1.
41. For 2    3,





  diverges and

  . We need lim

→∞




|  | = || lim  | | =  ||  1 for convergence, or
→∞



  converges. By Exercise 11.2.69,

converge for ||  2, the radius of convergence of ( +  )  is 2.


( +  )  diverges. Since both series

11.9 Representations of Functions as Power Series
1. If  () =

∞


  has radius of convergence 10, then  0 () =

=0

1
, and then use Equation (1) to represent the function as a sum of a power
1−



1
1 − 3



=

 
∞
∞
 1
2    

or, equivalently, 2
 . The series converges when    1,
+1
3 =0 3
3
=0 3

rS

2
2
=
3−
3

al

∞
∞


1
1
=
=
(−) =
(−1)  with |−|  1 ⇔ ||  1, so  = 1 and  = (−1 1).
1+
1 − (−) =0
=0

that is, when ||  3, so  = 3 and  = (−3 3).





   
∞
∞
∞

1
1
2
2+1
 


 

 2
=
=
=
(−1)  =
(−1) +1
=
−
2
2
2}
9+
9 1 + (3)
9 1 − {−(3)
9 =0
3
9 =0
9
9
=0

Fo

7.  () =

e

3. Our goal is to write the function in the form

5.  () =

 −1 also has radius of convergence 10 by

=1

Theorem 2.

series.  () =

∞


 = 3 and  = (−3 3).

 2
 
 1 ⇔ ||2  9 ⇔ ||  3, so
9



∞
∞
∞
∞
∞
∞
 
 
 +1
 
 
 
1
1+
= (1 + )
= (1 + )
 =
 +

=1+
 +
 = 1+2
 .
1−
1−
=0
=0
=0
=1
=1
=1

N

9.  () =

ot

   
   

 2
 2
1 ⇔
The geometric series converges when −
−

3
3 
=0
∞


The series converges when ||  1, so  = 1 and  = (−1 1).



∞
∞
 
 
1
−(1 − ) + 2
1+
=
= −1 + 2
= −1 + 2
 =1+2
 .
A second approach:  () =
1−
1−
1−
=0
=1

A third approach:
() =



1+
1
= (1 + )
= (1 + )(1 +  + 2 + 3 + · · · )
1−
1−

= (1 +  + 2 + 3 + · · · ) + ( + 2 + 3 + 4 + · · · ) = 1 + 2 + 22 + 23 + · · · = 1 + 2

∞


 .

=1

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.9

11.  () =

3


3
=
=
+
2 −  − 2
( − 2)( + 1)
−2
+1

REPRESENTATIONS OF FUNCTIONS AS POWER SERIES

¤

515

⇒ 3 = ( + 1) + ( − 2). Let  = 2 to get  = 1 and

 = −1 to get  = −1. Thus



∞
∞

1
1
1
1
1
1    
3
=
−
=
−
=−
−
(−)
2 −−2

−2
+1
−2 1 − (2)
1 − (−)
2 =0 2
=0




 
∞
∞


1 1
1


+1
−
(−1)
=
− 1(−1)  =
− +1 
2 2
2
=0
=0

We represented  as the sum of two geometric series; the first converges for  ∈ (−2 2) and the second converges for (−1 1).
Thus, the sum converges for  ∈ (−1 1) = .

=1

e




∞
1

−1
 
 
13. (a)  () =
=
(−1) 
[from Exercise 3]
=−
 1 + 
 =0
(1 + )2
∞
∞


(−1)+1 −1 [from Theorem 2(i)] =
(−1) ( + 1) with  = 1.
=
=0

al

In the last step, note that we decreased the initial value of the summation variable  by 1, and then increased each occurrence of  in the term by 1 [also note that (−1)+2 = (−1) ].


∞


1
1
1 
1  
=−
(−1) ( + 1)
=−
[from part (a)]
2  (1 + )2
2  =0
(1 + )3
∞
∞


(−1) ( + 1)−1 = 1
(−1) ( + 2)( + 1) with  = 1.
= −1
2
2

rS

(b)  () =

=1

=0

∞
2
1
1 
= 2 ·
= 2 ·
(−1) ( + 2)( + 1)
(1 + )3
(1 + )3
2 =0
∞
1 
=
(−1) ( + 2)( + 1)+2
2 =0

Fo

(c)  () =

[from part (b)]

ot

To write the power series with  rather than +2 , we will decrease each occurrence of  in the term by 2 and increase the initial value of the summation variable by 2. This gives us
1

=−
5−
5

N



15.  () = ln(5 − ) = −




1
=−
1 − 5
5

∞
1 
(−1) ()( − 1) with  = 1.
2 =2


∞
∞
∞

 
 
+1
1 
 =  −
=−
 ( + 1)

5 =0 5
=0 5
=1  5

 

Putting  = 0, we get  = ln 5. The series converges for |5|  1 ⇔ ||  5, so  = 5.
17. We know that

∞

1
1
(−4) . Differentiating, we get
=
=
1 + 4
1 − (−4) =0

∞
∞


−4
=
(−4) −1 =
(−4)+1 ( + 1) , so
2
(1 + 4)
=1
=0

 () =

∞
∞

−4
−
− 

·
=
=
(−4)+1 ( + 1) =
(−1) 4 ( + 1)+1
2
2
(1 + 4)
4 (1 + 4)
4 =0
=0

for |−4|  1

⇔

||  1 , so  = 1 .
4
4

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

516

¤

CHAPTER 11

19. By Example 5,

INFINITE SEQUENCES AND SERIES

∞

1
=
( + 1) . Thus,
(1 − )2
=0

 () =

∞
∞


1+
1

=
+
=
( + 1) +
( + 1)+1
2
2
2
(1 − )
(1 − )
(1 − )
=0
=0

=

∞


( + 1) +

=0

=1+

∞


[make the starting values equal]



=1
∞


[( + 1) + ] = 1 +

=1

21.  () =



=
2 + 16
16



1
1 − (−2 16)

∞


(2 + 1) =

=1



=

∞


(2 + 1) with  = 1.

=0

 2 
∞
∞
∞

 
1
1

 
=
(−1)  2 =
(−1) +1 2+1 .
−
16 =0
16
16 =0
16
16
=0




The series converges when −216  1 ⇔ 2  16 ⇔ ||  4, so  = 4. The partial sums are 1 =
,
16

3
5
7
9
, 3 = 2 + 3 , 4 = 3 − 4 , 5 = 4 + 5 ,    . Note that 1 corresponds to the first term of the infinite
2
16
16
16
16

e

2 = 1 −

Fo

rS

al

sum, regardless of the value of the summation variable and the value of the exponent.

ot

As  increases,  () approximates  better on the interval of convergence, which is (−4 4).







1+




= ln(1 + ) − ln(1 − ) =
+
=
+
1−
1+
1−
1 − (−)
1−


 ∞
∞

 
(−1)  +
  = [(1 −  + 2 − 3 + 4 − · · · ) + (1 +  + 2 + 3 + 4 + · · · )] 
=

N

23.  () = ln

=0

=



=0

(2 + 22 + 24 + · · · )  =



∞


22  =  +

=0

But  (0) = ln 1 = 0, so  = 0 and we have  () =
1

∞
 22+1
=0 2 + 1

∞
∞
 22+1

1 with  = 1. If  = ±1, then  () = ±2
,
2 + 1
2 + 1
=0
=0

which both diverge by the Limit Comparison Test with  =

1
.


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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.9

REPRESENTATIONS OF FUNCTIONS AS POWER SERIES

¤

517

2
23
25
, 2 = 1 +
, 3 = 2 +
, .
1
3
5

The partial sums are 1 =

As  increases,  () approximates  better on the interval of convergence, which is (−1 1).

∞


27. From Example 6, ln(1 + ) =

(−1)−1

=1



e


∞
∞
∞
 8 
 8+1
 8+2

1
1

. The series for
=·
=
( ) =

⇒
 =  + converges 8
8
8
1−
1−
1−
8 + 2
1 − 8
=0
=0
=0
  when 8   1 ⇔ ||  1, so  = 1 for that series and also the series for (1 − 8 ). By Theorem 2, the series for


 also has  = 1.
1 − 8
∞


+2 for ||  1, so 2 ln(1 + ) = and (−1)−1


=1

al

25.



1
 =
1 + 5

=



02

0



∞


Fo

∞
∞
  5 

1
1
=
=
(−1) 5
−
=
1 + 5
1 − (−5 )
=0
=0

(−1) 5  =  +

=0

∞


(−1)

=0

⇒

5+1
. Thus,
5 + 1

02

1
11
(02)11
6
(02)6
+
− ···
+
− · · · . The series is alternating, so if we use
 =  −
= 02 −
5
1+
6
11
6
11
0

ot

29.

(−1)

rS

+3
.  = 1 for the series for ln(1 + ), so  = 1 for the series representing
( + 3)
=1

2 ln(1 + ) as well. By Theorem 2, the series for 2 ln(1 + )  also has  = 1.
∞


2 ln(1 + )  =  +

N

the first two terms, the error is at most (02)1111 ≈ 19 × 10−9 . So  ≈ 02 − (02)66 ≈ 0199 989 to six decimal places.
31. We substitute 3 for  in Example 7, and find that



 arctan(3)  =

So





0



∞


(−1)

=0

(3)2+1
 =
2 + 1

01

 arctan(3)  =





∞


(−1)

=0

∞

32+1 2+2
32+1 2+3
 =  +
(−1)
2 + 1
(2 + 1)(2 + 3)
=0

33
33 5
35 7
37 9
−
+
−
+···
1·3
3·5
5·7
7·9

01
0

1
9
243
2187
= 3 −
+
−
+···.
10
5 × 105
35 × 107
63 × 109
The series is alternating, so if we use three terms, the error is at most


0

01

 arctan(3)  ≈

2187
≈ 35 × 10−8 . So
63 × 109

1
9
243
−
+
≈ 0000 983 to six decimal places.
103
5 × 105
35 × 107

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

518

¤

CHAPTER 11

INFINITE SEQUENCES AND SERIES

33. By Example 7, arctan  =  −

5
7
(02)3
(02)5
(02)7
3
+
−
+ · · · , so arctan 02 = 02 −
+
−
+···.
3
5
7
3
5
7

The series is alternating, so if we use three terms, the error is at most
Thus, to five decimal places, arctan 02 ≈ 02 −
35. (a) 0 () =

(02)7
≈ 0000 002.
7

(02)3
(02)5
+
≈ 0197 40.
3
5

∞
∞
∞
 (−1) 2 0
 (−1) 22−1
 (−1) 2(2 − 1)2−2
00
 0 () =
, and 0 () =
, so
2 (!)2
2 (!)2
2
22 (!)2
=0 2
=1
=1

00
0
2 0 () + 0 () + 2 0 () =

=

∞
∞
∞
 (−1) 2(2 − 1)2
 (−1) 22
 (−1) 2+2
+
+
22 (!)2
22 (!)2
22 (!)2
=1
=1
=0

∞
∞
∞
 (−1) 2(2 − 1)2
 (−1) 22
 (−1)−1 2
+
+
2−2 [( − 1)!]2
22 (!)2
22 (!)2
=1
=1
=1 2

∞
∞
∞
 (−1) 2(2 − 1)2
 (−1) 22
 (−1) (−1)−1 22 2 2
+
+
22 (!)2
22 (!)2
22 (!)2
=1
=1
=1


∞

2(2 − 1) + 2 − 22 2 2

=
(−1)
22 (!)2
=1
 2

∞

4 − 2 + 2 − 42 2
 =0
=
(−1)
22 (!)2
=1


 1
 1
 1 ∞
 (−1) 2
2
4
6
 =
1−
(b)
+
−
+ · · · 
0 ()  =
2
2
4
64
2304
=0 2 (!)
0
0
0

1
3
1
5
7
1
1
= −
+
−
+··· = 1 −
+
−
+ ···
3·4
5 · 64
7 · 2304
12
320
16,128
0

Fo

rS

al

e

=

∞
 
=0 !

⇒  0 () =

∞
∞
∞
 −1
 −1
 
=
=
= ()
!
( − 1)! =0 !
=1
=1

N

37. (a)  () =

ot

1
Since 16,128 ≈ 0000062, it follows from The Alternating Series Estimation Theorem that, correct to three decimal places,
1
1
1
 ()  ≈ 1 − 12 + 320 ≈ 0920.
0 0

(b) By Theorem 9.4.2, the only solution to the differential equation  () =  () is  () =  , but (0) = 1, so
 = 1 and  () =  .

Or: We could solve the equation  () =  () as a separable differential equation.


 +1


2
 +1 



2 
 = lim  
, then by the Ratio Test, lim 
·   = || lim
= ||  1 for
→∞   
→∞  ( + 1)2
→∞  + 1
2
 
 
∞
∞
   
  =  1 which is a convergent -series ( = 2  1), so the interval of convergence, so  = 1. When  = ±1,
 2 
2
=1
=1 

39. If  =

convergence for  is [−1 1]. By Theorem 2, the radii of convergence of  0 and  00 are both 1, so we need only check the endpoints.  () =

∞
 
2
=1 

⇒  0 () =

∞
∞
 −1
 
, and this series diverges for  = 1 (harmonic series)
=
2

=1
=0  + 1

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.10

TAYLOR AND MACLAURIN SERIES

and converges for  = −1 (Alternating Series Test), so the interval of convergence is [−1 1).  00 () = at both 1 and −1 (Test for Divergence) since lim

→∞

¤

∞
 −1 diverges =1  + 1


= 1 6= 0, so its interval of convergence is (−1 1).
+1

2+1
1
for ||  1. In particular, for  = √ , we
2 + 1
3
=0
 √ 2+1

 

∞
∞


1 3
1
1
1
1

√
=
, so
(−1)
(−1) have = tan−1 √
=
6
2 + 1
3
3
3 2 + 1
=0
=0

41. By Example 7, tan−1  =

∞


(−1)

∞
∞
√ 
(−1)
(−1)
6 
=2 3
.
= √

3 =0 (2 + 1)3
=0 (2 + 1)3

11.10 Taylor and Maclaurin Series
∞


=0

 ( − 5) ,  =

 () ()
 (8) (5)
, so 8 =
.
!
8!

e

1. Using Theorem 5 with

al

3. Since  () (0) = ( + 1)!, Equation 7 gives the Maclaurin series

rS

∞
∞
∞
  () (0) 
 ( + 1)! 

 =
 =
( + 1) . Applying the Ratio Test with  = ( + 1) gives us
!
!
=0
=0
=0




+1 
 +1 

 = lim  ( + 2)
 = || lim  + 2 = || · 1 = ||. For convergence, we must have ||  1, so the lim 
→∞   
→∞  ( + 1) 
→∞  + 1

5.

Fo

radius of convergence  = 1.


 () ()

 () (0)

0

(1 − )−2

1

2(1 − )−3

1

24(1 − )−5

24 3

6

+

120 4

24

+···

= 1 + 2 + 32 + 43 + 54 + · · · =

6

120(1 − )−6
.
.
.

24

120
.
.
.

N

3
4
.
.
.

 00 (0) 2  000 (0) 3  (4) (0) 4
 +
 +
 +···
2!
3!
4!

ot

6(1 − )

= 1 + 2 + 6 2 +
2

2

−4

2

7.

(1 − )−2 =  (0) +  0 (0) +



 () ()

 () (0)

0

sin 
 cos 

for convergence, so  = 1.

sin  =  (0) +  0 (0) +



2
3
4
5
.
.
.

2

− sin 

0

+

−3
0

=  −

5 cos 
.
.
.

5
.
.
.

=

∞


=0

 00 (0) 2  000 (0) 3
 +

2!
3!

 (4) (0) 4  (5) (0) 5
 +
 + ···
4!
5!

= 0 +  + 0 −

−3 cos 
 4 sin 

( + 1)

=0





+1 
 +1 

 = lim  ( + 2)
 = || lim  + 2 = || (1) = ||  1 lim 
→∞   
→∞  ( + 1) 
→∞  + 1

0

1

∞


3 3
5 5
 +0+
 +···
3!
5!

3 3 5 5 7 7
 +
 −
 + ···
3!
5!
7!

(−1)

519

2+1
2+1
(2 + 1)!



 2+3 2+3

 +1 

(2 + 1)! 

 2 2
 = lim   lim 
· 2+1 2+1  = lim
   →∞  (2 + 3)!
 →∞ (2 + 3)(2 + 2) = 0  1 for all , so  = ∞.
→∞


c
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520

¤

CHAPTER 11

INFINITE SEQUENCES AND SERIES

2 =

9.



 () ()

 () (0)

2

1

0
1



2 (ln 2)






+1 +1
 +1 



!
 = lim  (ln 2)

lim 
·
   →∞  ( + 1)!
  
→∞
(ln 2)

ln 2

2

2

2 (ln 2)

(ln 2)2

3

2 (ln 2)3

(ln 2)3

4
.
.
.

2 (ln 2)4
.
.
.

∞
∞
  () (0) 
 (ln 2) 
 =
 .
!
!
=0
=0

(ln 2)4
.
.
.

= lim

→∞

11.



 () ()

 () (0)

0

sinh 

0

1

cosh 

1

2

sinh 

0

3

cosh 

1

4
.
.
.

sinh 
.
.
.



0
.
.
.

1 if  is odd

e

al

rS

4 − 32 + 1

−1

2

6

43 − 6

−2

12 − 6

24

4

24

24

5

0

6
.
.
.

0
.
.
.

0

=

−1
−2
6
( − 1)0 +
( − 1)1 + ( − 1)2
0!
1!
2!

24
24
( − 1)3 +
( − 1)4
3!
4!
= −1 − 2( − 1) + 3( − 1)2 + 4( − 1)3 + ( − 1)4
+

A finite series converges for all , so  = ∞.

N

0
.
.
.

15.

() = ln  =


 () ()

 () (2)

0

ln 

ln 2

1

1

3
4
5
.
.
.

12

−1

−64

−624

=

∞
  () (2)
( − 2)
!
=0

ln 2
1
−1
2
( − 2)0 +
( − 2)1 +
( − 2)2 +
( − 2)3
0!
1! 21
2! 22
3! 23

−122

23

245
.
.
.

4
  () (1)
( − 1)
!
=0

ot

24

 () = 4 − 32 + 1 =

Fo

0

2

2+1
.
=0 (2 + 1)!
∞


 () () = 0 for  ≥ 5, so  has a finite series expansion about  = 1.

 () (1)

2

so sinh  =

= 0  1 for all , so  = ∞.

 () ()

3

0 if  is even

2+1
, then
(2 + 1)!


 2+3

 +1 

1
(2 + 1)! 
 = lim  
 = 2 · lim
·
lim 
→∞   
→∞  (2 + 3)!
→∞ (2 + 3)(2 + 2)
2+1 



2

(0) =



Use the Ratio Test to find . If  =

13.

1

()

(ln 2) ||
= 0  1 for all , so  = ∞.
+1

+

223

2425
.
.
.

= ln 2 +

∞


−6
24
( − 2)4 +
( − 2)5 + · · ·
4! 24
5! 25

(−1)+1

=1

= ln 2 +

∞


(−1)+1

=1

( − 1)!
( − 2)
! 2
1
( − 2)
 2

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.10

TAYLOR AND MACLAURIN SERIES

¤

521









+2

 +1 



| − 2|
( − 2)+1
 2

 = lim  (−1)
 = lim  (−1)( − 2)  = lim lim 
·
→∞   
→∞ 
( + 1) 2+1
(−1)+1 ( − 2)  →∞  ( + 1)2  →∞  + 1
2
=

| − 2|
 1 for convergence, so | − 2|  2 and  = 2.
2



 () ()

 () (3)

0

2

6

1

22

26

2

22 2

46

3

23 2

86

4
.
.
.

24 2
.
.
.

∞
  () (3)
( − 3)
!
=0

()= 2 =

17.

166
.
.
.

6
26
46
( − 3)0 +
( − 3)1 +
( − 3)2
0!
1!
2!

=

+
=

86
166
( − 3)3 +
( − 3)4 + · · ·
3!
4!

∞
 2 6
( − 3)
=0 !

19.

 () = cos  =
 () ()

 () ()

0

cos 

1

− sin 

−1
0

− cos 

0

cos 
.
.
.

4
.
.
.

1

sin 

3

= −1 +

−1
.
.
.

=

∞


( − )2
( − )4
( − )6
−
+
− ···
2!
4!
6!

(−1)+1

Fo

2

∞
  () ()
( − )
!
=0

rS



al

e



 +1 6

 +1 


 ( − 3)+1
!
 = lim  2
 = lim 2 | − 3| = 0  1 for all , so  = ∞.
·  6 lim 

→∞   
→∞ 
→∞  + 1
( + 1)!
2  ( − 3)

=0

( − )2
(2)!





2+2
 +1 
(2)!
 = lim | − |
·
lim 
→∞   
→∞
(2 + 2)!
| − |2

| − |2
= 0  1 for all , so  = ∞.
→∞ (2 + 2)(2 + 1)

ot

= lim







N

21. If  () = sin , then  (+1) () = ±+1 sin  or ±+1 cos . In each case,  (+1) () ≤  +1 , so by Formula 9

with  = 0 and  = +1 , | ()| ≤

+1
||+1
||+1 =
. Thus, | ()| → 0 as  → ∞ by Equation 10.
( + 1)!
( + 1)!

So lim  () = 0 and, by Theorem 8, the series in Exercise 7 represents sin  for all .
→∞

23. If  () = sinh , then for all ,  (+1) () = cosh  or sinh . Since |sinh |  |cosh | = cosh  for all , we have





 (+1) 


() ≤ cosh  for all . If  is any positive number and || ≤ , then  (+1) () ≤ cosh  ≤ cosh , so by

Formula 9 with  = 0 and  = cosh , we have | ()| ≤

cosh 
||+1 . It follows that | ()| → 0 as  → ∞ for
( + 1)!

|| ≤  (by Equation 10). But  was an arbitrary positive number. So by Theorem 8, the series represents sinh  for all .

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

522

¤

CHAPTER 11

INFINITE SEQUENCES AND SERIES



14
(−) = 1 + 1 (−) +
4

=0

∞
√

25. 4 1 −  = [1 + (−)]14 =

1
4

 3
−4
(−)2 +
2!

1
4

 3 7
−4 −4
(−)3 + · · ·
3!

∞
 (−1)−1 (−1) · [3 · 7 · · · · · (4 − 5)] 
1
=1− +

4
4 · !
=2

∞
 3 · 7 · · · · · (4 − 5) 
1
=1− −

4
4 · !
=2

and |−|  1

⇔

||  1, so  = 1.

 
∞
1
1
1  −3   
1
 −3
27.
=
=
=
. The binomial coefficient is
1+
8
2
8 =0 
2
(2 + )3
[2(1 + 2)]3

 
(−3)(−4)(−5) · · · · · (−3 −  + 1)
(−3)(−4)(−5) · · · · · [−( + 2)]
−3
=
=

!
!

∞


(−1)

=0

31.  =

∞
 
=0 !

2+1
(2 + 1)!

(−1)

=0



 () =  cos 1 2 =
2

=0

∞



⇒ cos 1 2 =
(−1)
2

2
(2)!

=0

∞


(−1)

∞

()2+1
2+1
=
2+1 ,  = ∞.
(−1)
(2 + 1)! =0
(2 + 1)!

(−1)

1

2
∞
2

4
=
, so
(−1) 2
(2)!
2 (2)!
=0

2

1
4+1 ,  = ∞.
22 (2)!

ot

∞


⇒ () = sin() =

∞
∞
∞
∞
∞
 (2)
 2 
 1 
 2 
 2 + 1 
=
, so () =  + 2 =
 +
 =
 ,
!
!
!
=0
=0
=0 !
=0 !
=0

⇒ 2 =

 = ∞.
33. cos  =

∞


rS

29. sin  =

al


∞
∞
 (−1) ( + 1)( + 2)
1  (−1) ( + 1)( + 2) 
1
 
=
= for    1 ⇔ ||  2, so  = 2.
8 =0
2
2
2+4
2
(2 + )3
=0

Fo

Thus,

(−1) · 2 · 3 · 4 · 5 · · · · · ( + 1)( + 2)
(−1) ( + 1)( + 2)
=
2 · !
2

e

=

=0

35. We must write the binomial in the form (1+ expression), so we’ll factor out a 4.

N

  

−12
∞
2



2

  −1
2
√
= 
=
1+
= 
=
2
2 4)
2 4
2
4
2 =0 
4
4+
4(1 + 
2 1+


 1  3   2 2  1  3  5   2 3
 1  2
−2 −2 −2
−2 −2



=
+
+···
1 + −2
+
2
4
2!
4
3!
4

=

=

∞
1 · 3 · 5 · · · · · (2 − 1) 2
  
(−1)
+

2
2 =1
2 · 4 · !

∞

1 · 3 · 5 · · · · · (2 − 1) 2+1

2
+
1 ⇔
(−1)
 and 3+1
2 =1
! 2
4

||
 1 ⇔ ||  2, so  = 2.
2





∞
∞
∞
 (−1) (2)2
 (−1) (2)2
 (−1)+1 22−1 2
1
1
1
1−
=
1−1−
=
,
37. sin  = (1 − cos 2) =
2
2
(2)!
2
(2)!
(2)!
=0
=1
=1
2

=∞

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.10
∞
(16) 

39. cos  =

(−1)

=0

() = cos(2 ) =
= 1 − 1 4 +
2

2
(2)!

TAYLOR AND MACLAURIN SERIES

¤

523

⇒

∞
∞
 (−1) (2 )2
 (−1) 4
=
(2)!
(2)!
=0
=0
1 8

24

−

1
12
720

+···

The series for cos  converges for all , so the same is true of the series for
 (), that is,  = ∞. Notice that, as  increases,  () becomes a better approximation to  ().

=0

∞
∞
 (−)



, so − =
=
, so
(−1)
!
!
!
=0
=0

 () = − =

∞


(−1)

=0

1 +1

!

=  − 2 + 1 3 − 1 4 +
2
6

1
6
120

−

+ ···



(−1)−1

=1


( − 1)!

al

=

∞


1 5

24

e

∞
(11) 

41.  =

rS

The series for  converges for all , so the same is true of the series

for (); that is,  = ∞. From the graphs of  and the first few Taylor

polynomials, we see that  () provides a closer fit to  () near 0 as  increases.

cos

  
∞

2

2
4
6
= radians and cos  =
=1−
+
−
+ · · · , so
(−1)
◦
180
36
(2)!
2!
4!
6!
=0

Fo

43. 5◦ = 5◦

(36)2
(36)4
(36)6
(36)2
(36)4

=1−
+
−
+ · · · . Now 1 −
≈ 099619 and adding
≈ 24 × 10−6
36
2!
4!
6!
2!
4!

N

ot

does not affect the fifth decimal place, so cos 5◦ ≈ 099619 by the Alternating Series Estimation Theorem.
 1  3  5 
 1  3 
√


  2 

− 2 − 2 − 2  2 3
− 2 − 2  2 2
2 = 1 + −2 −12 = 1 + − 1
45. (a) 1 1 − 
+
+ ···
− +
−
−
2
2!
3!
∞ 1 · 3 · 5 · · · · · (2 − 1)

=1+
2
2 · !
=1

∞
 1 · 3 · 5 · · · · · (2 − 1) 2+1
1
√
(b) sin−1  =
 =  +  +

(2 + 1)2 · !
1 − 2
=1
=+
∞
(16) 

47. cos  =

∞
 1 · 3 · 5 · · · · · (2 − 1) 2+1

(2 + 1)2 · !
=1

(−1)

=0

 cos(3 ) =

∞


(−1)

=0
∞
(16) 

49. cos  =

(−1)

=0



2
(2)!

⇒

6+1
(2)!

2
(2)!

cos(3 ) =

∞


(−1)

=0

⇒



since 0 = sin−1 0 = .
∞

(3 )2
6
=
(−1)
(2)!
(2)!
=0

 cos(3 )  =  +

⇒ cos  − 1 =

∞


(−1)

=0
∞


(−1)

=1

2
(2)!

∞

2 cos  − 1
 =  +
, with  = ∞.
(−1)

2 · (2)!
=1

⇒

⇒

6+2
, with  = ∞.
(6 + 2)(2)!

∞

2−1 cos  − 1
=
(−1)

(2)!
=1

⇒

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

524

CHAPTER 11

51. arctan  =

∞


(−1)

=0




INFINITE SEQUENCES AND SERIES
∞

2+1
2+4 for ||  1, so 3 arctan  = for ||  1 and
(−1)
2 + 1
2 + 1
=0

3 arctan   =  +

∞


(−1)

=0
12

3 arctan   =

∞


(−1)

=0

0

2+5
. Since
(2 + 1)(2 + 5)

1
2

 1, we have

(12)2+5
(12)5
(12)7
(12)9
(12)11
=
−
+
−
+ · · · . Now
(2 + 1)(2 + 5)
1·5
3·7
5·9
7 · 11

(12)5
(12)7
(12)9
(12)11
−
+
≈ 00059 and subtracting
≈ 63 × 10−6 does not affect the fourth decimal place,
1·5
3·7
5·9
7 · 11 so  12
0

3 arctan   ≈ 00059 by the Alternating Series Estimation Theorem.




 
∞
 12 4+1
12
4 
4  =  +
1+
( ) , so and hence, since 04  1,

 4 + 1
=0
=0

e

∞
√

53. 1 + 4 = (1 + 4 )12 =

04


∞

1 + 4  =

= 04 +

 1  3 
− 2 − 2 (04)13
+
3!
13

1
2

 1  3  5 
− 2 − 2 − 2 (04)17
+ ···
4!
17

(04)9
(04)13
5(04)17
(04)5
−
+
−
+···
10
72
208
2176

(04)5
(04)9
≈ 36 × 10−6  5 × 10−6 , so by the Alternating Series Estimation Theorem,  ≈ 04 +
≈ 040102
72
10

Fo

Now

1
2

rS



12 (04)4+1
=

4 + 1
=0
0
 1
1
1
− 2 (04)9
(04)5
(04)1
+ 2
+ 2
+
= (1)
0!
1! 5
2!
9


al

we have

(correct to five decimal places).

1 2
 − ( − 1 2 + 1 3 − 1 4 + 1 5 − · · · )
 − 1 3 + 1 4 − 1 5 + · · ·
 − ln(1 + )
2
3
4
5
3
4
5
= lim
= lim 2
2
2
→0
→0
→0


2

ot

55. lim

= lim ( 1 − 1  + 1 2 − 1 3 + · · · ) =
2
3
4
5

1
2

N

→0

since power series are continuous functions.

57. lim

→0



1
1
1
 − 3! 3 + 5! 5 − 7! 7 + · · · −  + 1 3 sin  −  + 1 3
6
6
= lim
→0
5
5


1 5
1
 − 7! 7 + · · ·
1
2
4
1
1
−
+
−··· =
=
= lim 5!
= lim
→0
→0 5!
5
7!
9!
5!
120

since power series are continuous functions.
2

59. From Equation 11, we have − = 1 −

2
4
6
2
4
+
−
+ · · · and we know that cos  = 1 −
+
− · · · from
1!
2!
3!
2!
4!



2
Equation 16. Therefore, − cos  = 1 − 2 + 1 4 − · · · 1 − 1 2 +
2
2
2

degree ≤ 4, we get − cos  = 1 − 1 2 +
2

1 4

24

1 4

24


− · · · . Writing only the terms with

− 2 + 1 4 + 1 4 + · · · = 1 − 3 2 +
2
2
2

25 4

24

+ ···.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.10

61.

 (15)

.
=
1 sin 
 − 1 3 + 120 5 − · · ·
6

1 + 1 2 +
6

TAYLOR AND MACLAURIN SERIES

1
5
120

− ···

1 3

6
1 3

6

(−1)−1

=1

67.

−

+···

+···
+···
···

e



∞

3
(35)
3
8
= ln 1 +
[from Table 1] = ln
=
(−1)−1
5

5
5
=1

 2+1
∞
 (−1) 
(−1)  2+1
4
=
= sin  =
4
2+1 (2 + 1)!
(2 + 1)!
=0 4
=0
∞


69. 3 +

+···

rS

∞


1
5
120
1 5

36

al

(−1)

−···

+···.

4


=0

65.



∞
 −4
4
=
= − , by (11).
!
!
=0

−

1
√ ,
2

by (15).

∞
∞
 3
 3
27
81
31
32
33
34
9
+
+
+ ··· =
+
+
+
+··· =
=
− 1 = 3 − 1, by (11).
2!
3!
4!
1!
2!
3!
4!
=1 !
=0 !

Fo

63.

∞


7
4
360

1
5
120


 − 1 3 +
6


From the long division above,
= 1 + 1 2 +
6
sin 

+···

7
5
360
7
5
360

 − 1 3 +
6

7
4
360

¤

71. If  is an th-degree polynomial, then () () = 0 for   , so its Taylor series at  is () =

ot

Put  −  = 1, so that  =  + 1. Then ( + 1) =


 () ()
.
!
=0

N

This is true for any , so replace  by : ( + 1) =


 () ()
!
=0


 () ()
( − ) .
!
=0


 000 ()  ≤    ⇒


 00 () −  00 () ≤  ( − ) ⇒  00 () ≤  00 () +  ( − ). Thus,   00 ()  ≤  [ 00 () +  ( − )]  ⇒

73. Assume that | 000 ()| ≤ , so  000 () ≤  for  ≤  ≤  + . Now




 0 () −  0 () ≤  00 ()( − ) + 1 ( − )2 ⇒  0 () ≤  0 () +  00 ()( − ) + 1  ( − )2
2
2

 0
 0
 ()  ≤   () +  00 ()( − ) + 1 ( − )2  ⇒
2


⇒

 () −  () ≤  0 ()( − ) + 1  00 ()( − )2 + 1 ( − )3 . So
2
6
 () −  () −  0 ()( − ) − 1  00 ()( − )2 ≤
2

1
6 (

− )3 . But

2 () =  () − 2 () =  () −  () −  0 ()( − ) − 1  00 ()( − )2 , so 2 () ≤ 1 ( − )3 .
2
6
A similar argument using  000 () ≥ − shows that 2 () ≥ − 1 ( − )3 . So |2 (2 )| ≤ 1  | − |3 .
6
6
Although we have assumed that   , a similar calculation shows that this inequality is also true if   . c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

525

526

¤

CHAPTER 11

INFINITE SEQUENCES AND SERIES

 
∞
  

75. (a) () =
=0 

 

−1 , so
⇒  () =

=1
∞


0

 
 
 
∞
∞
 
 

−1
−1
(1 + ) () = (1 + )
=
+



=1 
=1 
=1 
0

=

∞


=0

=

∞




∞



 
∞
 


( + 1) +

+1
=0 

( + 1)

=0



Replace  with  + 1 in the first series





∞

( − 1)( − 2) · · · ( −  + 1)( − ) 
( − 1)( − 2) · · · ( −  + 1) 
 +

()
( + 1)!
!
=0

∞
 ( + 1)( − 1)( − 2) · · · ( −  + 1)
[( − ) + ] 
( + 1)!
=0
 
∞
∞
 ( − 1)( − 2) · · · ( −  + 1) 
  
 = ()
 =
=
!
=0
=0 

()
.
1+

al

Thus, 0 () =

e

=

rS

(b) () = (1 + )− () ⇒

0 () = −(1 + )−−1 () + (1 + )− 0 ()
= −(1 + )−−1 () + (1 + )−

()
1+

[Product Rule]
[from part (a)]

Fo

= −(1 + )−−1 () + (1 + )−−1 () = 0

(c) From part (b) we see that () must be constant for  ∈ (−1 1), so () = (0) = 1 for  ∈ (−1 1).

ot

Thus, () = 1 = (1 + )− () ⇔ () = (1 + ) for  ∈ (−1 1).

1. (a)

N

11.11 Applications of Taylor Polynomials


 () ()

0

cos 

1
2
3

− sin 

− cos  sin 

 () (0)
1

1

0

1

−1

0

4

cos 

1

5

− sin 

0

6

− cos 

 ()

−1

1 − 1 2
2
1 − 1 2
2

1 − 1 2 +
2
1 − 1 2 +
2
1 − 1 2 +
2

1 4

24
1 4

24
1 4

24

−

1
6
720

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.11

APPLICATIONS OF TAYLOR POLYNOMIALS

(b)





4

2

0 = 1

2 = 3

4 = 5

6

07071

06916

07074

07071

0

1
1

−02337

00200

−1



1

−00009

−39348

01239

−12114

(c) As  increases,  () is a good approximation to  () on a larger and larger interval.
3.

 () ()

 () (2)

0

1

1
2
−1
4
1
4
−3
8

2

−1

23

−64

3
3 () =
=
=

3
  () (2)
( − 2)
!
=0
1
2

0!
1
2

−

1
4

1!

( − 2) +

1
4

2!

e

2

− 1 ( − 2) + 1 ( − 2)2 −
4
8

5.

 () ()

 () (2)

0

cos 

0

1

− sin 

−1

sin 

1

2

− cos 

3

0

3!

1
(
16

( − 2)3

− 2)3

Fo



3
8

( − 2)2 −

rS

1

al



3

  () (2) 
− 
2
!
=0



3
=− −  + 1 − 
2
6
2

N

7.

ot

3 () =

 () ()

 () (1)

0

ln 

0

1

1



1
2

2

−1

23

3
3 () =

−1
2

3
  () (1)
( − 1)
!
=0

=0+

−1
1
2
( − 1) +
( − 1)2 + ( − 1)3
1!
2!
3!

= ( − 1) − 1 ( − 1)2 + 1 ( − 1)3
2
3

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

527

528

¤

CHAPTER 11

INFINITE SEQUENCES AND SERIES

9.



 () ()

 () (0)

0

−2

0
−2

1

−2

1

−4

(1 − 2)

2

4( − 1)

3

−2

4(3 − 2)

3 () =

12

3
  () (0) 
 =
!
=0

0
1

· 1 + 1 1 +
1

−4 2

2

+

12 3

6

=  − 22 + 23

11. You may be able to simply find the Taylor polynomials for

 () = cot  using your CAS. We will list the values of  () (4)

e

for  = 0 to  = 5.



0

1

2

3

4

5

(4)

1

−2

4

−16

80

−512

rS

5

  () (4) 
− 
4
!
=0


2

3

=1−2 −  +2 −  − 8 −  +
4
4
3
4

al


()

5 () =

10
3


4
−  −
4

64
15


5
− 
4

Fo



For  = 2 to  = 5,  () is the polynomial consisting of all the terms up to and including the  −  term
4
(a)  () =

13.

1

1 −12

2

2

− 1 −32
4

= 2 + 1 ( − 4) −
4

2
1
4

1
− 32

N

3

3 −52

8

 () (4)

ot

0

 () ()
√




√
1
132
 ≈ 2 () = 2 + ( − 4) −
( − 4)2
4
2!

(b) |2 ()| ≤

1
(
64

− 4)2


| − 4|3 , where | 000 ()| ≤  Now 4 ≤  ≤ 42 ⇒
3!

| − 4| ≤ 02 ⇒ | − 4|3 ≤ 0008. Since  000 () is decreasing

on [4 42], we can take  = | 000 (4)| = 3 4−52 =
8
|2 ()| ≤

3
,
256

so

0008
3256
(0008) =
= 0000 015 625.
6
512

(c)
√
From the graph of |2 ()| = |  − 2 ()|, it seems that the error is less than 152 × 10−5 on [4 42].

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.11

(a)  () = 23 ≈ 3 () = 1 + 2 ( − 1) −
3

15.



 () ()

 () (1)

0

23

1

1

2 −13

3

2
3

2

− 2 −43
9

−2
9

3

8 −73
27 

8
27

4

APPLICATIONS OF TAYLOR POLYNOMIALS

− 56 −103
81

¤

529

29
827
( − 1)2 +
( − 1)3
2!
3!

= 1 + 2 ( − 1) − 1 ( − 1)2 +
3
9

4
81 (

− 1)3






| − 1|4 , where   (4) () ≤  . Now 08 ≤  ≤ 12 ⇒
4!




| − 1| ≤ 02 ⇒ | − 1|4 ≤ 00016. Since  (4) () is decreasing



 on [08 12], we can take  =   (4) (08) = 56 (08)−103 , so
81

(b) |3 ()| ≤

|3 ()| ≤

(c)

56
(08)−103
81

24

(00016) ≈ 0000 096 97.





From the graph of |3 ()| = 23 − 3 (), it seems that the

17.



0

()



()

(0)

sec 

1

1

sec  tan 

0

2

sec  (2 sec2  − 1)

1

2

sec  tan  (6 sec  − 1)





||3 , where   (3) () ≤  . Now −02 ≤  ≤ 02 ⇒ || ≤ 02 ⇒ ||3 ≤ (02)3 .
(b) |2 ()| ≤
3!

Fo

3

(a)  () = sec  ≈ 2 () = 1 + 1 2
2

rS



()

al

e

error is less than 0000 053 3 on [08 12].

 (3) () is an odd function and it is increasing on [0 02] since sec  and tan  are increasing on [0 02],

ot

N

(c)



 (3) (02)

 so   (3) () ≤  (3) (02) ≈ 1085 158 892. Thus, |2 ()| ≤
(02)3 ≈ 0001 447.
3!
From the graph of |2 ()| = |sec  − 2 ()|, it seems that the error is less than 0000 339 on [−02 02].

2

19.





()

()

2

0



1

 (2)

0
2

2

 (2 + 4 )

3

 (12 + 83 )

4

 (12 + 482 + 164 )

2
2

(0)

1

2

2



()

2
0

(a) () =  ≈ 3 () = 1 +
(b) |3 ()| ≤

2 2
 = 1 + 2
2!






||4 , where  (4) () ≤  . Now 0 ≤  ≤ 01 ⇒
4!

4 ≤ (01)4 , and letting  = 01 gives
|3 ()| ≤

001 (12 + 048 + 00016)
(01)4 ≈ 000006.
24

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

530

¤

CHAPTER 11

INFINITE SEQUENCES AND SERIES

(c)
 2



From the graph of |3 ()| =  − 3 (), it appears that the

error is less than 0000 051 on [0 01].

(a)  () =  sin  ≈ 4 () =

21.
()





()

0



()

 sin 

(0)

0

2
−4
1
( − 0)2 +
( − 0)4 = 2 − 4
2!
4!
6






||5 , where   (5) () ≤ . Now −1 ≤  ≤ 1 ⇒
5!




|| ≤ 1, and a graph of  (5) () shows that   (5) () ≤ 5 for −1 ≤  ≤ 1.

(b) |4 ()| ≤

2

2 cos  −  sin 

2

3

−3 sin  −  cos 

0

4

−4 cos  +  sin 

−4

5

5 sin  +  cos 

e

0

Thus, we can take  = 5 and get |4 ()| ≤

5
1
· 15 =
= 00416.
5!
24

al

sin  +  cos 

rS

1

(c)

From the graph of |4 ()| = | sin  − 4 ()|, it seems that the

Fo

error is less than 00082 on [−1 1].



3

 −  4 with
 −  + 3 (), where |3 ()| ≤
2
2
4!




 (4) 

 () = |cos | ≤  = 1. Now  = 80◦ = (90◦ − 10◦ ) = 2 − 18 = 4 radians, so the error is
9


1
24

  4
18



+

1
6

≈ 0000 039, which means our estimate would not be accurate to five decimal places. However,

N

  4 
≤
3
9


2

ot

23. From Exercise 5, cos  = −  −

  
3 = 4 , so we can use 4 4  ≤
9

1
120

  5
18

≈ 0000 001. Therefore, to five decimal places,

  3
  cos 80◦ ≈ − − 18 + 1 − 18 ≈ 017365.
6

25. All derivatives of  are  , so | ()| ≤

 (01) ≤


||+1 , where 0    01. Letting  = 01,
( + 1)!

01
(01)+1  000001, and by trial and error we find that  = 3 satisfies this inequality since
( + 1)!

3 (01)  00000046. Thus, by adding the four terms of the Maclaurin series for  corresponding to  = 0, 1, 2, and 3, we can estimate 01 to within 000001. (In fact, this sum is 110516 and 01 ≈ 110517.)

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.11

27. sin  =  −

APPLICATIONS OF TAYLOR POLYNOMIALS

¤

531

1 3
1
 + 5 − · · · . By the Alternating Series
3!
5!

Estimation Theorem, the error in the approximation


1 
1
sin  =  − 3 is less than  5   001 ⇔
 5! 
3!
 5
   120(001) ⇔ ||  (12)15 ≈ 1037. The curves

 =  − 1 3 and  = sin  − 001 intersect at  ≈ 1043, so
6

the graph confirms our estimate. Since both the sine function and the given approximation are odd functions, we need to check the estimate only for   0. Thus, the desired range of values for  is −1037    1037.
3
5
7
+
−
+ · · · . By the Alternating Series
3
5
7


Estimation Theorem, the error is less than − 1 7   005 ⇔
7
 7
   035 ⇔ ||  (035)17 ≈ 08607. The curves

al

 =  − 1 3 + 1 5 and  = arctan  + 005 intersect at
3
5

e

29. arctan  =  −

rS

 ≈ 09245, so the graph confirms our estimate. Since both the

arctangent function and the given approximation are odd functions, we need to check the estimate only for   0. Thus, the desired

Fo

range of values for  is −086    086.

31. Let () be the position function of the car, and for convenience set (0) = 0. The velocity of the car is () = 0 () and the

acceleration is () = 00 (), so the second degree Taylor polynomial is 2 () = (0) + (0) +

(0) 2
 = 20 + 2 . We
2

ot

estimate the distance traveled during the next second to be (1) ≈ 2 (1) = 20 + 1 = 21 m. The function 2 () would not be accurate over a full minute, since the car could not possibly maintain an acceleration of 2 ms2 for that long (if it did, its final

N

speed would be 140 ms ≈ 313 mih!).


−2 







33.  = 2 −
= 2 − 2
= 2 1− 1+
.

( + )2

 (1 + )2


We use the Binomial Series to expand (1 + )−2 :




  
 3
 
 2
 3
 2
2·3·4 




2·3 


−
+···
= 2 2
+4
− ···
+
−3
= 2 1− 1−2


2! 
3!





 


1
≈ 2 ·2
= 2 · 3



when  is much larger than ; that is, when  is far away from the dipole.
35. (a) If the water is deep, then 2 is large, and we know that tanh  → 1 as  → ∞. So we can approximate

tanh(2) ≈ 1, and so  2 ≈ (2) ⇔  ≈


(2).

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

532

¤

CHAPTER 11

INFINITE SEQUENCES AND SERIES

(b) From the table, the first term in the Maclaurin series of


2
 2
2
tanh
≈
, and so  2 ≈
·


2


√
⇔  ≈ .

 () ()

 () (0)

0

tanh  is , so if the water is shallow, we can approximate

tanh 

0

2

1
2
3

sech 

1

2

−2 sech  tanh 

2 sech2  (3 tanh2  − 1)

0
−2

(c) Since tanh  is an odd function, its Maclaurin series is alternating, so the error in the approximation

3

3
2
| 000 (0)| 2
1 2
2
≈ is less than the first neglected term, which is
=
. tanh 

3!

3 

3

3
1
1
3
1 2
2 ·
, so the error in the approximation  2 =  is less

=
If   10, then
3 
3
10
375
 3
·
≈ 00132.
2 375

e

than

⇒

 = . Now sec  = ( + ) ⇒  sec  =  + 

⇒

rS

 =  sec  −  =  sec() − .

al

37. (a)  is the length of the arc subtended by the angle , so  = 

(b) First we’ll find a Taylor polynomial 4 () for  () = sec  at  = 0.


 () (0)

sec 

1

sec  tan 

0

Fo

0

 () ()

1
2

2

sec (2 tan  + 1)
2

1

sec  tan (6 tan  + 5)

0

4

sec (24 tan4  + 28 tan2  + 5)

5

ot

3

N

1
5
5
Thus,  () = sec  ≈ 4 () = 1 + 2! ( − 0)2 + 4! ( − 0)4 = 1 + 1 2 + 24 4 . By part (a),
2

 2
 4 
2
4
54
5 
5
2
1
1 
+
+
.
− =+ · 2 + · 4 − =
 ≈ 1+
2 
24 
2

24

2
243

(c) Taking  = 100 km and  = 6370 km, the formula in part (a) says that
 =  sec() −  = 6370 sec(1006370) − 6370 ≈ 0785 009 965 44 km.
The formula in part (b) says that  ≈

54
5 · 1004
2
1002
+
+
=
≈ 0785 009 957 36 km.
3
2
24
2 · 6370
24 · 63703

The difference between these two results is only 0000 000 008 08 km, or 0000 008 08 m!
39. Using  () =  () +  () with  = 1 and  = , we have  () = 1 () + 1 (), where 1 is the first-degree Taylor

polynomial of  at . Because  =  ,  () =  ( ) +  0 ( )( −  ) + 1 (). But  is a root of  , so  () = 0 and we have 0 = ( ) +  0 ( )( −  ) + 1 (). Taking the first two terms to the left side gives us c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 11 REVIEW

¤

533

 ( )
1 ()
= 0
. By the formula for Newton’s
 0 ( )
 ( )


 1 () 

method, the left side of the preceding equation is +1 − , so |+1 − | =  0
  ( ) . Taylor’s Inequality gives us
 0 ( )( − ) − ( ) = 1 (). Dividing by  0 ( ), we get  −  −

|1 ()| ≤

| 00 ()|
| −  |2 . Combining this inequality with the facts | 00 ()| ≤  and | 0 ()| ≥  gives us
2!

|+1 − | ≤


| − |2 .
2

11 Review

1. (a) See Definition 11.1.1.

(c) The terms of the sequence { } approach 3 as  becomes large.

e

(b) See Definition 11.2.2.

al

(d) By adding sufficiently many terms of the series, we can make the partial sums as close to 3 as we like.

rS

2. (a) See the definition on page 721.

(b) A sequence is monotonic if it is either increasing or decreasing.

(c) By Theorem 11.1.12, every bounded, monotonic sequence is convergent.

(b) The -series
4. If



Fo

3. (a) See (4) in Section 11.2.

∞
 1 is convergent if   1.

=1 

 = 3, then lim  = 0 and lim  = 3.
→∞

→∞

ot

5. (a) Test for Divergence: If lim  does not exist or if lim  6= 0, then the series
→∞

→∞

∞

=1

 is divergent.

N

(b) Integral Test: Suppose  is a continuous, positive, decreasing function on [1 ∞) and let  =  (). Then the series
∞
∞
=1  is convergent if and only if the improper integral 1 ()  is convergent. In other words:
∞

(i) If 1  ()  is convergent, then ∞  is convergent.
=1
∞
∞
(ii) If 1  ()  is divergent, then =1  is divergent.



(c) Comparison Test: Suppose that  and  are series with positive terms.


(i) If  is convergent and  ≤  for all , then  is also convergent.


(ii) If  is divergent and  ≥  for all , then  is also divergent.


(d) Limit Comparison Test: Suppose that  and  are series with positive terms. If lim (  ) = , where  is a
→∞

finite number and   0, then either both series converge or both diverge.

(e) Alternating Series Test: If the alternating series

∞

−1

=1 (−1)

= 1 − 2 + 3 − 4 + 5 − 6 + · · · [  0]

satisfies (i) +1 ≤  for all  and (ii) lim  = 0, then the series is convergent.
→∞

c
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534

¤

CHAPTER 11 INFINITE SEQUENCES AND SERIES

(f ) Ratio Test:


∞
 +1 


(i) If lim 
   =   1, then the series =1  is absolutely convergent (and therefore convergent).
→∞




∞
 +1 


 =   1 or lim  +1  = ∞, then the series   is divergent.
(ii) If lim 
  
  
→∞
→∞
=1



 +1 
 = 1, the Ratio Test is inconclusive; that is, no conclusion can be drawn about the convergence or
(iii) If lim 
→∞   

divergence of .

(g) Root Test:




| | =   1, then the series ∞  is absolutely convergent (and therefore convergent).
=1
→∞





(ii) If lim
| | =   1 or lim
| | = ∞, then the series ∞  is divergent.
=1
→∞
→∞

(iii) If lim  | | = 1, the Root Test is inconclusive.
→∞



 is called absolutely convergent if the series of absolute values



| | is convergent.

al

6. (a) A series

e

(i) If lim

rS


(b) If a series  is absolutely convergent, then it is convergent.

(c) A series  is called conditionally convergent if it is convergent but not absolutely convergent.

7. (a) Use (3) in Section 11.3.

Fo

(b) See Example 5 in Section 11.4.

(c) By adding terms until you reach the desired accuracy given by the Alternating Series Estimation Theorem.
8. (a)

∞


=0

 ( − )

(b) Given the power series

∞


 ( − ) , the radius of convergence is:

ot

=0

(i) 0 if the series converges only when  = 

N

(ii) ∞ if the series converges for all , or

(iii) a positive number  such that the series converges if | − |   and diverges if | − |  .
(c) The interval of convergence of a power series is the interval that consists of all values of  for which the series converges.
Corresponding to the cases in part (b), the interval of convergence is: (i) the single point {}, (ii) all real numbers, that is, the real number line (−∞ ∞), or (iii) an interval with endpoints  −  and  +  which can contain neither, either, or both of the endpoints. In this case, we must test the series for convergence at each endpoint to determine the interval of convergence. 9. (a), (b) See Theorem 11.9.2.
10. (a)  () =

(b)


  () ()
( − )
!
=0

∞
  () ()
( − )
!
=0

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 11 REVIEW

(c)

∞
  () (0) 

!
=0

¤

535

[ = 0 in part (b)]

(d) See Theorem 11.10.8.
(e) See Taylor’s Inequality (11.10.9).
11. (a)–(f ) See Table 1 on page 786.
12. See the binomial series (11.10.17) for the expansion. The radius of convergence for the binomial series is 1.

1. False.

See Note 2 after Theorem 11.2.6.

3. True.

If lim  = , then as  → ∞, 2 + 1 → ∞, so 2+1 → .

5. False.

For example, take  = (−1)(6 ).






al




e

→∞






3



3

3

9. False.

rS

+1 
1

 
1 
1
= lim 
= lim 
= lim
7. False, since lim 
·
·
= 1.
→∞   
→∞  ( + 1)3
1  →∞  ( + 1)3 13  →∞ (1 + 1)3
See the note after Example 2 in Section 11.4.
See (9) in Section 11.1.

13. True.

By Theorem 11.10.5 the coefficient of 3 is

 000 (0)
1
=
3!
3

Fo

11. True.

⇒  000 (0) = 2.

Or: Use Theorem 11.9.2 to differentiate  three times.
For example, let  =  = (−1) . Then { } and { } are divergent, but   = 1, so {  } is convergent.

ot

15. False.

17. True by Theorem 11.6.3.

(−1)  is absolutely convergent and hence convergent.]

099999    = 09 + 09(01)1 + 09(01)2 + 09(01)3 + · · · =

N

19. True.



[

∞


(09)(01)−1 =

=1

09
= 1 by the formula
1 − 01

for the sum of a geometric series [ = 1 (1 − )] with ratio  satisfying ||  1.

21. True. A finite number of terms doesn’t affect convergence or divergence of a series.

1.



2 + 3
1 + 23



converges since lim

3. lim  = lim
→∞

→∞

→∞

2 + 3
23 + 1
1
= lim
= .
→∞ 13 + 2
1 + 23
2

3

= ∞, so the sequence diverges.
= lim
→∞ 12 + 1
1 + 2

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° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

536

¤

CHAPTER 11 INFINITE SEQUENCES AND SERIES



  sin  

1
≤
 , so | | → 0 as  → ∞. Thus, lim  = 0. The sequence { } is convergent.
→∞
2 + 1  2 + 1


5. | | = 



4 
4

3
3
1+
7.
. Then is convergent. Let  = 1 +





1
3
− 2
1 + 3

ln(1 + 3) H
12
= lim lim ln  = lim 4 ln(1 + 3) = lim
= lim
= 12, so
→∞
→∞
→∞
→∞
→∞ 1 + 3
1(4)
−1(42 )

4
3
1+
= 12 .
→∞


lim  = lim

→∞

9. We use induction, hypothesizing that −1    2. Note first that 1  2 =

1
3

(1 + 4) =

5
3

 2, so the hypothesis holds

for  = 2. Now assume that −1    2. Then  = 1 (−1 + 4)  1 ( + 4)  1 (2 + 4) = 2. So   +1  2,
3
3
3
and the induction is complete. To find the limit of the sequence, we note that  = lim  = lim +1

⇒

al

 = 1 ( + 4) ⇒  = 2.
3

∞
∞

 1



1
 3 = 2 , so converges by the Comparison Test with the convergent -series
[  = 2  1].
2
3 + 1


3 + 1
=1
=1 







 +1 
( + 1)3 5

lim
13. lim 
· 3
   = →∞
→∞
5+1




rS

11.

→∞

e

→∞


3
∞
 3
1
1
1
· =  1, so converges by the Ratio Test.
1+

→∞

5
5
=1 5

= lim

Fo

1
√
. Then  is continuous, positive, and decreasing on [2 ∞), so the Integral Test applies. ln 
 
 ln 
 ∞

 √ ln 

1
1
√
 ()  = lim
−12  = lim 2 
  = ln ,  =  = lim

→∞ 2 
→∞ ln 2
→∞
ln 2 ln 
2
 √

√
= lim 2 ln  − 2 ln 2 = ∞

15. Let  () =



so the series

series

1
√
diverges. ln 
=2 


 

∞

cos 3 
1
1
5
≤

=
, so
| | converges by comparison with the convergent geometric



1 + (12)
1 + (12)
(12)
6
=1

∞
  5  
=
6

=1



N




17. | | = 


∞


ot

→∞



5
6

∞


 1 . It follows that
 converges (by Theorem 3 in Section 11.6).
=1

 +1 
2 + 1
5 !
2
 = lim 1 · 3 · 5 · · · · · (2 − 1)(2 + 1) ·
19. lim 
= lim
=  1, so the series
→∞   
→∞
5+1 ( + 1)!
1 · 3 · 5 · · · · · (2 − 1) →∞ 5( + 1)
5
converges by the Ratio Test.

√
√
∞



−1
 0, { } is decreasing, and lim  = 0, so the series converges by the Alternating
21.  =
(−1)
→∞
+1
+1
=1
Series Test.

c
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CHAPTER 11 REVIEW

23. Consider the series of absolute values:

∞


−13 is a p-series with  =

=1

1
3

¤

537

≤ 1 and is therefore divergent. But if we apply the

∞

1
Alternating Series Test, we see that  = √  0, { } is decreasing, and lim  = 0, so the series
(−1)−1 −13
3
→∞

=1
∞


converges. Thus,

(−1)−1 −13 is conditionally convergent.

=1












Test,

∞
 (−1) ( + 1)3 is absolutely convergent.
22+1
=1

29.

∞


1 8
1
·
=
8 11
11

[tan−1 ( + 1) − tan−1 ] = lim 
→∞

=1

al

=

e

−1



∞
∞
∞
∞
∞
 (−3)−1
 (−3)−1
 (−3)−1
1
1  (−3)−1
1 
1
3
=
=
=
=
=
−
3 
23
8
8 =1 8−1
8 =1
8
8 1 − (−38)
=1
=1 (2 )
=1

rS

27.


 +2 3
( + 2)3+1
1 + (2) 3
3
22+1
=
· =
· →  1 as  → ∞, so by the Ratio
·
22+3
(−1) ( + 1)3   + 1 4
1 + (1) 4
4

+1

+1   (−1)
25. 

=

= lim [(tan−1 2 − tan−1 1) + (tan−1 3 − tan−1 2) + · · · + (tan−1 ( + 1) − tan−1 )]
→∞

= lim [tan−1 ( + 1) − tan−1 1] =
31. 1 −  +


2

−


4

=


4

Fo

→∞

∞
∞
∞

 (−)
 
2

3
4
(−1)
−
+
−··· =
=
= − since  = for all .
2!
3!
4!
!
!
=0
=0
=0 !

∞

∞
 (−)
1 
1  
( + − ) =
+
2
2 =0 ! =0 !

 

1
2
3
4
2
3
4
=
1++
+
+
+··· + 1 − +
−
+
−···
2
2!
3!
4!
2!
3!
4!


∞
 2
1
2
4
1
1
=
2+2·
+2·
+ · · · = 1 + 2 +
≥ 1 + 2 for all 
2
2!
4!
2
2
=2 (2)!

N

ot

33. cosh  =

35.

∞
 (−1)+1
1
1
1
1
1
1
1
+
−
+
−
+
−
+···.
=1−
5
32
243
1024
3125
7776
16,807
32,768
=1

Since 8 =

37.

∞


∞
7
 (−1)+1
 (−1)+1
1
1
=
≈
≈ 09721.
 0000031,
85
32,768
5
5
=1
=1

8

1
1
1
1
≈
≈ 018976224. To estimate the error, note that
  , so the remainder term is


2 + 5
5
=1 2 + 5
=1 2 + 5

8 =

∞

 1
1
159
= 64 × 10−7 geometric series with  =

=


1 − 15
=9 2 + 5
=9 5
∞


1
59

and  =

1
5


.

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538

¤

CHAPTER 11 INFINITE SEQUENCES AND SERIES





  + 1  
+1
1

 
= lim 1 +
= 1  0.
39. Use the Limit Comparison Test. lim 
 = lim
→∞ 
→∞
 →∞ 


Since




 

  + 1

| | is convergent, so is 
 , by the Limit Comparison Test.









 +1 
| + 2|+1
 | + 2|
| + 2|
 4

 = lim
=
 1 ⇔ | + 2|  4, so  = 4.
41. lim 
·
= lim
→∞
→∞  + 1
  →∞ ( + 1) 4+1 | + 2|
4
4
| + 2|  4 ⇔ −4   + 2  4 ⇔ −6    2. If  = −6, then the series

∞
 ( + 2) becomes  4
=1

series becomes the harmonic series





√
( − 3)+1
+3 
+3
 = 2 | − 3| lim
√
· 
= 2 | − 3|  1 ⇔ | − 3|  1 ,
2
→∞
2 ( − 3) 
+4
+4

+1

so  = 1 . | − 3| 
2

1
2

⇔ −1   − 3 
2

1
2

5

alternating series, so  =

7
2 2

45.

 () ()

0

sin 

1

cos 

3
4
.
.
.

− sin 

− cos  sin 
.
.
.



6
1
2
√
3
2
−1
2
√
− 23
1
2

∞
 2 ( − 3)
√
becomes
+3
=1

∞

 (−1)
√
, which is a convergent
≤ 1 , but for  = 5 , we get
2
+3
=0

N

2

 ()

   7 . For  = 7 , the series
2
2

ot




.

1
2

5
2

Fo

∞

 1
1
√
=
, which diverges  =
12
+3
=0
=3
∞


⇔

al




2
+1 
= lim 
43. lim 
→∞   
→∞ 

rS




∞
 1
, which diverges. Thus,  = [−6 2).
=1 

e

∞
∞
 (−4)
 (−1)
, the alternating harmonic series, which converges by the Alternating Series Test. When  = 2, the
=


=1 4
=1

.
.
.

 
 
 
00 
(4) 
  


 (3)
2
  6 

 3 
 4
6
6
+ 0
−
+
−
−
−
sin  = 
+
+
+ ···
6
6
6
2!
6
3!
6
4!
6
 √ 


1
1
3
1
 2
 4

1 
 3
=
+
− ··· +
+···
1−
−
−
−
−
−
2
2!
6
4!
6
2
6
3!
6
√

∞
∞
2
2+1
1 
1 
3 
1


=
(−1)
+
(−1)
−
−
2 =0
(2)!
6
2 =0
(2 + 1)!
6


47.

∞
∞


1
1
=
=
(−) =
(−1)  for ||  1 ⇒
1+
1 − (−) =0
=0

∞

2
=
(−1) +2 with  = 1.
1 +  =0

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CHAPTER 11 REVIEW

49.





1
 = − ln(4 − ) +  and
4−


1
1
 =
4−
4

ln(4 − ) = −

1
1
 =
1 − 4
4



¤

 ∞ 
∞
∞
   
 
1
1 
+1
 =
 =
+ . So

 ( + 1)
4 =0 4
4 =0 4
=0 4

∞
∞
∞

 
+1
+1
1 
+ . Putting  = 0, we get  = ln 4.
+ =−
+ =−

4 =0 4 ( + 1)
4+1 ( + 1)
=0
=1 4

Thus, () = ln(4 − ) = ln 4 −
Another solution:

∞
 
. The series converges for |4|  1

=1 4

⇔

||  4, so  = 4.

ln(4 − ) = ln[4(1 − 4)] = ln 4 + ln(1 − 4) = ln 4 + ln[1 + (−4)]
(−1)+1

=1

51. sin  =

∞
 (−1) 2+1
(2 + 1)!
=0

∞
∞

 
(−4)

[from Table 1] = ln 4 +
(−1)2+1  = ln 4 −
.


4
=1
=1 4

⇒ sin(4 ) =

∞
∞
 (−1) (4 )2+1
 (−1) 8+4
=
for all , so the radius of
(2 + 1)!
(2 + 1)!
=0
=0

rS

convergence is ∞.

e

∞


al

= ln 4 +


−14
1
1
1
1
1
= √ 
= 
14 = 2 1 − 16 
4
4
1
16 − 
16(1 − 16)
16 1 − 16 


 1  5  
 1  5  9


−4 −4
−4 −4 −4
1

 2
 3
1
1+ −
−
+
−
−
=
+
+···
2
4
16
2!
16
3!
16
∞
∞
 1 · 5 · 9 · · · · · (4 − 3) 
 1 · 5 · 9 · · · · · (4 − 3) 
1
1
 = +
+

 · ! · 16
2 =1
2·4
2 =1
26+1 !

 

 for −   1
16



||  16, so  = 16.

∞
∞
∞
∞
∞
 
 −1
 −1
 −1

1  
1
, so
=
=
= −1 +
= + and 
 =0 !
 =1 !
=0 !
=0 !
=1 !

N

55.  =

⇔

ot

=

Fo

53.  () = √
4

∞
 

 =  + ln || +
.

=1  · !

57. (a)


0
1
2
3
4
.
.
.

 () ()
12



1 −12

2
1 −32
−4
3 −52

8
− 15 −72
16

.
.
.

 () (1)
1

√
12
38
14
 ≈ 3 () = 1 +
( − 1) −
( − 1)2 +
( − 1)3
1!
2!
3!
= 1 + 1 ( − 1) − 1 ( − 1)2 +
2
8

1
(
16

− 1)3

1
2
−1
4
3
8
− 15
16

.
.
.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

539

540

¤

CHAPTER 11 INFINITE SEQUENCES AND SERIES

(b)

(c) |3 ()| ≤






| − 1|4 , where  (4) () ≤  with
4!

 (4) () = − 15 −72 . Now 09 ≤  ≤ 11 ⇒
16
−01 ≤  − 1 ≤ 01

⇒

( − 1)4 ≤ (01)4 ,

and letting  = 09 gives  =
|3 ()| ≤

15
, so
16(09)72

15
(01)4 ≈ 0000 005 648
16(09)72 4!
≈ 0000 006 = 6 × 10−6

(d)

√
From the graph of |3 ()| = |  − 3 ()|, it appears that

al

e

the error is less than 5 × 10−6 on [09 11].

2+1
3
5
7
3
5
7
= −
+
−
+ · · · , so sin  −  = −
+
−
+ · · · and
(2 + 1)!
3!
5!
7!
3!
5!
7!
=0


sin  − 
2
4
2
4
1
1
1
sin  − 
−
+ · · · . Thus, lim
−
+ ··· = − .
=− +
= lim − +
→0
→0
3
3!
5!
7!
3
6
120
5040
6

61.  () =

∞


=0

(−1)

 

⇒  (−) =

∞


=0

rS

∞


∞


Fo

59. sin  =

 (−) =

(−1)  

=0

(a) If  is an odd function, then  (−) = − () ⇒

∞


(−1)   =

=0

=0

−  . The coefficients of any power series

ot

are uniquely determined (by Theorem 11.10.5), so (−1)  = − .

∞


If  is even, then (−1) = 1, so  = −

⇒ 2 = 0 ⇒  = 0. Thus, all even coefficients are 0, that is,

N

0 = 2 = 4 = · · · = 0.

(b) If  is even, then  (−) =  () ⇒

∞


=0

If  is odd, then (−1) = −1, so − = 

(−1)   =

∞


=0

 

⇒ (−1)  =  .

⇒ 2 = 0 ⇒  = 0. Thus, all odd coefficients are 0,

that is, 1 = 3 = 5 = · · · = 0.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

PROBLEMS PLUS
1. It would be far too much work to compute 15 derivatives of . The key idea is to remember that  () (0) occurs in the

coefficient of  in the Maclaurin series of  . We start with the Maclaurin series for sin: sin  =  −
Then sin(3 ) = 3 −
 (15) (0) =

3
5
+
− ···.
3!
5!

15
1
9
 (15) (0)
+
− · · · , and so the coefficient of 15 is
= . Therefore,
3!
5!
15!
5!

15!
= 6 · 7 · 8 · 9 · 10 · 11 · 12 · 13 · 14 · 15 = 10,897,286,400.
5!

3. (a) From Formula 14a in Appendix D, with  =  = , we get tan 2 =

⇒

1 − tan2 
= cot  − tan . Replacing  by 1 , we get 2 cot  = cot 1  − tan 1 , or
2
2
2
tan 

e

2 cot 2 =

1 − tan2 
2 tan 
, so cot 2 =
2
1 − tan 
2 tan 


2−1

in place of , tan

∞
 1




= cot  − 2 cot −1 , so the th partial sum of tan  is


2
2
2
2
=1 2

rS

(b) From part (a) with

al

tan 1  = cot 1  − 2 cot .
2
2

tan(2) tan(4) tan(8) tan(2 )
+
+
+··· +
2
4
8
2

 
 

cot(2) cot(4) cot(8) cot(2) cot(4)
=
− cot  +
−
+
−
+ ···
2
4
2
8
4


cot(2 ) cot(2−1 ) cot(2 )
= − cot  +
−
[telescoping sum]
+

−1
2
2
2

Now

Fo

 =

cot(2 ) cos(2 )
2
1
1
cos(2 )
=
·
→ · 1 = as  → ∞ since 2 → 0
= 

)
2
2 sin(2

sin(2 )



N

ot

for  6= 0. Therefore, if  6= 0 and  6=  where  is any integer, then


∞
 1


1
1 tan  = lim  = lim − cot  +  cot  = − cot  +
→∞
→∞
2
2
2
2

=1

If  = 0, then all terms in the series are 0, so the sum is 0.

5. (a) At each stage, each side is replaced by four shorter sides, each of length
1
3

of the side length at the preceding stage. Writing 0 and 0 for the

number of sides and the length of the side of the initial triangle, we generate the table at right. In general, we have  = 3 · 4 and
 
 = 1 , so the length of the perimeter at the th stage of construction
3
 
  is  =   = 3 · 4 · 1 = 3 · 4 .
3
3
 
−1

4
4
(b)  = −1 = 4
. Since 4  1,  → ∞ as  → ∞.
3
3
3


0 = 3

0 = 1

1 = 3 · 4

1 = 13

2

2 = 132

3

3 = 133
.
.
.

2 = 3 · 4
3 = 3 · 4
.
.
.

(c) The area of each of the small triangles added at a given stage is one-ninth of the area of the triangle added at the preceding stage. Let  be the area of the original triangle. Then the area  of each of the small triangles added at stage  is c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

541

542

¤

CHAPTER 11 PROBLEMS PLUS

 =  ·

1

=  . Since a small triangle is added to each side at every stage, it follows that the total area  added to the
9
9

figure at the th stage is  = −1 ·  = 3 · 4−1 · curve is  =  + 1 + 2 + 3 + · · · =  +  · geometric series with common ratio

triangle with side 1 is  =


4−1
=  · 2−1 . Then the total area enclosed by the snowflake

9
3

1
42
43
4
+  · 3 +  · 5 +  · 7 + · · · . After the first term, this is a
3
3
3
3

3
8
 9
4
, so  =  +
. But the area of the original equilateral
=+ · =
9
3 5
5
1− 4
9

√
√
√
1
3
3

8
2 3
· 1 · sin =
. So the area enclosed by the snowflake curve is ·
=
.
2
3
4
5 4
5

7. (a) Let  = arctan  and  = arctan . Then, from Formula 14b in Appendix D,

tan(arctan ) − tan(arctan )
−
tan  − tan 
=
=
1 + tan  tan 
1 + tan(arctan ) tan(arctan )
1 + 

(b) From part (a) we have

120
1
− 239
119
1
+ 120 · 239
119

= arctan

28,561
28,441
28,561
28,441

rS

1 arctan 120 − arctan 239 = arctan
119

− since −   −  
2
1 + 

al

Now arctan  − arctan  =  −  = arctan(tan( − )) = arctan

e

tan( − ) =

1

(c) Replacing  by − in the formula of part (a), we get arctan  + arctan  = arctan

= arctan

1
5

= arctan 1 =


4

+
. So
1 − 

+1
5
5
5
5
= 2 arctan 12 = arctan 12 + arctan 12
1− 1 · 1
5 5

Fo



4 arctan 1 = 2 arctan 1 + arctan 1 = 2 arctan
5
5
5


.
2

5
12

+

5
12

1−

5
12

·

5
12

= arctan 120
119

ot

1
1
Thus, from part (b), we have 4 arctan 1 − arctan 239 = arctan 120 − arctan 239 =
5
119

5
7
9
11
3
+
−
+
−
+ · · · , so
3
5
7
9
11

N

(d) From Example 7 in Section 11.9 we have arctan  =  −


4.

arctan

1
1
1
1
1
1
1
+
−
+
−
+···
= −
5
5
3 · 53
5 · 55
7 · 57
9 · 59
11 · 511

This is an alternating series and the size of the terms decreases to 0, so by the Alternating Series Estimation Theorem, the sum lies between 5 and 6 , that is, 0197395560  arctan 1  0197395562.
5
(e) From the series in part (d) we get arctan

1
1
1
1
+
− · · · . The third term is less than
=
−
239
239
3 · 2393
5 · 2395

26 × 10−13 , so by the Alternating Series Estimation Theorem, we have, to nine decimal places,
1
1 arctan 239 ≈ 2 ≈ 0004184076. Thus, 0004184075  arctan 239  0004184077.
1
(f ) From part (c) we have  = 16 arctan 1 − 4 arctan 239 , so from parts (d) and (e) we have
5

16(0197395560) − 4(0004184077)    16(0197395562) − 4(0004184075) ⇒
3141592652    3141592692. So, to 7 decimal places,  ≈ 31415927. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

CHAPTER 11 PROBLEMS PLUS

∞


−1

=1

∞


1
, ||  1, and differentiate:
1−
=0
∞



∞
∞
 


1
1



=
=
 for ||  1 ⇒
 = 
−1 =
=
2
 =0
 1 − 
(1 − )
(1 − )2
=1
=1

9. We start with the geometric series

 =

for ||  1. Differentiate again:
∞


2 −1 =

=1
∞


3 −1 =

=1



(1 − )2 −  · 2(1 − )(−1)
+1
=
=
 (1 − )2
(1 − )4
(1 − )3

⇒

∞


2  =

=1

 2 + 
(1 − )3 (2 + 1) − (2 + )3(1 − )2 (−1)
2 + 4 + 1
=
=
3
6
 (1 − )
(1 − )
(1 − )4

2 + 
(1 − )3

⇒

⇒

3 + 42 + 
, ||  1. The radius of convergence is 1 because that is the radius of convergence for the
(1 − )4
=1
 geometric series we started with. If  = ±1, the series is 3 (±1) , which diverges by the Test For Divergence, so the
∞


3  =

1
2



 2

 −1
( + 1)( − 1)
= ln[( + 1)( − 1)] − ln 2
= ln
= ln
2
2

al



11. ln 1 −

e

interval of convergence is (−1 1).

= ln( + 1) + ln( − 1) − 2 ln  = ln( − 1) − ln  − ln  + ln( + 1)

−1
−1

− [ln  − ln( + 1)] = ln
− ln
.


+1








−1

1
Let  = ln − ln for  ≥ 2. Then ln 1 − 2 =


+1
=2
=2

 



1
2
2
3
−1

1

 = ln − ln
+ ln − ln
+ · · · + ln
− ln
= ln − ln
, so
2
3
3
4

+1
2
+1




∞


1
1
1
= ln − ln 1 = ln 1 − ln 2 − ln 1 = − ln 2. ln 1 − 2 = lim  = lim ln − ln
→∞
→∞

2
+1
2
=2

Fo

rS

= ln

The x-intercepts of the curve occur where sin  = 0 ⇔  = ,

ot

13. (a)

 an integer. So using the formula for disks (and either a CAS or

N

sin2  = 1 (1 − cos 2) and Formula 99 to evaluate the integral),
2

the volume of the nth bead is
 
 
 =  (−1) (−10 sin )2  =  (−1) −5 sin2  
=

250
(−(−1)5
101

− −5 )

(b) The total volume is


∞
0

−5 sin2   =

∞


 =

=1

250
101

∞


[−(−1)5 − −5 ] =

=1

Another method: If the volume in part (a) has been written as  = as a geometric series with  =

250
(1
101

250
101

[telescoping sum].

250 −5 5

(
101

− 1), then we recognize

− −5 ) and  = −5 

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

∞


=1



543

544

¤

CHAPTER 11 PROBLEMS PLUS

15. If  is the length of a side of the equilateral triangle, then the area is  =

1
2

·

√
3
2 

=

√

3 2
4 

and so 2 =

4
√ .
3

Let  be the radius of one of the circles. When there are  rows of circles, the figure shows that
=

√
√ 
√


√
3  +  + ( − 2)(2) +  + 3  =  2 − 2 + 2 3 , so  = 
.
2 + 3−1

The number of circles is 1 + 2 + · · · +  =

( + 1)
, and so the total area of the circles is
2

( + 1) 2
( + 1)
2
 =
 
√
2
2
2
4 + 3−1
√

4 3
( + 1)
( + 1)
 
=
√
√
2 = 
2 √
2
+ 3−1 2 3
4 + 3−1

 =

⇒


( + 1)

= 
√
2 √

+ 3−1 2 3

al

e

1 + 1


= 
 2 √ → √ as  → ∞
√
2 3
3 − 1  2 3
1+

17. As in Section 11.9 we have to integrate the function  by integrating series. Writing  = (ln  ) =  ln  and using the
∞
∞
 ( ln )
  (ln )
=
. As with power series, we can
!
!
=0
=0
 1
∞
 1
 (ln )
 =
 (ln ) . We integrate by parts
!
=0 ! 0

integrate this series term-by-term:



1

  =

∞


=0

0



−1

(ln )




0

1

 (ln )  = lim

→0+



1

 and  =

+1
:
+1

 (ln )  = lim


+1

→0+



1



+1
(ln )
+1

1


− lim

→0+





1


 (ln )−1 
+1

 (ln )−1 

ot

=0−



1

Fo

with  = (ln ) ,  =  , so  =

0

rS

Maclaurin series for  , we have  = (ln  ) =  ln  =

0

N

(where l’Hospital’s Rule was used to help evaluate the first limit). Further integration by parts gives
 1
 1

 (ln )  = −
 (ln )−1  and, combining these steps, we get
+1 0
0

 1
(−1) ! 1 
(−1) !
 (ln )  =
  =
⇒

( + 1) 0
( + 1)+1
0
 1
 1
∞
∞
∞
∞
 1
 1 (−1) !

 (−1)−1
(−1)
  =
 (ln )  =
=
=
.
+1
+1

=0 ! 0
=0 ! ( + 1)
=0 ( + 1)
=1
0
2+1
1 for ||  1. In particular, for  = √ , we
2 + 1
3
=0
 √ 2+1


 
∞
∞


1 3
1
1
1
1

√
=
=
, so
(−1)
(−1) have = tan−1 √
6
2 + 1
3
3
3 2 + 1
=0
=0


∞
∞
∞
∞
√ 
√


6 

(−1)
(−1)
(−1)
(−1)
= √
=2 3
=2 3 1+
= √ − 1.
⇒




3 =0 (2 + 1)3
2 3
=0 (2 + 1)3
=1 (2 + 1)3
=1 (2 + 1)3

19. By Table 1 in Section 11.10, tan−1  =

∞


(−1)

c
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CHAPTER 11 PROBLEMS PLUS

21. Let () denote the left-hand side of the equation 1 +

¤

545

2
3
4

+
+
+
+ · · · = 0. If  ≥ 0, then  () ≥ 1 and there are
2!
4!
6!
8!

2
4
6
8
+
−
+
− · · · = cos . The solutions of cos  = 0 for
2!
4!
6!
8!

2

  0 are given by  = − , where  is a positive integer. Thus, the solutions of  () = 0 are  = −
−  , where
2
2 no solutions of the equation. Note that  (−2 ) = 1 −

 is a positive integer.

23. Call the series . We group the terms according to the number of digits in their denominators:

=

1
1

+

1
2

+··· +

1

1
8

+

1
9





+



1

 11

+··· +

2

1
99




+



1

 111

+··· +

3

1
999




+···

∞
  9 −1
9 10 is a geometric series with  = 9 and  =

=1
∞


 

=1

25.  = 1 +

∞
  9 −1
9 10
=

=1

9
1 − 910

= 90.

 1. Therefore, by the Comparison Test,

rS

=

9
10

al

Now

e

Now in the group  , since we have 9 choices for each of the  digits in the denominator, there are 9 terms.
 9 −1
1
1
.
Furthermore, each term in  is less than 10−1 [except for the first term in 1 ]. So   9 · 10−1 = 9 10

6
9
4
7
10
2
5
8
3
+
+
+···,  =  +
+
+
+ ···,  =
+
+
+···.
3!
6!
9!
4!
7!
10!
2!
5!
8!

Fo

Use the Ratio Test to show that the series for , , and  have positive radii of convergence (∞ in each case), so
Theorem 11.9.2 applies, and hence, we may differentiate each of these series:
32
65
98
2
5
8

=
+
+
+ ··· =
+
+
+··· = 

3!
6!
9!
2!
5!
8!


3
6
9

4
7
10
=1+
+
+
+ · · · = , and
=+
+
+
+ · · · = .

3!
6!
9!

4!
7!
10!

ot

Similarly,

N

So 0 = ,  0 = , and 0 = . Now differentiate the left-hand side of the desired equation:
 3
( +  3 + 3 − 3) = 32 0 + 3 2  0 + 32 0 − 3(0  +  0  + 0 )

= 32  + 3 2  + 32  − 3(2 + 2  +  2 ) = 0

⇒

3 +  3 + 3 − 3 = . To find the value of the constant , we put  = 0 in the last equation and get
13 + 03 + 03 − 3(1 · 0 · 0) = 

⇒  = 1, so 3 +  3 + 3 − 3 = 1.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

N ot e

al

rS

Fo

APPENDIXES
A Numbers, Inequalities, and Absolute Values
1. |5 − 23| = |−18| = 18
3. |−| =  because   0.

√



5.  5 − 5 = −

√
√
√

5 − 5 = 5 − 5 because 5 − 5  0.

7. If   2,  − 2  0, so | − 2| = −( − 2) = 2 − .

if   −1

[since 2 + 1 ≥ 0 for all ].

⇔ 2  −4 ⇔   −2, so  ∈ (−2 ∞).

13. 2 + 7  3
15. 1 −  ≤ 2

− − 1

if  ≥ −1

⇔ − ≤ 1 ⇔  ≥ −1, so  ∈ [−1 ∞).

⇔ 9  3 ⇔ 3  , so  ∈ (3 ∞).

17. 2 + 1  5 − 8
19. −1  2 − 5  7
21. 0 ≤ 1 −   1

al



11. 2 + 1 = 2 + 1

if  + 1  0

+1

e

−( + 1)

=



rS



if  + 1 ≥ 0

+1

⇔ 4  2  12 ⇔ 2    6, so  ∈ (2 6).

Fo

9. | + 1| =



⇔ −1 ≤ −  0 ⇔ 1 ≥   0, so  ∈ (0 1].

⇔ 2  1 ⇔   1 , and
2


1
2 + 1 ≤ 3 + 2 ⇔ −1 ≤ . Thus,  ∈ −1 2 .

ot

23. 4  2 + 1 ≤ 3 + 2. So 4  2 + 1

N

25. ( − 1)( − 2)  0.

Case 1: (both factors are positive, so their product is positive)  − 1  0 ⇔   1, and  − 2  0 ⇔   2, so  ∈ (2 ∞).

Case 2: (both factors are negative, so their product is positive)  − 1  0 ⇔   1, and  − 2  0 ⇔   2, so  ∈ (−∞ 1).

Thus, the solution set is (−∞ 1) ∪ (2 ∞).
27. 22 +  ≤ 1

⇔ 22 +  − 1 ≤ 0 ⇔ (2 − 1) ( + 1) ≤ 0.

Case 1: 2 − 1 ≥ 0 ⇔  ≥ 1 , and  + 1 ≤ 0 ⇔  ≤ −1,
2
which is an impossible combination.


Case 2: 2 − 1 ≤ 0 ⇔  ≤ 1 , and  + 1 ≥ 0 ⇔  ≥ −1, so  ∈ −1 1 .
2
2


Thus, the solution set is −1 1 .
2
c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

547

548

¤

APPENDIX A NUMBERS, INEQUALITIES, AND ABSOLUTE VALUES

2

+ 1 +
2

 0. But since

2
 + 1 ≥ 0 for every real , the original inequality will be true for all real  as well.
2

29. 2 +  + 1  0

⇔ 2 +  +

1
4

+

3
4

0 ⇔

3
4

Thus, the solution set is (−∞ ∞).

√ 
√ 

⇔ 2 − 3  0 ⇔  − 3  + 3  0.
√
√
Case 1:   3 and   − 3, which is impossible.
√
√
Case 2:   3 and   − 3.
 √ √ 
Thus, the solution set is − 3 3 .

31. 2  3

Another method: 2  3 ⇔ || 

33. 3 − 2 ≤ 0

√
√
√
3 ⇔ − 3    3.

⇔ 2 ( − 1) ≤ 0. Since 2 ≥ 0 for all , the inequality is satisfied when  − 1 ≤ 0 ⇔  ≤ 1.

e

Thus, the solution set is (−∞ 1].



−1

+1

  −1

−

−

−

01

+

−

+

−1    0

+

−

−

+

+

+
−

+

Fo

1

−

( − 1)( + 1)

rS

Interval

al



⇔ 3 −   0 ⇔  2 − 1  0 ⇔ ( − 1)( + 1)  0. Construct a chart:

35. 3  

+

Since 3   when the last column is positive, the solution set is (−1 0) ∪ (1 ∞).
37. 1  4. This is clearly true for   0. So suppose   0. then 1  4

39.  =

5
(
9

1
4


 . Thus, the solution set is (−∞ 0) ∪ 4  ∞ .

ot

1  4 ⇔

1

⇔

− 32) ⇒  = 9  + 32. So 50 ≤  ≤ 95 ⇒ 50 ≤ 9  + 32 ≤ 95 ⇒ 18 ≤ 9  ≤ 63 ⇒
5
5
5

N

10 ≤  ≤ 35. So the interval is [10 35].

41. (a) Let  represent the temperature in degrees Celsius and  the height in km.  = 20 when  = 0 and  decreases by 10◦ C

for every km (1◦ C for each 100-m rise). Thus,  = 20 − 10 when 0 ≤  ≤ 12.
(b) From part (a),  = 20 − 10 ⇒ 10 = 20 − 

⇒  = 2 −  10. So 0 ≤  ≤ 5 ⇒ 0 ≤ 2 −  10 ≤ 5 ⇒

−2 ≤ − 10 ≤ 3 ⇒ −20 ≤ − ≤ 30 ⇒ 20 ≥  ≥ −30 ⇒ −30 ≤  ≤ 20. Thus, the range of temperatures (in ◦ C) to be expected is [−30 20].
43. |2| = 3

⇔ either 2 = 3 or 2 = −3 ⇔  =

45. | + 3| = |2 + 1|

3
2

or  = − 3 .
2

⇔ either  + 3 = 2 + 1 or  + 3 = − (2 + 1). In the first case,  = 2, and in the second case,

 + 3 = −2 − 1 ⇔ 3 = −4 ⇔  = − 4 . So the solutions are − 4 and 2.
3
3
47. By Property 5 of absolute values, ||  3

⇔ −3    3, so  ∈ (−3 3).

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

APPENDIX B COORDINATE GEOMETRY AND LINES

49. | − 4|  1

⇔ −1   − 4  1 ⇔ 3    5, so  ∈ (3 5).

51. | + 5| ≥ 2

¤

⇔  + 5 ≥ 2 or  + 5 ≤ −2 ⇔  ≥ −3 or  ≤ −7, so  ∈ (−∞ −7] ∪ [−3 ∞).
⇔ −04 ≤ 2 − 3 ≤ 04 ⇔ 26 ≤ 2 ≤ 34 ⇔ 13 ≤  ≤ 17, so  ∈ [13 17].

53. |2 − 3| ≤ 04

55. 1 ≤ || ≤ 4. So either 1 ≤  ≤ 4 or 1 ≤ − ≤ 4
57. ( − ) ≥ 
59.  +   

⇔  −  ≥




⇔  ≥

⇔ −1 ≥  ≥ −4. Thus,  ∈ [−4 −1] ∪ [1 4].

 + 
+=



⇔ ≥

 + 


−
[since   0]


⇔    −  ⇔  

61. |( + ) − 5| = |( − 2) + ( − 3)| ≤ | − 2| + | − 3|  001 + 004 = 005

√
√ √

()2 = 2 2 = 2 2 = || ||

( + )  .

al

65. || =

1
2

e

63. If    then  +    +  and  +    + . So 2   +   2. Dividing by 2, we get  

67. If 0    , then  ·    ·  and  ·    ·  [using Rule 3 of Inequalities]. So 2    2 and hence 2  2 .

rS

69. Observe that the sum, difference and product of two integers is always an integer. Let the rational numbers be represented

by  =  and  =  (where , ,  and  are integers with  6= 0,  6= 0). Now  +  = but  +  and  are both integers, so

 + 
=  +  is a rational number by definition. Similarly,



 − 
 


− = is a rational number. Finally,  ·  =
· = but  and  are both integers, so



 


Fo

− =


 + 

+ =
,




ot


=  ·  is a rational number by definition.


B Coordinate Geometry and Lines

N

1. Use the distance formula with 1 (1  1 ) = (1 1) and 2 (2  2 ) = (4 5) to get

|1 2 | =

3. The distance from (6 −2) to (−1 3) is
5. The distance from (2 5) to (4 −7) is


√
√
(4 − 1)2 + (5 − 1)2 = 32 + 42 = 25 = 5



√
−1 − 6)2 + [3 − (−2)]2 = (−7)2 + 52 = 74.



√
√
(4 − 2)2 + (−7 − 5)2 = 22 + (−12)2 = 148 = 2 37.

7. The slope  of the line through  (1 5) and (4 11) is  =

11 − 5
6
= = 2.
4−1
3

9. The slope  of the line through  (−3 3) and (−1 −6) is  =

−6 − 3
9
=− .
−1 − (−3)
2



√
(−4 − 0)2 + (3 − 2)2 = (−4)2 + 12 = 17 and


√
|| = [−4 − (−3)]2 + [3 − (−1)]2 = (−1)2 + 42 = 17, so the triangle has two sides of equal length, and is

11. Using (0 2), (−3 −1), and (−4 3), we have || =

isosceles.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

549

¤

550

APPENDIX B COORDINATE GEOMETRY AND LINES

13. Using (−2 9), (4 6), (1 0), and (−5 3), we have

|| =
|| =
|| =
|| =



√
√ √
√
[4 − (−2)]2 + (6 − 9)2 = 62 + (−3)2 = 45 = 9 5 = 3 5,



√
√ √
√
(1 − 4)2 + (0 − 6)2 = (−3)2 + (−6)2 = 45 = 9 5 = 3 5,



√
√ √
√
(−5 − 1)2 + (3 − 0)2 = (−6)2 + 32 = 45 = 9 5 = 3 5, and


√
√ √
√
√
[−2 − (−5)]2 + (9 − 3)2 = 32 + 62 = 45 = 9 5 = 3 5. So all sides are of equal length and we have a

rhombus. Moreover,  =
 =

6−9
0−6
3−0
1
1
= − ,  =
= 2,  =
= − , and
4 − (−2)
2
1−4
−5 − 1
2

9−3
= 2, so the sides are perpendicular. Thus, , , , and  are vertices of a square.
−2 − (−5)

1
10 − 4
1−7
7 − 10
= , the slope of  is
= −3, and the slope of  is
= −3. So  is parallel to  and
−1 − 5
2
5−7
1 − (−1)

al

is

1
4−1
= , the slope of 
7−1
2

e

15. For the vertices (1 1), (7 4), (5 10), and (−1 7), the slope of the line segment  is

 is parallel to . Hence  is a parallelogram.

⇔

 = 0 or  = 0. The graph consists

of the coordinate axes.

ot

Fo

-intercept 3. The line does not have a slope.

19.  = 0

rS

17. The graph of the equation  = 3 is a vertical line with

21. By the point-slope form of the equation of a line, an equation of the line through (2 −3) with slope 6 is

23.  − 7 =

2
(
3

N

 − (−3) = 6( − 2) or  = 6 − 15.
− 1) or  = 2  +
3

19
3

25. The slope of the line through (2 1) and (1 6) is  =

 − 1 = −5( − 2) or  = −5 + 11.

6−1
= −5, so an equation of the line is
1−2

27. By the slope-intercept form of the equation of a line, an equation of the line is  = 3 − 2.
29. Since the line passes through (1 0) and (0 −3), its slope is  =

Another method: From Exercise 61,

−3 − 0
= 3, so an equation is  = 3 − 3.
0−1



+
= 1 ⇒ −3 +  = −3 ⇒  = 3 − 3.
1
−3

31. The line is parallel to the -axis, so it is horizontal and must have the form  = . Since it goes through the point

( ) = (4 5), the equation is  = 5. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

APPENDIX B COORDINATE GEOMETRY AND LINES

¤

551

33. Putting the line  + 2 = 6 into its slope-intercept form gives us  = − 1  + 3, so we see that this line has slope − 1 . Thus,
2
2

we want the line of slope − 1 that passes through the point (1 −6):  − (−6) = − 1 ( − 1) ⇔  = − 1  −
2
2
2
35. 2 + 5 + 8 = 0

11
.
2

⇔  = − 2  − 8 . Since this line has slope − 2 , a line perpendicular to it would have slope 5 , so the
5
5
5
2

required line is  − (−2) = 5 [ − (−1)] ⇔  = 5  + 1 .
2
2
2
37.  + 3 = 0

⇔  = − 1 , so the slope is − 1 and the
3
3

39.  = −2 is a horizontal line with slope 0 and

-intercept is 0.

41. 3 − 4 = 12

⇔  = 3  − 3, so the slope is
4

and

43. {( ) |   0}

ot

Fo

rS

the -intercept is −3.

3
4

al

e

-intercept −2.

45. {( ) |   0} = {( ) |   0 and   0}

N

∪ {( ) |   0 and   0}

47.





( )  || ≤ 2 = {( ) | −2 ≤  ≤ 2}

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

552

¤

APPENDIX B COORDINATE GEOMETRY AND LINES

49. {( ) | 0 ≤  ≤ 4  ≤ 2}

51. {( ) | 1 +  ≤  ≤ 1 − 2}

53. Let  (0 ) be a point on the -axis. The distance from  to (5 −5) is



(5 − 0)2 + (−5 − )2 = 52 + ( + 5)2 . The



distance from  to (1 1) is (1 − 0)2 + (1 − )2 = 12 + ( − 1)2 . We want these distances to be

 equal: 52 + ( + 5)2 = 12 + ( − 1)2 ⇔ 52 + ( + 5)2 = 12 + ( − 1)2 ⇔

25 + ( 2 + 10 + 25) = 1 + (2 − 2 + 1) ⇔ 12 = −48 ⇔  = −4. So the desired point is (0 −4).
2

(b) Using the midpoint formula from Exercise 54 with (−1 6) and (8 −12), we get





3 + 15
2

= (4 9).

−1 + 8 6 + (−12)

2
2



=

7
2


 −3 .

⇔  = 2 − 4 ⇒ 1 = 2 and 6 − 2 = 10 ⇔ 2 = 6 − 10 ⇔  = 3 − 5 ⇒ 2 = 3.

rS

57. 2 −  = 4



e

1+7

al

55. (a) Using the midpoint formula from Exercise 54 with (1 3) and (7 15), we get

Since 1 6= 2 , the two lines are not parallel. To find the point of intersection: 2 − 4 = 3 − 5 ⇔  = 1 ⇒
 = −2. Thus, the point of intersection is (1 −2).

of  is



1 + 7 4 + (−2)
2 
2



−2 − 4
7−1

= −1, so its perpendicular bisector has slope 1. The midpoint

Fo

59. With (1 4) and (7 −2), the slope of segment  is

= (4 1), so an equation of the perpendicular bisector is  − 1 = 1( − 4) or  =  − 3.

61. (a) Since the -intercept is , the point ( 0) is on the line, and similarly since the -intercept is , (0 ) is on the line. Hence,


+ = ⇔


N



+ = 1.



−0


= − . Substituting into  =  +  gives  = −  +  ⇔
0−



ot

the slope of the line is  =

(b) Letting  = 6 and  = −8 gives



+
= 1 ⇔ −8 + 6 = −48 [multiply by −48] ⇔ 6 = 8 − 48 ⇔
6
−8

3 = 4 − 24 ⇔  = 4  − 8.
3

C Graphs of Second-Degree Equations
1. An equation of the circle with center (3 −1) and radius 5 is ( − 3)2 + ( + 1)2 = 52 = 25.
3. The equation has the form 2 +  2 = 2 . Since (4 7) lies on the circle, we have 42 + 72 = 2

⇒ 2 = 65. So the

required equation is 2 +  2 = 65.
5. 2 +  2 − 4 + 10 + 13 = 0

⇔ 2 − 4 +  2 + 10 = −13 ⇔

(2 − 4 + 4) + ( 2 + 10 + 25) = −13 + 4 + 25 = 16 ⇔ ( − 2)2 + ( + 5)2 = 42 . Thus, we have a circle with center (2 −5) and radius 4. c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

APPENDIX C GRAPHS OF SECOND-DEGREE EQUATIONS

7. 2 +  2 +  = 0


 2
 +  + 1 + 2 =
4

⇔

and radius 1 .
2

1
4

⇔

¤

2

 2


 + 1 +  2 = 1 . Thus, we have a circle with center − 1  0
2
2
2





1
1
⇔ 2 2 − 1  + 16 + 2 2 + 1  + 16 = 1 + 1 + 1 ⇔
2
2
8
8










1 2
1 2
5
1 2
1 2
5
2  − 4 + 2  + 4 = 4 ⇔  − 4 +  + 4 = 8 . Thus, we have a circle with center 1  − 1 and
4
4

9. 22 + 2 2 −  +  = 1

radius

√
5
√
2 2

=

√
10
4 .

11.  = −2 . Parabola

⇔

2
2
+
= 1. Ellipse
16
4

⇔

2
2
−
= 1. Hyperbola
25
16

al

e

13. 2 + 4 2 = 16

17. 42 +  2 = 1

2
+  2 = 1. Ellipse
14

⇔

Fo

rS

15. 162 − 25 2 = 400

21. 9 2 − 2 = 9

⇔ 2 −

2
= 1. Hyperbola
9

N

ot

19.  =  2 − 1. Parabola with vertex at (−1 0)

23.  = 4. Hyperbola

553

25. 9( − 1)2 + 4( − 2)2 = 36
2

2

⇔

( − 1)
( − 2)
+
= 1. Ellipse centered at (1 2)
4
9

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

554

¤

APPENDIX C GRAPHS OF SECOND-DEGREE EQUATIONS





2

27.  = 2 − 6 + 13 = 2 − 6 + 9 + 4 = ( − 3) + 4.

29.  = 4 −  2 = − 2 + 4. Parabola with vertex at (4 0)

31. 2 + 4 2 − 6 + 5 = 0

33.  = 3 and  = 2 intersect where 3 = 2

Parabola with vertex at (3 4)

2

⇔

2

⇔

0 =  − 3 = ( − 3), that is, at (0 0) and (3 9).
2

( − 6 + 9) + 4 = −5 + 9 = 4 ⇔
2

rS

al

e

( − 3)
+  2 = 1. Ellipse centered at (3 0)
4

35. The parabola must have an equation of the form  = ( − 1)2 − 1. Substituting  = 3 and  = 3 into the equation gives

Fo

3 = (3 − 1)2 − 1, so  = 1, and the equation is  = ( − 1)2 − 1 = 2 − 2. Note that using the other point (−1 3) would have given the same value for , and hence the same equation.


( ) | 2 +  2 ≤ 1





39. ( ) |  ≥ 2 − 1

N

ot

37.

D Trigonometry
1. 210◦ = 210◦
5. 900◦ = 900◦
9.

5
12

rad =

5
12







180◦

180◦

180◦







=

7
6

rad

3. 9◦ = 9◦


180◦

7. 4 rad = 4

= 5 rad





=

180◦


11. − 3 rad = − 3
8
8

= 75◦

13. Using Formula 3,  =  = 36 ·




12




20



rad
= 720◦

180◦




= −675◦

= 3 cm.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

APPENDIX D TRIGONOMETRY

15. Using Formula 3,  =  =

1
15

17.

=

2
3

rad =

2
3

19.




◦

180


=

 120 ◦


¤

555

≈ 382◦ .
21.

From the diagram we see that a point on the terminal side is  (−1 1).
√
Therefore, taking  = −1,  = 1,  = 2 in the definitions of the

23.

al

e

1
1
trigonometric ratios, we have sin 3 = √2 , cos 3 = − √2 ,
4
4
√
√ tan 3 = −1, csc 3 = 2, sec 3 = − 2, and cot 3 = −1.
4
4
4
4

From the diagram we see that a point on the terminal side is  (0 1).

25.

Therefore taking  = 0,  = 1,  = 1 in the definitions of the

rS

trigonometric ratios, we have sin 9 = 1, cos 9 = 0, tan 9 =  is
2
2
2

undefined since  = 0, csc 9 = 1, sec 9 =  is undefined since
2
2
 = 0, and cot

9
2

= 0.

Fo

 √ 
Using Figure 8 we see that a point on the terminal side is  − 3 1 .
√
Therefore taking  = − 3,  = 1,  = 2 in the definitions of the

27.

3
5

5
6

=−

√
3
,
2

√
1
2 tan 5 = − √3 , csc 5 = 2, sec 5 = − √3 , and cot 5 = − 3.
6
6
6
6

⇒  = 3,  = 5, and  =

N

29. sin  =  =

ot

trigonometric ratios, we have sin 5 = 1 , cos
6
2


2 −  2 = 4 (since 0   


2 ).

Therefore taking  = 4,  = 3,  = 5 in

the definitions of the trigonometric ratios, we have cos  = 4 , tan  = 3 , csc  = 5 , sec  = 5 , and cot  = 4 .
5
4
3
4
3
31.

⇒  is in the second quadrant, where  is negative and  is positive. Therefore
√
√
√
sec  =  = −15 = − 3 ⇒  = 3,  = −2, and  = 2 − 2 = 5. Taking  = −2,  = 5, and  = 3 in the
2

2



definitions of the trigonometric ratios, we have sin  =

√

5
,
3

cos  = − 2 , tan  = −
3

√
5
,
2

csc  =

3
√ ,
5

2 and cot  = − √5 .

33.     2 means that  is in the third or fourth quadrant where  is negative. Also since cot  =  = 3 which is


√
positive,  must also be negative. Therefore cot  =  = 3 ⇒  = −3,  = −1, and  = 2 +  2 = 10. Taking
1
√
 = −3,  = −1 and  = 10 in the definitions of the trigonometric ratios, we have sin  = − √1 , cos  = − √3 ,
10
10
√
√
10
tan  = 1 , csc  = − 10, and sec  = − 3 .
3

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

556

¤

APPENDIX D TRIGONOMETRY

35. sin 35◦ =


10

37. tan


8

2
5

=

⇒  = 10 sin 35◦ ≈ 573576 cm
⇒  = 8 tan 2 ≈ 2462147 cm
5
(a) From the diagram we see that sin  =

39.


−


= , and sin(−) =
= − = − sin .





(b) Again from the diagram we see that cos  =



= = cos(−).



41. (a) Using (12a) and (13a), we have

= 1 [sin  cos  + cos  sin  + sin  cos  − cos  sin ] = 1 (2 sin  cos ) = sin  cos .
2
2

e

1
2 [sin( + ) + sin( − )]

1
[cos( + ) + cos( − )]
2

al

(b) This time, using (12b) and (13b), we have

= 1 [cos  cos  − sin  sin  + cos  cos  + sin  sin ] = 1 (2 cos  cos ) = cos  cos .
2
2

1
2 [cos(

rS

(c) Again using (12b) and (13b), we have

− ) − cos( + )] = 1 [cos  cos  + sin  sin  − cos  cos  + sin  sin ]
2
= 1 (2 sin  sin ) = sin  sin 
2

2


+  = sin  cos  + cos  sin  = 1 · cos  + 0 · sin  = cos .
2
2

45. Using (6), we have sin  cot  = sin  ·

cos 
= cos . sin 

1
1 − cos2  sin2  sin 
− cos  [by (6)] =
=
[by (7)] = sin  = tan  sin  [by (6)] cos  cos  cos  cos  cos2  cos2  + sin2  cos2 
1
[by (6)] =
+
2
2
cos sin  sin2  cos2 

N

49. cot2  + sec2  =

ot

47. sec  − cos  =

Fo



43. Using (12a), we have sin

=

(1 − sin2 )(1 − sin2 ) + sin2 
1 − sin2  + sin4 
[by (7)] =
2
2 sin  cos sin2  cos2 

=

cos2  + sin4 
1
sin2 
= csc2  + tan2  [by (6)]
[ (7)] =
+
2
2
2 cos2  sin  cos sin 

51. Using (14a), we have tan 2 = tan( + ) =

2 tan  tan  + tan 
=
.
1 − tan  tan 
1 − tan2 

53. Using (15a) and (16a),

sin  sin 2 + cos  cos 2 = sin  (2 sin  cos ) + cos  (2 cos2  − 1) = 2 sin2  cos  + 2 cos3  − cos 
= 2(1 − cos2 ) cos  + 2 cos3  − cos  [by (7)]
= 2 cos  − 2 cos3  + 2 cos3  − cos  = cos 
Or: sin  sin 2 + cos  cos 2 = cos (2 − ) [by 13(b)] = cos  c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

APPENDIX D TRIGONOMETRY

55.

sin  sin 
1 + cos  sin  (1 + cos ) sin  (1 + cos )
=
·
=
=
1 − cos 
1 − cos  1 + cos 
1 − cos2  sin2 

[by (7)]

1 cos 
1 + cos 
=
+
= csc  + cot  [by (6)] sin  sin  sin 

=
57. Using (12a),

sin 3 + sin  = sin(2 + ) + sin  = sin 2 cos  + cos 2 sin  + sin 
[by (16a)]

= sin 2 cos  + (2 cos2  − 1) sin  + sin 
2

= sin 2 cos  + 2 cos  sin  − sin  + sin  = sin 2 cos  + sin 2 cos 

[by (15a)]

= 2 sin 2 cos 
1
3

we can label the opposite side as having length 1,

the hypotenuse as having length 3, and use the Pythagorean Theorem
√
to get that the adjacent side has length 8. Then, from the diagram,
√

8
.
3

Similarly we have that sin  = 3 . Now use (12a):
5
1
3

4
5

·

+

√
8
3

·

3
5

=

4
15

+

√
3 8
15

=

√
4+6 2
.
15

rS

sin( + ) = sin  cos  + cos  sin  =

al

cos  =

e

59. Since sin  =

61. Using (13b) and the values for cos  and sin  obtained in Exercise 59, we have

cos( − ) = cos  cos  + sin  sin  =

√

8
3

·

4
5

+

1
3

·

3
5

=

√
8 2+3
15

65. 2 cos  − 1 = 0

⇔ cos  =

⇔ sin2  =

1
2

1
2

 5
3, 3

⇒ =

1
⇔ sin  = ± √2

ot

67. 2 sin2  = 1

Fo

63. Using (15a) and the values for sin  and cos  obtained in Exercise 59, we have sin 2 = 2 sin  cos  = 2 ·

69. Using (15a), we have sin 2 = cos 
 3
2, 2

N

2 sin  − 1 = 0 ⇒  =
71. sin  = tan 

1−

3
5

·

4
5

=

24
.
25

for  ∈ [0 2].
⇒ =

 3 5 7
4, 4 , 4 , 4 .

⇔ 2 sin  cos  − cos  = 0 ⇔ cos (2 sin  − 1) = 0 ⇔ cos  = 0 or

or sin  =

1
2

⇒ =

⇔ sin  − tan  = 0 ⇔ sin  −

1
1
= 0 ⇒  = 0, , 2 or 1 = cos  cos 


6

or

5
6 .

Therefore, the solutions are  =

  5 3
6, 2, 6 , 2 .



1 sin 
= 0 ⇔ sin  1 −
= 0 ⇔ sin  = 0 or cos  cos 

⇒ cos  = 1 ⇒  = 0, 2. Therefore the solutions

are  = 0, , 2.
73. We know that sin  =
5
6

1
2

when  =


6

or

5
,
6

and from Figure 13(a), we see that sin  ≤

1
2

⇒ 0≤≤


6

or

≤  ≤ 2 for  ∈ [0 2].

75. tan  = −1 when  =

0≤

 3
4, 4



3 7
, 4,
4

and tan  = 1 when  =

5
4 ,

7
4

and


4

or

5
.
4

From Figure 14(a) we see that −1  tan   1 ⇒

  ≤ 2.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

557

558

¤

APPENDIX D TRIGONOMETRY



77.  = cos  −

and shift it


3


3


. We start with the graph of  = cos 

79.  =

1
3



tan  −  . We start with the graph of
2

 = tan , shift it

units to the right.

1
3


2

units to the right and compress it to

of its original vertical size.

al

e

81.  = |sin |. We start with the graph of  = sin  and reflect the parts below the -axis about the -axis.

83. From the figure in the text, we see that  =  cos ,  =  sin , and from the distance formula we have that the


( − )2 + ( − 0)2

⇒

rS

distance  from ( ) to ( 0) is  =

2 = ( cos  − )2 + ( sin )2 = 2 cos2  − 2 cos  + 2 + 2 sin2 
= 2 + 2 (cos2  + sin2 ) − 2 cos  = 2 + 2 − 2 cos 

[by (7)]

Fo

85. Using the Law of Cosines, we have 2 = 12 + 12 − 2(1)(1) cos ( − ) = 2 [1 − cos( − )]. Now, using the distance

formula, 2 = ||2 = (cos  − cos )2 + (sin  − sin )2 . Equating these two expressions for 2 , we get
2[1 − cos( − )] = cos2  + sin2  + cos2  + sin2  − 2 cos  cos  − 2 sin  sin 

⇒ cos( − ) = cos  cos  + sin  sin .

ot

1 − cos( − ) = 1 − cos  cos  − sin  sin 

⇒

87. In Exercise 86 we used the subtraction formula for cosine to prove the addition formula for cosine. Using that formula with

N








 =  − ,  = , we get cos  −  +  = cos  −  cos  − sin  −  sin  ⇒
2
2
2
2





 cos  − ( − ) = cos  −  cos  − sin  −  sin . Now we use the identities given in the problem,
2
2
2




cos  −  = sin  and sin  −  = cos , to get sin( − ) = sin  cos  − cos  sin .
2
2

89. Using the formula from Exercise 88, the area of the triangle is 1 (10)(3) sin 107◦ ≈ 1434457 cm2 .
2

E Sigma Notation
1.

5
√
√
√
√
√
√
= 1+ 2+ 3+ 4+ 5

3.

=1

5.

9.

4
 2 − 1
1
3
5
7
= −1 + + + +
2 + 1
3
5
7
9
=0
−1

=0

(−1) = 1 − 1 + 1 − 1 + · · · + (−1)−1

6


3 = 34 + 35 + 36

=4

7.




=1

10 = 110 + 210 + 310 + · · · + 10

11. 1 + 2 + 3 + 4 + · · · + 10 =

10




=1

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

APPENDIX E SIGMA NOTATION

13.

19
 
1
2
3
4
19
+ + + + ··· +
=
2
3
4
5
20 =1  + 1

17. 1 + 2 + 4 + 8 + 16 + 32 =

5


15. 2 + 4 + 6 + 8 + · · · + 2 =

2

19.  + 2 + 3 + · · · +  =

=0

21.

8





¤

559

2

=1






=1

(3 − 2) = [3(4) − 2] + [3(5) − 2] + [3(6) − 2] + [3(7) − 2] + [3(8) − 2] = 10 + 13 + 16 + 19 + 22 = 80

=4

23.

6


3+1 = 32 + 33 + 34 + 35 + 36 + 37 = 9 + 27 + 81 + 243 + 729 + 2187 = 3276

=1

(For a more general method, see Exercise 47.)
25.

20


(−1) = −1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 = 0

=1

27.

4


(2 + 2 ) = (1 + 0) + (2 + 1) + (4 + 4) + (8 + 9) + (16 + 16) = 61

=1

31.







=1

=2·

(2 + 3 + 4) =

=1

33.




( + 1)
[by Theorem 3(c)] = ( + 1)
2




2 + 3

=1




+

=1




4=

al

2 = 2

3( + 1)
( + 1)(2 + 1)
+
+ 4
6
2

rS




=1

= 1 [(23 + 32 + ) + (92 + 9) + 24] = 1 (23 + 122 + 34) = 1 (2 + 6 + 17)
6
6
3



( + 1)( + 2) =

=1

(2 + 3 + 2) =

=1




2 + 3

=1




+

=1




2=

=1

3( + 1)
( + 1)(2 + 1)
+
+ 2
6
2

Fo

29.

e

=0

( + 1)
( + 1)
[(2 + 1) + 9] + 2 =
( + 5) + 2
6
3


= [( + 1)( + 5) + 6] = (2 + 6 + 11)
3
3
2





 3
 3 

( + 1)
( + 1)
− 2
35.
( −  − 2) =
 −
−
2=
−
2
2
=1
=1
=1
=1

ot

=

N

= 1 ( + 1)[( + 1) − 2] − 2 = 1 ( + 1)( + 2)( − 1) − 2
4
4

= 1 [( + 1)( − 1)( + 2) − 8] = 1 [(2 − 1)( + 2) − 8] = 1 (3 + 22 −  − 10)
4
4
4

37. By Theorem 2(a) and Example 3,




=

=1

39.







1 = .

=1



[( + 1)4 − 4 ] = (24 − 14 ) + (34 − 24 ) + (44 − 34 ) + · · · + ( + 1)4 − 4

=1

= ( + 1)4 − 14 = 4 + 43 + 62 + 4

On the other hand,



[( + 1)4 − 4 ] =

=1




(43 + 62 + 4 + 1) = 4

=1




=1

3 + 6




=1

= 4 + ( + 1)(2 + 1) + 2( + 1) + 

2 + 4




=1

+




1

=1




 3

where  =
=1

3

2

2

3

2

= 4 + 2 + 3 +  + 2 + 2 +  = 4 + 2 + 5 + 4

[continued] c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

560

¤

APPENDIX E SIGMA NOTATION

Thus, 4 + 43 + 62 + 4 = 4 + 23 + 52 + 4, from which it follows that
2

( + 1)
.
4 = 4 + 23 + 2 = 2 (2 + 2 + 1) = 2 ( + 1)2 and  =
2

 
 
 



4
 − ( − 1)4 = 14 − 04 + 24 − 14 + 34 − 24 + · · · + 4 − ( − 1)4 = 4 − 0 = 4

41. (a)

=1

100 


(b)

=1

 
 
 



5 − 5−1 = 51 − 50 + 52 − 51 + 53 − 52 + · · · + 5100 − 599 = 5100 − 50 = 5100 − 1


 
 
 



99
 1
1
1 1
1 1
1 1
1
1
1
1
97
−
=
−
+
−
+
−
+··· +
−
= −
=

+1
3
4
4
5
5
6
99
100
3
100
300
=3

(c)




(d)

( − −1 ) = (1 − 0 ) + (2 − 1 ) + (3 − 2 ) + · · · + ( − −1 ) =  − 0

=1

e



2


3

al


 2
45. lim
→∞ =1 

 2




1  2
1 ( + 1)(2 + 1)
1

1
1
= lim
1+
2+
= 1 (1)(2) =
= lim 3
 = lim 3
6
→∞  =1
→∞ 
→∞ 6

6




 





 16 3 20
16  3 20 
2
= lim
+5
 + 2  = lim
 + 2

→∞ =1 4
→∞ 4 =1


 =1




16 2 ( + 1)2
4( + 1)2
20 ( + 1)
10( + 1)
+ 2
= lim
+
→∞ 4
→∞
4

2
2
2

 
2


1
1
= 4 · 1 + 10 · 1 = 14
= lim 4 1 +
+ 10 1 +
→∞






=1



Fo

= lim

47. Let  =

rS


 1
→∞ =1 

43. lim

−1 =  +  + 2 + · · · + −1 . Multiplying both sides by  gives us

ot

 =  + 2 + · · · + −1 +  . Subtracting the first equation from the second, we find

49.




N

( − 1) =  −  = ( − 1), so  =
(2 + 2 ) = 2

=1




=1

+




=1

2 · 2−1 = 2

( − 1)
−1

[since  6= 1].

2(2 − 1)
( + 1)
+
= 2+1 + 2 +  − 2.
2
2−1

For the first sum we have used Theorems 2(a) and 3(c), and for the second, Exercise 47 with  =  = 2.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

1
3

APPENDIX G GRAPHING CALCULATORS AND COMPUTERS

G Graphing Calculators and Computers
1.  () =

√
3 − 52

(a) [−5 5] by [−5 5]

(b) [0 10] by [0 2]

(c) [0 10] by [0 10]

(There is no graph shown.)

The most appropriate graph is produced in viewing rectangle (c).

al

an appropriate viewing rectangle should include the minimum point.

e

3. Since the graph of  () = 2 − 36 + 32 is a parabola opening upward,

Completing the square, we get () = ( − 18)2 − 292, and so the

rS

minimum point is (18 −292).

⇒ 50 ≥ 02 ⇒  ≤ 250, so the domain of the
√
root function () = 50 − 02 is (−∞ 250].

ot

Fo

5. 50 − 02 ≥ 0

7. The graph of  () = 3 − 225 is symmetric with respect to the origin.

N

Since  () = 3 − 225 = (2 − 225) = ( + 15)( − 15), there are -intercepts at 0, −15, and 15.  (20) = 3500.

9. The period of () = sin(1000) is

2
1000

≈ 00063 and its range is

[−1 1]. Since  () = sin2 (1000) is the square of , its range is
[0 1] and a viewing rectangle of [−001 001] by [0 11] seems appropriate. c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

561

562

¤

APPENDIX G GRAPHING CALCULATORS AND COMPUTERS

11. The domain of  =

√
√
 is  ≥ 0, so the domain of () = sin  is [0 ∞)

and the range is [−1 1]. With a little trial-and-error experimentation, we find that an Xmax of 100 illustrates the general shape of  , so an appropriate viewing rectangle is [0 100] by [−15 15].
13. The first term, 10 sin , has period 2 and range [−10 10]. It will be the dominant term in any “large” graph of

 = 10 sin  + sin 100, as shown in the first figure. The second term, sin 100, has period

2
100

=


50

and range [−1 1].

It causes the bumps in the first figure and will be the dominant term in any “small” graph, as shown in the view near the

al

e

origin in the second figure.

rS

15. (a) The first figure shows the "big

picture" for  () = ( − 10)3 2− .
The second figure shows a maximum

Fo

near  = 10.

(b) You need more than one window because no single window can show what the function looks like globally and the detail

ot

of the function near  = 10.

17. We must solve the given equation for  to obtain equations for the upper and

N

lower halves of the ellipse.

42 + 2 2 = 1 ⇔ 2 2 = 1 − 42

1 − 42
=±
2

⇔ 2 =

1 − 42
2

⇔

19. From the graph of  = 32 − 6 + 1 and  = 023 − 225 in the viewing rectangle [−1 3] by [−25 15], it is difficult to

see if the graphs intersect. If we zoom in on the fourth quadrant, we see the graphs do not intersect.

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

APPENDIX G GRAPHING CALCULATORS AND COMPUTERS

¤

563

21. We see that the graphs of  () = 4 −  and () = 1 intersect twice.

The -coordinates of these points (which are the solutions of the equations) are approximately −072 and 122. Alternatively, we could find these values by finding the zeros of () = 4 −  − 1.

23. We see that the graphs of  () = tan  and () =

√
1 − 2 intersect

once. Using an intersect feature or zooming in, we find this value to be

e

approximately 065. Alternatively, we could find this value by finding the
√
positive zero of () = tan  − 1 − 2 .

al

Note: After producing the graph on a TI-84 Plus, we can find the approximate value 0.65 by using the following keystrokes:
. The “.6” is just a guess for 065.

Fo

rS

25. () = 3 10 is larger than  () = 102 whenever   100.

ot

27. We see from the graphs of  = |tan  − | and  = 001 that there are two

solutions to the equation  = |tan  − | = 001 for −2    2:
 ≈ −031 and  ≈ 031. The condition |tan  − |  001 holds for any

N

 lying between these two values, that is, −031    031.

29. (a) The root functions  =

=

√
√
4
 and  = 6 

√
,

(b) The root functions  = ,
√
√
 = 3  and  = 5 

(c) The root functions  =
√
√
 = 4  and  = 5 

√
√
,  = 3 ,

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

564

¤

APPENDIX H COMPLEX NUMBERS

(d) • For any , the th root of 0 is 0 and the th root of 1 is 1; that is, all th root functions pass through the points (0 0) and (1 1).
• For odd , the domain of the th root function is R, while for even , it is { ∈ R |  ≥ 0}.
√
√
• Graphs of even root functions look similar to that of , while those of odd root functions resemble that of 3 .
√
• As  increases, the graph of   becomes steeper near 0 and flatter for   1.
31.  () = 4 + 2 + . If   −15, there are three humps: two minimum points

and a maximum point. These humps get flatter as  increases, until at  = −15 two of the humps disappear and there is only one minimum point. This single hump then moves to the right and approaches the origin as  increases.
33.  =  2− . As  increases, the maximum of the

e

function moves further from the origin, and gets

al

larger. Note, however, that regardless of , the

rS

function approaches 0 as  → ∞.

35.  2 = 3 + 2 . If   0, the loop is to the right of the origin, and if  is positive,

it is to the left. In both cases, the closer  is to 0, the larger the loop is. (In the

Fo

limiting case,  = 0, the loop is “infinite,” that is, it doesn’t close.) Also, the larger || is, the steeper the slope is on the loopless side of the origin.

37. The graphing window is 95 pixels wide and we want to start with  = 0 and end with  = 2. Since there are 94 “gaps”

Thus, the -values that the calculator actually plots are  = 0 + 2 · ,
94
 2

 2 where  = 0, 1, 2,    , 93, 94. For  = sin 2, the actual points plotted by the calculator are 94 ·  sin 2 · 94 ·  for



 = 0, 1,    , 94 For  = sin 96, the points plotted are 2 ·  sin 96 · 2 ·  for  = 0, 1,    , 94 But
94
94

2−0
94 .

N

ot

between pixels, the distance between pixels is

 sin 96 ·

2
94






·  = sin 94 · 2 ·  + 2 · 2 ·  = sin 2 + 2 · 2 · 
94
94
94


[by periodicity of sine]  = 0, 1,    , 94
= sin 2 · 2 · 
94

So the -values, and hence the points, plotted for  = sin 96 are identical to those plotted for  = sin 2.
Note: Try graphing  = sin 94. Can you see why all the -values are zero?

H Complex Numbers
1. (5 − 6) + (3 + 2) = (5 + 3) + (−6 + 2) = 8 + (−4) = 8 − 4
3. (2 + 5)(4 − ) = 2(4) + 2(−) + (5)(4) + (5)(−) = 8 − 2 + 20 − 52 = 8 + 18 − 5(−1)

= 8 + 18 + 5 = 13 + 18 c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

APPENDIX H COMPLEX NUMBERS

¤

565

5. 12 + 7 = 12 − 7
7.

1 + 4
1 + 4 3 − 2
3 − 2 + 12 − 8(−1)
11
10
11 + 10
=
·
=
=
+ 
=
3 + 2
3 + 2 3 − 2
32 + 22
13
13
13

9.

1
1
1−
1−
1−
1
1
=
·
=
=
= − 
1+
1+ 1−
1 − (−1)
2
2
2

11. 3 = 2 ·  = (−1) = −

√
√
−25 = 25  = 5

√
√
122 + (−5)2 = 144 + 25 = 169 = 13

19. 42 + 9 = 0


√
02 + (−4)2 = 16 = 4

⇔ 42 = −9 ⇔ 2 = − 9
4

21. By the quadratic formula, 2 + 2 + 5 = 0

23. By the quadratic formula,  2 +  + 2 = 0



⇔  = ± − 9 = ± 9  = ± 3 .
4
4
2

⇔ =

−2 ±

−1 ±

⇔ =


√
(−3)2 + 32 = 3 2 and tan  =
√ 

Therefore, −3 + 3 = 3 2 cos 3 +  sin 3 .
4
4
√
32 + 42 = 5 and tan  =



 
3 + 4 = 5 cos tan−1 4 +  sin tan−1 4 .
3
3

27. For  = 3 + 4,  =

4
3


√
√
12 − 4(1)(2)
7
−1 ± −7
1
=
=− ±
.
2(1)
2
2
2

3
−3

= −1 ⇒  =

⇒  = tan−1

√ 
√
2
3 + ,  =
3 + 12 = 2 and tan  =

ot

29. For  =


√
22 − 4(1)(5)
−2 ± −16
−2 ± 4
=
=
= −1 ± 2.
2(1)
2
2

Fo

25. For  = −3 + 3,  =

e

17. −4 = 0 − 4 = 0 + 4 = 4 and |−4| =

al

15. 12 − 5 = 12 + 15 and |12 − 15| =

rS

13.

1
√
3

4
3

⇒ =


6

3
4

(since  lies in the second quadrant).

(since  lies in the first quadrant). Therefore,



⇒  = 2 cos  +  sin  .
6
6

√
√


3 ,  = 2 and tan  = 3 ⇒  =  ⇒  = 2 cos  +  sin  .
3
3
3





 
Therefore,  = 2 · 2 cos  +  +  sin  +  = 4 cos  +  sin  ,
6
3
6
3
2
2
 







 = 2 cos  −  +  sin  −  = cos −  +  sin −  , and 1 = 1 + 0 = 1(cos 0 +  sin 0) ⇒
2
6
3
6
3
6
6
 



 



1 = 1 cos 0 −  +  sin 0 −  = 1 cos −  +  sin −  . For 1, we could also use the formula that precedes
2
6
6
2
6
6


Example 5 to obtain 1 = 1 cos  −  sin  .
2
6
6

N

For  = 1 +

31. For  = 2

 √ 
√
2
2 3 + (−2)2 = 4 and tan  =
3 − 2,  =

−2
√
2 3

1
= − √3

⇒  = −
6

⇒

√
 



1
 = 4 cos −  +  sin −  . For  = −1 + ,  = 2, tan  = −1 = −1 ⇒  = 3 ⇒
6
6
4
√   
√ 
√ 


 


3
3
3
3
= 4 2 cos 7 +  sin 7 ,
 = 2 cos 4 +  sin 4 . Therefore,  = 4 2 cos − 6 + 4 +  sin − 6 + 4
12
12
√ 
 



 




4
4
 = √2 cos −  − 3 +  sin −  − 3 = √2 cos − 11 +  sin − 11 = 2 2 cos 13 +  sin 13 , and
6
4
6
4
12
12
12
12
 





1 = 1 cos −  −  sin −  = 1 cos  +  sin  .
4
6
6
4
6
6 c ° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

566

¤

APPENDIX H COMPLEX NUMBERS

33. For  = 1 + ,  =

√
2 and tan  =

(1 + )20 =

1
1

=1 ⇒ =


4

⇒ =

√ 

2 cos  +  sin  . So by De Moivre’s Theorem,
4
4

20


√ 
2 cos  +  sin 
= (212 )20 cos 204·  +  sin 204·  = 210 (cos 5 +  sin 5)
4
4

= 210 [−1 + (0)] = −210 = −1024
35. For  = 2

 √ 
√
√
2
2 3 + 22 = 16 = 4 and tan  =
3 + 2,  =

2
√
2 3

=

1
√
3

⇒ =


6


⇒  = 4 cos


6

So by De Moivre’s Theorem,

 √
√
5  
5


 √
= 45 cos 5 +  sin 5 = 1024 − 23 + 1  = −512 3 + 512.
2 3 + 2 = 4 cos  +  sin 
6
6
6
6
2

37. 1 = 1 + 0 = 1 (cos 0 +  sin 0). Using Equation 3 with  = 1,  = 8, and  = 0, we have

rS

al



1
1
0 = 1(cos 0 +  sin 0) = 1, 1 = 1 cos  +  sin  = √2 + √2 ,
4
4




1
1
2 = 1 cos  +  sin  = , 3 = 1 cos 3 +  sin 3 = − √2 + √2 ,
2
2
4
4


1
1
4 = 1(cos  +  sin ) = −1, 5 = 1 cos 5 +  sin 5 = − √2 − √2 ,
4
4




1
1
6 = 1 cos 3 +  sin 3 = −, 7 = 1 cos 7 +  sin 7 = √2 − √2 
2
2
4
4

e

 



0 + 2
0 + 2


+  sin
= cos
+  sin
, where  = 0 1 2     7.
 = 118 cos
8
8
4
4


+  sin  . Using Equation 3 with  = 1,  = 3, and  =
2
 



+ 2
+ 2
 = 113 cos 2
+  sin 2
, where  = 0 1 2.
3
3
 √

0 = cos  +  sin  = 23 + 1 
6
6
2



2


,
2

we have

Fo

39.  = 0 +  = 1 cos

ot

√


1 = cos 5 +  sin 5 = − 23 + 1 
6
6
2


9
9
2 = cos 6 +  sin 6 = −

2,

we have 2 = cos  +  sin  = 0 + 1 = .
2
2

N

41. Using Euler’s formula (6) with  =
43. Using Euler’s formula (6) with  =

√


3

1
, we have 3 = cos +  sin = +
.
3
3
3
2
2

45. Using Equation 7 with  = 2 and  = , we have 2+ = 2  = 2 (cos  +  sin ) = 2 (−1 + 0) = −2 .
47. Take  = 1 and  = 3 in De Moivre’s Theorem to get

[1(cos  +  sin )]3 = 13 (cos 3 +  sin 3)
(cos  +  sin )3 = cos 3 +  sin 3 cos3  + 3(cos2 )( sin ) + 3(cos )( sin )2 + ( sin )3 = cos 3 +  sin 3 cos3  + (3 cos2  sin ) − 3 cos  sin2  − (sin3 ) = cos 3 +  sin 3
(cos3  − 3 sin2  cos ) + (3 sin  cos2  − sin3 ) = cos 3 +  sin 3
Equating real and imaginary parts gives cos 3 = cos3  − 3 sin2  cos 

and sin 3 = 3 sin  cos2  − sin3 .

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.


+  sin  .
6

APPENDIX H COMPLEX NUMBERS

49.  () =  = (+) = + =  (cos  +  sin ) =  cos  + ( sin )

⇒

 0 () = ( cos )0 + ( sin )0
= ( cos  −  sin ) + ( sin  +  cos )
= [ (cos  +  sin )] + [ (− sin  +  cos )]
=  + [ (2 sin  +  cos )]

N

ot

Fo

rS

al

e

=  + [ (cos  +  sin )] =  +  = ( + ) = 

c
° 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

¤

567

N ot e

al

rS

Fo

N ot e

al

rS

Fo

N ot e

al

rS

Fo

N ot e

al

rS

Fo

N ot e

al

rS

Fo

N ot e

al

rS

Fo

N ot e

al

rS

Fo

N ot e

al

rS

Fo

N ot e

al

rS

Fo

N ot e

al

rS

Fo

N ot e

al

rS

Fo

N ot e

al

rS

Fo

N ot e

al

rS

Fo