Calculus for Everyone

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CALCULUS MADE EASY

MACMILLAN AND CO., Limited
LONDON : BOMBAY : CALCUTTA MELBOURNE

THE MACMILLAN COMPANY
NEW YORK : BOSTON : CHICAGO DALLAS : SAN FRANCISCO

THE MACMILLAN CO. OF CANADA, Ltd.
TORONTO

CALCULUS MADE EASY:
BEING A VERY-SIMPLEST INTRODUCTION TO THOSE BEAUTIFUL METHODS OF RECKONING WHICH ARE GENERALLY CALLED BY THE TERRIFYING NAMES OF THE

DIFFERENTIAL CALCULUS
AND THE

INTEGRAL CALCULUS.
BY

F. R. S.

SECOND EDITION, ENLARGED

MACMILLAN AND CO., LIMITED ST. MARTIN’S STREET, LONDON 

COPYRIGHT. First Edition 1910. Reprinted 1911 (twice), 1912, 1913. Second Edition 1914.

What one fool can do, another can. (Ancient Simian Proverb.)

PREFACE TO THE SECOND EDITION.
The surprising success of this work has led the author to add a considerable number of worked examples and exercises. Advantage has also been taken to enlarge certain parts where experience showed that further explanations would be useful. The author acknowledges with gratitude many valuable suggestions and letters received from teachers, students, and—critics. October, 1914.

CONTENTS.

Chapter

Page

Prologue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I. To deliver you from the Preliminary Terrors II. On Different Degrees of Smallness . . . . . . . . . . . III. On Relative Growings . . . . . . . . . . . . . . . . . . . . . . . . . . IV. Simplest Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V. Next Stage. What to do with Constants . . . . . . VI. Sums, Differences, Products and Quotients . . . VII. Successive Differentiation . . . . . . . . . . . . . . . . . . . . . VIII. When Time Varies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . IX. Introducing a Useful Dodge . . . . . . . . . . . . . . . . . . . X. Geometrical Meaning of Differentiation . . . . . . XI. Maxima and Minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

ix 1 3 9 17 25 34 48 52 66 75 91

XII. Curvature of Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 XIII. Other Useful Dodges . . . . . . . . . . . . . . . . . . . . . . . . . . 118 XIV. On true Compound Interest and the Law of Organic Growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 vii

CALCULUS MADE EASY

viii
Page

Chapter

XV. How to deal with Sines and Cosines . . . . . . . . . . . 162 XVI. Partial Differentiation . . . . . . . . . . . . . . . . . . . . . . . . 172 XVII. Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180 XVIII. Integrating as the Reverse of Differentiating 189 XIX. On Finding Areas by Integrating . . . . . . . . . . . . . . 204 XX. Dodges, Pitfalls, and Triumphs . . . . . . . . . . . . . . . . 224 XXI. Finding some Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 232 Table of Standard Forms . . . . . . . . . . . . . . . . . . . . . . . . 249 Answers to Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . 252

PROLOGUE.
Considering how many fools can calculate, it is surprising that it should be thought either a diﬃcult or a tedious task for any other fool to learn how to master the same tricks. Some calculus-tricks are quite easy. Some are enormously diﬃcult. The fools who write the textbooks of advanced mathematics—and they are mostly clever fools—seldom take the trouble to show you how easy the easy calculations are. On the contrary, they seem to desire to impress you with their tremendous cleverness by going about it in the most diﬃcult way. Being myself a remarkably stupid fellow, I have had to unteach myself the diﬃculties, and now beg to present to my fellow fools the parts that are not hard. Master these thoroughly, and the rest will follow. What one fool can do, another can.

CHAPTER I.
TO DELIVER YOU FROM THE PRELIMINARY TERRORS.
The preliminary terror, which chokes oﬀ most ﬁfth-form boys from even attempting to learn how to calculate, can be abolished once for all by simply stating what is the meaning—in common-sense terms—of the two principal symbols that are used in calculating. These dreadful symbols are: (1) d which merely means “a little bit of.” Thus dx means a little bit of x; or du means a little bit of u. Ordinary mathematicians think it more polite to say “an element of,” instead of “a little bit of.” Just as you please. But you will ﬁnd that these little bits (or elements) may be considered to be indeﬁnitely small. (2) sum of.” Thus dx means the sum of all the little bits of x; or dt means the sum of all the little bits of t. Ordinary mathematicians call this symbol “the integral of.” Now any fool can see that if x is considered as made up of a lot of little bits, each of which is called dx, if you add them all up together you get the sum of all the dx’s, (which is the which is merely a long S, and may be called (if you like) “the

CALCULUS MADE EASY

2

same thing as the whole of x). The word “integral” simply means “the whole.” If you think of the duration of time for one hour, you may (if you like) think of it as cut up into 3600 little bits called seconds. The whole of the 3600 little bits added up together make one hour. When you see an expression that begins with this terrifying symbol, you will henceforth know that it is put there merely to give you instructions that you are now to perform the operation (if you can) of totalling up all the little bits that are indicated by the symbols that follow. That’s all.

CHAPTER II.
ON DIFFERENT DEGREES OF SMALLNESS.
We shall ﬁnd that in our processes of calculation we have to deal with small quantities of various degrees of smallness. We shall have also to learn under what circumstances we may consider small quantities to be so minute that we may omit them from consideration. Everything depends upon relative minuteness. Before we ﬁx any rules let us think of some familiar cases. There are 60 minutes in the hour, 24 hours in the day, 7 days in the week. There are therefore 1440 minutes in the day and 10080 minutes in the week. Obviously 1 minute is a very small quantity of time compared with a whole week. Indeed, our forefathers considered it small as compared with an hour, and called it “one min`te,” meaning a minute u fraction—namely one sixtieth—of an hour. When they came to require still smaller subdivisions of time, they divided each minute into 60 still smaller parts, which, in Queen Elizabeth’s days, they called “second min`tes” (i.e. small quantities of the second order of minuteu ness). Nowadays we call these small quantities of the second order of smallness “seconds.” But few people know why they are so called. Now if one minute is so small as compared with a whole day, how

CALCULUS MADE EASY

4

much smaller by comparison is one second! Again, think of a farthing as compared with a sovereign: it is barely worth more than
1 1000

part. A farthing more or less is of precious little

importance compared with a sovereign: it may certainly be regarded as a small quantity. But compare a farthing with £1000: relatively to this greater sum, the farthing is of no more importance than a negligible quantity in the wealth of a millionaire. Now if we ﬁx upon any numerical fraction as constituting the proportion which for any purpose we call relatively small, we can easily state other fractions of a higher degree of smallness. Thus if, for the purpose of time,
1 60 1 1000

of a

farthing would be to a sovereign. Even a golden sovereign is relatively

be called a small fraction, then

1 60

of

1 60

(being a

small fraction of a small fraction) may be regarded as a small quantity of the second order of smallness.∗ Or, if for any purpose we were to take 1 per cent. (i.e. small fraction, then 1 per cent. of 1 per cent. (i.e. small fraction of the second order of smallness; 1 per cent. of 1 per cent. Lastly, suppose that for some very precise purpose we should regard
1 1,000,000 1 ) 100

as a

1 ) would be a 10,000 1 and 1,000,000 would be

a small fraction of the third order of smallness, being 1 per cent. of

as “small.” Thus, if a ﬁrst-rate chronometer is not to lose

or gain more than half a minute in a year, it must keep time with an accuracy of 1 part in 1, 051, 200. Now if, for such a purpose, we
∗

The mathematicians talk about the second order of “magnitude” (i.e. great-

ness) when they really mean second order of smallness. This is very confusing to beginners.

DIFFERENT DEGREES OF SMALLNESS
1 1,000,000 1 1,000,000

5 of

regard

(or one millionth) as a small quantity, then
1 1,000,000,000,000

1 , 1,000,000

that is

(or one billionth) will be a small quantity

of the second order of smallness, and may be utterly disregarded, by comparison. Then we see that the smaller a small quantity itself is, the more negligible does the corresponding small quantity of the second order become. Hence we know that in all cases we are justiﬁed in neglecting the small quantities of the second—or third (or higher)—orders, if only we take the small quantity of the ﬁrst order small enough in itself. But, it must be remembered, that small quantities if they occur in our expressions as factors multiplied by some other factor, may become important if the other factor is itself large. Even a farthing becomes important if only it is multiplied by a few hundred. Now in the calculus we write dx for a little bit of x. These things such as dx, and du, and dy, are called “diﬀerentials,” the diﬀerential of x, or of u, or of y, as the case may be. [You read them as dee-eks, or dee-you, or dee-wy.] If dx be a small bit of x, and relatively small of itself, it does not follow that such quantities as x · dx, or x2 dx, or ax dx are negligible. But dx × dx would be negligible, being a small quantity of the second order. A very simple example will serve as illustration. Let us think of x as a quantity that can grow by a small amount so as to become x + dx, where dx is the small increment added by growth. The square of this is x2 + 2x · dx + (dx)2 . The second term is not negligible because it is a ﬁrst-order quantity; while the third term is of the second order of smallness, being a bit of, a bit of x2 . Thus if we

CALCULUS MADE EASY
1 60

6

took dx to mean numerically, say, be
2 60

of x, then the second term would
1 3600

of x2 , whereas the third term would be
1 1000

of x2 . This last term
2 1000

is clearly less important than the second. But if we go further and take dx to mean only of x, then the second term will be
1 1,000,000

of x2 , while

the third term will be only

of x2 .

x

x

Fig. 1.

Geometrically this may be depicted as follows: Draw a square (Fig. 1) the side of which we will take to represent x. Now suppose the square to grow by having a bit dx added to its size each way. The enlarged square is made up of the original square x2 , the two (or together 2x · dx), and the little square at the top right-hand corner of x—about
1 . 5

rectangles at the top and on the right, each of which is of area x · dx

which is (dx)2 . In Fig. 2 we have taken dx as quite a big fraction But suppose we had taken it only
1 10,000 1 —about 100

the

thickness of an inked line drawn with a ﬁne pen. Then the little corner square will have an area of only itself small enough. Let us consider a simile. of x2 , and be practically invisible. Clearly (dx)2 is negligible if only we consider the increment dx to be

DIFFERENT DEGREES OF SMALLNESS

7

x dx

dx dx x · dx (dx)2

x

x

x2

x · dx

x

dx

Fig. 2.

Fig. 3.

Suppose a millionaire were to say to his secretary: next week I will give you a small fraction of any money that comes in to me. Suppose that the secretary were to say to his boy: I will give you a small fraction of what I get. Suppose the fraction in each case to be
1 100

part. Now

if Mr. Millionaire received during the next week £1000, the secretary would receive £10 and the boy 2 shillings. Ten pounds would be a small quantity compared with £1000; but two shillings is a small small quantity indeed, of a very secondary order. But what would be the disproportion if the fraction, instead of being
1 1000 1 , 100

had been settled at

part? Then, while Mr. Millionaire got his £1000, Mr. Secretary

would get only £1, and the boy less than one farthing! The witty Dean Swift∗ once wrote:
“So, Nat’ralists observe, a Flea “Hath smaller Fleas that on him prey. “And these have smaller Fleas to bite ’em, “And so proceed ad inﬁnitum.”
∗

On Poetry: a Rhapsody (p. 20), printed 1733—usually misquoted.

CALCULUS MADE EASY

8

An ox might worry about a ﬂea of ordinary size—a small creature of the ﬁrst order of smallness. But he would probably not trouble himself about a ﬂea’s ﬂea; being of the second order of smallness, it would be negligible. Even a gross of ﬂeas’ ﬂeas would not be of much account to the ox.

CHAPTER III.
ON RELATIVE GROWINGS.
All through the calculus we are dealing with quantities that are growing, and with rates of growth. We classify all quantities into two classes: constants and variables. Those which we regard as of ﬁxed value, and call constants, we generally denote algebraically by letters from the beginning of the alphabet, such as a, b, or c; while those which we consider as capable of growing, or (as mathematicians say) of “varying,” we denote by letters from the end of the alphabet, such as x, y, z, u, v, w, or sometimes t. Moreover, we are usually dealing with more than one variable at once, and thinking of the way in which one variable depends on the other: for instance, we think of the way in which the height reached by a projectile depends on the time of attaining that height. Or we are asked to consider a rectangle of given area, and to enquire how any increase in the length of it will compel a corresponding decrease in the breadth of it. Or we think of the way in which any variation in the slope of a ladder will cause the height that it reaches, to vary. Suppose we have got two such variables that depend one on the other. An alteration in one will bring about an alteration in the other, because of this dependence. Let us call one of the variables x, and the

CALCULUS MADE EASY

10

other that depends on it y. Suppose we make x to vary, that is to say, we either alter it or imagine it to be altered, by adding to it a bit which we call dx. We are thus causing x to become x + dx. Then, because x has been altered, y will have altered also, and will have become y + dy. Here the bit dy may be in some cases positive, in others negative; and it won’t (except by a miracle) be the same size as dx. Take two examples. (1) Let x and y be respectively the base and the height of a rightangled triangle (Fig. 4), of which the slope of the other side is ﬁxed dy y 30◦ x dx

y

Fig. 4.

at 30◦ . If we suppose this triangle to expand and yet keep its angles the same as at ﬁrst, then, when the base grows so as to become x + dx, the height becomes y + dy. Here, increasing x results in an increase of y. The little triangle, the height of which is dy, and the base of which is dx, is similar to the original triangle; and it is obvious that the value dy y of the ratio is the same as that of the ratio . As the angle is 30◦ dx x it will be seen that here dy 1 = . dx 1.73

ON RELATIVE GROWINGS

11

(2) Let x represent, in Fig. 5, the horizontal distance, from a wall, of the bottom end of a ladder, AB, of ﬁxed length; and let y be the
B

y

O

x

A

Fig. 5.

height it reaches up the wall. Now y clearly depends on x. It is easy to see that, if we pull the bottom end A a bit further from the wall, the top end B will come down a little lower. Let us state this in scientiﬁc language. If we increase x to x + dx, then y will become y − dy; that is, when x receives a positive increment, the increment which results to y is negative. Yes, but how much? Suppose the ladder was so long that when the bottom end A was 19 inches from the wall the top end B reached just 15 feet from the ground. Now, if you were to pull the bottom end out 1 inch more, how much would the top end come down? Put it all into inches: x = 19 inches, y = 180 inches. Now the increment of x which we call dx, is 1 inch: or x + dx = 20 inches.

CALCULUS MADE EASY

12

we work out the height by Euclid I. 47, then we shall be able to ﬁnd how much dy will be. The length of the ladder is (180)2 + (19)2 = 181 inches. Clearly then, the new height, which is y − dy, will be such that (y − dy)2 = (181)2 − (20)2 = 32761 − 400 = 32361, √ y − dy = 32361 = 179.89 inches. Now y is 180, so that dy is 180 − 179.89 = 0.11 inch. making dy a decrease of 0.11 inch. And the ratio of dy to dx may be stated thus: 0.11 dy =− . dx 1 It is also easy to see that (except in one particular position) dy will be of a diﬀerent size from dx. Now right through the diﬀerential calculus we are hunting, hunting, hunting for a curious thing, a mere ratio, namely, the proportion which dy bears to dx when both of them are indeﬁnitely small. dy when dx y and x are related to each other in some way, so that whenever x varies It should be noted here that we can only ﬁnd this ratio y does vary also. For instance, in the ﬁrst example just taken, if the base x of the triangle be made longer, the height y of the triangle becomes greater also, and in the second example, if the distance x of the foot of the ladder from the wall be made to increase, the height y So we see that making dx an increase of 1 inch has resulted in

How much will y be diminished? The new height will be y − dy. If

ON RELATIVE GROWINGS

13

reached by the ladder decreases in a corresponding manner, slowly at ﬁrst, but more and more rapidly as x becomes greater. In these cases the relation between x and y is perfectly deﬁnite, it can be expressed y = tan 30◦ and x2 + y 2 = l2 (where l is the mathematically, being x dy length of the ladder) respectively, and has the meaning we found in dx each case. If, while x is, as before, the distance of the foot of the ladder from the wall, y is, instead of the height reached, the horizontal length of the wall, or the number of bricks in it, or the number of years since it was built, any change in x would naturally cause no change whatever dy has no meaning whatever, and it is not possible in y; in this case dx to ﬁnd an expression for it. Whenever we use diﬀerentials dx, dy, dz, etc., the existence of some kind of relation between x, y, z, etc., is implied, and this relation is called a “function” in x, y, z, etc.; the two y expressions given above, for instance, namely = tan 30◦ and x2 +y 2 = x l2 , are functions of x and y. Such expressions contain implicitly (that is, contain without distinctly showing it) the means of expressing either x in terms of y or y in terms of x, and for this reason they are called implicit functions in x and y; they can be respectively put into the forms y y = x tan 30◦ or x = tan 30◦ √ y = l2 − x2 or x = l2 − y 2 .

and

These last expressions state explicitly (that is, distinctly) the value of x in terms of y, or of y in terms of x, and they are for this reason called explicit functions of x or y. For example x2 + 3 = 2y − 7 is an

CALCULUS MADE EASY

14

x2 + 10 (explicit implicit function in x and y; it may be written y = 2 √ function of x) or x = 2y − 10 (explicit function of y). We see that an explicit function in x, y, z, etc., is simply something the value of which changes when x, y, z, etc., are changing, either one at the time or several together. Because of this, the value of the explicit function is called the dependent variable, as it depends on the value of the other variable quantities in the function; these other variables are called the independent variables because their value is not determined from the value assumed by the function. For example, if u = x2 sin θ, x and θ are the independent variables, and u is the dependent variable. Sometimes the exact relation between several quantities x, y, z either is not known or it is not convenient to state it; it is only known, or convenient to state, that there is some sort of relation between these variables, so that one cannot alter either x or y or z singly without aﬀecting the other quantities; the existence of a function in x, y, z is then indicated by the notation F (x, y, z) (implicit function) or by x = F (y, z), y = F (x, z) or z = F (x, y) (explicit function). Sometimes the letter f or φ is used instead of F , so that y = F (x), y = f (x) and y = φ(x) all mean the same thing, namely, that the value of y depends on the value of x in some way which is not stated. dy “the diﬀerential coeﬃcient of y with respect We call the ratio dx to x.” It is a solemn scientiﬁc name for this very simple thing. But we are not going to be frightened by solemn names, when the things themselves are so easy. Instead of being frightened we will simply pronounce a brief curse on the stupidity of giving long crack-jaw names; and, having relieved our minds, will go on to the simple thing itself,

ON RELATIVE GROWINGS

15

namely the ratio

dy . dx In ordinary algebra which you learned at school, you were always

hunting after some unknown quantity which you called x or y; or sometimes there were two unknown quantities to be hunted for simultaneously. You have now to learn to go hunting in a new way; the fox being now neither x nor y. Instead of this you have to hunt for this curious dy dy . The process of ﬁnding the value of is called “difcub called dx dx ferentiating.” But, remember, what is wanted is the value of this ratio when both dy and dx are themselves indeﬁnitely small. The true value of the diﬀerential coeﬃcient is that to which it approximates in the limiting case when each of them is considered as inﬁnitesimally minute. dy . Let us now learn how to go in quest of dx

CALCULUS MADE EASY

16

NOTE TO CHAPTER III. How to read Diﬀerentials. It will never do to fall into the schoolboy error of thinking that dx means d times x, for d is not a factor—it means “an element of” or “a bit of” whatever follows. One reads dx thus: “dee-eks.” In case the reader has no one to guide him in such matters it may here be simply said that one reads diﬀerential coeﬃcients in the following way. The diﬀerential coeﬃcient dy is read “dee-wy by dee-eks,” or “dee-wy over dee-eks.” dx du is read “dee-you by dee-tee.” So also dt Second diﬀerential coeﬃcients will be met with later on. They are like this: d2 y ; which is read “dee-two-wy over dee-eks-squared,” dx2 and it means that the operation of diﬀerentiating y with respect to x has been (or has to be) performed twice over. Another way of indicating that a function has been diﬀerentiated is by putting an accent to the symbol of the function. Thus if y = F (x), which means that y is some unspeciﬁed function of x (see p. 13), we may d F (x) write F (x) instead of . Similarly, F (x) will mean that the dx original function F (x) has been diﬀerentiated twice over with respect to x.

CHAPTER IV.
SIMPLEST CASES.
Now let us see how, on ﬁrst principles, we can diﬀerentiate some simple algebraical expression. Case 1. Let us begin with the simple expression y = x2 . Now remember that the fundamental notion about the calculus is the idea of growing. Mathematicians call it varying. Now as y and x2 are equal to one another, it is clear that if x grows, x2 will also grow. And if x2 grows, then y will also grow. What we have got to ﬁnd out is the proportion between the growing of y and the growing of x. In other words our task is to ﬁnd out the ratio between dy and dx, or, in brief, to ﬁnd the value dy . of dx Let x, then, grow a little bit bigger and become x + dx; similarly, y will grow a bit bigger and will become y+dy. Then, clearly, it will still be true that the enlarged y will be equal to the square of the enlarged x. Writing this down, we have: y + dy = (x + dx)2 .

CALCULUS MADE EASY

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Doing the squaring we get: y + dy = x2 + 2x · dx + (dx)2 . What does (dx)2 mean? Remember that dx meant a bit—a little bit—of x. Then (dx)2 will mean a little bit of a little bit of x; that is, as explained above (p. 4), it is a small quantity of the second order of smallness. It may therefore be discarded as quite inconsiderable in comparison with the other terms. Leaving it out, we then have: y + dy = x2 + 2x · dx. Now y = x2 ; so let us subtract this from the equation and we have left dy = 2x · dx. Dividing across by dx, we ﬁnd dy = 2x. dx Now this ∗ is what we set out to ﬁnd. The ratio of the growing of y to the growing of x is, in the case before us, found to be 2x. dy is the result of diﬀerentiating y with respect to x. Difdx ferentiating means ﬁnding the diﬀerential coeﬃcient. Suppose we had some other
∗

N.B.—This ratio

function of x, as, for example, u = 7x2 + 3. Then if we were told to diﬀerentiate this du d(7x2 + 3) with respect to x, we should have to ﬁnd , or, what is the same thing, . dx dx On the other hand, we may have a case in which time was the independent variable
1 (see p. 14), such as this: y = b + 2 at2 . Then, if we were told to diﬀerentiate it, that

means we must ﬁnd its diﬀerential coeﬃcient with respect to t. So that then our d(b + 1 at2 ) dy 2 business would be to try to ﬁnd , that is, to ﬁnd . dt dt

SIMPLEST CASES

19

Numerical example. Suppose x = 100 and ∴ y = 10, 000. Then let x grow till it becomes 10, 201. But if we agree that we may ignore small quantities of the 101 (that is, let dx = 1). Then the enlarged y will be 101 × 101 =

second order, 1 may be rejected as compared with 10, 000; so we may round oﬀ the enlarged y to 10, 200. y has grown from 10, 000 to 10, 200; the bit added on is dy, which is therefore 200. 200 dy = = 200. According to the algebra-working of the previous dx 1 dy paragraph, we ﬁnd = 2x. And so it is; for x = 100 and 2x = 200. dx But, you will say, we neglected a whole unit. Well, try again, making dx a still smaller bit. Try dx =
1 . 10

Then x + dx = 100.1, and

(x + dx)2 = 100.1 × 100.1 = 10, 020.01. Now the last ﬁgure 1 is only one-millionth part of the 10, 000, and is utterly negligible; so we may take 10, 020 without the little decimal dy 20 at the end. And this makes dy = 20; and = = 200, which is dx 0.1 still the same as 2x. Case 2. Try diﬀerentiating y = x3 in the same way. We let y grow to y + dy, while x grows to x + dx. Then we have y + dy = (x + dx)3 .

Doing the cubing we obtain y + dy = x3 + 3x2 · dx + 3x(dx)2 + (dx)3 .

CALCULUS MADE EASY

20

Now we know that we may neglect small quantities of the second and third orders; since, when dy and dx are both made indeﬁnitely small, (dx)2 and (dx)3 will become indeﬁnitely smaller by comparison. So, regarding them as negligible, we have left: y + dy = x3 + 3x2 · dx. But y = x3 ; and, subtracting this, we have: dy = 3x2 · dx,

and Case 3.

dy = 3x2 . dx

Try diﬀerentiating y = x4 . Starting as before by letting both y and x grow a bit, we have: y + dy = (x + dx)4 . Working out the raising to the fourth power, we get y + dy = x4 + 4x3 dx + 6x2 (dx)2 + 4x(dx)3 + (dx)4 . Then striking out the terms containing all the higher powers of dx, as being negligible by comparison, we have y + dy = x4 + 4x3 dx. Subtracting the original y = x4 , we have left dy = 4x3 dx, and dy = 4x3 . dx

SIMPLEST CASES

21

Now all these cases are quite easy. Let us collect the results to see if we can infer any general rule. Put them in two columns, the values of y dy in the other: thus in one and the corresponding values found for dx y x2 x3 x4 dy dx 2x 3x2 4x3

Just look at these results: the operation of diﬀerentiating appears to have had the eﬀect of diminishing the power of x by 1 (for example in the last case reducing x4 to x3 ), and at the same time multiplying by a number (the same number in fact which originally appeared as the power). Now, when you have once seen this, you might easily conjecture how the others will run. You would expect that diﬀerentiating x5 would give 5x4 , or diﬀerentiating x6 would give 6x5 . If you hesitate, try one of these, and see whether the conjecture comes right. Try y = x5 . Then y + dy = (x + dx)5 = x5 + 5x4 dx + 10x3 (dx)2 + 10x2 (dx)3 + 5x(dx)4 + (dx)5 . Neglecting all the terms containing small quantities of the higher orders, we have left y + dy = x5 + 5x4 dx,

CALCULUS MADE EASY

22

and subtracting

y = x5 leaves us dy = 5x4 dx, dy = 5x4 , exactly as we supposed. dx

whence

Following out logically our observation, we should conclude that if we want to deal with any higher power,—call it n—we could tackle it in the same way. Let y = xn ,

then, we should expect to ﬁnd that dy = nx(n−1) . dx For example, let n = 8, then y = x8 ; and diﬀerentiating it would dy = 8x7 . give dx And, indeed, the rule that diﬀerentiating xn gives as the result nxn−1 is true for all cases where n is a whole number and positive. [Expanding (x + dx)n by the binomial theorem will at once show this.] But the question whether it is true for cases where n has negative or fractional values requires further consideration. Case of a negative power. Let y = x−2 . Then proceed as before: y + dy = (x + dx)−2 = x−2 1 + dx x
−2

.

SIMPLEST CASES

23

Expanding this by the binomial theorem (see p. 137), we get = x−2 1 − 2 dx 2(2 + 1) + x 1×2 dx x
2

− etc.

= x−2 − 2x−3 · dx + 3x−4 (dx)2 − 4x−5 (dx)3 + etc. So, neglecting the small quantities of higher orders of smallness, we have: y + dy = x−2 − 2x−3 · dx. Subtracting the original y = x−2 , we ﬁnd dy = −2x−3 dx,

dy = −2x−3 . dx

And this is still in accordance with the rule inferred above. Case of a fractional power. Let y = x 2 . Then, as before, y + dy = (x + dx) = x =
1 2 1 2 1

dx 1+ x

1 2

√ 1 dx 1 (dx)2 √ + terms with higher x+ √ − 2 x 8x x powers of dx.
1

Subtracting the original y = x 2 , and neglecting higher powers we have left: dy = 1 dx 1 1 √ = x− 2 · dx, 2 x 2

CALCULUS MADE EASY

24

and

1 1 dy = x− 2 . Agreeing with the general rule. dx 2 Summary. Let us see how far we have got. We have arrived at the

following rule: To diﬀerentiate xn , multiply by the power and reduce the power by one, so giving us nxn−1 as the result. Exercises I. (See p. 252 for Answers.)

Diﬀerentiate the following: (1) y = x13 (3) y = x2a (5) z = (7) u = (9) y = √ 3
5

(2) y = x− 2 (4) u = t2.4 (6) y =

3

u 1 x8

√ 3 −5 x

(8) y = 2xa 1 xm

√ q 3 x

(10) y =

n

You have now learned how to diﬀerentiate powers of x. How easy it is!

CHAPTER V.
NEXT STAGE. WHAT TO DO WITH CONSTANTS.
In our equations we have regarded x as growing, and as a result of x being made to grow y also changed its value and grew. We usually think of x as a quantity that we can vary; and, regarding the variation of x as a sort of cause, we consider the resulting variation of y as an eﬀect. In other words, we regard the value of y as depending on that of x. Both x and y are variables, but x is the one that we operate upon, and y is the “dependent variable.” In all the preceding chapter we have been trying to ﬁnd out rules for the proportion which the dependent variation in y bears to the variation independently made in x. Our next step is to ﬁnd out what eﬀect on the process of diﬀerentiating is caused by the presence of constants, that is, of numbers which don’t change when x or y change their values. Added Constants. Let us begin with some simple case of an added constant, thus: Let y = x3 + 5.

Just as before, let us suppose x to grow to x+dx and y to grow to y+dy.

CALCULUS MADE EASY

26

Then:

y + dy = (x + dx)3 + 5 = x3 + 3x2 dx + 3x(dx)2 + (dx)3 + 5.

Neglecting the small quantities of higher orders, this becomes y + dy = x3 + 3x2 · dx + 5. Subtract the original y = x3 + 5, and we have left: dy = 3x2 dx. dy = 3x2 . dx So the 5 has quite disappeared. It added nothing to the growth of x, and does not enter into the diﬀerential coeﬃcient. If we had put 7, or 700, or any other number, instead of 5, it would have disappeared. So if we take the letter a, or b, or c to represent any constant, it will simply disappear when we diﬀerentiate. If the additional constant had been of negative value, such as −5 or −b, it would equally have disappeared. Multiplied Constants. Take as a simple experiment this case: Let y = 7x2 . Then on proceeding as before we get: y + dy = 7(x + dx)2 = 7{x2 + 2x · dx + (dx)2 } = 7x2 + 14x · dx + 7(dx)2 .

WHAT TO DO WITH CONSTANTS

27

Then, subtracting the original y = 7x2 , and neglecting the last term, we have dy = 14x · dx.

dy = 14x. dx

Let us illustrate this example by working out the graphs of the dy equations y = 7x2 and = 14x, by assigning to x a set of successive dx values, 0, 1, 2, 3, etc., and ﬁnding the corresponding values of y and dy . of dx These values we tabulate as follows: x y dy dx 0 0 0 1 7 14 2 28 28 3 63 42 4 112 56 5 175 70 −1 7 −14 −2 28 −28 −3 63 −42

Now plot these values to some convenient scale, and we obtain the two curves, Figs. 6 and 6a. Carefully compare the two ﬁgures, and verify by inspection that the height of the ordinate of the derived curve, Fig. 6a, is proportional to the slope of the original curve,∗ Fig. 6, at the corresponding value of x. To the left of the origin, where the original curve slopes negatively (that is, downward from left to right) the corresponding ordinates of the derived curve are negative. Now if we look back at p. 18, we shall see that simply diﬀerentiating x2 gives us 2x. So that the diﬀerential coeﬃcient of 7x2 is just
∗

See p. 76 about slopes of curves.

CALCULUS MADE EASY

28 dy dx

y 200
150

150
100

100
50

50 x

−3 −2 −1

0

1

2

3

4

5

x

−3 −2 −1 0

1

2

3

4

5

−50

Fig. 6.—Graph of y = 7x2 .

Fig. 6a.—Graph of

dy = 14x. dx

7 times as big as that of x2 . If we had taken 8x2 , the diﬀerential coeﬃcient would have come out eight times as great as that of x2 . If we put y = ax2 , we shall get dy = a × 2x. dx

dy = a × nxn−1 . dx So that any mere multiplication by a constant reappears as a mere If we had begun with y = axn , we should have had multiplication when the thing is diﬀerentiated. And, what is true about multiplication is equally true about division: for if, in the example above, we had taken as the constant the same
1 7 1 7

instead of 7, we should have had

come out in the result after diﬀerentiation.

Some Further Examples. The following further examples, fully worked out, will enable you to master completely the process of diﬀerentiation as applied to ordinary

WHAT TO DO WITH CONSTANTS

29

algebraical expressions, and enable you to work out by yourself the examples given at the end of this chapter. x5 3 − . (1) Diﬀerentiate y = 7 5 3 is an added constant and vanishes (see p. 25). 5 We may then write at once dy 1 = × 5 × x5−1 , dx 7 dy 5 = x4 . dx 7

or

√ 1√ a. (2) Diﬀerentiate y = a x − 2 √ 1√ The term a vanishes, being an added constant; and as a x, in 2 1 the index form, is written ax 2 , we have
1 1 dy 1 a = a × × x 2 −1 = × x− 2 , dx 2 2 dy a or = √ . dx 2 x √ (3) If ay + bx = by − ax + (x + y) a2 − b2 ,

ﬁnd the diﬀerential coeﬃcient of y with respect to x. As a rule an expression of this kind will need a little more knowledge than we have acquired so far; it is, however, always worth while to try whether the expression can be put in a simpler form. First we must try to bring it into the form y = some expression involving x only. The expression may be written √ (a − b)y + (a + b)x = (x + y) a2 − b2 .

CALCULUS MADE EASY

30

Squaring, we get (a − b)2 y 2 + (a + b)2 x2 + 2(a + b)(a − b)xy = (x2 + y 2 + 2xy)(a2 − b2 ), which simpliﬁes to (a − b)2 y 2 + (a + b)2 x2 = x2 (a2 − b2 ) + y 2 (a2 − b2 ); or that is hence y= [(a − b)2 − (a2 − b2 )]y 2 = [(a2 − b2 ) − (a + b)2 ]x2 , 2b(b − a)y 2 = −2b(b + a)x2 ; a+b x and a−b dy = dx a+b . a−b

(4) The volume of a cylinder of radius r and height h is given by the formula V = πr2 h. Find the rate of variation of volume with the radius when r = 5.5 in. and h = 20 in. If r = h, ﬁnd the dimensions of the cylinder so that a change of 1 in. in radius causes a change of 400 cub. in. in the volume. The rate of variation of V with regard to r is dV = 2πrh. dr If r = 5.5 in. and h = 20 in. this becomes 690.8. It means that a change of radius of 1 inch will cause a change of volume of 690.8 cub. inch. This can be easily veriﬁed, for the volumes with r = 5 and r = 6 are 1570 cub. in. and 2260.8 cub. in. respectively, and 2260.8 − 1570 = 690.8. Also, if r = h, dV = 2πr2 = 400 and r = h = dr 400 = 7.98 in. 2π

WHAT TO DO WITH CONSTANTS

31

(5) The reading θ of a F´ry’s Radiation pyrometer is related to the e Centigrade temperature t of the observed body by the relation θ = θ1 t t1
4

,

where θ1 is the reading corresponding to a known temperature t1 of the observed body. Compare the sensitiveness of the pyrometer at temperatures 800◦ C., 1000◦ C., 1200◦ C., given that it read 25 when the temperature was 1000◦ C. The sensitiveness is the rate of variation of the reading with the dθ . The formula may be written temperature, that is dt θ= and we have 25t4 θ1 4 t = , t4 10004 1

dθ 100t3 t3 = = . dt 10004 10, 000, 000, 000 dθ = 0.0512, 0.1 and 0.1728 dt

When t = 800, 1000 and 1200, we get respectively.

The sensitiveness is approximately doubled from 800◦ to 1000◦ , and becomes three-quarters as great again up to 1200◦ . Exercises II. (See p. 252 for Answers.)

Diﬀerentiate the following: (1) y = ax3 + 6. (2) y = 13x 2 − c.
3

CALCULUS MADE EASY
1 1 1 1

32

(3) y = 12x 2 + c 2 . (5) u = az n − 1 . c

(4) y = c 2 x 2 . (6) y = 1.18t2 + 22.4.

Make up some other examples for yourself, and try your hand at diﬀerentiating them. (7) If lt and l0 be the lengths of a rod of iron at the temperatures t◦ C. and 0◦ C. respectively, then lt = l0 (1+0.000012t). Find the change of length of the rod per degree Centigrade. (8) It has been found that if c be the candle power of an incandescent electric lamp, and V be the voltage, c = aV b , where a and b are constants. Find the rate of change of the candle power with the voltage, and calculate the change of candle power per volt at 80, 100 and 120 volts in the case of a lamp for which a = 0.5 × 10−10 and b = 6. (9) The frequency n of vibration of a string of diameter D, length L and speciﬁc gravity σ, stretched with a force T , is given by n= 1 DL gT . πσ

Find the rate of change of the frequency when D, L, σ and T are varied singly. (10) The greatest external pressure P which a tube can support without collapsing is given by P = 2E 1 − σ2 t3 , D3

WHAT TO DO WITH CONSTANTS

33

where E and σ are constants, t is the thickness of the tube and D is its diameter. (This formula assumes that 4t is small compared to D.) Compare the rate at which P varies for a small change of thickness and for a small change of diameter taking place separately. (11) Find, from ﬁrst principles, the rate at which the following vary with respect to a change in radius: (a) the circumference of a circle of radius r; (b) the area of a circle of radius r; (c) the lateral area of a cone of slant dimension l; (d ) the volume of a cone of radius r and height h; (e) the area of a sphere of radius r; (f ) the volume of a sphere of radius r. (12) The length L of an iron rod at the temperature T being given by L = lt 1 + 0.000012(T − t) , where lt is the length at the temperature t, ﬁnd the rate of variation of the diameter D of an iron tyre suitable for being shrunk on a wheel, when the temperature T varies.

CHAPTER VI.
SUMS, DIFFERENCES, PRODUCTS AND QUOTIENTS.
We have learned how to diﬀerentiate simple algebraical functions such as x2 + c or ax4 , and we have now to consider how to tackle the sum of two or more functions. For instance, let y = (x2 + c) + (ax4 + b); what will its dy be? How are we to go to work on this new job? dx The answer to this question is quite simple: just diﬀerentiate them,

one after the other, thus: dy = 2x + 4ax3 . (Ans.) dx If you have any doubt whether this is right, try a more general case, working it by ﬁrst principles. And this is the way. Let y = u+v, where u is any function of x, and v any other function of x. Then, letting x increase to x + dx, y will increase to y + dy; and u will increase to u + du; and v to v + dv. And we shall have: y + dy = u + du + v + dv.

SUMS, DIFFERENCES, PRODUCTS

35

Subtracting the original y = u + v, we get dy = du + dv, and dividing through by dx, we get: du dv dy = + . dx dx dx This justiﬁes the procedure. You diﬀerentiate each function separately and add the results. So if now we take the example of the preceding paragraph, and put in the values of the two functions, we shall have, using the notation shown (p. 16), dy d(x2 + c) d(ax4 + b) = + dx dx dx 3 = 2x + 4ax , exactly as before. If there were three functions of x, which we may call u, v and w, so that y = u + v + w; then du dv dw dy = + + . dx dx dx dx

As for subtraction, it follows at once; for if the function v had itself had a negative sign, its diﬀerential coeﬃcient would also be negative; so that by diﬀerentiating y = u − v,

we should get

dy du dv = − . dx dx dx

CALCULUS MADE EASY

36

But when we come to do with Products, the thing is not quite so simple. Suppose we were asked to diﬀerentiate the expression y = (x2 + c) × (ax4 + b), what are we to do? The result will certainly not be 2x × 4ax3 ; for it into that product. is easy to see that neither c × ax4 , nor x2 × b, would have been taken Now there are two ways in which we may go to work. First way. Do the multiplying ﬁrst, and, having worked it out, then diﬀerentiate. Accordingly, we multiply together x2 + c and ax4 + b. This gives ax6 + acx4 + bx2 + bc. Now diﬀerentiate, and we get: dy = 6ax5 + 4acx3 + 2bx. dx Second way. Go back to ﬁrst principles, and consider the equation y = u × v; where u is one function of x, and v is any other function of x. Then, if x grows to be x + dx; and y to y + dy; and u becomes u + du, and v becomes v + dv, we shall have: y + dy = (u + du) × (v + dv) = u · v + u · dv + v · du + du · dv.

SUMS, DIFFERENCES, PRODUCTS

37

therefore in the limit may be discarded, leaving

Now du · dv is a small quantity of the second order of smallness, and y + dy = u · v + u · dv + v · du.

Then, subtracting the original y = u · v, we have left dy = u · dv + v · du; and, dividing through by dx, we get the result: dv du dy =u +v . dx dx dx This shows that our instructions will be as follows: To diﬀerentiate the product of two functions, multiply each function by the diﬀerential coeﬃcient of the other, and add together the two products so obtained. You should note that this process amounts to the following: Treat u as constant while you diﬀerentiate v; then treat v as constant while you dy diﬀerentiate u; and the whole diﬀerential coeﬃcient will be the sum dx of these two treatments. Now, having found this rule, apply it to the concrete example which was considered above. We want to diﬀerentiate the product (x2 + c) × (ax4 + b). Call (x2 + c) = u; and (ax4 + b) = v.

CALCULUS MADE EASY

38

Then, by the general rule just established, we may write: d(ax4 + b) d(x2 + c) dy = (x2 + c) + (ax4 + b) dx dx dx = (x2 + c) 4ax3 + (ax4 + b) 2x = 4ax5 + 4acx3 dy = 6ax5 + 4acx3 dx exactly as before. Lastly, we have to diﬀerentiate quotients. bx5 + c Think of this example, y = 2 . In such a case it is no use to x +a try to work out the division beforehand, because x2 + a will not divide into bx5 + c, neither have they any common factor. So there is nothing for it but to go back to ﬁrst principles, and ﬁnd a rule. So we will put y= u ; v + 2ax5 + 2bx, + 2bx,

where u and v are two diﬀerent functions of the independent variable x. Then, when x becomes x + dx, y will become y + dy; and u will become u + du; and v will become v + dv. So then y + dy = u + du . v + dv

QUOTIENTS

39

Now perform the algebraic division, thus: v + dv u + du u · dv u+ v du − u du u · dv + − v v v2

u · dv v du · dv du + v − u · dv du · dv − v v u · dv u · dv · dv − − v v2 − du · dv u · dv · dv + . v v2

As both these remainders are small quantities of the second order, they may be neglected, and the division may stop here, since any further remainders would be of still smaller magnitudes. So we have got: y + dy = which may be written = u v · du − u · dv + . v v2 u du u · dv ; + − v v v2

CALCULUS MADE EASY

40

Now subtract the original y = dy =

u , and we have left: v

whence

v · du − u · dv ; v2 du dv v −u dy = dx 2 dx . dx v

This gives us our instructions as to how to diﬀerentiate a quotient of two functions. Multiply the divisor function by the diﬀerential coefﬁcient of the dividend function; then multiply the dividend function by the diﬀerential coeﬃcient of the divisor function; and subtract. Lastly divide by the square of the divisor function. bx5 + c , Going back to our example y = 2 x +a write and bx5 + c = u; x2 + a = v.

Then d(bx5 + c) d(x2 + a) − (bx5 + c) dy dx dx = 2 + a)2 dx (x 2 4 (x + a)(5bx ) − (bx5 + c)(2x) = , (x2 + a)2 dy 3bx6 + 5abx4 − 2cx = . (Answer.) dx (x2 + a)2 (x2 + a) The working out of quotients is often tedious, but there is nothing diﬃcult about it. Some further examples fully worked out are given hereafter.

DIFFERENTIATION

41

(1) Diﬀerentiate y = Being a constant,

a2 vanishes, and we have b2

a2 a 3 a2 x − x+ 2. b2 b b

a a2 dy = 2 × 3 × x3−1 − × 1 × x1−1 . dx b b But x1−1 = x0 = 1; so we get: dy 3a 2 a2 = 2x − . dx b b √ √ √ 3b 3 a (2) Diﬀerentiate y = 2a bx3 − − 2 ab. x Putting x in the index form, we get √ 3 √ √ y = 2a bx 2 − 3b 3 ax−1 − 2 ab. Now √ √ 3 dy = 2a b × 3 × x 2 −1 − 3b 3 a × (−1) × x−1−1 ; 2 dx √ √ 3b 3 a dy . = 3a bx + dx x2 1 4.4 − √ − 27◦ . 5 2 θ θ 2 1 −3 This may be written: z = 1.8 θ − 4.4 θ− 5 − 27◦ . (3) Diﬀerentiate z = 1.8 3 The 27◦ vanishes, and we have
2 1 dz 2 = 1.8 × − 3 × θ− 3 −1 − 4.4 × − 1 θ− 5 −1 ; 5 dθ 5 6 dz = −1.2 θ− 3 + 0.88 θ− 5 ; dθ 0.88 1.2 dz = √ −√ . 5 6 3 5 dθ θ θ

or,

or, or,

CALCULUS MADE EASY

42

A direct way of doing this will be explained later (see p. 66); but we can nevertheless manage it now without any diﬃculty. Developing the cube, we get v = 27t6 − 32.4t5 + 39.96t4 − 23.328t3 − 13.32t2 − 3.6t + 1; hence dv = 162t5 − 162t4 + 159.84t3 − 69.984t2 + 26.64t − 3.6. dt (5) Diﬀerentiate y = (2x − 3)(x + 1)2 . d (x + 1)(x + 1) dy d(2x − 3) = (2x − 3) + (x + 1)2 dx dx dx d(x + 1) d(x + 1) = (2x − 3) (x + 1) + (x + 1) dx dx d(2x − 3) + (x + 1)2 dx = 2(x + 1) (2x − 3) + (x + 1) = 2(x + 1)(3x − 2); or, more simply, multiply out and then diﬀerentiate. (6) Diﬀerentiate y = 0.5x3 (x − 3). dy d(x − 3) d(x3 ) = 0.5 x3 + (x − 3) dx dx dx = 0.5 x3 + (x − 3) × 3x2 = 2x3 − 4.5x2 . Same remarks as for preceding example. √ 1 1 (7) Diﬀerentiate w = θ + θ+√ . θ θ

(4) Diﬀerentiate v = (3t2 − 1.2t + 1)3 .

DIFFERENTIATION

43

This may be written w = (θ + θ−1 )(θ 2 + θ− 2 ).
−2 −1 2 1 d(θ + θ 1 dw ) ) −1 d(θ + θ 2 + θ− 2 ) = (θ + θ ) + (θ dθ dθ dθ 1 1 1 −1 1 − 2 1 −3 2 ) + (θ 2 + θ − 2 )(1 − θ −2 ) = (θ + θ )( 2 θ − 2 θ
1 3 1 5 1 1 3 1 1 1 1

= 1 (θ 2 + θ− 2 − θ− 2 − θ− 2 ) + (θ 2 + θ− 2 − θ− 2 − θ− 2 ) 2 √ 1 1 1 =3 θ−√ +1 √ −√ . 2 2 θ θ5 θ3

5

This, again, could be obtained more simply by multiplying the two factors ﬁrst, and diﬀerentiating afterwards. This is not, however, always possible; see, for instance, p. 170, example 8, in which the rule for diﬀerentiating a product must be used. a √ . (8) Diﬀerentiate y = 1 + a x + a2 x d(1 + ax 2 + a2 x) (1 + ax + a x) × 0 − a dy dx √ = dx (1 + a x + a2 x)2
1 2 1

2

=−

a( 1 ax− 2 + a2 ) 2 (1 + ax 2 + a2 x)2
1

1

.

x2 . (9) Diﬀerentiate y = 2 x +1 dy (x2 + 1) 2x − x2 × 2x 2x = = 2 . dx (x2 + 1)2 (x + 1)2 √ a+ x √ . (10) Diﬀerentiate y = a− x

CALCULUS MADE EASY
1 1 1

44

In the indexed form, y =
1 1

a + x2 a − x2

.
1 1 1

(a − x 2 )( 1 x− 2 ) − (a + x 2 )(− 1 x− 2 ) a − x2 + a + x2 dy 2 2 = = 1 ; 1 1 dx (a − x 2 )2 2(a − x 2 )2 x 2 a dy √ 2√ . = hence dx (a − x) x (11) Diﬀerentiate Now
3

√ 3 1 − a t2 √ . θ= 2 1 + a t3 2 1 − at 3 θ= 3 . 1 + at 2
1 2 1

(1 + at 2 )(− 2 at− 3 ) − (1 − at 3 ) × 3 at 2 dθ 3 2 = 3 dt 2 )2 (1 + at √ √ 4a 6 5a2 t7 − √ − 9a 2 t 3 t √ . = 2 3 2 6(1 + a t ) (12) A reservoir of square cross-section has sides sloping at an angle of 45◦ with the vertical. The side of the bottom is 200 feet. Find an expression for the quantity pouring in or out when the depth of water varies by 1 foot; hence ﬁnd, in gallons, the quantity withdrawn hourly when the depth is reduced from 14 to 10 feet in 24 hours. The volume of a frustum of pyramid of height H, and of bases A √ H and a, is V = (A + a + Aa). It is easily seen that, the slope being 3 45◦ , if the depth be h, the length of the side of the square surface of the water is 200 + 2h feet, so that the volume of water is h 4h3 [2002 + (200 + 2h)2 + 200(200 + 2h)] = 40, 000h + 400h2 + . 3 3

DIFFERENTIATION

45

dV = 40, 000 + 800h + 4h2 = cubic feet per foot of depth variation. dh dV The mean level from 14 to 10 feet is 12 feet, when h = 12, = dh 50, 176 cubic feet. Gallons per hour corresponding to a change of depth of 4 ft. in 4 × 50, 176 × 6.25 = 52, 267 gallons. 24 hours = 24 (13) The absolute pressure, in atmospheres, P , of saturated steam 5 40 + t ◦ at the temperature t C. is given by Dulong as being P = 140 as long as t is above 80◦ . Find the rate of variation of the pressure with the temperature at 100◦ C. Expand the numerator by the binomial theorem (see p. 137). P = 1 (405 + 5 × 404 t + 10 × 403 t2 + 10 × 402 t3 + 5 × 40t4 + t5 ); 1405 1 dP = dt 537, 824 × 105

hence

(5 × 404 + 20 × 403 t + 30 × 402 t2 + 20 × 40t3 + 5t4 ),

when t = 100 this becomes 0.036 atmosphere per degree Centigrade change of temperature. Exercises III. (1) Diﬀerentiate (a) u = 1 + x + x2 x3 + + ···. 1×2 1×2×3 (b) y = ax2 + bx + c. (c) y = (x + a)2 . (d ) y = (x + a)3 . (See the Answers on p. 253.)

CALCULUS MADE EASY

46

dw . dt (3) Find the diﬀerential coeﬃcient of
1 (2) If w = at − 2 bt2 , ﬁnd

y = (x + (4) Diﬀerentiate

√

−1) × (x −

√

−1).

y = (197x − 34x2 ) × (7 + 22x − 83x3 ). dx . dy (6) Diﬀerentiate y = 1.3709x × (112.6 + 45.202x2 ). (5) If x = (y + 3) × (y + 5), ﬁnd Find the diﬀerential coeﬃcients of (7) y = (9) y = 2x + 3 . 3x + 2 ax + b . cx + d (8) y = (10) y = 1 + x + 2x2 + 3x3 . 1 + x + 2x2 xn + a . x−n + b

(11) The temperature t of the ﬁlament of an incandescent electric lamp is connected to the current passing through the lamp by the relation C = a + bt + ct2 . Find an expression giving the variation of the current corresponding to a variation of temperature. (12) The following formulae have been proposed to express the relation between the electric resistance R of a wire at the temperature

DIFFERENTIATION

47

t◦ C., and the resistance R0 of that same wire at 0◦ Centigrade, a, b, c being constants. R = R0 (1 + at + bt2 ). √ R = R0 (1 + at + b t). R = R0 (1 + at + bt2 )−1 . Find the rate of variation of the resistance with regard to temperature as given by each of these formulae. (13) The electromotive-force E of a certain type of standard cell has been found to vary with the temperature t according to the relation E = 1.4340 1 − 0.000814(t − 15) + 0.000007(t − 15)2 volts. Find the change of electromotive-force per degree, at 15◦ , 20◦ and 25◦ . (14) The electromotive-force necessary to maintain an electric arc of length l with a current of intensity i has been found by Mrs. Ayrton to be E = a + bl + where a, b, c, k are constants. Find an expression for the variation of the electromotive-force (a) with regard to the length of the arc; (b) with regard to the strength of the current. c + kl , i

CHAPTER VII.
SUCCESSIVE DIFFERENTIATION.
Let us try the eﬀect of repeating several times over the operation of diﬀerentiating a function (see p. 13). Begin with a concrete case. Let y = x5 . First diﬀerentiation, 5x4 . = 20x3 . = 60x2 . = 120x.

Second diﬀerentiation, 5 × 4x3 Third diﬀerentiation, 5 × 4 × 3x2 Fourth diﬀerentiation, 5 × 4 × 3 × 2x Fifth diﬀerentiation, Sixth diﬀerentiation,

5 × 4 × 3 × 2 × 1 = 120. = 0.

There is a certain notation, with which we are already acquainted (see p. 14), used by some writers, that is very convenient. This is to employ the general symbol f (x) for any function of x. Here the symbol f ( ) is read as “function of,” without saying what particular function is meant. So the statement y = f (x) merely tells us that y is a function of x, it may be x2 or axn , or cos x or any other complicated function of x.

SUCCESSIVE DIFFERENTIATION

49

The corresponding symbol for the diﬀerential coeﬃcient is f (x), dy . This is called the “derived function” which is simpler to write than dx of x. Suppose we diﬀerentiate over again, we shall get the “second derived function” or second diﬀerential coeﬃcient, which is denoted by f (x); and so on. Now let us generalize. Let y = f (x) = xn . First diﬀerentiation, Second diﬀerentiation, Third diﬀerentiation, Fourth diﬀerentiation, f (x) = nxn−1 . f (x) = n(n − 1)xn−2 . f (x) = n(n − 1)(n − 2)xn−3 . f (x) = n(n − 1)(n − 2)(n − 3)xn−4 . etc., etc. But this is not the only way of indicating successive diﬀerentiations. For, if the original function be once diﬀerentiating gives y = f (x);

dy = f (x); dx dy d dx twice diﬀerentiating gives = f (x); dx d2 y d2 y and this is more conveniently written as , or more usually . (dx)2 dx2 d3 y Similarly, we may write as the result of thrice diﬀerentiating, = dx3 f (x).

CALCULUS MADE EASY

50

Examples. Now let us try y = f (x) = 7x4 + 3.5x3 − 1 x2 + x − 2. 2 dy dx d2 y dx2 d3 y dx3 d4 y dx4 d5 y dx5 = f (x) = 28x3 + 10.5x2 − x + 1, = f (x) = 84x2 + 21x − 1, = f (x) = 168x + 21, = f (x) = 168, =f (x) = 0.

In a similar manner if y = φ(x) = 3x(x2 − 4), dy = 3 x × 2x + (x2 − 4) × 1 = 3(3x2 − 4), dx d2 y φ (x) = 2 = 3 × 6x = 18x, dx d3 y φ (x) = 3 = 18, dx d4 y φ (x) = 4 = 0. dx φ (x) =

Exercises IV. (See page 253 for Answers.) dy d2 y Find and 2 for the following expressions: dx dx (1) y = 17x + 12x2 . (2) y = x2 + a . x+a

(3) y = 1 +

x x2 x3 x4 + + + . 1 1×2 1×2×3 1×2×3×4

SUCCESSIVE DIFFERENTIATION

51

(4) Find the 2nd and 3rd derived functions in the Exercises III. (p. 45), No. 1 to No. 7, and in the Examples given (p. 40), No. 1 to No. 7.

CHAPTER VIII.
WHEN TIME VARIES.
Some of the most important problems of the calculus are those where time is the independent variable, and we have to think about the values of some other quantity that varies when the time varies. Some things grow larger as time goes on; some other things grow smaller. The distance that a train has got from its starting place goes on ever increasing as time goes on. Trees grow taller as the years go by. Which is growing at the greater rate; a plant 12 inches high which in one month becomes 14 inches high, or a tree 12 feet high which in a year becomes 14 feet high? In this chapter we are going to make much use of the word rate. Nothing to do with poor-rate, or water-rate (except that even here the word suggests a proportion—a ratio—so many pence in the pound). Nothing to do even with birth-rate or death-rate, though these words suggest so many births or deaths per thousand of the population. When a motor-car whizzes by us, we say: What a terriﬁc rate! When a spendthrift is ﬂinging about his money, we remark that that young man is living at a prodigious rate. What do we mean by rate? In both these cases we are making a mental comparison of something that is happening, and the length of time that it takes to happen. If the

WHEN TIME VARIES

53

motor-car ﬂies past us going 10 yards per second, a simple bit of mental arithmetic will show us that this is equivalent—while it lasts—to a rate of 600 yards per minute, or over 20 miles per hour. Now in what sense is it true that a speed of 10 yards per second is the same as 600 yards per minute? Ten yards is not the same as 600 yards, nor is one second the same thing as one minute. What we mean by saying that the rate is the same, is this: that the proportion borne between distance passed over and time taken to pass over it, is the same in both cases. Take another example. A man may have only a few pounds in his possession, and yet be able to spend money at the rate of millions a year—provided he goes on spending money at that rate for a few minutes only. Suppose you hand a shilling over the counter to pay for some goods; and suppose the operation lasts exactly one second. Then, during that brief operation, you are parting with your money at the rate of 1 shilling per second, which is the same rate as £3 per minute, or £180 per hour, or £4320 per day, or £1, 576, 800 per year! If you have £10 in your pocket, you can go on spending money at the rate of a million a year for just 5 1 minutes. 4 It is said that Sandy had not been in London above ﬁve minutes when “bang went saxpence.” If he were to spend money at that rate all day long, say for 12 hours, he would be spending 6 shillings an hour, or £3. 12s. per day, or £21. 12s. a week, not counting the Sawbbath. Now try to put some of these ideas into diﬀerential notation. Let y in this case stand for money, and let t stand for time. If you are spending money, and the amount you spend in a short

CALCULUS MADE EASY

54

dy time dt be called dy, the rate of spending it will be , or rather, should dt dy be written with a minus sign, as − , because dy is a decrement, not an dt increment. But money is not a good example for the calculus, because it generally comes and goes by jumps, not by a continuous ﬂow—you may earn £200 a year, but it does not keep running in all day long in a thin stream; it comes in only weekly, or monthly, or quarterly, in lumps: and your expenditure also goes out in sudden payments. A more apt illustration of the idea of a rate is furnished by the speed of a moving body. From London (Euston station) to Liverpool is 200 miles. If a train leaves London at 7 o’clock, and reaches Liverpool at 11 o’clock, you know that, since it has travelled 200 miles in 4 hours, its average rate must have been 50 miles per hour; because
200 4

=

50 . 1

Here you are really making a mental comparison between the distance passed over and the time taken to pass over it. You are dividing one by the other. If y is the whole distance, and t the whole time, clearly y the average rate is . Now the speed was not actually constant all the t way: at starting, and during the slowing up at the end of the journey, the speed was less. Probably at some part, when running downhill, the speed was over 60 miles an hour. If, during any particular element of time dt, the corresponding element of distance passed over was dy, dy then at that part of the journey the speed was . The rate at which dt one quantity (in the present instance, distance) is changing in relation to the other quantity (in this case, time) is properly expressed, then, by stating the diﬀerential coeﬃcient of one with respect to the other. A velocity, scientiﬁcally expressed, is the rate at which a very small distance in any given direction is being passed over; and may therefore

WHEN TIME VARIES

55

be written

dy . dt But if the velocity v is not uniform, then it must be either increasing v=

or else decreasing. The rate at which a velocity is increasing is called the acceleration. If a moving body is, at any particular instant, gaining an additional velocity dv in an element of time dt, then the acceleration a at that instant may be written a= but dv is itself d dv ; dt

dy . Hence we may put dt d a= dy dt ; dt

and this is usually written a =

d2 y ; dt2 or the acceleration is the second diﬀerential coeﬃcient of the distance, with respect to time. Acceleration is expressed as a change of velocity in unit time, for instance, as being so many feet per second per second; the notation used being feet ÷ second2 . When a railway train has just begun to move, its velocity v is small;

but it is rapidly gaining speed—it is being hurried up, or accelerated, d2 y by the eﬀort of the engine. So its 2 is large. When it has got up its dt d2 y top speed it is no longer being accelerated, so that then 2 has fallen dt to zero. But when it nears its stopping place its speed begins to slow down; may, indeed, slow down very quickly if the brakes are put on,

CALCULUS MADE EASY

56

and during this period of deceleration or slackening of pace, the value d2 y dv , that is, of 2 will be negative. of dt dt To accelerate a mass m requires the continuous application of force. The force necessary to accelerate a mass is proportional to the mass, and it is also proportional to the acceleration which is being imparted. Hence we may write for the force f , the expression f = ma; or or f =m dv ; dt d2 y f =m 2. dt

The product of a mass by the speed at which it is going is called its momentum, and is in symbols mv. If we diﬀerentiate momentum with d(mv) for the rate of change of momentum. respect to time we shall get dt dv But, since m is a constant quantity, this may be written m , which dt we see above is the same as f . That is to say, force may be expressed either as mass times acceleration, or as rate of change of momentum. Again, if a force is employed to move something (against an equal and opposite counter-force), it does work ; and the amount of work done is measured by the product of the force into the distance (in its own direction) through which its point of application moves forward. So if a force f moves forward through a length y, the work done (which we may call w) will be w = f × y; where we take f as a constant force. If the force varies at diﬀerent parts of the range y, then we must ﬁnd an expression for its value from

WHEN TIME VARIES

57

point to point. If f be the force along the small element of length dy, the amount of work done will be f × dy. But as dy is only an element then an element of work will be dw; and we have dw = f × dy; which may be written dw = ma · dy; dw = m of length, only an element of work will be done. If we write w for work,

or or

d2 y · dy; dt2 dv dw = m · dy. dt Further, we may transpose the expression and write dw = f. dy This gives us yet a third deﬁnition of force; that if it is being used

to produce a displacement in any direction, the force (in that direction) is equal to the rate at which work is being done per unit of length in that direction. In this last sentence the word rate is clearly not used in its time-sense, but in its meaning as ratio or proportion. Sir Isaac Newton, who was (along with Leibnitz) an inventor of the methods of the calculus, regarded all quantities that were varying as ﬂowing; and the ratio which we nowadays call the diﬀerential coeﬃcient he regarded as the rate of ﬂowing, or the ﬂuxion of the quantity in question. He did not use the notation of the dy and dx, and dt (this was due to Leibnitz), but had instead a notation of his own. If y was a quantity

CALCULUS MADE EASY

58

that varied, or “ﬂowed,” then his symbol for its rate of variation (or “ﬂuxion”) was y. If x was the variable, then its ﬂuxion was called x. ˙ ˙ The dot over the letter indicated that it had been diﬀerentiated. But this notation does not tell us what is the independent variable with dy respect to which the diﬀerentiation has been eﬀected. When we see dt dy we know that y is to be diﬀerentiated with respect to t. If we see dx we know that y is to be diﬀerentiated with respect to x. But if we see merely y, we cannot tell without looking at the context whether this is ˙ dy dy dy or or , or what is the other variable. So, therefore, to mean dx dt dz this ﬂuxional notation is less informing than the diﬀerential notation, and has in consequence largely dropped out of use. But its simplicity gives it an advantage if only we will agree to use it for those cases exclusively where time is the independent variable. In that case y will ˙ 2 du dx dy and u will mean ˙ ; and x will mean 2 . ¨ mean dt dt dt Adopting this ﬂuxional notation we may write the mechanical equations considered in the paragraphs above, as follows: distance velocity acceleration force work Examples. (1) A body moves so that the distance x (in feet), which it travels from a certain point O, is given by the relation x = 0.2t2 + 10.4, where t is the time in seconds elapsed since a certain instant. Find the velocity x, v = x, ˙ a = v = x, ˙ ¨ f = mv = m¨, ˙ x w = x × m¨. x

WHEN TIME VARIES

59

and acceleration 5 seconds after the body began to move, and also ﬁnd the corresponding values when the distance covered is 100 feet. Find also the average velocity during the ﬁrst 10 seconds of its motion. (Suppose distances and motion to the right to be positive.) Now v=x= ˙ dx = 0.4t; dt x = 0.2t2 + 10.4 and a = x = ¨ d2 x = 0.4 = constant. dt2

When t = 0, x = 10.4 and v = 0. The body started from a point 10.4 feet to the right of the point O; and the time was reckoned from the instant the body started. When t = 5, v = 0.4 × 5 = 2 ft./sec.; a = 0.4 ft./sec2 . v = 0.4 × 21.17 = 8.468 ft./sec. When t = 10, distance travelled = 0.2 × 102 + 10.4 − 10.4 = 20 ft. Average velocity =
20 10

When x = 100, 100 = 0.2t2 + 10.4, or t2 = 448, and t = 21.17 sec.;

= 2 ft./sec.

(It is the same velocity as the velocity at the middle of the interval, t = 5; for, the acceleration being constant, the velocity has varied uniformly from zero when t = 0 to 4 ft./sec. when t = 10.) (2) In the above problem let us suppose x = 0.2t2 + 3t + 10.4. dx v=x= ˙ = 0.4t + 3; dt d2 x a = x = 2 = 0.4 = constant. ¨ dt

When t = 0, x = 10.4 and v = 3 ft./sec, the time is reckoned from the instant at which the body passed a point 10.4 ft. from the point O,

CALCULUS MADE EASY

60

its velocity being then already 3 ft./sec. To ﬁnd the time elapsed since The body began moving 7.5 sec. before time was begun to be observed; 5 seconds after this gives t = −2.5 and v = 0.4 × −2.5 + 3 = 2 ft./sec. When x = 100 ft., 100 = 0.2t2 + 3t + 10.4; or t2 + 15t − 448 = 0;
3 it began moving, let v = 0; then 0.4t + 3 = 0, t = − .4 = −7.5 sec.

hence t = 14.95 sec., v = 0.4 × 14.95 + 3 = 8.98 ft./sec.

To ﬁnd the distance travelled during the 10 ﬁrst seconds of the

motion one must know how far the body was from the point O when it started. When t = −7.5, x = 0.2 × (−7.5)2 − 3 × 7.5 + 10.4 = −0.85 ft., that is 0.85 ft. to the left of the point O. Now, when t = 2.5, x = 0.2 × 2.52 + 3 × 2.5 + 10.4 = 19.15. So, in 10 seconds, the distance travelled was 19.15 + 0.85 = 20 ft., and the average velocity =
20 10

= 2 ft./sec.

(3) Consider a similar problem when the distance is given by x = 0.2t2 − 3t + 10.4. Then v = 0.4t − 3, a = 0.4 = constant. When in the direction opposite to its motion in the previous cases. As the acceleration is positive, however, we see that this velocity will decrease t = 0, x = 10.4 as before, and v = −3; so that the body was moving

WHEN TIME VARIES

61

as time goes on, until it becomes zero, when v = 0 or 0.4t − 3 = 0; or after the body started, t = 12.5, and v = 0.4 × 12.5 − 3 = 2 ft./sec. When x = 100, 100 = 0.2t2 − 3t + 10.4, and or t2 − 15t − 448 = 0,

t = 7.5 sec. After this, the velocity becomes positive; and 5 seconds

t = 29.95; v = 0.4 × 29.95 − 3 = 8.98 ft./sec.

us that the body moves back to 0.85 ft. beyond the point O before it stops. Ten seconds later t = 17.5 and x = 0.2 × 17.52 − 3 × 17.5 + 10.4 = 19.15. The distance travelled = .85 + 19.15 = 20.0, and the average velocity is again 2 ft./sec. 3t2 + 10.4; v = 0.6t2 − 6t; a = 1.2t − 6. The acceleration is no more constant. When t = 0, x = 10.4, v = 0, a = −6. The body is at rest, but just (4) Consider yet another problem of the same sort with x = 0.2t3 −

When v is zero, x = 0.2 × 7.52 − 3 × 7.5 + 10.4 = −0.85, informing

ready to move with a negative acceleration, that is to gain a velocity towards the point O. (5) If we have x = 0.2t3 − 3t + 10.4, then v = 0.6t2 − 3, and a = 1.2t. When t = 0, x = 10.4; v = −3; a = 0. The body is moving towards the point O with a velocity of 3 ft./sec.,

and just at that instant the velocity is uniform.

CALCULUS MADE EASY

62

We see that the conditions of the motion can always be at once ascertained from the time-distance equation and its ﬁrst and second derived functions. In the last two cases the mean velocity during the ﬁrst 10 seconds and the velocity 5 seconds after the start will no more be the same, because the velocity is not increasing uniformly, the acceleration being no longer constant. (6) The angle θ (in radians) turned through by a wheel is given by θ = 3 + 2t − 0.1t3 , where t is the time in seconds from a certain instant; ﬁnd the angular velocity ω and the angular acceleration α, (a) after at rest, and how many revolutions has it performed up to that instant? Writing for the acceleration ˙ dθ = 2 − 0.3t2 , ω=θ= dt
2 ¨ dθ α = θ = 2 = −0.6t. dt

1 second; (b) after it has performed one revolution. At what time is it

When t = 0, θ = 3; ω = 2 rad./sec.; α = 0. When t = 1, ω = 2 − 0.3 = 1.7 rad./sec.; α = −0.6 rad./sec2 .

This is a retardation; the wheel is slowing down. After 1 revolution θ = 2π = 6.28; 6.28 = 3 + 2t − 0.1t3 .

values of t for which θ = 6.28; these are 2.11 and 3.03 (there is a third negative value).

By plotting the graph, θ = 3 + 2t − 0.1t3 , we can get the value or

WHEN TIME VARIES

63

When t = 2.11, θ = 6.28; ω = 2 − 1.34 = 0.66 rad./sec.;

α = −1.27 rad./sec2 . When t = 3.03, θ = 6.28;

ω = 2 − 2.754 = −0.754 rad./sec.; α = −1.82 rad./sec2 .

The velocity is reversed. The wheel is evidently at rest between or when t = 2.58 sec., it has performed these two instants; it is at rest when ω = 0, that is when 0 = 2 − 0.3t3 , 3 + 2 × 2.58 − 0.1 × 2.583 θ = = 1.025 revolutions. 2π 6.28 Exercises V. (See page 255 for Answers.) dy d2 y and 2 . dt dt d2 y = 2b + 12ct2 . dt2

(1) If y = a + bt2 + ct4 ; ﬁnd Ans.

dy = 2bt + 4ct3 ; dt

(2) A body falling freely in space describes in t seconds a space s, in feet, expressed by the equation s = 16t2 . Draw a curve showing the relation between s and t. Also determine the velocity of the body at the following times from its being let drop: t = 2 seconds; t = 4.6 seconds; t = 0.01 second. ˙ ¨ (3) If x = at − 1 gt2 ; ﬁnd x and x. 2

CALCULUS MADE EASY

64

(4) If a body move according to the law s = 12 − 4.5t + 6.2t2 , ﬁnd its velocity when t = 4 seconds; s being in feet. (5) Find the acceleration of the body mentioned in the preceding example. Is the acceleration the same for all values of t? (6) The angle θ (in radians) turned through by a revolving wheel is connected with the time t (in seconds) that has elapsed since starting; by the law θ = 2.1 − 3.2t + 4.8t2 . Find the angular velocity (in radians per second) of that wheel when
1 12

seconds have elapsed. Find also its angular acceleration. (7) A slider moves so that, during the ﬁrst part of its motion, its

distance s in inches from its starting point is given by the expression s = 6.8t3 − 10.8t; t being in seconds.

Find the expression for the velocity and the acceleration at any time; and hence ﬁnd the velocity and the acceleration after 3 seconds. (8) The motion of a rising balloon is such that its height h, in miles, √ 1 is given at any instant by the expression h = 0.5 + 10 3 t − 125; t being in seconds. Find an expression for the velocity and the acceleration at any time. Draw curves to show the variation of height, velocity and acceleration during the ﬁrst ten minutes of the ascent.

WHEN TIME VARIES

65

(9) A stone is thrown downwards into water and its depth p in metres at any instant t seconds after reaching the surface of the water is given by the expression p= 4 + 0.8t − 1. 4 + t2

Find an expression for the velocity and the acceleration at any time. Find the velocity and acceleration after 10 seconds. (10) A body moves in such a way that the spaces described in the time t from starting is given by s = tn , where n is a constant. Find the value of n when the velocity is doubled from the 5th to the 10th second; ﬁnd it also when the velocity is numerically equal to the acceleration at the end of the 10th second.

CHAPTER IX.
INTRODUCING A USEFUL DODGE.
Sometimes one is stumped by ﬁnding that the expression to be diﬀerentiated is too complicated to tackle directly. Thus, the equation y = (x2 + a2 ) 2 is awkward to a beginner. Now the dodge to turn the diﬃculty is this: Write some symbol, such as u, for the expression x2 + a2 ; then the equation becomes y = u2 , which you can easily manage; for 3 1 dy = u2 . du 2 Then tackle the expression u = x 2 + a2 , and diﬀerentiate it with respect to x, du = 2x. dx
3 3

INTRODUCING A USEFUL DODGE

67

Then all that remains is plain sailing; for that is, dy du dy = × ; dx du dx dy 3 1 = u 2 × 2x dx 2 1 = 3 (x2 + a2 ) 2 × 2x 2 = 3x(x2 + a2 ) 2 ; and so the trick is done. By and bye, when you have learned how to deal with sines, and cosines, and exponentials, you will ﬁnd this dodge of increasing usefulness.
1

Examples. Let us practise this dodge on a few examples. √ (1) Diﬀerentiate y = a + x. Let a + x = u.
1 1 1 du dy = 1; y = u 2 ; = 1 u− 2 = 1 (a + x)− 2 . 2 2 dx du dy dy du 1 = × = √ . dx du dx 2 a+x

(2) Diﬀerentiate y = √ Let a + x2 = u.

1 . a + x2

1 3 du dy 1 = 2x; y = u− 2 ; = − 2 u− 2 . dx du dy dy du x = × =− . dx du dx (a + x2 )3

CALCULUS MADE EASY

68

(3) Diﬀerentiate y =
2 4

m − nx +

2 3

p x3
4

a

.

Let m − nx 3 + px− 3 = u.
1 7 du = − 2 nx− 3 − 4 px− 3 ; 3 3 dx dy y = ua ; = aua−1 . du a−1 1 7 2 p dy dy du 2 ( 3 nx− 3 + 4 px− 3 ). = × = −a m − nx 3 + 4 3 dx du dx x3

(4) Diﬀerentiate y = √ Let u = x3 − a2 .

x3

1 . − a2

1 3 1 dy du = 3x2 ; y = u− 2 ; = − (x3 − a2 )− 2 . dx du 2 2 dy dy du 3x = × =− . dx du dx 2 (x3 − a2 )3

1−x . 1+x 1 (1 − x) 2 Write this as y = 1 . (1 + x) 2 (5) Diﬀerentiate y = d(1 − x) 2 1 d(1 + x) 2 (1 + x) − (1 − x) 2 dy dx dx = . dx 1+x
1 2 1 1

product.)

(We may also write y = (1 − x) 2 (1 + x)− 2 and diﬀerentiate as a Proceeding as in example (1) above, we get d(1 − x) 2 1 =− √ ; dx 2 1−x
1

1

1

and

d(1 + x) 2 1 = √ . dx 2 1+x

1

INTRODUCING A USEFUL DODGE

69

Hence (1 + x) 2 (1 − x) 2 dy √ √ =− − dx 2(1 + x) 1 − x 2(1 + x) 1 + x √ 1 1−x √ ; =− √ − 2 1 + x 1 − x 2 (1 + x)3 dy 1 √ =− . dx (1 + x) 1 − x2 (6) Diﬀerentiate y = We may write this x3 . 1 + x2 y = x 2 (1 + x2 )− 2 ; d (1 + x2 )− 2 1 3 dy 3 1 = 2 x 2 (1 + x2 )− 2 + x 2 × . dx dx Diﬀerentiating (1 + x2 )− 2 , as shown in example (2) above, we get d (1 + x2 )− 2 =− dx so that √ dy 3 x = √ − dx 2 1 + x2 √ √ x5 (1 + x2 )3
1 1 1 3 1 1 1

or

x (1 + x2 )3 √ 2

;

=

x(3 + x2 ) (1 + x2 )3

.

(7) Diﬀerentiate y = (x + √ Let x + x2 + x + a = u.

x2 + x + a)3 .

y = u3 ;

d (x2 + x + a) 2 du =1+ . dx dx √ dy and = 3u2 = 3 x + x2 + x + a du

1

2

.

CALCULUS MADE EASY

70

Now let (x2 + x + a) 2 = v and (x2 + x + a) = w. dw dx dv dx du Hence dx dy dx = 2x + 1; = v = w2;
1 1 dv = 1 w− 2 . 2 dw

1

1 dv dw 1 × = 2 (x2 + x + a)− 2 (2x + 1). dw dx 2x + 1 =1+ √ , 2 x2 + x + a dy du = × du dx 2 √ 2x + 1 = 3 x + x2 + x + a 1+ √ 2 x2 + x + a

.

(8) Diﬀerentiate y = We get
1

a2 + x 2 a2 − x 2

3

a2 − x 2 . a2 + x 2
1 1

y=

2 2 − d (a2 + x2 ) 6 1 d (a − x ) 6 dy = (a2 + x2 ) 6 + . 1 dx dx (a2 − x2 ) 6 dx

(a2 − x ) (a2 + x )

(a2 + x2 ) 2 (a2 − x2 ) 3
1 2 2 1 2 3

1

= (a2 + x2 ) 6 (a2 − x2 )− 6 .
1 1

Let u = (a2 − x2 )− 6 and v = (a2 − x2 ). du 1 7 dv = − v− 6 ; = −2x. dv 6 dx 7 du du dv 1 = × = x(a2 − x2 )− 6 . dx dv dx 3 u = v− 6 ;
1

1

Let w = (a2 + x2 ) 6 and z = (a2 + x2 ). dw 1 5 dz = z− 6 ; = 2x. dz 6 dx 5 dw dw dz 1 = × = x(a2 + x2 )− 6 . dx dz dx 3 w = z6;
1

1

INTRODUCING A USEFUL DODGE

71

Hence
1 x x dy = (a2 + x2 ) 6 7 + 1 5 ; dx 3(a2 − x2 ) 6 3(a2 − x2 ) 6 (a2 + x2 ) 6

or

x dy = dx 3

6

a2 + x 2 + (a2 − x2 )7

1
6

(a2 − x2 )(a2 + x2 )5 ]

.

(9) Diﬀerentiate y n with respect to y 5 . d(y n ) ny n−1 n = 5−1 = y n−5 . 5) d(y 5y 5 (10) Find the ﬁrst and second diﬀerential coeﬃcients x of y = (a − x)x. b dy x d (a − x)x = dx b dx Let (a − x)x
1 2 1 2

+

(a − x)x . b
1

= u and let (a − x)x = w; then u = w 2 .

du 1 1 1 1 = w− 2 = . 1 = dw 2 2 (a − x)x 2w 2 dw = a − 2x. dx du dw du a − 2x × = = . dw dx dx 2 (a − x)x Hence x(a − 2x) dy = + dx 2b (a − x)x (a − x)x x(3a − 4x) = . b 2b (a − x)x

CALCULUS MADE EASY

72

Now d2 y = dx2 = 2b (a − x)x (3a − 8x) − 3a − 12ax + 8x
2 2

(3ax − 4x2 )b(a − 2x) (a − x)x

4b2 (a − x)x .

4b(a − x) (a − x)x

(We shall need these two last diﬀerential coeﬃcients later on. See Ex. X. No. 11.) Exercises VI. √ (See page 255 for Answers.) √

Diﬀerentiate the following: (1) y = x2 + 1. 1 . a+x x 2 − a2 . x2 (2) y = x 2 + a2 .

(3) y = √ (5) y = (7) y = √

a . (4) y = √ a − x2

√ 3 4 x +a (6) y = √ . 2 3 x +a

a2 + x 2 . (a + x)2

(8) Diﬀerentiate y 5 with respect to y 2 . √ 1 − θ2 (9) Diﬀerentiate y = . 1−θ The process can be extended to three or more diﬀerential coeﬃdy dy dz dv = × × . cients, so that dx dz dv dx

INTRODUCING A USEFUL DODGE

73

Examples. (1) If z = 3x4 ; v = We have

√ dv 7 ; y = 1 + v, ﬁnd . 2 z dx

dv 1 14 dz dy ; = √ = − 3; = 12x3 . dv dz z dx 2 1+v dy 168x3 28 √ √ =− =− . dx (2 1 + v)z 3 3x5 9x8 + 7 t 1 7x2 dv (2) If t = √ ; x = t3 + ; v = √ , ﬁnd . 3 2 dθ x−1 5 θ 7x(5x − 6) dv = 3 ; dx 3 (x − 1)4 dx 1 = 3t2 + 2 ; dt

dt 1 =− √ . dθ 10 θ3

Hence

an expression in which x must be replaced by its value, and t by its value in terms of θ. √ √ 3a2 x 1 − θ2 1 dφ (3) If θ = √ ; ω = ; and φ = 3 − √ , ﬁnd . 1+θ dx ω 2 x3 We get θ = 3a2 x− 2 ; √ 1−θ 1 ; and φ = 3 − √ ω −1 . 1+θ 2 2 dθ 3a dω 1 √ =− √ ; =− 3 dx dθ (1 + θ) 1 − θ2 2 x
1

1 7x(5x − 6)(3t2 + 2 ) dv √ =− , dθ 30 3 (x − 1)4 θ3

ω=

(see example 5, p. 68); and

dφ 1 =√ . dω 2ω 2 1 3a2 dθ 1 √ =√ × × √ . dx 2 × ω 2 (1 + θ) 1 − θ2 2 x3 Replace now ﬁrst ω, then θ by its value. So that

CALCULUS MADE EASY

74

Exercises VII.

You can now successfully try the following. (See

page 256 for Answers.) dw 1 , ﬁnd . 2 v dx √ √ 1 dv (2) If y = 3x2 + 2; z = 1 + y; and v = √ , ﬁnd . dx 3 + 4z x3 1 du (3) If y = √ ; z = (1 + y)2 ; and u = √ , ﬁnd . dx 1+z 3
1 (1) If u = 2 x3 ; v = 3(u + u2 ); and w =

CHAPTER X.
GEOMETRICAL MEANING OF DIFFERENTIATION.
It is useful to consider what geometrical meaning can be given to the diﬀerential coeﬃcient. In the ﬁrst place, any function of x, such, for example, as x2 , or familiar with the process of curve-plotting.
R Y Q dx P y

√ x,

or ax + b, can be plotted as a curve; and nowadays every schoolboy is

dy

O

x

dx

X

Fig. 7.

Let P QR, in Fig. 7, be a portion of a curve plotted with respect to the axes of coordinates OX and OY . Consider any point Q on this curve, where the abscissa of the point is x and its ordinate is y. Now observe how y changes when x is varied. If x is made to increase by

CALCULUS MADE EASY

76

a small increment dx, to the right, it will be observed that y also (in this particular curve) increases by a small increment dy (because this particular curve happens to be an ascending curve). Then the ratio of dy to dx is a measure of the degree to which the curve is sloping up between the two points Q and T . As a matter of fact, it can be seen on the ﬁgure that the curve between Q and T has many diﬀerent slopes, so that we cannot very well speak of the slope of the curve between Q and T . If, however, Q and T are so near each other that the small portion QT of the curve is practically straight, then it is true to say that dy is the slope of the curve along QT . The straight line QT the ratio dx produced on either side touches the curve along the portion QT only, and if this portion is indeﬁnitely small, the straight line will touch the curve at practically one point only, and be therefore a tangent to the curve. This tangent to the curve has evidently the same slope as QT , so dy that is the slope of the tangent to the curve at the point Q for which dx dy the value of is found. dx We have seen that the short expression “the slope of a curve” has no precise meaning, because a curve has so many slopes—in fact, every small portion of a curve has a diﬀerent slope. “The slope of a curve at a point” is, however, a perfectly deﬁned thing; it is the slope of a very small portion of the curve situated just at that point; and we have seen that this is the same as “the slope of the tangent to the curve at that point.” Observe that dx is a short step to the right, and dy the corresponding short step upwards. These steps must be considered as short as

MEANING OF DIFFERENTIATION

77

possible—in fact indeﬁnitely short,—though in diagrams we have to represent them by bits that are not inﬁnitesimally small, otherwise they could not be seen. We shall hereafter make considerable use of this circumstance that dy represents the slope of the curve at any point. dx
Y dy dx

O

X

Fig. 8.

If a curve is sloping up at 45◦ at a particular point, as in Fig. 8, dy dy and dx will be equal, and the value of = 1. dx dy If the curve slopes up steeper than 45◦ (Fig. 9), will be greater dx than 1. dy If the curve slopes up very gently, as in Fig. 10, will be a fraction dx smaller than 1. For a horizontal line, or a horizontal place in a curve, dy = 0, and dy therefore = 0. dx If a curve slopes downward, as in Fig. 11, dy will be a step down, dy will have and must therefore be reckoned of negative value; hence dx negative sign also.

CALCULUS MADE EASY
Y Y

78

dy dx

dy dx

O

X

O

X

Fig. 9.

Fig. 10.

If the “curve” happens to be a straight line, like that in Fig. 12, the dy will be the same at all points along it. In other words its value of dx slope is constant. If a curve is one that turns more upwards as it goes along to the dy right, the values of will become greater and greater with the indx creasing steepness, as in Fig. 13. If a curve is one that gets ﬂatter and ﬂatter as it goes along, the dy values of will become smaller and smaller as the ﬂatter part is dx
Y Q dx dy

y

O

x

dx

X

Fig. 11.

MEANING OF DIFFERENTIATION
Y dy dx dy dx dy dx

79

O

X

Fig. 12.
Y

dy dx dy dy dx O X dx

Fig. 13.

reached, as in Fig. 14. If a curve ﬁrst descends, and then goes up again, as in Fig. 15, dy presenting a concavity upwards, then clearly will ﬁrst be negative, dx with diminishing values as the curve ﬂattens, then will be zero at the point where the bottom of the trough of the curve is reached; and from dy this point onward will have positive values that go on increasing. In dx such a case y is said to pass by a minimum. The minimum value of y is not necessarily the smallest value of y, it is that value of y corresponding to the bottom of the trough; for instance, in Fig. 28 (p. 99), the value of y corresponding to the bottom of the trough is 1, while y takes

CALCULUS MADE EASY
Y Y

80

y min. O X O X

Fig. 14.

Fig. 15.

elsewhere values which are smaller than this. The characteristic of a minimum is that y must increase on either side of it. N.B.—For the particular value of x that makes y a minimum, the dy = 0. value of dx dy If a curve ﬁrst ascends and then descends, the values of will be dx positive at ﬁrst; then zero, as the summit is reached; then negative, as the curve slopes downwards, as in Fig. 16. In this case y is said to pass by a maximum, but the maximum value of y is not necessarily the greatest value of y. In Fig. 28, the maximum of y is 2 1 , but this is 3 by no means the greatest value y can have at some other point of the curve. N.B.—For the particular value of x that makes y a maximum, the dy value of = 0. dx dy will If a curve has the peculiar form of Fig. 17, the values of dx always be positive; but there will be one particular place where the dy slope is least steep, where the value of will be a minimum; that is, dx less than it is at any other part of the curve.

MEANING OF DIFFERENTIATION
Y Y

81

y max.

O

X

O

X

Fig. 16.

Fig. 17.

dy will be negative dx in the upper part, and positive in the lower part; while at the nose of dy the curve where it becomes actually perpendicular, the value of will dx be inﬁnitely great. If a curve has the form of Fig. 18, the value of
Y dx dy

Q

dy dx O X

Fig. 18.

dy measures the steepness of a curve dx at any point, let us turn to some of the equations which we have already Now that we understand that learned how to diﬀerentiate.

CALCULUS MADE EASY

82

(1) As the simplest case take this: y = x + b. It is plotted out in Fig. 19, using equal scales for x and y. If we put x = 0, then the corresponding ordinate will be y = b; that is to say, the “curve” crosses the y-axis at the height b. From here it ascends at 45◦ ;
Y Y

dy dx b O X b O X

Fig. 19.

Fig. 20.

for whatever values we give to x to the right, we have an equal y to ascend. The line has a gradient of 1 in 1. Now diﬀerentiate y = x + b, by the rules we have already learned dy (pp. 21 and 25 ante), and we get = 1. dx The slope of the line is such that for every little step dx to the right, we go an equal little step dy upward. And this slope is constant—always the same slope. (2) Take another case: y = ax + b.

MEANING OF DIFFERENTIATION

83

We know that this curve, like the preceding one, will start from a height b on the y-axis. But before we draw the curve, let us ﬁnd its dy = a. The slope will be constant, slope by diﬀerentiating; which gives dx at an angle, the tangent of which is here called a. Let us assign to a some numerical value—say 1 . Then we must give it such a slope that 3 it ascends 1 in 3; or dx will be 3 times as great as dy; as magniﬁed in Fig. 21. So, draw the line in Fig. 20 at this slope.

Fig. 21.

(3) Now for a slightly harder case. Let y = ax2 + b.

Again the curve will start on the y-axis at a height b above the origin. Now diﬀerentiate. [If you have forgotten, turn back to p. 25; or, rather, don’t turn back, but think out the diﬀerentiation.] dy = 2ax. dx This shows that the steepness will not be constant: it increases as x increases. At the starting point P , where x = 0, the curve (Fig. 22) has no steepness—that is, it is level. On the left of the origin, where x dy has negative values, will also have negative values, or will descend dx from left to right, as in the Figure.

CALCULUS MADE EASY
Y

84

R

Q P b O X

Fig. 22.

Let us illustrate this by working out a particular instance. Taking the equation y = 1 x2 + 3, 4 and diﬀerentiating it, we get dy = 1 x. 2 dx Now assign a few successive values, say from 0 to 5, to x; and calculate dy from the the corresponding values of y by the ﬁrst equation; and of dx second equation. Tabulating results, we have: x y dy dx 0 3 0 1 31 4
1 2

2 4 1

3 51 4 11 2

4 7 2

5 91 4 21 2

Then plot them out in two curves, Figs. 23 and 24, in Fig. 23 plotting dy the values of y against those of x and in Fig. 24 those of against dx those of x. For any assigned value of x, the height of the ordinate in the second curve is proportional to the slope of the ﬁrst curve.

MEANING OF DIFFERENTIATION y 9 8 7 6 5 dy dx

85

x2 +

3

=

1 x2 4

4

1

b −3 −2 −1 0 1 2 3 4 5 x

5 4 3 2 −3 −2 −1 1

y

dy = 12x dx
0 1 2 3 4 5 x

Fig. 23.

Fig. 24.

If a curve comes to a sudden cusp, as in Fig. 25, the slope at that point suddenly changes from a slope upward to a slope downward. In
Y

O

X

Fig. 25.

dy will clearly undergo an abrupt change from a positive to dx a negative value. that case The following examples show further applications of the principles just explained.

CALCULUS MADE EASY

86

(4) Find the slope of the tangent to the curve y= 1 + 3, 2x

at the point where x = −1. Find the angle which this tangent makes with the curve y = 2x2 + 2. The slope of the tangent is the slope of the curve at the point where dy of the curve for they touch one another (see p. 76); that is, it is the dx dy 1 dy 1 that point. Here = − 2 and for x = −1, = − , which is the dx 2x dx 2 slope of the tangent and of the curve at that point. The tangent, being dy = a, a straight line, has for equation y = ax + b, and its slope is dx 1 1 hence a = − . Also if x = −1, y = + 3 = 2 1 ; and as the 2 2 2(−1) tangent passes by this point, the coordinates of the point must satisfy the equation of the tangent, namely 1 y = − x + b, 2 1 1 so that 2 2 = − × (−1) + b and b = 2; the equation of the tangent is 2 1 therefore y = − x + 2. 2 Now, when two curves meet, the intersection being a point common to both curves, its coordinates must satisfy the equation of each one of the two curves; that is, it must be a solution of the system of simultaneous equations formed by coupling together the equations of the curves. Here the curves meet one another at points given by the solution of   y = 2x2 + 2,  y = − 1 x + 2 or 2x2 + 2 = − 1 x + 2; 2 2

MEANING OF DIFFERENTIATION

87

that is,

1 x(2x + 2 ) = 0.

the curve y = 2x2 + 2 at any point is

1 This equation has for its solutions x = 0 and x = − 4 . The slope of

dy = 4x. dx For the point where x = 0, this slope is zero; the curve is horizontal. For the point where dy 1 = −1; x=− , 4 dx hence the curve at that point slopes downwards to the right at such an angle θ with the horizontal that tan θ = 1; that is, at 45◦ to the horizontal. the right and makes with the horizontal an angle φ such that tan φ = 1 ; 2 that is, an angle of 26◦ 34 . It follows that at the ﬁrst point the curve cuts the straight line at an angle of 26◦ 34 , while at the second it cuts it at an angle of 45◦ − 26◦ 34 = 18◦ 26 . (5) A straight line is to be drawn, through a point whose coordinates
1 The slope of the straight line is − 2 ; that is, it slopes downwards to

are x = 2, y = −1, as tangent to the curve y = x2 − 5x + 6. Find the coordinates of the point of contact. that is, 2x − 5. The slope of the tangent must be the same as the dy of the curve; dx

The equation of the straight line is y = ax+b, and as it is satisﬁed for dy the values x = 2, y = −1, then −1 = a×2+b; also, its = a = 2x−5. dx The x and the y of the point of contact must also satisfy both the equation of the tangent and the equation of the curve.

CALCULUS MADE EASY

88

We have then         y = x2 − 5x + 6, y = ax + b, (i) (ii) (iii) (iv)

  −1 = 2a + b,      a = 2x − 5, four equations in a, b, x, y. Equations (i) and (ii) give x2 − 5x + 6 = ax + b. Replacing a and b by their value in this, we get x2 − 5x + 6 = (2x − 5)x − 1 − 2(2x − 5),

which simpliﬁes to x2 − 4x + 3 = 0, the solutions of which are: x = 3 two points of contact are then x = 1, y = 2, and x = 3, y = 0.

and x = 1. Replacing in (i), we get y = 0 and y = 2 respectively; the

Note.—In all exercises dealing with curves, students will ﬁnd it extremely instructive to verify the deductions obtained by actually plotting the curves. Exercises VIII. (See page 256 for Answers.)

at points corresponding to diﬀerent values of x, the angle of its slope.

3 (1) Plot the curve y = 4 x2 −5, using a scale of millimetres. Measure

Find, by diﬀerentiating the equation, the expression for slope; and see, from a Table of Natural Tangents, whether this agrees with the measured angle.

MEANING OF DIFFERENTIATION

89

(2) Find what will be the slope of the curve y = 0.12x3 − 2, at the particular point that has as abscissa x = 2. (3) If y = (x − a)(x − b), show that at the particular point of the dy = 0, x will have the value 1 (a + b). curve where 2 dx dy of the equation y = x3 + 3x; and calculate the (4) Find the dx dy numerical values of for the points corresponding to x = 0, x = 1 , 2 dx x = 1, x = 2. (5) In the curve to which the equation is x2 + y 2 = 4, ﬁnd the values of x at those points where the slope = 1. (6) Find the slope, at any point, of the curve whose equation is y2 x2 + 2 = 1; and give the numerical value of the slope at the place 32 2 where x = 0, and at that where x = 1. (7) The equation of a tangent to the curve y = 5 − 2x + 0.5x3 , being

of the form y = mx + n, where m and n are constants, ﬁnd the value for abscissa. (8) At what angle do the two curves y = 3.5x2 + 2 and y = x2 − 5x + 9.5 cut one another?

of m and n if the point where the tangent touches the curve has x = 2

which x = 3 and x = 4. Find the coordinates of the point of intersection of the tangents and their mutual inclination.

√ (9) Tangents to the curve y = ± 25 − x2 are drawn at points for

CALCULUS MADE EASY

90

point. What are the coordinates of the point of contact, and what is the value of b?

(10) A straight line y = 2x − b touches a curve y = 3x2 + 2 at one

CHAPTER XI.
MAXIMA AND MINIMA.
One of the principal uses of the process of diﬀerentiating is to ﬁnd out under what conditions the value of the thing diﬀerentiated becomes a maximum, or a minimum. This is often exceedingly important in engineering questions, where it is most desirable to know what conditions will make the cost of working a minimum, or will make the eﬃciency a maximum. Now, to begin with a concrete case, let us take the equation y = x2 − 4x + 7. By assigning a number of successive values to x, and ﬁnding the corresponding values of y, we can readily see that the equation represents a curve with a minimum. x y 0 7 1 4 2 3 3 4 4 7 5 12

These values are plotted in Fig. 26, which shows that y has apparently a minimum value of 3, when x is made equal to 2. But are you sure that the minimum occurs at 2, and not at 2 1 or at 1 3 ? 4 4

CALCULUS MADE EASY

92

Y 7 6 5 4 3 2 1 O 1 2 3 4 X

Fig. 26.

Of course it would be possible with any algebraic expression to work out a lot of values, and in this way arrive gradually at the particular value that may be a maximum or a minimum.
Y 4 3 2 1 −1 O −1 −2 −3 −4 1 2 3 4 X

Fig. 27.

Here is another example: Let y = 3x − x2 .

MAXIMA AND MINIMA

93

Calculate a few values thus: x y −1 −4 0 0 1 2 2 2 3 0 4 −4 5 −10

Plot these values as in Fig. 27. It will be evident that there will be a maximum somewhere between x = 1 and x = 2; and the thing looks as if the maximum value of y ought
1 to be about 2 1 . Try some intermediate values. If x = 1 4 , y = 2.187; if 4

x = 1 1 , y = 2.25; if x = 1.6, y = 2.24. How can we be sure that 2.25 is 2
1 the real maximum, or that it occurs exactly when x = 1 2 ?

Now it may sound like juggling to be assured that there is a way by which one can arrive straight at a maximum (or minimum) value without making a lot of preliminary trials or guesses. And that way depends on diﬀerentiating. Look back to an earlier page (78) for the remarks about Figs. 14 and 15, and you will see that whenever a curve gets either to its maximum or to its minimum height, at that point its dy = 0. Now this gives us the clue to the dodge that is wanted. When dx there is put before you an equation, and you want to ﬁnd that value of x that will make its y a minimum (or a maximum), ﬁrst diﬀerentiate dy it, and having done so, write its as equal to zero, and then solve dx for x. Put this particular value of x into the original equation, and you will then get the required value of y. This process is commonly called “equating to zero.” To see how simply it works, take the example with which this chapter opens, namely y = x2 − 4x + 7.

CALCULUS MADE EASY

94

Diﬀerentiating, we get:

dy = 2x − 4. dx Now equate this to zero, thus: 2x − 4 = 0. Solving this equation for x, we get: 2x = 4, x = 2. Now, we know that the maximum (or minimum) will occur exactly when x = 2. Putting the value x = 2 into the original equation, we get y = 22 − (4 × 2) + 7 =4−8+7 = 3. Now look back at Fig. 26, and you will see that the minimum occurs when x = 2, and that this minimum of y = 3. Try the second example (Fig. 24), which is y = 3x − x2 .

Diﬀerentiating, Equating to zero,

dy = 3 − 2x. dx

3 − 2x = 0, whence
1 x = 12;

MAXIMA AND MINIMA

95

and putting this value of x into the original equation, we ﬁnd:
1 y = 4 2 − (1 1 × 1 1 ), 2 2

y = 21. 4 This gives us exactly the information as to which the method of trying a lot of values left us uncertain. Now, before we go on to any further cases, we have two remarks to dy to zero, you feel at ﬁrst (that make. When you are told to equate dx is if you have any wits of your own) a kind of resentment, because you dy has all sorts of diﬀerent values at diﬀerent parts of the know that dx curve, according to whether it is sloping up or down. So, when you are suddenly told to write dy = 0, dx you resent it, and feel inclined to say that it can’t be true. Now you will have to understand the essential diﬀerence between “an equation,” and “an equation of condition.” Ordinarily you are dealing with equations that are true in themselves, but, on occasions, of which the present are examples, you have to write down equations that are not necessarily true, but are only true if certain conditions are to be fulﬁlled; and you write them down in order, by solving them, to ﬁnd the conditions which make them true. Now we want to ﬁnd the particular value that x has when the curve is neither sloping up nor sloping down, that is, at the dy dy particular place where = 0. So, writing = 0 does not mean that dx dx it always is = 0; but you write it down as a condition in order to see dy how much x will come out if is to be zero. dx

CALCULUS MADE EASY

96

The second remark is one which (if you have any wits of your own) you will probably have already made: namely, that this much-belauded process of equating to zero entirely fails to tell you whether the x that you thereby ﬁnd is going to give you a maximum value of y or a minimum value of y. Quite so. It does not of itself discriminate; it ﬁnds for you the right value of x but leaves you to ﬁnd out for yourselves whether the corresponding y is a maximum or a minimum. Of course, if you have plotted the curve, you know already which it will be. For instance, take the equation: 1 y = 4x + . x Without stopping to think what curve it corresponds to, diﬀerentiate it, and equate to zero: 1 dy = 4 − x−2 = 4 − 2 = 0; dx x 1 x = 2;

whence

and, inserting this value, y=4 will be either a maximum or else a minimum. But which? You will hereafter be told a way, depending upon a second diﬀerentiation, (see Chap. XII., p. 109). But at present it is enough if you will simply try any other value of x diﬀering a little from the one found, and see whether with this altered value the corresponding value of y is less or greater than that already found.

MAXIMA AND MINIMA

97

Try another simple problem in maxima and minima. Suppose you were asked to divide any number into two parts, such that the product was a maximum? How would you set about it if you did not know the trick of equating to zero? I suppose you could worry it out by the rule of try, try, try again. Let 60 be the number. You can try cutting it into two parts, and multiplying them together. Thus, 50 times 10 is 500; 52 times 8 is 416; 40 times 20 is 800; 45 times 15 is 675; 30 times 30 is 900. This looks like a maximum: try varying it. 31 times 29 is 899, which is not so good; and 32 times 28 is 896, which is worse. So it seems that the biggest product will be got by dividing into two equal halves. Now see what the calculus tells you. Let the number to be cut into and the product will be x(n − x) or nx − x2 . So we write y = nx − x2 . Now diﬀerentiate and equate to zero; dy = n − 2x = 0 dx n = x. 2 two parts be called n. Then if x is one part, the other will be n − x,

Solving for x, we get

So now we know that whatever number n may be, we must divide it into two equal parts if the product of the parts is to be a maximum;
1 and the value of that maximum product will always be = 4 n2 .

This is a very useful rule, and applies to any number of factors, so that if m + n + p = a constant number, m × n × p is a maximum when m = n = p.

CALCULUS MADE EASY

98

Test Case. Let us at once apply our knowledge to a case that we can test. Let y = x2 − x;

and let us ﬁnd whether this function has a maximum or minimum; and if so, test whether it is a maximum or a minimum. Diﬀerentiating, we get dy = 2x − 1. dx Equating to zero, we get 2x − 1 = 0, whence or 2x = 1,
1 x = 2. 1 That is to say, when x is made = 2 , the corresponding value of y

will be either a maximum or a minimum. Accordingly, putting x = in the original equation, we get
1 y = ( 2 )2 − 1 , 2

1 2

or

y = −1. 4 Is this a maximum or a minimum? To test it, try putting x a little

1 bigger than 2 ,—say make x = 0.6. Then

y = (0.6)2 − 0.6 = 0.36 − 0.6 = −0.24, which is higher up than −0.25; showing that y = −0.25 is a minimum. Plot the curve for yourself, and verify the calculation.

MAXIMA AND MINIMA

99

Further Examples. A most interesting example is aﬀorded by a curve that has both a maximum and a minimum. Its equation is:
1 y = 3 x3 − 2x2 + 3x + 1.

Now

dy = x2 − 4x + 3. dx
Y 6 5 4 3 2 −1 1 1 O −1 −2 −3 −4 2 3

4

5

X

Fig. 28.

Equating to zero, we get the quadratic, x2 − 4x + 3 = 0; and solving the quadratic gives us two roots, viz.  x = 3  x = 1.

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Now, when x = 3, y = 1; and when x = 1, y = 2 1 . The ﬁrst of 3 these is a minimum, the second a maximum. The curve itself may be plotted (as in Fig. 28) from the values calculated, as below, from the original equation. x y −1 −4 1 3 0 1 1 21 3 2
2 13

3 1

4 21 3

5 72 3

6 19

A further exercise in maxima and minima is aﬀorded by the following example: The equation to a circle of radius r, having its centre C at the point whose coordinates are x = a, y = b, as depicted in Fig. 29, is: (y − b)2 + (x − a)2 = r2 .

Y

r
C y b x O a X

Fig. 29.

This may be transformed into y= r2 − (x − a)2 + b.

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Now we know beforehand, by mere inspection of the ﬁgure, that when x = a, y will be either at its maximum value, b + r, or else at its minimum value, b − r. But let us not take advantage of this knowledge; minimum, by the process of diﬀerentiating and equating to zero. 1 dy = dx 2 which reduces to dy = dx a−x . 1 r2 − (x − a)2 × (2a − 2x), let us set about ﬁnding what value of x will make y a maximum or a

r2 − (x − a)2 a−x

Then the condition for y being maximum or minimum is: r2 − (x − a)2 the only condition to give zero is x = a. Inserting this value in the original equation for the circle, we ﬁnd y= √ r2 + b; = 0.

Since no value whatever of x will make the denominator inﬁnite,

and as the root of r2 is either +r or −r, we have two resulting values of y,  y = b + r  y = b − r.

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The ﬁrst of these is the maximum, at the top; the second the minimum, at the bottom. If the curve is such that there is no place that is a maximum or minimum, the process of equating to zero will yield an impossible result. For instance: Let Then y = ax3 + bx + c.

dy = 3ax2 + b. dx Equating this to zero, we get 3ax2 + b = 0, −b −b , and x = , which is impossible. 3a 3a Therefore y has no maximum nor minimum. x2 = A few more worked examples will enable you to thoroughly master this most interesting and useful application of the calculus. (1) What are the sides of the rectangle of maximum area inscribed in a circle of radius R? If one side be called x, the other side = (diagonal)2 − x2 ;

and as the diagonal of the rectangle is necessarily a diameter, the other √ side = 4R2 − x2 . √ Then, area of rectangle S = x 4R2 − x2 , √ √ d 4R2 − x2 dS d(x) =x× + 4R2 − x2 × . dx dx dx √ If you have forgotten how to diﬀerentiate 4R2 − x2 , here is a hint: √ dy dw and ; ﬁght it out, write 4R2 − x2 = w and y = w, and seek dw dx and only if you can’t get on refer to page 66.

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You will get √ x 4R2 − 2x2 dS = x × −√ + 4R2 − x2 = √ . dx 4R2 − x2 4R2 − x2

For maximum or minimum we must have

4R2 − 2x2 √ = 0; 4R2 − x2 √ that is, 4R2 − 2x2 = 0 and x = R 2. √ √ The other side = 4R2 − 2R2 = R 2; the two sides are equal; the ﬁgure is a square the side of which is equal to the diagonal of the square constructed on the radius. In this case it is, of course, a maximum with which we are dealing. (2) What is the radius of the opening of a conical vessel the sloping side of which has a length l when the capacity of the vessel is greatest? √ If R be the radius and H the corresponding height, H = l2 − R2 . √ H l 2 − R2 2 2 Volume V = πR × = πR × . 3 3 Proceeding as in the previous problem, we get dV R 2πR √ 2 = πR2 × − √ + l − R2 dR 3 3 l 2 − R2 2πR(l2 − R2 ) − πR3 √ = =0 3 l 2 − R2
2 , 3

for maximum or minimum. obviously.

Or, 2πR(l2 − R2 ) − πR2 = 0, and R = l

for a maximum,

(3) Find the maxima and minima of the function y= 4−x x + . 4−x x

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We get (4 − x) − (−x) −x − (4 − x) dy = + =0 dx (4 − x)2 x2 for maximum or minimum; or 4 4 − 2 = 0 and x = 2. 2 (4 − x) x There is only one value, hence only one maximum or minimum. For x = 2, for x = 1.5, for x = 2.5, y = 2, y = 2.27, y = 2.27;

it is therefore a minimum. (It is instructive to plot the graph of the function.) √ (4) Find the maxima and minima of the function y = 1 − x. (It will be found instructive to plot the graph.) dy 1 1 = √ − √ =0 dx 2 1+x 2 1−x for maximum or minimum. √ √ Hence 1 + x = 1 − x and x = 0, the only solution For x = 0, y = 2. For x = ±0.5, y = 1.932, so this is a maximum. x2 − 5 . 2x − 4 Diﬀerentiating gives at once (see example No. 1, p. 67) √ 1+x +

(5) Find the maxima and minima of the function y=

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We have

for maximum or minimum; or

(2x − 4) × 2x − (x2 − 5)2 dy = =0 dx (2x − 4)2 2x2 − 8x + 10 = 0; (2x − 4)2

or x2 − 4x + 5 = 0; which has for solutions x=
5 2

±

√ −1.

These being imaginary, there is no real value of x for which hence there is neither maximum nor minimum. (6) Find the maxima and minima of the function (y − x2 )2 = x5 . This may be written y = x2 ± x 2 . dy 5 3 = 2x ± 2 x 2 = 0 for maximum or minimum; dx
5

dy = 0; dx

5 that is, x(2± 2 x 2 ) = 0, which is satisﬁed for x = 0, and for 2± 5 x 2 = 0, 2

1

1

that is for x =

16 . 25

So there are two solutions.
2 2

x = +0.5, y = 0.25 ±

Taking ﬁrst x = 0. If x = −0.5, y = 0.25 ±

(.5)5 . On one side y is imaginary; that is,

−(.5)5 , and if

there is no value of y that can be represented by a graph; the latter is therefore entirely on the right side of the axis of y (see Fig. 30). On plotting the graph it will be found that the curve goes to the origin, as if there were a minimum there; but instead of continuing beyond, as it should do for a minimum, it retraces its steps (forming

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what is called a “cusp”). There is no minimum, therefore, although the dy = 0. It is necessary condition for a minimum is satisﬁed, namely dx therefore always to check by taking one value on either side. y 0.3

0.2

0.1

0

0.2

0.4

0.6

0.8 1.0

x

Fig. 30.

Now, if we take x =

16 25

= 0.64. If x = 0.64, y = 0.7373 and

y = 0.0819; if x = 0.6, y becomes 0.6389 and 0.0811; and if x = 0.7, y becomes 0.8996 and 0.0804. This shows that there are two branches of the curve; the upper one does not pass through a maximum, but the lower one does. (7) A cylinder whose height is twice the radius of the base is increasing in volume, so that all its parts keep always in the same proportion to each other; that is, at any instant, the cylinder is similar to the original cylinder. When the radius of the base is r feet, the surface area is increasing at the rate of 20 square inches per second; at what

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rate is its volume then increasing? Area = S = 2(πr2 ) + 2πr × 2r = 6πr2 . Volume = V = πr2 × 2r = 2πr3 . dS = 12πr, dr dV = 6πr2 , dr 20 , 12πr

dS = 12πr dr = 20,

dr =

dV = 6πr2 dr = 6πr2 ×

20 = 10r. 12πr

The volume changes at the rate of 10r cubic inches. Make other examples for yourself. There are few subjects which oﬀer such a wealth for interesting examples. Exercises IX. (See page 257 for Answers.)

(1) What values of x will make y a maximum and a minimum, if x2 ? y= x+1 (2) What value of x will make y a maximum in the equation y = x ? 2 + x2 a (3) A line of length p is to be cut up into 4 parts and put together as a rectangle. Show that the area of the rectangle will be a maximum if each of its sides is equal to 1 p. 4 (4) A piece of string 30 inches long has its two ends joined together and is stretched by 3 pegs so as to form a triangle. What is the largest triangular area that can be enclosed by the string?

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(5) Plot the curve corresponding to the equation y= also ﬁnd 10 10 + ; x 8−x

dy , and deduce the value of x that will make y a minimum; dx and ﬁnd that minimum value of y. (6) If y = x5 − 5x, ﬁnd what values of x will make y a maximum or

a minimum.

(7) What is the smallest square that can be inscribed in a given square? (8) Inscribe in a given cone, the height of which is equal to the radius of the base, a cylinder (a) whose volume is a maximum; (b) whose lateral area is a maximum; (c) whose total area is a maximum. (9) Inscribe in a sphere, a cylinder (a) whose volume is a maximum; (b) whose lateral area is a maximum; (c) whose total area is a maximum. (10) A spherical balloon is increasing in volume. If, when its radius is r feet, its volume is increasing at the rate of 4 cubic feet per second, at what rate is its surface then increasing? (11) Inscribe in a given sphere a cone whose volume is a maximum. (12) The current C given by a battery of N similar voltaic cells is n×E C = , where E, R, r, are constants and n is the number of rn2 R+ N cells coupled in series. Find the proportion of n to N for which the current is greatest.

CHAPTER XII.
CURVATURE OF CURVES.
Returning to the process of successive diﬀerentiation, it may be asked: Why does anybody want to diﬀerentiate twice over? We know that when the variable quantities are space and time, by diﬀerentiating twice over we get the acceleration of a moving body, and that in the dy means the slope of geometrical interpretation, as applied to curves, dx 2 dy mean in this case? Clearly it means the the curve. But what can dx2 rate (per unit of length x) at which the slope is changing—in brief, it is a measure of the curvature of the slope.
Y
Y

O

X

O

X

Fig. 31.

Fig. 32.

Suppose a slope constant, as in Fig. 31.

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dy is of constant value. dx Suppose, however, a case in which, like Fig. 32, the slope itself is dy d d2 y dx getting greater upwards, then , that is, 2 , will be positive. dx dx If the slope is becoming less as you go to the right (as in Fig. 14, Here, p. 80), or as in Fig. 33, then, even though the curve may be going d2 y will upward, since the change is such as to diminish its slope, its dx2 be negative.
Y

O

X

Fig. 33.

It is now time to initiate you into another secret—how to tell whether the result that you get by “equating to zero” is a maximum or a minimum. The trick is this: After you have diﬀerentiated (so as to get the expression which you equate to zero), you then diﬀerentiate a second time, and look whether the result of the second diﬀerentiation d2 y is positive or negative. If comes out positive, then you know that dx2 d2 y the value of y which you got was a minimum; but if comes out dx2 negative, then the value of y which you got must be a maximum. That’s

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the rule. The reason of it ought to be quite evident. Think of any curve that has a minimum point in it (like Fig. 15, p. 80), or like Fig. 34, where the point of minimum y is marked M , and the curve is concave upwards. To the left of M the slope is downward, that is, negative, and is getting less negative. To the right of M the slope has become upward, and is getting more and more upward. Clearly the change of slope as the
Y Y M

M y min. O X O

y max.

x

x

X

Fig. 34.

Fig. 35.

d2 y curve passes through M is such that is positive, for its operation, dx2 as x increases toward the right, is to convert a downward slope into an upward one. Similarly, consider any curve that has a maximum point in it (like Fig. 16, p. 81), or like Fig. 35, where the curve is convex, and the maximum point is marked M . In this case, as the curve passes through M from left to right, its upward slope is converted into a downward or d2 y negative slope, so that in this case the “slope of the slope” is dx2 negative. Go back now to the examples of the last chapter and verify in this

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way the conclusions arrived at as to whether in any particular case there is a maximum or a minimum. You will ﬁnd below a few worked out examples. (1) Find the maximum or minimum of (a) y = 4x2 − 9x − 6; (b) y = 6 + 9x − 4x2 ;

and ascertain if it be a maximum or a minimum in each case. (a) dy dx d2 y dx2 dy (b) dx d2 y dx2 = 8x − 9 = 0; = 8;
1 x = 18,

and y = −11.065.

it is +; hence it is a minimum.
1 x = 18;

= 9 − 8x = 0; = −8;

and y = +11.065.

it is −; hence it is a maximum.

(2) Find the maxima and minima of the function y = x3 − 3x + 16. dy = 3x2 − 3 = 0; x2 = 1; and x = ±1. dx d2 y = 6x; for x = 1; it is +; dx2 hence x = 1 corresponds to a minimum y = 14. For x = −1 it is −; hence x = −1 corresponds to a maximum y = +18. x−1 (3) Find the maxima and minima of y = 2 . x +2

dy (x2 + 2) × 1 − (x − 1) × 2x 2x − x2 + 2 = = = 0; dx (x2 + 2)2 (x2 + 2)2

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or x2 − 2x − 2 = 0, whose solutions are x = +2.73 and x = −0.73. (x2 + 2)2 × (2x − 2) − (x2 − 2x − 2)(4x3 + 8x) d2 y =− dx2 (x2 + 2)4 5 4 3 2x − 6x − 8x − 8x2 − 24x + 8 =− . (x2 + 2)4 The denominator is always positive, so it is suﬃcient to ascertain the sign of the numerator. If we put x = 2.73, the numerator is negative; the maximum, y = 0.183. y = −0.683. If we put x = −0.73, the numerator is positive; the minimum, (4) The expense C of handling the products of a certain factory

varies with the weekly output P according to the relation C = aP + b + d, where a, b, c, d are positive constants. For what output will c+P the expense be least? b dC =a− = 0 for maximum or minimum; dP (c + P )2 hence a = b and P = ± (c + P )2 b − c. a b − c. a

As the output cannot be negative, P = + Now d2 C b(2c + 2P ) =+ , 2 dP (c + P )4

which is positive for all the values of P ; hence P = + to a minimum.

b −c corresponds a

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(5) The total cost per hour C of lighting a building with N lamps of a certain kind is C=N Cl EP Ce + t 1000 ,

where E is the commercial eﬃciency (watts per candle), P is the candle power of each lamp, t is the average life of each lamp in hours, Cl = cost of renewal in pence per hour of use, Ce = cost of energy per 1000 watts per hour. Moreover, the relation connecting the average life of a lamp with the commercial eﬃciency at which it is run is approximately t = mE n , where m and n are constants depending on the kind of lamp. Find the commercial eﬃciency for which the total cost of lighting will be least. We have Cl −n P Ce E + E , m 1000 dC P Ce nCl −(n+1) = − E =0 dE 1000 m C=N

for maximum or minimum. E n+1 = 1000 × nCl mP Ce and E = n+1 1000 × nCl . mP Ce

This is clearly for minimum, since d2 C nCl −(n+2) = (n + 1) E , 2 dE m

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which is positive for a positive value of E. For a particular type of 16 candle-power lamps, Cl = 17 pence, Ce = 5 pence; and it was found that m = 10 and n = 3.6. E=
4.6

1000 × 3.6 × 17 = 2.6 watts per candle-power. 10 × 16 × 5 (You are advised to plot the graph of any numerical

Exercises X.

example.) (See p. 258 for the Answers.) (1) Find the maxima and minima of y = x3 + x2 − 10x + 8. dy d2 y b x − cx2 , ﬁnd expressions for , and for , also a dx dx2 ﬁnd the value of x which makes y a maximum or a minimum, and show (2) Given y = whether it is maximum or minimum. (3) Find how many maxima and how many minima there are in the curve, the equation to which is y =1− x2 x4 + ; 2 24

and how many in that of which the equation is y =1− x 2 x4 x6 + − . 2 24 720

(4) Find the maxima and minima of y = 2x + 1 + 5 . x2

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(5) Find the maxima and minima of y= x2 3 . +x+1

(6) Find the maxima and minima of y= 5x . 2 + x2

(7) Find the maxima and minima of y= x2 3x x + + 5. −3 2

(8) Divide a number N into two parts in such a way that three times the square of one part plus twice the square of the other part shall be a minimum. (9) The eﬃciency u of an electric generator at diﬀerent values of output x is expressed by the general equation: u= x ; a + bx + cx2

where a is a constant depending chieﬂy on the energy losses in the iron and c a constant depending chieﬂy on the resistance of the copper parts. Find an expression for that value of the output at which the eﬃciency will be a maximum. (10) Suppose it to be known that consumption of coal by a certain steamer may be represented by the formula y = 0.3+0.001v 3 ; where y is the number of tons of coal burned per hour and v is the speed expressed in nautical miles per hour. The cost of wages, interest on capital, and depreciation of that ship are together equal, per hour, to the cost of

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1 ton of coal. What speed will make the total cost of a voyage of 1000 nautical miles a minimum? And, if coal costs 10 shillings per ton, what will that minimum cost of the voyage amount to? (11) Find the maxima and minima of y=± x 6 x(10 − x).

(12) Find the maxima and minima of y = 4x3 − x2 − 2x + 1.

CHAPTER XIII.
OTHER USEFUL DODGES.
Partial Fractions. We have seen that when we diﬀerentiate a fraction we have to perform a rather complicated operation; and, if the fraction is not itself a simple one, the result is bound to be a complicated expression. If we could split the fraction into two or more simpler fractions such that their sum is equivalent to the original fraction, we could then proceed by diﬀerentiating each of these simpler expressions. And the result of diﬀerentiating would be the sum of two (or more) diﬀerentials, each one of which is relatively simple; while the ﬁnal expression, though of course it will be the same as that which could be obtained without resorting to this dodge, is thus obtained with much less eﬀort and appears in a simpliﬁed form. Let us see how to reach this result. Try ﬁrst the job of adding two fractions together to form a resultant fraction. Take, for example, 2 1 and . Every schoolboy can add these the two fractions x+1 x−1 3x + 1 together and ﬁnd their sum to be 2 . And in the same way he can x −1 add together three or more fractions. Now this process can certainly be reversed: that is to say, that if this last expression were given, it

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is certain that it can somehow be split back again into its original components or partial fractions. Only we do not know in every case that may be presented to us how we can so split it. In order to ﬁnd this out we shall consider a simple case at ﬁrst. But it is important to bear in mind that all which follows applies only to what are called “proper” algebraic fractions, meaning fractions like the above, which have the numerator of a lesser degree than the denominator; that is, those in which the highest index of x is less in the numerator than in x2 + 2 , the denominator. If we have to deal with such an expression as 2 x −1 3 we can simplify it by division, since it is equivalent to 1 + 2 ; and x −1 3 is a proper algebraic fraction to which the operation of splitting 2−1 x into partial fractions can be applied, as explained hereafter. Case I. If we perform many additions of two or more fractions the denominators of which contain only terms in x, and no terms in x2 , x3 , or any other powers of x, we always ﬁnd that the denominator of the ﬁnal resulting fraction is the product of the denominators of the fractions which were added to form the result. It follows that by factorizing the denominator of this ﬁnal fraction, we can ﬁnd every one of the denominators of the partial fractions of which we are in search. 3x + 1 Suppose we wish to go back from 2 to the components which x −1 1 2 we know are and . If we did not know what those compox+1 x−1 nents were we can still prepare the way by writing: 3x + 1 3x + 1 = = + , x2 − 1 (x + 1)(x − 1) x+1 x−1 leaving blank the places for the numerators until we know what to put

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there. We always may assume the sign between the partial fractions to be plus, since, if it be minus, we shall simply ﬁnd the corresponding numerator to be negative. Now, since the partial fractions are proper fractions, the numerators are mere numbers without x at all, and we can call them A, B, C . . . as we please. So, in this case, we have: A B 3x + 1 = + . 2−1 x x+1 x−1 If now we perform the addition of these two partial fractions, we A(x − 1) + B(x + 1) 3x + 1 get ; and this must be equal to . (x + 1)(x − 1) (x + 1)(x − 1) And, as the denominators in these two expressions are the same, the numerators must be equal, giving us: 3x + 1 = A(x − 1) + B(x + 1). Now, this is an equation with two unknown quantities, and it would seem that we need another equation before we can solve them and ﬁnd A and B. But there is another way out of this diﬃculty. The equation must be true for all values of x; therefore it must be true for such values of x as will cause x − 1 and x + 1 to become zero, that is for x = 1 and for x = −1 respectively. If we make x = 1, we get 4 = (A×0)+(B ×2),

so that A = 1. Replacing the A and B of the partial fractions by these 1 2 new values, we ﬁnd them to become and ; and the thing is x+1 x−1 done. 4x2 + 2x − 14 . The As a farther example, let us take the fraction 3 x + 3x2 − x − 3 denominator becomes zero when x is given the value 1; hence x − 1 is a factor of it, and obviously then the other factor will be x2 + 4x + 3;

so that B = 2; and if we make x = −1, we get −2 = (A×−2)+(B ×0),

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121

and this can again be decomposed into (x + 1)(x + 3). So we may write the fraction thus: A B C 4x2 + 2x − 14 = + + , 3 + 3x2 − x − 3 x x+1 x−1 x+3 making three partial factors. Proceeding as before, we ﬁnd 4x2 + 2x − 14 = A(x − 1)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x − 1). Now, if we make x = 1, we get: −8 = (A × 0) + B(2 × 4) + (C × 0); If x = −1, we get: −12 = A(−2 × 2) + (B × 0) + (C × 0); If x = −3, we get: 16 = (A × 0) + (B × 0) + C(−2 × −4); So then the partial fractions are: 1 2 3 − + , x+1 x−1 x+3 which is far easier to diﬀerentiate with respect to x than the complicated expression from which it is derived. whence C = 2. whence A = 3. that is, B = −1.

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Case II. If some of the factors of the denominator contain terms in x2 , and are not conveniently put into factors, then the corresponding numerator may contain a term in x, as well as a simple number; and hence it becomes necessary to represent this unknown numerator not by the symbol A but by Ax + B; the rest of the calculation being made as before. −x2 − 3 . Try, for instance: (x2 + 1)(x + 1) Ax + B C −x2 − 3 = 2 + ; 2 + 1)(x + 1) (x x +1 x+1 −x2 − 3 = (Ax + B)(x + 1) + C(x2 + 1). Putting x = −1, we get −4 = C × 2; and C = −2; hence and −x2 − 3 = (Ax + B)(x + 1) − 2x2 − 2; x2 − 1 = Ax(x + 1) + B(x + 1).

hence

Putting x = 0, we get −1 = B; x2 − 1 = Ax(x + 1) − x − 1; or x2 + x = Ax(x + 1);

and

x + 1 = A(x + 1),

so that A = 1, and the partial fractions are: x−1 2 − . 2+1 x x+1 Take as another example the fraction x3 − 2 . (x2 + 1)(x2 + 2)

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We get Ax + B Cx + D x3 − 2 = 2 + 2 2 + 1)(x2 + 2) (x x +1 x +2 2 (Ax + B)(x + 2) + (Cx + D)(x2 + 1) = . (x2 + 1)(x2 + 2) In this case the determination of A, B, C, D is not so easy. It will be simpler to proceed as follows: Since the given fraction and the fraction found by adding the partial fractions are equal, and have identical denominators, the numerators must also be identically the same. In such a case, and for such algebraical expressions as those with which we are dealing here, the coeﬃcients of the same powers of x are equal and of same sign. Hence, since x3 − 2 = (Ax + B)(x2 + 2) + (Cx + D)(x2 + 1) = (A + C)x3 + (B + D)x2 + (2A + C)x + 2B + D, we have 1 = A + C; 0 = B + D (the coeﬃcient of x2 in the left expression being zero); 0 = 2A + C; and −2 = 2B + D. Here are four equations, from which we readily obtain A = −1; B = −2; C = 2; 2(x + 1) x + 2 − 2 . This method D = 0; so that the partial fractions are 2 x +2 x +1 can always be used; but the method shown ﬁrst will be found the quickest in the case of factors in x only. Case III. When, among the factors of the denominator there are some which are raised to some power, one must allow for the possible existence of partial fractions having for denominator the several powers of that factor up to the highest. For instance, in splitting the

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3x2 − 2x + 1 we must allow for the possible existence of a (x + 1)2 (x − 2) denominator x + 1 as well as (x + 1)2 and (x − 2). fraction It maybe thought, however, that, since the numerator of the fraction the denominator of which is (x + 1)2 may contain terms in x, we must allow for this in writing Ax + B for its numerator, so that Ax + B C D 3x2 − 2x + 1 = + + . 2 (x − 2) 2 (x + 1) (x + 1) x+1 x−2 If, however, we try to ﬁnd A, B, C and D in this case, we fail, because we get four unknowns; and we have only three relations connecting them, yet 3x2 − 2x + 1 x−1 1 1 = + + . 2 (x − 2) 2 (x + 1) (x + 1) x+1 x−2 But if we write 3x2 − 2x + 1 A B C + = + , 2 (x − 2) 2 (x + 1) (x + 1) x+1 x−2 we get 3x2 − 2x + 1 = A(x − 2) + B(x + 1)(x − 2) + C(x + 1)2 , which gives C = 1 for x = 2. Replacing C by its value, transposing, gathering like terms and dividing by x − 2, we get −2x = A + B(x + 1), which gives A = −2 for x = −1. Replacing A by its value, we get 2x = −2 + B(x + 1). Hence B = 2; so that the partial fractions are: 2 2 1 − + , 2 x + 1 (x + 1) x−2

OTHER USEFUL DODGES

125

x−1 1 1 + stated above as being the fractions + 2 x + 1 (x + 1) x−2 3x2 − 2x + 1 from which was obtained. The mystery is cleared if we (x + 1)2 (x − 2) x−1 1 observe that can itself be split into the two fractions − 2 (x + 1) x+1 2 , so that the three fractions given are really equivalent to (x + 1)2 instead of 1 1 2 1 2 2 1 + − + = − + , 2 2 x + 1 x + 1 (x + 1) x−2 x + 1 (x + 1) x−2 which are the partial fractions obtained. We see that it is suﬃcient to allow for one numerical term in each numerator, and that we always get the ultimate partial fractions. When there is a power of a factor of x2 in the denominator, however, the corresponding numerators must be of the form Ax+B; for example, (2x2 which gives 3x − 1 = (Ax + B)(x + 1) + (Cx + D)(x + 1)(2x2 − 1) + E(2x2 − 1)2 . like terms, and dividing by x + 1, we get For x = −1, this gives E = −4. Replacing, transposing, collecting 16x3 − 16x2 + 3 = 2Cx3 + 2Dx2 + x(A − C) + (B − D). A − 8 = 0 and A = 8, and ﬁnally, B − D = 3 or B = −5. So that we (8x − 5) 8(x − 1) 4 + − . 2 − 1)2 2−1 (2x 2x x+1 Hence 2C = 16 and C = 8; 2D = −16 and D = −8; A − C = 0 or Ax + B Cx + D E 3x − 1 = + 2 + , 2 (x + 1) 2 − 1)2 − 1) (2x 2x − 1 x+1

obtain as the partial fractions:

CALCULUS MADE EASY

126

It is useful to check the results obtained. The simplest way is to replace x by a single value, say +1, both in the given expression and in the partial fractions obtained. Whenever the denominator contains but a power of a single factor, a very quick method is as follows: 4x + 1 , let x + 1 = z; then x = z − 1. Taking, for example, (x + 1)3 Replacing, we get 4(z − 1) + 1 4z − 3 4 3 = = 2 − 3. z3 z3 z z The partial fractions are, therefore, 4 3 − . 2 (x + 1) (x + 1)3

Application to diﬀerentiation. Let it be required to diﬀerentiate 5 − 4x y= 2 ; we have 6x + 7x − 3 dy (6x2 + 7x − 3) × 4 + (5 − 4x)(12x + 7) =− dx (6x2 + 7x − 3)2 2 24x − 60x − 23 = . (6x2 + 7x − 3)2 If we split the given expression into 1 2 − , 3x − 1 2x + 3 we get, however, 3 4 dy =− + , 2 dx (3x − 1) (2x + 3)2

OTHER USEFUL DODGES

127

which is really the same result as above split into partial fractions. But the splitting, if done after diﬀerentiating, is more complicated, as will easily be seen. When we shall deal with the integration of such expressions, we shall ﬁnd the splitting into partial fractions a precious auxiliary (see p. 228). Exercises XI. (See page 259 for Answers.)

Split into fractions: (1) (3) (5) 3x + 5 . (x − 3)(x + 4) x2 3x + 5 . + x − 12 (2) (4) (6) 3x − 4 . (x − 1)(x − 2) x2 x+1 . − 7x + 12

x−8 . (2x + 3)(3x − 2) x2 − 3x + 1 . (x − 1)(x + 2)(x − 3)

x2 − 13x + 26 . (x − 2)(x − 3)(x − 4)

(7) (8) (9) (11) (13)

5x2 + 7x + 1 . (2x + 1)(3x − 2)(3x + 1) x2 . x3 − 1 (10) (12)

x4 + 1 . x3 + 1 x . (x − 1)(x − 2)2 x+3 . (x + 2)2 (x − 1)

5x2 + 6x + 4 . (x + 1)(x2 + x + 1) x . (x2 − 1)(x + 1)

(14)

CALCULUS MADE EASY

128

(15)

3x2 + 2x + 1 . (x + 2)(x2 + x + 1)2 7x2 + 9x − 1 . (3x − 2)4

(16)

5x2 + 8x − 12 . (x + 4)3 x2 . (x3 − 8)(x − 2)

(17)

(18)

Diﬀerential of an Inverse Function. Consider the function (see p. 13) y = 3x; it can be expressed in the y form x = ; this latter form is called the inverse function to the one 3 originally given. dy y dx 1 If y = 3x, = 3; if x = , = , and we see that dx 3 dy 3 1 dy = dx dx dy Consider y = 4x2 , y2 x= , 2
1

or

dy dx × = 1. dx dy

dy = 8x; the inverse function is dx and dx 1 1 1 = √ = = . dy 4 y 4 × 2x 8x dy dx × = 1. dx dy

Here again

It can be shown that for all functions which can be put into the inverse form, one can always write dy dx × = 1 or dx dy dy 1 = . dx dx dy

OTHER USEFUL DODGES

129

It follows that, being given a function, if it be easier to diﬀerentiate the inverse function, this may be done, and the reciprocal of the diﬀerential coeﬃcient of the inverse function gives the diﬀerential coeﬃcient of the given function itself. As an example, suppose that we wish to diﬀerentiate y =
2

3 We have seen one way of doing this, by writing u = − 1, and ﬁnding x du dy and . This gives du dx 3 dy =− . dx 3 2x2 −1 x If we had forgotten how to proceed by this method, or wished to check our result by some other way of obtaining the diﬀerential coefﬁcient, or for any other reason we could not use the ordinary method, 3 . we can proceed as follows: The inverse function is x = 1 + y2 dx 3 × 2y 6y =− =− ; 2 )2 dy (1 + y (1 + y 2 )2 hence 3 −1 2 2 dy (1 + y ) 1 3 x =− = =− =− . dx dx 6y 3 3 6× 2 −1 2x2 −1 dy x x 1 Let us take as an other example y = √ . 3 θ+5 1 The inverse function is θ = 3 − 5 or θ = y −3 − 5, and y 1+ dθ = −3y −4 = −3 3 (θ + 5)4 . dy
2

3 − 1. x

CALCULUS MADE EASY

130

It follows that erwise.

dy =− dx 3

1 (θ + 5)4

, as might have been found oth-

We shall ﬁnd this dodge most useful later on; meanwhile you are advised to become familiar with it by verifying by its means the results obtained in Exercises I. (p. 24), Nos. 5, 6, 7; Examples (p. 67), Nos. 1, 2, 4; and Exercises VI. (p. 72), Nos. 1, 2, 3 and 4. You will surely realize from this chapter and the preceding, that in many respects the calculus is an art rather than a science: an art only to be acquired, as all other arts are, by practice. Hence you should work many examples, and set yourself other examples, to see if you can work them out, until the various artiﬁces become familiar by use.

CHAPTER XIV.
ON TRUE COMPOUND INTEREST AND THE LAW OF ORGANIC GROWTH.
Let there be a quantity growing in such a way that the increment of its growth, during a given time, shall always be proportional to its own magnitude. This resembles the process of reckoning interest on money at some ﬁxed rate; for the bigger the capital, the bigger the amount of interest on it in a given time. Now we must distinguish clearly between two cases, in our calculation, according as the calculation is made by what the arithmetic books call “simple interest,” or by what they call “compound interest.” For in the former case the capital remains ﬁxed, while in the latter the interest is added to the capital, which therefore increases by successive additions. (1) At simple interest. Consider a concrete case. Let the capital at start be £100, and let the rate of interest be 10 per cent. per annum. Then the increment to the owner of the capital will be £10 every year. Let him go on drawing his interest every year, and hoard it by putting it by in a stocking, or locking it up in his safe. Then, if he goes on for 10 years, by the end of that time he will have received 10 increments

CALCULUS MADE EASY

132

of £10 each, or £100, making, with the original £100, a total of £200 in all. His property will have doubled itself in 10 years. If the rate of interest had been 5 per cent., he would have had to hoard for 20 years to double his property. If it had been only 2 per cent., he would have had to hoard for 50 years. It is easy to see that if the value of the yearly 1 interest is of the capital, he must go on hoarding for n years in order n to double his property. y Or, if y be the original capital, and the yearly interest is , then, n at the end of n years, his property will be y+n y = 2y. n

(2) At compound interest. As before, let the owner begin with a capital of £100, earning interest at the rate of 10 per cent. per annum; but, instead of hoarding the interest, let it be added to the capital each year, so that the capital grows year by year. Then, at the end of one year, the capital will have grown to £110; and in the second year (still at 10%) this will earn £11 interest. He will start the third year with £121, and the interest on that will be £12. 2s.; so that he starts the fourth year with £133. 2s., and so on. It is easy to work it out, and ﬁnd that at the end of the ten years the total capital will have grown to £259. 7s. 6d. In fact, we see that at the end of each year, each pound will have earned
1 10

of a pound, and therefore, if
11 ; 10

this is always added on, each year multiplies the capital by

and if

continued for ten years (which will multiply by this factor ten times over) will multiply the original capital by 2.59374. Let us put this into 1 symbols. Put y0 for the original capital; for the fraction added on at n

ON TRUE COMPOUND INTEREST

133

each of the n operations; and yn for the value of the capital at the end of the nth operation. Then yn = y0 1 + 1 n n .

But this mode of reckoning compound interest once a year, is really not quite fair; for even during the ﬁrst year the £100 ought to have been growing. At the end of half a year it ought to have been at least £105, and it certainly would have been fairer had the interest for the second half of the year been calculated on £105. This would be equivalent to calling it 5% per half-year; with 20 operations, therefore, at each of which the capital is multiplied by
21 . 20

If reckoned this way, by the end

of ten years the capital would have grown to £265. 6s. 7d.; for (1 +
1 20 ) 20

= 2.653.

But, even so, the process is still not quite fair; for, by the end of the ﬁrst month, there will be some interest earned; and a half-yearly reckoning assumes that the capital remains stationary for six months at a time. Suppose we divided the year into 10 parts, and reckon a one-percent. interest for each tenth of the year. We now have 100 operations lasting over the ten years; or yn = £100 1 + which works out to £270. 9s. 7 1 d. 2 Even this is not ﬁnal. Let the ten years be divided into 1000 periods, each of then yn = £100 1 +
1000 1 1000 1 100 1 100 100

;

of a year; the interest being

1 10

per cent. for each such period; ;

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134

which works out to £271. 13s. 10d. Go even more minutely, and divide the ten years into 10, 000 parts, each
1 1000

of a year, with interest at

1 100

of 1 per cent. Then
10,000

yn = £100 1 +
1 which amounts to £271. 16s. 3 2 d.

1 10,000

;

Finally, it will be seen that what we are trying to ﬁnd is in reality n 1 the ultimate value of the expression 1 + , which, as we see, is n greater than 2; and which, as we take n larger and larger, grows closer and closer to a particular limiting value. However big you make n, the value of this expression grows nearer and nearer to the ﬁgure 2.71828 . . . a number never to be forgotten. Let us take geometrical illustrations of these things. In Fig. 36, OP stands for the original value. OT is the whole time during which the value is growing. It is divided into 10 periods, in each of which dy there is an equal step up. Here is a constant; and if each step up dx 1 is 10 of the original OP , then, by 10 such steps, the height is doubled. If we had taken 20 steps, each of half the height shown, at the end the 1 height would still be just doubled. Or n such steps, each of of the n original height OP , would suﬃce to double the height. This is the case of simple interest. Here is 1 growing till it becomes 2. In Fig. 37, we have the corresponding illustration of the geometrical 1 progression. Each of the successive ordinates is to be 1 + , that is, n n+1 times as high as its predecessor. The steps up are not equal, n

ON TRUE COMPOUND INTEREST

135

U

P

2

1

O

1

2

3

4

5

6

7

8

9

T

Fig. 36.

because each step up is now

1 of the ordinate at that part of the curve. n 1 If we had literally 10 steps, with 1 + 10 for the multiplying factor, the ﬁnal total would be (1 +
1 10 ) 10

or 2.594 times the original 1. But if 1 only we take n suﬃciently large (and the corresponding suﬃciently n n 1 small), then the ﬁnal value 1 + to which unity will grow will n be 2.71828.
U

2.7182 P 1 O 1 2 3 4 5 6 7 8 9 T

Fig. 37.

Epsilon. To this mysterious number 2.7182818 etc., the mathemati-

CALCULUS MADE EASY

136 (pronounced ep-

cians have assigned as a symbol the Greek letter

silon). All schoolboys know that the Greek letter π (called pi ) stands for 3.141592 etc.; but how many of them know that epsilon means 2.71828? Yet it is an even more important number than π! What, then, is epsilon? Suppose we were to let 1 grow at simple interest till it became 2; then, if at the same nominal rate of interest, and for the same time, we were to let 1 grow at true compound interest, instead of simple, it would grow to the value epsilon. This process of growing proportionately, at every instant, to the magnitude at that instant, some people call a logarithmic rate of growing. Unit logarithmic rate of growth is that rate which in unit time will cause 1 to grow to 2.718281. It might also be called the organic rate of growing: because it is characteristic of organic growth (in certain circumstances) that the increment of the organism in a given time is proportional to the magnitude of the organism itself. If we take 100 per cent. as the unit of rate, and any ﬁxed period as the unit of time, then the result of letting 1 grow arithmetically at unit rate, for unit time, will be 2, while the result of letting 1 grow logarithmically at unit rate, for the same time, will be 2.71828 . . . . A little more about Epsilon. We have seen that we require to know n 1 , when n becomes what value is reached by the expression 1 + n indeﬁnitely great. Arithmetically, here are tabulated a lot of values (which anybody can calculate out by the help of an ordinary table of logarithms) got by assuming n = 2; n = 5; n = 10; and so on, up to

THE LAW OF ORGANIC GROWTH

137

n = 10, 000. (1 + 1 )2 2 (1 + 1 )5 5 (1 + (1 + (1 + (1 + (1 +
1 10 ) 10 1 20 ) 20 1 100 ) 100 1 )1000 1000 1 )10,000 10,000

= 2.25. = 2.488. = 2.594. = 2.653. = 2.705. = 2.7169. = 2.7181.

It is, however, worth while to ﬁnd another way of calculating this immensely important ﬁgure. Accordingly, we will avail ourselves of the binomial theorem, and n 1 expand the expression 1 + in that well-known way. n The binomial theorem gives the rule that (a + b)n = an + n an−2 b2 an−1 b + n(n − 1) 1! 2! n−3 3 a b + n(n − 1)(n − 2) + etc. 3!

Putting a = 1 and b = 1 1+ n n 1 , we get n 1 n−1 1 (n − 1)(n − 2) + 2! n 3! n2 1 (n − 1)(n − 2)(n − 3) + + etc. 4! n3

=1+1+

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138

Now, if we suppose n to become indeﬁnitely great, say a billion, or equal to n; and then the series becomes =1+1+ a billion billions, then n − 1, n − 2, and n − 3, etc., will all be sensibly 1 1 1 + + + etc. . . . 2! 3! 4!

By taking this rapidly convergent series to as many terms as we please, we can work out the sum to any desired point of accuracy. Here is the working for ten terms: 1.000000 dividing by 1 dividing by 2 dividing by 3 dividing by 4 dividing by 5 dividing by 6 dividing by 7 dividing by 8 dividing by 9 1.000000 0.500000 0.166667 0.041667 0.008333 0.001389 0.000198 0.000025 0.000002

Total 2.718281 is incommensurable with 1, and resembles π in being an interminable non-recurrent decimal. The Exponential Series. We shall have need of yet another series. Let us, again making use of the binomial theorem, expand the exnx 1 pression 1 + , which is the same as x when we make n indeﬁn

THE LAW OF ORGANIC GROWTH

139

nitely great. 1 x nx−1

=1

nx

+ nx

1 n

1 + nx(nx − 1) 1nx−3 1 n

nx−2

1 n

2

1!

2!
3

+ etc. 3! 1 n2 x2 − nx 1 n3 x3 − 3n2 x2 + 2nx =1+x+ · + · + etc. 2! n2 3! n3 x 3x2 2x x2 − x3 − + 2 n + n n + etc. =1+x+ 2! 3! But, when n is made indeﬁnitely great, this simpliﬁes down to the following: x2 x 3 x4 + + + etc. . . . 2! 3! 4! This series is called the exponential series. x + nx(nx − 1)(nx − 2)

=1+x+

The great reason why is regarded of importance is that

x

possesses

a property, not possessed by any other function of x, that when you diﬀerentiate it its value remains unchanged ; or, in other words, its diﬀerential coeﬃcient is the same as itself. This can be instantly seen by diﬀerentiating it with respect to x, thus: d( x ) 2x 3x2 4x3 =0+1+ + + dx 1·2 1·2·3 1·2·3·4 5x4 + + etc. 1·2·3·4·5 x2 x3 x4 =1+x+ + + + etc., 1·2 1·2·3 1·2·3·4

or

which is exactly the same as the original series.

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140

Now we might have gone to work the other way, and said: Go to; let us ﬁnd a function of x, such that its diﬀerential coeﬃcient is the same as itself. Or, is there any expression, involving only powers of x, which is unchanged by diﬀerentiation? Accordingly; let us assume as a general expression that y = A + Bx + Cx2 + Dx3 + Ex4 + etc., (in which the coeﬃcients A, B, C, etc. will have to be determined), and diﬀerentiate it. dy = B + 2Cx + 3Dx2 + 4Ex3 + etc. dx Now, if this new expression is really to be the same as that from B A which it was derived, it is clear that A must = B; that C = = ; 2 1·2 A D A C = ; that E = = , etc. that D = 3 1·2·3 4 1·2·3·4 The law of change is therefore that y =A 1+ x x2 x3 x4 + + + + etc. . 1 1·2 1·2·3 1·2·3·4 x x2 x3 x4 + + + + etc. 1 1·2 1·2·3 1·2·3·4

If, now, we take A = 1 for the sake of further simplicity, we have y =1+

Diﬀerentiating it any number of times will give always the same series over again. If, now, we take the particular case of A = 1, and evaluate the

THE LAW OF ORGANIC GROWTH

141

series, we shall get simply when x = 1, when x = 2, when x = 3, and therefore when x = x, y = (2.718281 etc.)x ; that is, y = x y = 2.718281 etc.; y = (2.718281 etc.)2 ; y = (2.718281 etc.)3 ;

that is, y = ; that is, y = that is, y =
2 3

; ;

,

thus ﬁnally demonstrating that x =1+

x2 x3 x4 x + + + + etc. 1 1·2 1·2·3 1·2·3·4 x [Note.—How to read exponentials. For the beneﬁt of those who have no tutor at hand it may be of use to state that So pt is read as

“epsilon to the eksth power ;” or some people read it “exponential eks.” is read “epsilon to the pee-teeth-power ” or “exponential pee tee.”
−2

Take some similar expressions:—Thus,

is read “epsilon to the minus
−ax

two power ” or “exponential minus two.” Of course it follows that respect to y. Also ax y

is read “epsilon to the

minus ay-eksth” or “exponential minus ay-eks.”] remains unchanged if diﬀerentiated with , which is equal to ( a )x , will, when diﬀerentiated ax with respect to x, be a

, because a is a constant.

Natural or Naperian Logarithms. Another reason why is important is because it was made by Napier, the inventor of logarithms, the basis of his system. If y is the value of

CALCULUS MADE EASY x 142

, then x is the logarithm, to the base , of y. Or, if y= x ,

then

x = log y.

The two curves plotted in Figs. 38 and 39 represent these equations. The points calculated are:   x For Fig. 38  y   y For Fig. 39  x y 8 7 6 5 4 3 2 1 O 0.5 1

0 1 1 0

0.5 1.65 2 0.69 y 8 7 6 5 4 3 2 1

1 2.71 3 1.10

1.5 4.50 4 1.39

2 7.39 8 2.08

=

ǫx

y

1.5

2

x

x

=

log

ǫ

y

O

1

2

x

Fig. 38.

Fig. 39.

It will be seen that, though the calculations yield diﬀerent points for plotting, yet the result is identical. The two equations really mean the same thing.

THE LAW OF ORGANIC GROWTH

143

As many persons who use ordinary logarithms, which are calculated to base 10 instead of base , are unfamiliar with the “natural” logarithms, it may be worth while to say a word about them. The ordinary rule that adding logarithms gives the logarithm of the product still holds good; or log a + log b = log ab. Also the rule of powers holds good; n × log a = log an . But as 10 is no longer the basis, one cannot multiply by 100 or 1000 by merely adding 2 or 3 to the index. One can change the natural logarithm to the ordinary logarithm simply by multiplying it by 0.4343; or log10 x = 0.4343 × log x, and conversely, log x = 2.3026 × log10 x.

Exponential and Logarithmic Equations. Now let us try our hands at diﬀerentiating certain expressions that contain logarithms or exponentials. Take the equation: y = log x. First transform this into y = x, y whence, since the diﬀerential of

with regard to y is the original

function unchanged (see p. 139), dx = dy

y

,

CALCULUS MADE EASY

144

A Useful Table of “Naperian Logarithms” (Also called Natural Logarithms or Hyperbolic Logarithms)
Number

log 0.0000 0.0953 0.1823 0.4055 0.5306 0.6931 0.7885 0.9163 0.9933 1.0296 1.0986 1.2528 1.3863 1.5041 1.6094

Number

log 1.7918 1.9459 2.0794 2.1972 2.3026 2.9957 3.9120 4.6052 5.2983 6.2146 6.9078 7.6009 8.5172 9.2103 9.9035

1 1.1 1.2 1.5 1.7 2.0 2.2 2.5 2.7 2.8 3.0 3.5 4.0 4.5 5.0

6 7 8 9 10 20 50 100 200 500 1, 000 2, 000 5, 000 10, 000 20, 000

THE LAW OF ORGANIC GROWTH

145

and, reverting from the inverse to the original function, 1 1 1 dy = = y = . dx dx x dy Now this is a very curious result. It may be written d(log x) = x−1 . dx Note that x−1 is a result that we could never have got by the rule for diﬀerentiating powers. That rule (page 24) is to multiply by the power, and reduce the power by 1. Thus, diﬀerentiating x3 gave us 3x2 ; and diﬀerentiating x2 gave 2x1 . But diﬀerentiating x0 does not give us have to come back to this curious fact that diﬀerentiating log x gives 1 us when we reach the chapter on integrating. x Now, try to diﬀerentiate y = log (x + a), that is we have d(x + a) = dy y y

x−1 or 0 × x−1 , because x0 is itself = 1, and is a constant. We shall

= x + a; y , since the diﬀerential of

remains

y

.

dx = y = x + a; dy hence, reverting to the original function (see p. 128), we get dy 1 1 = = . dx dx x+a dy This gives

CALCULUS MADE EASY

146

Next try

y = log10 x.

First change to natural logarithms by multiplying by the modulus 0.4343. This gives us y = 0.4343 log x; whence 0.4343 dy = . dx x

The next thing is not quite so simple. Try this: y = ax . Taking the logarithm of both sides, we get log y = x log a, or Since x= log y 1 = × log y. log a log a

1 is a constant, we get log a dx 1 1 1 = × = x ; dy log a y a × log a

hence, reverting to the original function. dy 1 = = ax × log a. dx dx dy

THE LAW OF ORGANIC GROWTH

147

We see that, since dx dy × = 1 and dy dx dx 1 1 = × , dy y log a 1 dy × = log a. y dx

We shall ﬁnd that whenever we have an expression such as log y = 1 dy = the diﬀerential coeﬃcient of a function of x, we always have y dx the function of x, so that we could have written at once, from log y = x log a, 1 dy = log a and y dx dy = ax log a. dx

Let us now attempt further examples. Examples. (1) y =
−ax

. Let −ax = z; then y = z z

. dy = −a dx
−ax

dy = dx Or thus:

;

dz = −a; dx

hence

.

log y = −ax; (2) y = x2 3

1 dy = −a; y dx z dy = −ay = −a dx . dy 2x = dx 3 x2 3

−ax

.

. Let

x2 = z; then y = 3 z dy = dz Or thus:

;

dz 2x = ; dx 3

.

x2 log y = ; 3

1 dy 2x = ; y dx 3

dy 2x = dx 3

x2 3

.

CALCULUS MADE EASY
2x x+1

148

(3) y =

. 2x 1 dy 2(x + 1) − 2x , = ; x+1 y dx (x + 1)2 2x 2 dy x+1 . = dx (x + 1)2

log y = hence

2x Check by writing = z. x+1 √ 1 x2 +a . log y = (x2 + a) 2 . (4) y = 1 dy x = 1 y dx (x2 + a) 2
1

and

√ 2 x × x +a dy = 1 . dx (x2 + a) 2
1

For if (x2 + a) 2 = u and x2 + a = v, u = v 2 , du 1 dv = 1; = 2x; dv dx 2v 2 √ Check by writing x2 + a = z. du x = 1 . dx (x2 + a) 2

(5) y = log(a + x3 ). Let (a + x3 ) = z; then y = log z. dz dy 3x2 . = 3x2 ; hence = dx dx a + x3 √ √ (6) y = log {3x2 + a + x2 }. Let 3x2 + a + x2 = z; then y = dy 1 = ; dz z dz x = 6x + √ ; dx x2 + a x √ 6x + √ 2+a dy x(1 + 6 x2 + a) x √ √ √ = = . dx 3x2 + a + x2 (3x2 + x2 + a) x2 + a dy 1 = ; dz z

log z.

THE LAW OF ORGANIC GROWTH

149

√ (7) y = (x + 3)2 x − 2. log y = 2 log (x + 3) + 1 log (x − 2). 2 1 dy 2 1 = + ; y dx (x + 3) 2(x − 2) √ 2 1 dy = (x + 3)2 x − 2 + dx x + 3 2(x − 2)
2

.

(8) y = (x2 + 3)3 (x3 − 2) 3 . log y = 3 log (x2 + 3) + 2 log (x3 − 2); 3 2x 2 3x2 6x 2x2 1 dy =3 2 + = 2 + 3 . y dx (x + 3) 3 x3 − 2 x +3 x −2 du 1 = ; dz z dz = 2x; dx du 2x = 2 . dx x +3

For if y = log (x2 + 3), let x2 + 3 = z and u = log z.

Similarly, if v = log (x3 − 2),

dv 3x2 = 3 and dx x −2 6x 2x2 + 3 x2 + 3 x − 2 .

2 dy = (x2 + 3)3 (x3 − 2) 3 dx √ 2 2 x +a √ (9) y = 3 . x3 − a

log y =

and

1 1 log (x2 + a) − log (x3 − a). 2 3 1 dy 1 2x 1 3x2 x x2 = − = 2 − 3 y dx 2 x2 + a 3 x3 − a x +a x −a √ 2 2 dy x +a x x2 = √ − 3 . 3 3 dx x − a x2 + a x − a

CALCULUS MADE EASY

150

(10) y =

1 log x dy = dx log x × 0 − 1 × log x
1

2

1 x =−

1 . x log2 x
1

(11) y =

3

log x = (log x) 3 . Let z = log x; y = z 3 . dz 1 = ; dx x dy 1 . = 3 dx 3x log2 x

dy 1 2 = z− 3 ; dz 3 (12) y = 1 ax ax .

1 dy = −ax × ax log a − a log ax . y dx ax dy 1 and =− (x × ax+1 log a + a log ax ). x dx a Try now the following exercises. Exercises XII. (See page 260 for Answers.) ax log y = ax(log 1 − log ax ) = −ax log ax .

(1) Diﬀerentiate y = b( u = at2 + 2 log t.

−

−ax

).

(2) Find the diﬀerential coeﬃcient with respect to t of the expression d(log y) . dt 1 abx (4) Show that if y = · , b log a dw (5) If w = pv n , ﬁnd . dv Diﬀerentiate (3) If y = nt , ﬁnd

dy = abx . dx

THE LAW OF ORGANIC GROWTH

151 .

(6) y = log xn . (8) y = (3x2 + 1)
−5x

(7) y = 3 .

x − x−1

(9) y = log (xa + a).

√ (10) y = (3x2 − 1)( x + 1). (11) y = log (x + 3) . x+3 (12) y = ax × xa .

(13) It was shown by Lord Kelvin that the speed of signalling through a submarine cable depends on the value of the ratio of the external diameter of the core to the diameter of the enclosed copper wire. If this ratio is called y, then the number of signals s that can be sent per minute can be expressed by the formula s = ay 2 log 1 ; y

where a is a constant depending on the length and the quality of the materials. Show that if these are given, s will be a maximum if y = √ 1÷ . (14) Find the maximum or minimum of y = x3 − log x. (15) Diﬀerentiate y = log (ax x ). (16) Diﬀerentiate y = (log ax)3 .

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152

The Logarithmic Curve. Let us return to the curve which has its successive ordinates in geometrical progression, such as that represented by the equation y = bpx . We can see, by putting x = 0, that b is the initial height of y. Then when x = 1, y = bp; x = 2, y = bp2 ; x = 3, y = bp3 , etc.

Also, we see that p is the numerical value of the ratio between the height of any ordinate and that of the next preceding it. In Fig. 40, we
6 have taken p as 5 ; each ordinate being 6 5

as high as the preceding one.

Y

b O 1 2 3 4 5 6 X

log b O 1 2 3 4 5 6 X

Fig. 40.

log y

Fig. 41.

If two successive ordinates are related together thus in a constant ratio, their logarithms will have a constant diﬀerence; so that, if we should plot out a new curve, Fig. 41, with values of log y as ordinates, it would be a straight line sloping up by equal steps. In fact, it follows

THE LAW OF ORGANIC GROWTH

153

from the equation, that log y = log b + x · log p, whence log y − log b = x · log p.

Now, since log p is a mere number, and may be written as log p = a, it follows that y = ax, b and the equation takes the new form log y=b ax .

The Die-away Curve. If we were to take p as a proper fraction (less than unity), the curve would obviously tend to sink downwards, as in Fig. 42, where each successive ordinate is
3 4

of the height of the preceding one. y = bpx ;

The equation is still

but since p is less than one, log p will be a negative quantity, and may be written −a; so that p = takes the form
−a

, and now our equation for the curve
−ax

y=b

.

The importance of this expression is that, in the case where the independent variable is time, the equation represents the course of a

CALCULUS MADE EASY
Y

154

b

O

1

2

3

4

5

6

X

Fig. 42.

great many physical processes in which something is gradually dying away. Thus, the cooling of a hot body is represented (in Newton’s celebrated “law of cooling”) by the equation θt = θ0
−at

;

where θ0 is the original excess of temperature of a hot body over that of its surroundings, θt the excess of temperature at the end of time t, and a is a constant—namely, the constant of decrement, depending on the amount of surface exposed by the body, and on its coeﬃcients of conductivity and emissivity, etc. A similar formula, Qt = Q0
−at

,

is used to express the charge of an electriﬁed body, originally having a charge Q0 , which is leaking away with a constant of decrement a; which constant depends in this case on the capacity of the body and on the resistance of the leakage-path. Oscillations given to a ﬂexible spring die out after a time; and the dying-out of the amplitude of the motion may be expressed in a similar

THE LAW OF ORGANIC GROWTH

155

way. In fact
−at

serves as a die-away factor for all those phenomena in

which the rate of decrease is proportional to the magnitude of that dy is proportional which is decreasing; or where, in our usual symbols, dt at every moment to the value that y has at that moment. For we have only to inspect the curve, Fig. 42 above, to see that, at every part of it, dy is proportional to the height y; the curve becoming ﬂatter the slope dx as y grows smaller. In symbols, thus y=b or
−ax

log y = log b − ax log dy =b dx

and, diﬀerentiating, hence

1 dy = −a; y dx
−ax

= log b − ax,

× (−a) = −ay;

or, in words, the slope of the curve is downward, and proportional to y and to the constant a. We should have got the same result if we had taken the equation in the form y = bpx ; for then But giving us as before. dy = bpx × log p. dx log p = −a;

dy = y × (−a) = −ay, dx

CALCULUS MADE EASY

156
−at

The Time-constant. In the expression for the “die-away factor”

,

the quantity a is the reciprocal of another quantity known as “the timeconstant,” which we may denote by the symbol T . Then the die-away factor will be written ; and it will be seen, by making t = T that the 1 meaning of T or of is that this is the length of time which it takes a for the original quantity (called θ0 or Q0 in the preceding instances) to 1 die away th part—that is to 0.3678—of its original value. The values of x t −T

and

−x

are continually required in diﬀerent

branches of physics, and as they are given in very few sets of mathematical tables, some of the values are tabulated on p. 157 for convenience. As an example of the use of this table, suppose there is a hot body cooling, and that at the beginning of the experiment (i.e. when t = 0) it is 72◦ hotter than the surrounding objects, and if the time-constant of its cooling is 20 minutes (that is, if it takes 20 minutes for its excess 1 of temperature to fall to part of 72◦ ), then we can calculate to what it will have fallen in any given time t. For instance, let t be 60 minutes. t Then = 60 ÷ 20 = 3, and we shall have to ﬁnd the value of −3 , T and then multiply the original 72◦ by this. The table shows that −3 is 0.0498. So that at the end of 60 minutes the excess of temperature will have fallen to 72◦ × 0.0498 = 3.586◦ .

THE LAW OF ORGANIC GROWTH

157

x 0.00 0.10 0.20 0.50 0.75 0.90 1.00 1.10 1.20 1.25 1.50 1.75 2.00 2.50 3.00 3.50 4.00 4.50 5.00 5.50 6.00 7.50 10.00

x

−x

1−

−x

1.0000 1.1052 1.2214 1.6487 2.1170 2.4596 2.7183 3.0042 3.3201 3.4903 4.4817 5.755 7.389 12.182 20.086 33.115 54.598 90.017 148.41 244.69 403.43 1808.04 22026.5

1.0000 0.9048 0.8187 0.6065 0.4724 0.4066 0.3679 0.3329 0.3012 0.2865 0.2231 0.1738 0.1353 0.0821 0.0498 0.0302 0.0183 0.0111 0.0067 0.0041 0.00248 0.00055 0.000045

0.0000 0.0952 0.1813 0.3935 0.5276 0.5934 0.6321 0.6671 0.6988 0.7135 0.7769 0.8262 0.8647 0.9179 0.9502 0.9698 0.9817 0.9889 0.9933 0.9959 0.99752 0.99947 0.999955

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158

Further Examples. (1) The strength of an electric current in a conductor at a time t secs. after the application of the electromotive force producing it is E Rt 1− −L . given by the expression C = R L The time constant is . R Rt If E = 10, R = 1, L = 0.01; then when t is very large the term − L E becomes 1, and C = = 10; also R L = T = 0.01. R Its value at any time may be written: C = 10 − 10 t − 0.01

,

the time-constant being 0.01. This means that it takes 0.01 sec. for the 1 0 variable term to fall by = 0.3678 of its initial value 10 − 0.01 = 10. t To ﬁnd the value of the current when t = 0.001 sec., say, = 0.1, T −0.1 = 0.9048 (from table). 9.048, and the actual current is 10 − 9.048 = 0.952. Similarly, at the end of 0.1 sec., t = 10; T
−10

It follows that, after 0.001 sec., the variable term is 0.9048 × 10 =

= 0.000045;

the variable term is 10 × 0.000045 = 0.00045, the current being 9.9995. thickness l cm. of some transparent medium is I = I0
−Kl

(2) The intensity I of a beam of light which has passed through a , where I0 is

the initial intensity of the beam and K is a “constant of absorption.”

THE LAW OF ORGANIC GROWTH

159

This constant is usually found by experiments. If it be found, for instance, that a beam of light has its intensity diminished by 18% in passing through 10 cms. of a certain transparent medium, this means that 82 = 100 ×
−K×10

or

−10K

= 0.82, and from the table one sees

that 10K = 0.20 very nearly; hence K = 0.02. To ﬁnd the thickness that will reduce the intensity to half its value, one must ﬁnd the value of l which satisﬁes the equality 50 = 100× or 0.5 =
−0.02l −0.02l

,

. It is found by putting this equation in its logarithmic log 0.5 = −0.02 × l × log ,

form, namely,

which gives l= −0.3010 = 34.7 centimetres nearly. −0.02 × 0.4343

(3) The quantity Q of a radio-active substance which has not yet undergone transformation is known to be related to the initial quantity Q0 of the substance by the relation Q = Q0
−λt

, where λ is a constant and

t the time in seconds elapsed since the transformation began. For “Radium A,” if time is expressed in seconds, experiment shows substance. (This time is called the “mean life” of the substance.) We have 0.5 =
−0.00385t

that λ = 3.85 × 10−3 . Find the time required for transforming half the .

log 0.5 = −0.00385t × log ; and t = 3 minutes very nearly.

CALCULUS MADE EASY

160

Exercises XIII.

(See page 260 for Answers.) t −T

(1) Draw the curve y = b various values from 0 to 20.

; where b = 12, T = 8, and t is given

(2) If a hot body cools so that in 24 minutes its excess of temperature has fallen to half the initial amount, deduce the time-constant, and ﬁnd how long it will be in cooling down to 1 per cent. of the original excess.
−2t

(3) Plot the curve y = 100(1 −

).

(4) The following equations give very similar curves: (i) y = ax ; x+b x (ii) y = a(1 − − b ); a x (iii) y = ◦ arc tan . 90 b Draw all three curves, taking a = 100 millimetres; b = 30 millimetres. (5) Find the diﬀerential coeﬃcient of y with respect to x, if (a) y = xx ; (b) y = ( x )x ; (c) y = xx .

(6) For “Thorium A,” the value of λ is 5; ﬁnd the “mean life,” that is, the time taken by the transformation of a quantity Q of “Thorium A” equal to half the initial quantity Q0 in the expression Q = Q0 t being in seconds.
−λt

;

THE LAW OF ORGANIC GROWTH

161

V0 = 20, is discharging through a resistance of 10, 000 ohms. Find the potential V after (a) 0.1 second; (b) 0.01 second; assuming that the fall of potential follows the rule V = V0 t − KR

(7) A condenser of capacity K = 4 × 10−6 , charged to a potential

.

(8) The charge Q of an electriﬁed insulated metal sphere is reduced from 20 to 16 units in 10 minutes. Find the coeﬃcient µ of leakage, Hence ﬁnd the time taken by half the charge to leak away. (9) The damping on a telephone line can be ascertained from the relation i = i0
−βl

if Q = Q0 ×

−µt

; Q0 being the initial charge and t being in seconds.

, where i is the strength, after t seconds, of a telephonic

current of initial strength i0 ; l is the length of the line in kilometres, and β is a constant. For the Franco-English submarine cable laid in 1910, β = 0.0114. Find the damping at the end of the cable (40 kilometres), and the length along which i is still 8% of the original current (limiting value of very good audition). (10) The pressure p of the atmosphere at an altitude h kilometres is given by p = p0
−kh

; p0 being the pressure at sea-level (760 millimetres).

The pressures at 10, 20 and 50 kilometres being 199.2, 42.2, 0.32 respectively, ﬁnd k in each case. Using the mean value of k, ﬁnd the percentage error in each case. (11) Find the minimum or maximum of y = xx . (12) Find the minimum or maximum of y = x x . (13) Find the minimum or maximum of y = xa x .
1 1

CHAPTER XV.
HOW TO DEAL WITH SINES AND COSINES.
Greek letters being usual to denote angles, we will take as the usual letter for any variable angle the letter θ (“theta”). Let us consider the function y = sin θ.

dθ

y θ O

Fig. 43.

d(sin θ) ; or, in other dθ words, if the angle θ varies, we have to ﬁnd the relation between the What we have to investigate is the value of increment of the sine and the increment of the angle, both increments being indeﬁnitely small in themselves. Examine Fig. 43, wherein, if the

SINES AND COSINES

163

radius of the circle is unity, the height of y is the sine, and θ is the angle. Now, if θ is supposed to increase by the addition to it of the small angle dθ—an element of angle—the height of y, the sine, will be increased by a small element dy. The new height y + dy will be the sine of the new angle θ + dθ, or, stating it as an equation, y + dy = sin(θ + dθ); and subtracting from this the ﬁrst equation gives dy = sin(θ + dθ) − sin θ. The quantity on the right-hand side is the diﬀerence between two sines, and books on trigonometry tell us how to work this out. For they tell us that if M and N are two diﬀerent angles, sin M − sin N = 2 cos we may write θ + dθ + θ θ + dθ − θ · sin , 2 2 dy = 2 cos(θ + 1 dθ) · sin 1 dθ. 2 2 dy = 2 cos But if we regard dθ as indeﬁnitely small, then in the limit we may
1 neglect 2 dθ by comparison with θ, and may also take sin 1 dθ as being 2

M −N M +N · sin . 2 2

If, then, we put M = θ + dθ for one angle, and N = θ for the other,

or,

the same as 1 dθ. The equation then becomes: 2 dy = 2 cos θ × 1 dθ; 2 and, ﬁnally, dy = cos θ. dθ dy = cos θ · dθ,

CALCULUS MADE EASY

164

The accompanying curves, Figs. 44 and 45, show, plotted to scale, dy = cos θ, for the corresponding values the values of y = sin θ, and dθ of θ. y 1 0.5 θ
30◦ 60◦ 90◦ 120◦ 150◦ 180◦ 270◦ 360◦

−0.5 −1

Fig. 44.

y 1 0.5 θ
30◦ 60◦ 90◦ 180◦ 270◦ 360◦

−0.5 −1

Fig. 45.

Take next the cosine. Let y = cos θ.

SINES AND COSINES

165

Now cos θ = sin Therefore

π −θ . 2 π −θ 2 = cos π − θ × d(−θ), 2 π − θ × (−dθ), = cos 2

dy = d sin

dy π = − cos −θ . dθ 2 And it follows that dy = − sin θ. dθ

Lastly, take the tangent. Let y = tan θ, dy = tan(θ + dθ) − tan θ. Expanding, as shown in books on trigonometry, tan(θ + dθ) = whence tan θ + tan dθ ; 1 − tan θ · tan dθ tan θ + tan dθ − tan θ dy = 1 − tan θ · tan dθ (1 + tan2 θ) tan dθ . = 1 − tan θ · tan dθ

Now remember that if dθ is indeﬁnitely diminished, the value of tan dθ becomes identical with dθ, and tan θ · dθ is negligibly small

CALCULUS MADE EASY

166

compared with 1, so that the expression reduces to dy = so that or (1 + tan2 θ) dθ , 1

dy = 1 + tan2 θ, dθ dy = sec2 θ. dθ

Collecting these results, we have: y sin θ cos θ tan θ dy dθ cos θ − sin θ sec2 θ

Sometimes, in mechanical and physical questions, as, for example, in simple harmonic motion and in wave-motions, we have to deal with angles that increase in proportion to the time. Thus, if T be the time of one complete period, or movement round the circle, then, since the angle all round the circle is 2π radians, or 360◦ , the amount of angle moved through in time t, will be t θ = 2π , in radians, T t θ = 360 , in degrees. T

or

If the frequency, or number of periods per second, be denoted by n, 1 then n = , and we may then write: T θ = 2πnt.

SINES AND COSINES

167

Then we shall have y = sin 2πnt. If, now, we wish to know how the sine varies with respect to time, we must diﬀerentiate with respect, not to θ, but to t. For this we must resort to the artiﬁce explained in Chapter IX., p. 66, and put dy dθ dy = · . dt dθ dt Now dθ will obviously be 2πn; so that dt dy = cos θ × 2πn dt = 2πn · cos 2πnt. Similarly, it follows that d(cos 2πnt) = −2πn · sin 2πnt. dt Second Diﬀerential Coeﬃcient of Sine or Cosine. We have seen that when sin θ is diﬀerentiated with respect to θ it becomes cos θ; and that when cos θ is diﬀerentiated with respect to θ it becomes − sin θ; or, in symbols, d2 (sin θ) = − sin θ. dθ2 So we have this curious result that we have found a function such that if we diﬀerentiate it twice over, we get the same thing from which we started, but with the sign changed from + to −.

CALCULUS MADE EASY

168

The same thing is true for the cosine; for diﬀerentiating cos θ gives us − sin θ, and diﬀerentiating − sin θ gives us − cos θ; or thus: d2 (cos θ) = − cos θ. dθ2 Sines and cosines are the only functions of which the second diﬀerential coeﬃcient is equal (and of opposite sign to) the original function.

Examples. With what we have so far learned we can now diﬀerentiate expressions of a more complex nature. (1) y = arc sin x. If y is the arc whose sine is x, then x = sin y. dx = cos y. dy Passing now from the inverse function to the original one, we get dy 1 1 = = . dx dx cos y dy Now hence cos y = dy 1 =√ , dx 1 − x2 1 − sin2 y = √ 1 − x2 ;

a rather unexpected result. (2) y = cos3 θ. This is the same thing as y = (cos θ)3 .

SINES AND COSINES

169

Let cos θ = v; then y = v 3 ;

dy = 3v 2 . dv

dv = − sin θ. dθ dy dv dy = × = −3 cos2 θ sin θ. dθ dv dθ (3) y = sin(x + a). Let x + a = v; then y = sin v. dy = cos v; dv (4) y = log sin θ. Let sin θ = v; y = log v. dy 1 dv = ; = cos θ; dv v dθ dy 1 = × cos θ = cot θ. dθ sin θ (5) y = cot θ = cos θ . sin θ dy − sin2 θ − cos2 θ = dθ sin2 θ = −(1 + cot2 θ) = − cosec2 θ. (6) y = tan 3θ. Let 3θ = v; y = tan v; dy = sec2 v. dv dy = 3 sec2 3θ. dθ dv = 1 and dx dy = cos(x + a). dx

dv = 3; dθ (7) y =

√ 1 1 + 3 tan2 θ; y = (1 + 3 tan2 θ) 2 .

CALCULUS MADE EASY

170

Let 3 tan2 θ = v. y = (1 + v) 2 ;
1

dy 1 = √ (see p. 67); dv 2 1+v

dv = 6 tan θ sec2 θ dθ (for, if tan θ = u, v = 3u2 ; hence hence dv = 6u; du du = sec2 θ; dθ

dv = 6(tan θ sec2 θ) dθ dy 6 tan θ sec2 θ = √ . dθ 2 1 + 3 tan2 θ

(8) y = sin x cos x. dy = sin x(− sin x) + cos x × cos x dx = cos2 x − sin2 x. Exercises XIV. (See page 261 for Answers.)

(1) Diﬀerentiate the following: (i) y = A sin θ − (ii) y = sin2 θ; (iii) y = sin3 θ; π . 2 and y = sin 2θ.

and y = sin 3θ.

(2) Find the value of θ for which sin θ × cos θ is a maximum. 1 (3) Diﬀerentiate y = cos 2πnt. 2π

SINES AND COSINES

171

dy . dx (5) Diﬀerentiate y = log cos x. (4) If y = sin ax , ﬁnd (6) Diﬀerentiate y = 18.2 sin(x + 26◦ ). (7) Plot the curve y = 100 sin(θ − 15◦ ); and show that the slope of

the curve at θ = 75◦ is half the maximum slope. dy (8) If y = sin θ · sin 2θ, ﬁnd . dθ (9) If y = a · tanm (θn ), ﬁnd the diﬀerential coeﬃcient of y with respect to θ. (10) Diﬀerentiate y = x sin2 x.

(11) Diﬀerentiate the three equations of Exercises XIII. (p. 160), No. 4, and compare their diﬀerential coeﬃcients, as to whether they are equal, or nearly equal, for very small values of x, or for very large values of x, or for values of x in the neighbourhood of x = 30. (12) Diﬀerentiate the following: (i) y = sec x. (iii) y = arc tan x. √ (v) y = tan x × 3 sec x. (13) Diﬀerentiate y = sin(2θ + 3)2.3 . (14) Diﬀerentiate y = θ3 + 3 sin(θ + 3) − 3sin θ − 3θ . (15) Find the maximum or minimum of y = θ cos θ. (ii) y = arc cos x. (iv) y = arc sec x.

CHAPTER XVI.
PARTIAL DIFFERENTIATION.
We sometimes come across quantities that are functions of more than one independent variable. Thus, we may ﬁnd a case where y depends on two other variable quantities, one of which we will call u and the other v. In symbols y = f (u, v). Take the simplest concrete case. Let y = u × v.

What are we to do? If we were to treat v as a constant, and diﬀerentiate with respect to u, we should get dyv = v du; or if we treat u as a constant, and diﬀerentiate with respect to v, we should have: dyu = u dv. The little letters here put as subscripts are to show which quantity has been taken as constant in the operation.

PARTIAL DIFFERENTIATION

173

Another way of indicating that the diﬀerentiation has been performed only partially, that is, has been performed only with respect to one of the independent variables, is to write the diﬀerential coeﬃcients with Greek deltas, like ∂, instead of little d. In this way ∂y = v, ∂u ∂y = u. ∂v If we put in these values for v and u respectively, we shall have  ∂y   dyv = du, ∂u which are partial diﬀerentials.  ∂y  dyu = dv,  ∂v But, if you think of it, you will observe that the total variation of y depends on both these things at the same time. That is to say, if both are varying, the real dy ought to be written dy = ∂y ∂y du + dv; ∂u ∂v

and this is called a total diﬀerential. In some books it is written dy = dy dy du + dv. du dv Example (1). Find the partial diﬀerential coeﬃcients of the expression w = 2ax2 + 3bxy + 4cy 3 . The answers are:  ∂w  = 4ax + 3by.   ∂x ∂w  = 3bx + 12cy 2 .  ∂y

CALCULUS MADE EASY

174

The ﬁrst is obtained by supposing y constant, the second is obtained by supposing x constant; then dw = (4ax + 3by) dx + (3bx + 12cy 2 ) dy. Example (2). Let z = xy . Then, treating ﬁrst y and then x as constant, we get in the usual way  ∂z  y−1  = yx ,  ∂x ∂z  = xy × log x,  ∂y so that dz = yxy−1 dx + xy log x dy. Example (3). A cone having height h and radius of base r has
1 volume V = 3 πr2 h. If its height remains constant, while r changes,

the ratio of change of volume, with respect to radius, is diﬀerent from ratio of change of volume with respect to height which would occur if the height were varied and the radius kept constant, for  2π  ∂V = rh, ∂r 3 ∂V π  = r2 .  ∂h 3 The variation when both the radius and the height change is given 2π π by dV = rh dV + r2 dh. 3 3 Example (4). In the following example F and f denote two arbitrary functions of any form whatsoever. For example, they may be sine-functions, or exponentials, or mere algebraic functions of the two

PARTIAL DIFFERENTIATION

175

independent variables, t and x. This being understood, let us take the expression y = F (x + at) + f (x − at), or, where Then y = F (w) + f (v); w = x + at, ∂F (w) ∂w ∂f (v) ∂v ∂y = · + · ∂x ∂w ∂x ∂v ∂x = F (w) · 1 + f (v) · 1 and v = x − at.

(where the ﬁgure 1 is simply the coeﬃcient of x in w and v); and Also ∂ 2y = F (w) + f (v). ∂x2 ∂y ∂F (w) ∂w ∂f (v) ∂v = · + · ∂t ∂w ∂t ∂v ∂t = F (w) · a − f (v)a; ∂ 2y = F (w)a2 + f (v)a2 ; ∂t2 ∂ 2y ∂ 2y = a2 2 . ∂t2 ∂x

and whence

This diﬀerential equation is of immense importance in mathematical physics. Maxima and Minima of Functions of two Independent Variables. Example (5). Let us take up again Exercise IX., p. 107, No. 4. Let x and y be the length of two of the portions of the string. The third is 30 − (x + y), and the area of the triangle is A =

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s(s − x)(s − y)(s − 30 + x + y), where s is the half perimeter, 15, so √ that A = 15P , where P = (15 − x)(15 − y)(x + y − 15) = xy 2 + x2 y − 15x2 − 15y 2 − 45xy + 450x + 450y − 3375. Clearly A is maximum when P is maximum. dP = ∂P ∂P dx + dy. ∂x ∂y

For a maximum (clearly it will not be a minimum in this case), one must have simultaneously ∂P = 0 and ∂x that is, ∂P = 0; ∂y

 2xy − 30x + y 2 − 45y + 450 = 0, 2xy − 30y + x2 − 45x + 450 = 0.

An immediate solution is x = y. If we now introduce this condition in the value of P , we ﬁnd P = (15 − x)2 (2x − 15) = 2x3 − 75x2 + 900x − 3375. For maximum or minimum, x = 15 or x = 10. dP = 6x2 − 150x + 900 = 0, which gives dx

Clearly x = 15 gives minimum area; x = 10 gives the maximum, for d2 P = 12x − 150, which is +30 for x = 15 and −30 for x = 10. dx2

PARTIAL DIFFERENTIATION

177

Example (6). Find the dimensions of an ordinary railway coal truck with rectangular ends, so that, for a given volume V the area of sides and ﬂoor together is as small as possible. The truck is a rectangular box open at the top. Let x be the length V . The surface area is S = and y be the width; then the depth is xy 2V 2V + . xy + x y dS = ∂S ∂S dx + dy = ∂x ∂y y− 2V x2 dx + x − 2V y2 dy.

For minimum (clearly it won’t be a maximum here), y− 2V = 0, x2 x− 2V = 0. y2

4V , Here also, an immediate solution is x = y, so that S = x2 + x dS 4V = 2x − 2 = 0 for minimum, and dx x √ 3 x = 2V .

Exercises XV.

(See page 263 for Answers.)

x3 y − 2x3 y − 2y 2 x + with respect 3 3 to x alone, and with respect to y alone. (1) Diﬀerentiate the expression (2) Find the partial diﬀerential coeﬃcients with respect to x, y and z, of the expression x2 yz + xy 2 z + xyz 2 + x2 y 2 z 2 .

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(3) Let r2 = (x − a)2 + (y − b)2 + (z − c)2 . Find the value of ∂ 2r . ∂z 2 (4) Find the total diﬀerential of y = uv .

∂ 2r ∂ 2r ∂r ∂r ∂r + + . Also ﬁnd the value of + + ∂x ∂y ∂z ∂x2 ∂y 2

(5) Find the total diﬀerential of y = u3 sin v; of y = (sin x)u ; and of log u y= . v (6) Verify that the sum of three quantities x, y, z, whose product is a constant k, is maximum when these three quantities are equal. (7) Find the maximum or minimum of the function u = x + 2xy + y. (8) The post-oﬃce regulations state that no parcel is to be of such a size that its length plus its girth exceeds 6 feet. What is the greatest volume that can be sent by post (a) in the case of a package of rectangular cross section; (b) in the case of a package of circular cross section. (9) Divide π into 3 parts such that the continued product of their sines may be a maximum or minimum. x+y (10) Find the maximum or minimum of u = (11) Find maximum and minimum of

xy

.

u = y + 2x − 2 log y − log x.

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179

(12) A telpherage bucket of given capacity has the shape of a horizontal isosceles triangular prism with the apex underneath, and the opposite face open. Find its dimensions in order that the least amount of iron sheet may be used in its construction.

CHAPTER XVII.
INTEGRATION.
The great secret has already been revealed that this mysterious symbol , which is after all only a long S, merely means “the sum of,” or (the Greek Sigma), which is also a sign of summation. There is generally used to indicate the is generally “the sum of all such quantities as.” It therefore resembles that other symbol is this diﬀerence, however, in the practice of mathematical men as to the use of these signs, that while sum of a number of ﬁnite quantities, the integral sign

used to indicate the summing up of a vast number of small quantities of indeﬁnitely minute magnitude, mere elements in fact, that go to make up the total required. Thus dy = y, and dx = x. Any one can understand how the whole of anything can be conceived of as made up of a lot of little bits; and the smaller the bits the more of them there will be. Thus, a line one inch long may be conceived as made up of 10 pieces, each
1 100 1 10

of an inch long; or of 100 parts, each part being
1 1,000,000

of an inch long; or of 1, 000, 000 parts, each of which is

of

an inch long; or, pushing the thought to the limits of conceivability, it may be regarded as made up of an inﬁnite number of elements each of which is inﬁnitesimally small. Yes, you will say, but what is the use of thinking of anything that

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181

way? Why not think of it straight oﬀ, as a whole? The simple reason is that there are a vast number of cases in which one cannot calculate the bigness of the thing as a whole without reckoning up the sum of a lot of small parts. The process of “integrating” is to enable us to calculate totals that otherwise we should be unable to estimate directly. Let us ﬁrst take one or two simple cases to familiarize ourselves with this notion of summing up a lot of separate parts. Consider the series:
1 1+ 2 + 1 + 1 + 4 8 1 16

+

1 32

+

1 64

+ etc.

Here each member of the series is formed by taking it half the value of the preceding. What is the value of the total if we could go on to an inﬁnite number of terms? Every schoolboy knows that the answer is 2. Think of it, if you like, as a line. Begin with one inch; add a half

1

1/2

1/4

1/8

Fig. 46.

inch, add a quarter; add an eighth; and so on. If at any point of the operation we stop, there will still be a piece wanting to make up the whole 2 inches; and the piece wanting will always be the same size as the last piece added. Thus, if after having put together 1, 1 , and 1 , we 2 4 stop, there will be will still be
1 64 1 4

wanting. If we go on till we have added

1 , 64

there

wanting. The remainder needed will always be equal to

the last term added. By an inﬁnite number of operations only should we reach the actual 2 inches. Practically we should reach it when we

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182

got to pieces so small that they could not be drawn—that would be after about 10 terms, for the eleventh term is
1 . 1024

If we want to go

so far that not even a Whitworth’s measuring machine would detect it, we should merely have to go to about 20 terms. A microscope would not show even the 18th term! So the inﬁnite number of operations is no such dreadful thing after all. The integral is simply the whole lot. But, as we shall see, there are cases in which the integral calculus enables us to get at the exact total that there would be as the result of an inﬁnite number of operations. In such cases the integral calculus gives us a rapid and easy way of getting at a result that would otherwise require an interminable lot of elaborate working out. So we had best lose no time in learning how to integrate. Slopes of Curves, and the Curves themselves. Let us make a little preliminary enquiry about the slopes of curves. For we have seen that diﬀerentiating a curve means ﬁnding an expression for its slope (or for its slopes at diﬀerent points). Can we perform the reverse process of reconstructing the whole curve if the slope (or slopes) are prescribed for us? Go back to case (2) on p. 82. Here we have the simplest of curves, a sloping line with the equation y = ax + b.

We know that here b represents the initial height of y when x = 0, dy , is the “slope” of the line. The line and that a, which is the same as dx

INTEGRATION

183

Y

b O X

Fig. 47. dy dx

has a constant slope. All along it the elementary triangles

have the same proportion between height and base. Suppose we were to take the dx’s, and dy’s of ﬁnite magnitude, so that 10 dx’s made up one inch, then there would be ten little triangles like

Now, suppose that we were ordered to reconstruct the “curve,” dy starting merely from the information that = a. What could we dx do? Still taking the little d’s as of ﬁnite size, we could draw 10 of them, all with the same slope, and then put them together, end to end, like this: And, as the slope is the same for all, they would join to make, as dy in Fig. 48, a sloping line sloping with the correct slope = a. And dx whether we take the dy’s and dx’s as ﬁnite or inﬁnitely small, as they y are all alike, clearly = a, if we reckon y as the total of all the dy’s, x and x as the total of all the dx’s. But whereabouts are we to put this

CALCULUS MADE EASY
Y

184

c

O

X

Fig. 48.

sloping line? Are we to start at the origin O, or higher up? As the only information we have is as to the slope, we are without any instructions as to the particular height above O; in fact the initial height is undetermined. The slope will be the same, whatever the initial height. Let us therefore make a shot at what may be wanted, and start the sloping line at a height C above O. That is, we have the equation y = ax + C. It becomes evident now that in this case the added constant means the particular value that y has when x = 0. Now let us take a harder case, that of a line, the slope of which is not constant, but turns up more and more. Let us assume that the upward slope gets greater and greater in proportion as x grows. In

INTEGRATION

185

dy = ax. dx Or, to give a concrete case, take a = 1 , so that 5 dy = 1 x. 5 dx Then we had best begin by calculating a few of the values of the slope at diﬀerent values of x, and also draw little diagrams of them.

symbols this is:

When

x = 0, x = 1, x = 2, x = 3,

dy = 0, dx dy = 0.2, dx dy = 0.4, dx dy = 0.6, dx dy = 0.8, dx dy = 1.0. dx

x = 4,

x = 5,

Now try to put the pieces together, setting each so that the middle of its base is the proper distance to the right, and so that they ﬁt together at the corners; thus (Fig. 49). The result is, of course, not a smooth curve: but it is an approximation to one. If we had taken bits half as long, and twice as numerous, like Fig. 50, we should have a

CALCULUS MADE EASY
Y

186
P

O

1

2

3

4

5

X

Fig. 49.

better approximation. But for a perfect curve we ought to take each dx and its corresponding dy inﬁnitesimally small, and inﬁnitely numerous.
Y P

O

1

2

3

4

5

X

Fig. 50.

Then, how much ought the value of any y to be? Clearly, at any point P of the curve, the value of y will be the sum of all the little dy’s from 0 up to that level, that is to say, dy = y. And as each dy is
1 x 5 1 equal to 5 x · dx, it follows that the whole y will be equal to the sum of

all such bits as 1 x · dx, or, as we should write it, 5 Now if x had been constant,
1 x 5

· dx.

· dx would have been the same

INTEGRATION

187

as

1 x 5

dx, or

1 2 x. 5

But x began by being 0, and increases to the
1 x dx 5

particular value of x at the point P , so that its average value from 0 to
1 that point is 2 x. Hence

=

1 2 x; 10

or y =

1 2 x. 10

But, as in the previous case, this requires the addition of an undetermined constant C, because we have not been told at what height above the origin the curve will begin, when x = 0. So we write, as the equation of the curve drawn in Fig. 51, y=
1 2 x 10

+ C.

Y

y

C

x

Fig. 51.

Exercises XVI.

(See page 264 for Answers.)
2 3 1 +1+6+ 3 1 12

(1) Find the ultimate sum of

+

1 24

+ etc.

and ﬁnd its sum to 8 terms.

1 1 1 (2) Show that the series 1 − 2 + 1 − 4 + 1 − 6 + 1 etc., is convergent, 3 5 7

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188

x2 x3 x4 + − + etc., ﬁnd log 1.3. 2 3 4 (4) Following a reasoning similar to that explained in this chapter, (3) If log (1 + x) = x − ﬁnd y, (a) if (5) If dy dy 1 = 4 x; (b) if = cos x. dx dx

dy = 2x + 3, ﬁnd y. dx

CHAPTER XVIII.
INTEGRATING AS THE REVERSE OF DIFFERENTIATING.
Differentiating is the process by which when y is given us (as a dy . function of x), we can ﬁnd dx Like every other mathematical operation, the process of diﬀerentiady = 4x3 ; tion may be reversed; thus, if diﬀerentiating y = x4 gives us dx dy if one begins with = 4x3 one would say that reversing the process dx would yield y = x4 . But here comes in a curious point. We should get dy = 4x3 if we had begun with any of the following: x4 , or x4 + a, dx or x4 + c, or x4 with any added constant. So it is clear that in working dy backwards from to y, one must make provision for the possibility dx of there being an added constant, the value of which will be undetermined until ascertained in some other way. So, if diﬀerentiating xn dy yields nxn−1 , going backwards from = nxn−1 will give us y = xn +C; dx where C stands for the yet undetermined possible constant. Clearly, in dealing with powers of x, the rule for working backwards will be: Increase the power by 1, then divide by that increased power, and add the undetermined constant.

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190

So, in the case where

dy = xn , dx 1 xn+1 + C. n+1

working backwards, we get y=

If diﬀerentiating the equation y = axn gives us dy = anxn−1 , dx it is a matter of common sense that beginning with dy = anxn−1 , dx and reversing the process, will give us y = axn . So, when we are dealing with a multiplying constant, we must simply put the constant as a multiplier of the result of the integration. dy Thus, if = 4x2 , the reverse process gives us y = 4 x3 . 3 dx But this is incomplete. For we must remember that if we had started with y = axn + C, where C is any constant quantity whatever, we should equally have found dy = anxn−1 . dx So, therefore, when we reverse the process we must always remember

to add on this undetermined constant, even if we do not yet know what its value will be.

HOW TO INTEGRATE

191

This process, the reverse of diﬀerentiating, is called integrating; for it consists in ﬁnding the value of the whole quantity y when you are dy . Hitherto we have as much as given only an expression for dy or for dx possible kept dy and dx together as a diﬀerential coeﬃcient: henceforth we shall more often have to separate them. If we begin with a simple case, dy = x2 . dx We may write this, if we like, as dy = x2 dx. Now this is a “diﬀerential equation” which informs us that an element of y is equal to the corresponding element of x multiplied by x2 . Now, what we want is the integral; therefore, write down with the proper symbol the instructions to integrate both sides, thus: dy = x2 dx.

[Note as to reading integrals: the above would be read thus: “Integral dee-wy equals integral eks-squared dee-eks.”] We haven’t yet integrated: we have only written down instructions to integrate—if we can. Let us try. Plenty of other fools can do it—why not we also? The left-hand side is simplicity itself. The sum of all the bits of y is the same thing as y itself. So we may at once put: y= x2 dx.

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192

But when we come to the right-hand side of the equation we must remember that what we have got to sum up together is not all the dx’s, but all such terms as x2 dx; and this will not be the same as x2 dx, because x2 is not a constant. For some of the dx’s will be multiplied by big values of x2 , and some will be multiplied by small values of x2 , according to what x happens to be. So we must bethink ourselves as to what we know about this process of integration being the reverse of diﬀerentiation. Now, our rule for this reversed process—see p. 189 ante—when dealing with xn is “increase the power by one, and divide by the same number as this increased power.” That is to say, x2 dx will
1 be changed∗ to 3 x3 . Put this into the equation; but don’t forget to add

the “constant of integration” C at the end. So we get:
1 y = 3 x3 + C.

You have actually performed the integration. How easy! Let us try another simple case. dy = ax12 , dx

Let

where a is any constant multiplier. Well, we found when diﬀerentiating (see p. 27) that any constant factor in the value of y reappeared
∗

You may ask, what has become of the little dx at the end? Well, remember

that it was really part of the diﬀerential coeﬃcient, and when changed over to the right-hand side, as in the x2 dx, serves as a reminder that x is the independent variable with respect to which the operation is to be eﬀected; and, as the result of the product being totalled up, the power of x has increased by one. You will soon become familiar with all this.

HOW TO INTEGRATE

193

dy . In the reversed process of integrating, it dx will therefore also reappear in the value of y. So we may go to work as unchanged in the value of before, thus dy = ax12 · dx, dy = dy = a y =a× So that is done. How easy! We begin to realize now that integrating is a process of ﬁnding our way back, as compared with diﬀerentiating. If ever, during diﬀerentiating, we have found any particular expression—in this example ax12 —we can ﬁnd our way back to the y from which it was derived. The contrast between the two processes may be illustrated by the following remark due to a well-known teacher. If a stranger were set down in Trafalgar Square, and told to ﬁnd his way to Euston Station, he might ﬁnd the task hopeless. But if he had previously been personally conducted from Euston Station to Trafalgar Square, it would be comparatively easy to him to ﬁnd his way back to Euston Station. Integration of the Sum or Diﬀerence of two Functions. ax12 · dx, x12 dx,
1 13 x 13

+ C.

Let then

dy = x2 + x3 , dx dy = x2 dx + x3 dx.

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194

There is no reason why we should not integrate each term separately: for, as may be seen on p. 34, we found that when we diﬀerentiated the sum of two separate functions, the diﬀerential coeﬃcient was simply the sum of the two separate diﬀerentiations. So, when we work backwards, integrating, the integration will be simply the sum of the two separate integrations. Our instructions will then be: dy = = (x2 + x3 ) dx x2 dx + x3 dx

y = 1 x3 + 1 x4 + C. 3 4 If either of the terms had been a negative quantity, the corresponding term in the integral would have also been negative. So that diﬀerences are as readily dealt with as sums. How to deal with Constant Terms. Suppose there is in the expression to be integrated a constant term— dy = xn + b. dx This is laughably easy. For you have only to remember that when dy you diﬀerentiated the expression y = ax, the result was = a. Hence, dx when you work the other way and integrate, the constant reappears such as this:

EASIEST INTEGRATIONS

195

multiplied by x. So we get dy = xn dx + b · dx, dy = y= xn dx + b dx, 1 xn+1 + bx + C. n+1

Here are a lot of examples on which to try your newly acquired powers.

Examples. (1) Given

dy = 24x11 . Find y. dx (2) Find (a + b)(x + 1) dx. x dx + dx

Ans. y = 2x12 + C. It is (a + b) (x + 1) dx

or (a + b)

or (a + b)

x2 + x + C. 2
3

du 1 (3) Given = gt 2 . Find u. dt dy (4) = x3 − x2 + x. Find y. dx dy = (x3 − x2 + x) dx or dy = x3 dx − x2 dx + x dx; and y=

Ans. u = 2 gt 2 + C. 3

x3 dx −

x2 dx +

x dx;

1 y = 1 x4 − 3 x3 + 1 x2 + C. 4 2

(5) Integrate 9.75x2.25 dx.

Ans. y = 3x3.25 + C.

CALCULUS MADE EASY

196

All these are easy enough. Let us try another case.

Let

dy = ax−1 . dx Proceeding as before, we will write dy = ax−1 · dx, dy = a x−1 dx.

Well, but what is the integral of x−1 dx? If you look back amongst the results of diﬀerentiating x2 and x3 and xn , etc., you will ﬁnd we never got x−1 from any one of them as dy . We got 3x2 from x3 ; we got 2x from x2 ; we got 1 the value of dx from x1 (that is, from x itself); but we did not get x−1 from x0 , for two very good reasons. First, x0 is simply = 1, and is a constant, and could not have a diﬀerential coeﬃcient. Secondly, even if it could be diﬀerentiated, its diﬀerential coeﬃcient (got by slavishly following the zero value! Therefore when we now come to try to integrate x−1 dx, we see that it does not come in anywhere in the powers of x that are given by the rule: xn dx = It is an exceptional case. Well; but try again. Look through all the various diﬀerentials obtained from various functions of x, and try to ﬁnd amongst them x−1 . dy A suﬃcient search will show that we actually did get = x−1 as the dx result of diﬀerentiating the function y = log x (see p. 145). 1 xn+1 . n+1 usual rule) would be 0 × x−1 , and that multiplication by zero gives it

EASIEST INTEGRATIONS

197

Then, of course, since we know that diﬀerentiating log x gives us x−1 , we know that, by reversing the process, integrating dy = x−1 dx will give us y = log x. But we must not forget the constant factor a that was given, nor must we omit to add the undetermined constant of integration. This then gives us as the solution to the present problem, y = a log x + C. N.B.—Here note this very remarkable fact, that we could not have integrated in the above case if we had not happened to know the corresponding diﬀerentiation. If no one had found out that diﬀerentiating log x gave x−1 , we should have been utterly stuck by the problem how to integrate x−1 dx. Indeed it should be frankly admitted that this is one of the curious features of the integral calculus:—that you can’t integrate anything before the reverse process of diﬀerentiating something else has yielded that expression which you want to integrate. No one, even to-day, is able to ﬁnd the general integral of the expression, dy 2 = a−x , dx because a−x has never yet been found to result from diﬀerentiating anything else. Another simple case. Find (x + 1)(x + 2) dx. On looking at the function to be integrated, you remark that it is the product of two diﬀerent functions of x. You could, you think, integrate (x+1) dx by itself, or (x+2) dx by itself. Of course you could. But what to do with a product? None of the diﬀerentiations you have
2

CALCULUS MADE EASY

198

learned have yielded you for the diﬀerential coeﬃcient a product like this. Failing such, the simplest thing is to multiply up the two functions, and then integrate. This gives us (x2 + 3x + 2) dx. And this is the same as x2 dx + 3x dx + 2 dx.

And performing the integrations, we get
1 3 x 3

+ 3 x2 + 2x + C. 2

Some other Integrals. Now that we know that integration is the reverse of diﬀerentiation, we may at once look up the diﬀerential coeﬃcients we already know, and see from what functions they were derived. This gives us the following integrals ready made: x−1 (p. 145); x−1 dx = log x + C.

1 (p. 145); x+a x 1 dx = log (x + a) + C. x+a x (p. 139);

dx dx

=

x

+ C.
−x

−x

−x

=−

+C

SOME OTHER INTEGRALS

199

(for if y = − sin x

1

, x

dy =− dx

x

×0−1×
2x

x

=

−x

).

(p. 164);

sin x dx

= − cos x + C.

cos x (p. 163);

cos x dx = sin x + C.

Also we may deduce the following: log x; (for if y = x log x − x, log10 x; ax cos ax; (p. 146); log x dx = x(log x − 1) + C x dy = + log x − 1 = log x). dx x log10 x dx = 0.4343x(log x − 1) + C. ax dx ax + C. log a 1 cos ax dx = sin ax + C a =

dy (for if y = sin ax, = a cos ax; hence to get cos ax one must diﬀerendx 1 tiate y = sin ax). a sin ax; 1 sin ax dx = − cos ax + C. a

Try also cos2 θ; a little dodge will simplify matters: cos 2θ = cos2 θ − sin2 θ = 2 cos2 θ − 1; hence
1 cos2 θ = 2 (cos 2θ + 1),

CALCULUS MADE EASY

200

and

cos2 θ dθ = = =

1 2 1 2

(cos 2θ + 1) dθ cos 2θ dθ +
1 2

dθ.

sin 2θ θ + + C. (See also p. 225). 4 2

See also the Table of Standard Forms on pp. 249–251. You should make such a table for yourself, putting in it only the general functions which you have successfully diﬀerentiated and integrated. See to it that it grows steadily! On Double and Triple Integrals. In many cases it is necessary to integrate some expression for two or more variables contained in it; and in that case the sign of integration appears more than once. Thus, f (x, y, ) dx dy means that some function of the variables x and y has to be integrated for each. It does not matter in which order they are done. Thus, take the function x2 + y 2 . Integrating it with respect to x gives us: (x2 + y 2 ) dx = 1 x3 + xy 2 . 3 Now, integrate this with respect to y:
1 ( 3 x3 + xy 2 ) dy = 1 x3 y + 1 xy 3 , 3 3

DOUBLE INTEGRALS

201

to which of course a constant is to be added. If we had reversed the order of the operations, the result would have been the same. In dealing with areas of surfaces and of solids, we have often to integrate both for length and breadth, and thus have integrals of the form u · dx dy, where u is some property that depends, at each point, on x and on y. This would then be called a surface-integral. It indicates that the value of all such elements as u · dx · dy (that is to say, of the value of u over a little rectangle dx long and dy broad) has to be summed up over the whole length and whole breadth. Similarly in the case of solids, where we deal with three dimensions. Consider any element of volume, the small cube whose dimensions are dx dy dz. If the ﬁgure of the solid be expressed by the function f (x, y, z), then the whole solid will have the volume-integral, volume = f (x, y, z) · dx · dy · dz.

Naturally, such integrations have to be taken between appropriate limits∗ in each dimension; and the integration cannot be performed unless one knows in what way the boundaries of the surface depend on x, y, and z. If the limits for x are from x1 to x2 , those for y from y1 to y2 , and those for z from z1 to z2 , then clearly we have z2 y2 y1 x2 x1

volume = z1 ∗

f (x, y, z) · dx · dy · dz.

See p. 206 for integration between limits.

CALCULUS MADE EASY

202

There are of course plenty of complicated and diﬃcult cases; but, in general, it is quite easy to see the signiﬁcance of the symbols where they are intended to indicate that a certain integration has to be performed over a given surface, or throughout a given solid space. Exercises XVII. (1) Find (2) Find (4) Find (6) Find (7) If (See p. 264 for the Answers.)

y dx when y 2 = 4ax. 3 dx. x4 (x2 + a) dx. (4x3 + 3x2 + 2x + 1) dx. (3) Find 1 3 x dx. a
7

(5) Integrate 5x− 2 .

dy ax bx2 cx3 = + + ; ﬁnd y. dx 2 3 4 x2 + a x+a dx. (9) Find (x + 3)3 dx.

(8) Find (10) Find (11) Find (12) Find (13) Find

(x + 2)(x − a) dx. √ √ ( x + 3 x)3a2 dx.
1 (sin θ − 2 )

dθ . 3 (14) Find sin2 θ dθ.

cos2 aθ dθ.

SIMPLE INTEGRATIONS

203
3x

(15) Find (17) Find

sin2 aθ dθ. dx . 1+x

(16) Find (18) Find

dx.

dx . 1−x

CHAPTER XIX.
ON FINDING AREAS BY INTEGRATING.
One use of the integral calculus is to enable us to ascertain the values of areas bounded by curves. Let us try to get at the subject bit by bit.

Y Q

B

P A y1 M O x1 x2 N y2

Fig. 52.
Let AB (Fig. 52) be a curve, the equation to which is known. That is, y in this curve is some known function of x. Think of a piece of the curve from the point P to the point Q.

FINDING AREAS BY INTEGRATING

205

Let a perpendicular P M be dropped from P , and another QN from the point Q. Then call OM = x1 and ON = x2 , and the ordinates P M = y1 and QN = y2 . We have thus marked out the area P QN M that lies beneath the piece P Q. The problem is, how can we calculate the value of this area? The secret of solving this problem is to conceive the area as being divided up into a lot of narrow strips, each of them being of the width dx. The smaller we take dx, the more of them there will be between x1 and x2 . Now, the whole area is clearly equal to the sum of the areas of all such strips. Our business will then be to discover an expression for the area of any one narrow strip, and to integrate it so as to add together all the strips. Now think of any one of the strips. It will be like this: being bounded between two vertical sides, with a ﬂat bottom dx, and with a slightly curved sloping top. Suppose we take its average height as being y; then, as its width is dx, its area will be y dx. And seeing that we may take the width as narrow as we please, if we only take it narrow enough its average height will be the same as the height at the middle of it. Now let us call the unknown value of the whole area S, meaning surface. The area of one strip will be simply a bit of the whole area, and may therefore be called dS. So we may write area of 1 strip = dS = y · dx. If then we add up all the strips, we get total area S = dS = y dx.

CALCULUS MADE EASY

206

the particular case, when we know what the value of y is as a function of x.

So then our ﬁnding S depends on whether we can integrate y ·dx for

For instance, if you were told that for the particular curve in question y = b + ax2 , no doubt you could put that value into the expression and say: then I must ﬁnd (b + ax2 ) dx. That is all very well; but a little thought will show you that something more must be done. Because the area we are trying to ﬁnd is not the area under the whole length of the curve, but only the area limited on the left by P M , and on the right by QN , it follows that we must do something to deﬁne our area between those ‘limits.’ This introduces us to a new notion, namely that of integrating between limits. We suppose x to vary, and for the present purpose we do not require any value of x below x1 (that is OM ), nor any value of x above x2 (that is ON ). When an integral is to be thus deﬁned between two limits, we call the lower of the two values the inferior limit, and the upper value the superior limit. Any integral so limited we designate as a deﬁnite integral, by way of distinguishing it from a general integral to which no limits are assigned. In the symbols which give instructions to integrate, the limits are marked by putting them at the top and bottom respectively of the sign of integration. Thus the instruction x=x2 x=x1

y · dx

the superior limit x2 .

will be read: ﬁnd the integral of y · dx between the inferior limit x1 and

FINDING AREAS BY INTEGRATING

207

Sometimes the thing is written more simply x2 x1

y · dx.

Well, but how do you ﬁnd an integral between limits, when you have got these instructions? Look again at Fig. 52 (p. 204). Suppose we could ﬁnd the area under the larger piece of curve from A to Q, that is from x = 0 to x = x2 , naming the area AQN O. Then, suppose we could ﬁnd the area under the smaller piece from A to P , that is from x = 0 to x = x1 , namely the area AP M O. If then we were to subtract the smaller area from the larger, we should have left as a remainder the area P QN M , which is what we want. Here we have the clue as to what to do; the deﬁnite integral between the two limits is the diﬀerence between the integral worked out for the superior limit and the integral worked out for the lower limit. Let us then go ahead. First, ﬁnd the general integral thus: y dx, and, as y = b + ax2 is the equation to the curve (Fig. 52), (b + ax2 ) dx is the general integral which we must ﬁnd. Doing the integration in question by the rule (p. 193), we get a bx + x3 + C; 3 and this will be the whole area from 0 up to any value of x that we may assign.

CALCULUS MADE EASY

208

Therefore, the larger area up to the superior limit x2 will be a bx2 + x3 + C; 3 2 and the smaller area up to the inferior limit x1 will be a bx1 + x3 + C. 3 1 Now, subtract the smaller from the larger, and we get for the area S the value, a area S = b(x2 − x1 ) + (x3 − x3 ). 1 3 2 This is the answer we wanted. Let us give some numerical values.

Suppose b = 10, a = 0.06, and x2 = 8 and x1 = 6. Then the area S is equal to 10(8 − 6) + 0.06 3 (8 − 63 ) 3 = 20 + 0.02(512 − 216) = 20 + 5.92 = 25.92. Let us here put down a symbolic way of stating what we have ascertained about limits: x=x2 x=x1

= 20 + 0.02 × 296

y dx = y2 − y1 ,

where y2 is the integrated value of y dx corresponding to x2 , and y1 that corresponding to x1 .

FINDING AREAS BY INTEGRATING

209

All integration between limits requires the diﬀerence between two values to be thus found. Also note that, in making the subtraction the added constant C has disappeared. Examples. (1) To familiarize ourselves with the process, let us take a case of which we know the answer beforehand. Let us ﬁnd the area of the triangle (Fig. 53), which has base x = 12 and height y = 4. We know beforehand, from obvious mensuration, that the answer will come 24. y 4 O 12 x

Fig. 53.
Now, here we have as the “curve” a sloping line for which the equation is y= The area in question will be x=12 x=0 x=12

x . 3

y · dx =

x=0

x · dx. 3

x dx (p. 192), and putting down the value of the general 3 integral in square brackets with the limits marked above and below, we Integrating

CALCULUS MADE EASY

210

get 1 1 2 · x area = 3 2 x=12 x=12

+C x=0 x2 +C 6 x=0 122 02 = − 6 6 144 = 24. Ans. = 6 = Let us satisfy ourselves about this rather surprising dodge of calculation, by testing it on a simple example. Get some squared paper, preferably some that is ruled in little squares of one-eighth inch or oney 5 4 3 2 1 O 3 6 9 12 x

Fig. 54.

tenth inch each way. On this squared paper plot out the graph of the equation, y= The values to be plotted will be: x y 0 0 3 1 6 2 9 3 12 4 x . 3

FINDING AREAS BY INTEGRATING

211

The plot is given in Fig. 54. Now reckon out the area beneath the curve by counting the little squares below the line, from x = 0 as far as x = 12 on the right. There are 18 whole squares and four triangles, each of which has an area equal to 1 1 squares; or, in total, 24 squares. Hence 24 is the numerical value 2 x of the integral of dx between the lower limit of x = 0 and the higher 3 limit of x = 12. As a further exercise, show that the value of the same integral between the limits of x = 3 and x = 15 is 36.
Y

b a

a

O

x

X

Fig. 55.

(2) Find the area, between limits x = x1 and x = 0, of the curve b . y= x+a x=x1 x=x1

Area = x=0 y · dx = x1 0

x=0

b dx x+a

= b log (x + a)

+C

= b log (x1 + a) − log (0 + a)

CALCULUS MADE EASY

212

x1 + a . Ans. a N.B.—Notice that in dealing with deﬁnite integrals the constant C = b log always disappears by subtraction. Let it be noted that this process of subtracting one part from a larger to ﬁnd the diﬀerence is really a common practice. How do you ﬁnd the area of a plane ring (Fig. 56), the outer radius of which is r2

r2

r1

Fig. 56.

and the inner radius is r1 ? You know from mensuration that the area
2 2 of the outer circle is πr2 ; then you ﬁnd the area of the inner circle, πr1 ;

then you subtract the latter from the former, and ﬁnd area of ring
2 2 = π(r2 − r1 ); which may be written

π(r2 + r1 )(r2 − r1 ) = mean circumference of ring × width of ring. (3) Here’s another case—that of the die-away curve (p. 153). Find

the area between x = 0 and x = a, of the curve (Fig. 57) whose equation is y=b Area = b x=0 −x

.
−x

x=a

· dx.

FINDING AREAS BY INTEGRATING

213

The integration (p. 198) gives =b − =b −
−x a 0 −a

= b(1 −
Y

−a

− (− ).

−0

)

p

p1 b p2 O
O a X

v1 v2

v

Fig. 57.

Fig. 58.

(4) Another example is aﬀorded by the adiabatic curve of a perfect gas, the equation to which is pv n = c, where p stands for pressure, v for volume, and n is of the value 1.42 (Fig. 58). Find the area under the curve (which is proportional to the work done in suddenly compressing the gas) from volume v2 to volume v1 . Here we have v=v2 area = v=v1 cv −n · dv v 2 1 v 1−n 1−n v1 1 1−n =c (v 1−n − v1 ) 1−n 2 −c 1 1 = − 0.42 . 0.42 0.42 v2 v1

=c

CALCULUS MADE EASY

214

An Exercise. Prove the ordinary mensuration formula, that the area A of a circle whose radius is R, is equal to πR2 .

dr r R

Fig. 59.

Consider an elementary zone or annulus of the surface (Fig. 59), of breadth dr, situated at a distance r from the centre. We may consider the entire surface as consisting of such narrow zones, and the whole area A will simply be the integral of all such elementary zones from centre to margin, that is, integrated from r = 0 to r = R. We have therefore to ﬁnd an expression for the elementary area dA of the narrow zone. Think of it as a strip of breadth dr, and of a length that is the periphery of the circle of radius r, that is, a length of 2πr. Then we have, as the area of the narrow zone, dA = 2πr dr. Hence the area of the whole circle will be: r=R r=R

A=

dA = r=0 2πr · dr = 2π

r=0

r · dr.

FINDING AREAS BY INTEGRATING
1 Now, the general integral of r · dr is 2 r2 . Therefore,

215

A = 2π or whence Another Exercise. A = 2π

1 2 r=R r r=0 ; 2 1 2 R 2 1 − 2 (0)2 ;

A = πR2 .

Let us ﬁnd the mean ordinate of the positive part of the curve y = x − x2 , which is shown in Fig. 60. To ﬁnd the mean ordinate, we
Y M
1/4

N O 1

Fig. 60.

shall have to ﬁnd the area of the piece OM N , and then divide it by the length of the base ON . But before we can ﬁnd the area we must ascertain the length of the base, so as to know up to what limit we are to integrate. At N the ordinate y has zero value; therefore, we must look at the equation and see what value of x will make y = 0. Now, clearly, if x is 0, y will also be 0, the curve passing through the origin O; but also, if x = 1, y = 0; so that x = 1 gives us the position of the point N .

CALCULUS MADE EASY

216

Then the area wanted is x=1 = x=0 (x − x2 ) dx
1 − 3 x3 1 3 1 0

= =

1 2 x 2 1 2

−

− [0 − 0]

1 = 6.

But the base length is 1.
1 Therefore, the average ordinate of the curve = 6 .

[N.B.—It will be a pretty and simple exercise in maxima and minima to ﬁnd by diﬀerentiation what is the height of the maximum ordinate. It must be greater than the average.] The mean ordinate of any curve, over a range from x = 0 to x = x1 , is given by the expression, 1 x=x1 y · dx. x1 x=0 One can also ﬁnd in the same way the surface area of a solid of mean y = revolution. Example. of the surface generated by the curve between x = 0 and x = 6. The curve y = x2 − 5 is revolving about the axis of x. Find the area A point on the curve, the ordinate of which is y, describes a circumference of length 2πy, and a narrow belt of the surface, of width dx, corresponding to this point, has for area 2πy dx. The total area is x=6 x=6

2π x=0 y dx = 2π x=0 (x2 − 5) dx = 2π

x3 − 5x 3

6 0

= 6.28 × 42 = 263.76.

FINDING AREAS BY INTEGRATING

217

Areas in Polar Coordinates. When the equation of the boundary of an area is given as a function of the distance r of a point of it from a ﬁxed point O (see Fig. 61) called the pole, and of the angle which r makes with the positive horizontal
B A

dθ

r

θ O X

Fig. 61.

direction OX, the process just explained can be applied just as easily, with a small modiﬁcation. Instead of a strip of area, we consider a small triangle OAB, the angle at O being dθ, and we ﬁnd the sum of all the little triangles making up the required area. AB The area of such a small triangle is approximately × r or 2 r dθ × r; hence the portion of the area included between the curve and 2 two positions of r corresponding to the angles θ1 and θ2 is given by
1 2 θ=θ2 θ=θ1

r2 dθ.

Examples. (1) Find the area of the sector of 1 radian in a circumference of radius a inches.

CALCULUS MADE EASY

218

The polar equation of the circumference is evidently r = a. The area is
1 2 θ=θ2 θ=θ1

a2 dθ =

a2 2

θ=1

dθ = θ=0 a2 . 2

(2) Find the area of the ﬁrst quadrant of the curve (known as “Pascal’s Snail”), the polar equation of which is r = a(1 + cos θ). Area = =
1 2 θ= π 2

a2 (1 + cos θ)2 dθ (1 + 2 cos θ + cos2 θ) dθ π 2

a 2

2

θ=0 θ= π 2 θ=0

a θ sin 2θ = θ + 2 sin θ + + 2 2 4 2 a (3π + 8) . = 8

2

0

Volumes by Integration. What we have done with the area of a little strip of a surface, we can, of course, just as easily do with the volume of a little strip of a solid. We can add up all the little strips that make up the total solid, and ﬁnd its volume, just as we have added up all the small little bits that made up an area to ﬁnd the ﬁnal area of the ﬁgure operated upon.

Examples. (1) Find the volume of a sphere of radius r. A thin spherical shell has for volume 4πx2 dx (see Fig. 59, p. 214); summing up all the concentric shells which make up the sphere, we

FINDING AREAS BY INTEGRATING

219

have volume sphere =

x=r x=0

x3 4πx dx = 4π 3
2

r

= 4 πr3 . 3
0

y

dx x O

y

x

Fig. 62.

We can also proceed as follows: a slice of the sphere, of thickness dx, has for volume πy 2 dx (see Fig. 62). Also x and y are related by the expression y 2 = r 2 − x2 . x=r Hence volume sphere = 2 x=0 π(r2 − x2 ) dx x=r x=r

= 2π x=0 r2 dx − r 0

x2 dx x=0 x3 = 2π r2 x − 3

4π 3 = r . 3

(2) Find the volume of the solid generated by the revolution of the curve y 2 = 6x about the axis of x, between x = 0 and x = 4.

CALCULUS MADE EASY

220

The volume of a strip of the solid is πy 2 dx. x=4 x=4

Hence

volume = x=0 πy dx = 6π x=0 2

x dx

= 6π

x2 2

4

= 48π = 150.8.
0

On Quadratic Means. In certain branches of physics, particularly in the study of alternating electric currents, it is necessary to be able to calculate the quadratic mean of a variable quantity. By “quadratic mean” is denoted the square root of the mean of the squares of all the values between the limits considered. Other names for the quadratic mean of any quantity are its “virtual” value, or its “r.m.s.” (meaning root-mean-square) value. The French term is valeur eﬃcace. If y is the function under consideration, and the quadratic mean is to be taken between the limits of x = 0 and x = l; then the quadratic mean is expressed as
2

1 l

l

y 2 dx.
0

Examples. (1) To ﬁnd the quadratic mean of the function y = ax (Fig. 63). l Here the integral is
0

1 a2 x2 dx, which is 3 a2 l3 .

Dividing by l and taking the square root, we have 1 quadratic mean = √ al. 3

FINDING AREAS BY INTEGRATING
Y

221

y

O l

X

Fig. 63.

Here the arithmetical mean is 1 al; and the ratio of quadratic to 2 2 arithmetical mean (this ratio is called the form-factor ) is √ = 1.155. 3 (2) To ﬁnd the quadratic mean of the function y = xa . x=l l2a+1 The integral is x2a dx, that is . 2a + 1 x=0 Hence quadratic mean =
2

l2a . 2a + 1 x (3) To ﬁnd the quadratic mean of the function y = a 2 . x=l The integral is x=0 (a 2 )2 dx, that is x=0 x

x=l

ax dx, ,

or which is al − 1 . log a

ax log a

x=l x=0

Hence the quadratic mean is Exercises XVIII.

2

al − 1 . l log a

(See p. 265 for Answers.)

x = 6, and the mean ordinates between these limits.

(1) Find the area of the curve y = x2 + x − 5 between x = 0 and

CALCULUS MADE EASY

222

√ (2) Find the area of the parabola y = 2a x between x = 0 and x = a. Show that it is two-thirds of the rectangle of the limiting ordinate and of its abscissa. (3) Find the area of the positive portion of a sine curve and the mean ordinate. (4) Find the area of the positive portion of the curve y = sin2 x, and ﬁnd the mean ordinate. (5) Find the area included between the two branches of the curve of the lower branch of the curve (see Fig. 30, p. 106). y = x2 ± x 2 from x = 0 to x = 1, also the area of the positive portion (6) Find the volume of a cone of radius of base r, and of height h. (7) Find the area of the curve y = x3 − log x between x = 0 and √ 1 + x2 , as it
5

x = 1.

(8) Find the volume generated by the curve y = revolves about the axis of x, between x = 0 and x = 4.

(9) Find the volume generated by a sine curve revolving about the axis of x. Find also the area of its surface. (10) Find the area of the portion of the curve xy = a included between x = 1 and x = a. Find the mean ordinate between these limits. (11) Show that the quadratic mean of the function y = sin x, between √ 2 the limits of 0 and π radians, is . Find also the arithmetical mean 2 of the same function between the same limits; and show that the formfactor is = 1.11.

FINDING AREAS BY INTEGRATING

223

(12) Find the arithmetical and quadratic means of the function x2 + 3x + 2, from x = 0 to x = 3. (13) Find the quadratic mean and the arithmetical mean of the function y = A1 sin x + A1 sin 3x. (14) A certain curve has the equation y = 3.42
0.21x

. Find the area

included between the curve and the axis of x, from the ordinate at x = 2 to the ordinate at x = 8. Find also the height of the mean ordinate of the curve between these points. (15) Show that the radius of a circle, the area of which is twice the area of a polar diagram, is equal to the quadratic mean of all the values of r for that polar diagram. (16) Find the volume generated by the curve y = ± x 6 x(10 − x)

rotating about the axis of x.

CHAPTER XX.
DODGES, PITFALLS, AND TRIUMPHS.
Dodges. A great part of the labour of integrating things consists in licking them into some shape that can be integrated. The books—and by this is meant the serious books—on the Integral Calculus are full of plans and methods and dodges and artiﬁces for this kind of work. The following are a few of them. Integration by Parts. This name is given to a dodge, the formula for which is u dx = ux − that if in any case x du + C.

It is useful in some cases that you can’t tackle directly, for it shows x du can be found, then u dx can also be found. The formula can be deduced as follows. From p. 37, we have, d(ux) = u dx + x du, which may be written u(dx) = d(ux) − x du, which by direct integration gives the above expression.

DODGES, PITFALLS, AND TRIUMPHS

225

Examples. (1) Find du = dw, while Write u = w, and for sin w · dw write dx. We shall then have Putting these into the formula, we get sin w · dw = − cos w = x. w · sin w dw.

w · sin w dw = w(− cos w) −

− cos w dw

= −w cos w + sin w + C. (2) Find Write then and x x x

x

dx. u = x, du = dx, dx = x =x x x x

dx = dv; v= x ,

− x x

dx (by the formula) x −

=

(x − 1) + C.

(3) Try

cos2 θ dθ. u = cos θ, cos θ dθ = dv. v = sin θ, sin2 θ dθ

Hence

du = − sin θ dθ, cos2 θ dθ = cos θ sin θ + =

2 cos θ sin θ + (1 − cos2 θ) dθ 2 sin 2θ = + dθ − cos2 θ dθ. 2

CALCULUS MADE EASY

226

Hence and

2

sin 2θ +θ 2 sin 2θ θ + + C. cos2 θ dθ = 4 2 cos2 θ dθ =

(4) Find Write then

x2 sin x dx. x2 = u, du = 2x dx, sin x dx = dv; v = − cos x, x cos x dx.

x2 sin x dx = −x2 cos x + 2 Now ﬁnd

x cos x dx, integrating by parts (as in Example 1 above): x cos x dx = x sin x + cos x + C.

Hence x2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + C = 2 x sin x + cos x 1 − (5) Find Write then √ 1 − x2 dx. √ 1 − x2 , dx = dv; (see Chap. IX., p. 66) x2 2 +C .

u=

du = − √

x dx 1 − x2

DODGES, PITFALLS, AND TRIUMPHS

227

and x = v; so that √ √ 1 − x2 dx = x 1 − x2 + x2 dx √ . 1 − x2 x2 dx √ . 1 − x2 x2 dx √ , and we 1 − x2

Here we may use a little dodge, for we can write √ 1 − x2 dx = (1 − x2 ) dx √ = 1 − x2 dx √ − 1 − x2

Adding these two last equations, we get rid of have 2 √ √ 1 − x2 dx = x 1 − x2 + √

dx . 1 − x2

dx Do you remember meeting √ ? it is got by diﬀerentiating 1 − x2 y = arc sin x (see p. 168); hence its integral is arc sin x, and so √ √ x 1 − x2 1 2 dx = 1−x + 2 arc sin x + C. 2 You can try now some exercises by yourself; you will ﬁnd some at the end of this chapter. Substitution. This is the same dodge as explained in Chap. IX., p. 66. Let us illustrate its application to integration by a few examples. √ (1) 3 + x dx. Let replace 3 + x = u,
1 3

dx = du;
3

2 2 u 2 du = 3 u 2 = 3 (3 + x) 2 .

CALCULUS MADE EASY

228

(2) Let so that

dx . x + −x x = u,

dx = x + −x

du du = x , and dx = x ; dx du du = = x ( x + −x ) 1 u u+ u

du . u2 + 1

du is the result of diﬀerentiating arc tan x. 1 + u2 Hence the integral is arc tan x . dx dx dx √ . = = (3) 2 + 2x + 3 2 + 2x + 1 + 2 x x (x + 1)2 + ( 2)2 Let x + 1 = u, dx = du;

then the integral becomes diﬀerentiating u =

1 u arc tan . a a 1 x+1 Hence one has ﬁnally √ arc tan √ for the value of the given 2 2 integral. Formulæ of Reduction are special forms applicable chieﬂy to binomial and trigonometrical expressions that have to be integrated, and have to be reduced into some form of which the integral is known. Rationalization, and Factorization of Denominator are dodges applicable in special cases, but they do not admit of any short or general explanation. Much practice is needed to become familiar with these preparatory processes. The following example shows how the process of splitting into partial fractions, which we learned in Chap. XIII., p. 118, can be made use of in integration.

du du √ ; but 2 is the result of u + a2 u2 + ( 2)2

DODGES, PITFALLS, AND TRIUMPHS

229

dx 1 ; if we split 2 into partial fracx2 + 2x + 3 x + 2x + 3 tions, this becomes (see p. 230): Take again dx √ x + 1 + −2 √ 1 x + 1 − −2 √ . = √ log 2 −2 x + 1 + −2 Notice that the same integral can be expressed sometimes in more than one way (which are equivalent to one another). Pitfalls. A beginner is liable to overlook certain points that a practised hand would avoid; such as the use of factors that are equivalent to either zero or inﬁnity, and the occurrence of indeterminate quantities such as 0 . There is no golden rule that will meet every possible case. 0 Nothing but practice and intelligent care will avail. An example of a pitfall which had to be circumvented arose in Chap. XVIII., p. 189, when we came to the problem of integrating x−1 dx. Triumphs. By triumphs must be understood the successes with which the calculus has been applied to the solution of problems otherwise intractable. Often in the consideration of physical relations one is able to build up an expression for the law governing the interaction of the parts or of the forces that govern them, such expression being naturally in the form of a diﬀerential equation, that is an equation containing diﬀerential coeﬃcients with or without other algebraic quantities. And when such a diﬀerential equation has been found, one can get no further until it has been integrated. Generally it is much easier to state the appropriate diﬀerential equation than to solve it:— the real trouble begins then only when one wants to integrate, unless 1 √ 2 −2 dx √ − x + 1 − −2

CALCULUS MADE EASY

230

indeed the equation is seen to possess some standard form of which the integral is known, and then the triumph is easy. The equation which results from integrating a diﬀerential equation is called∗ its “solution”; and it is quite astonishing how in many cases the solution looks as if it had no relation to the diﬀerential equation of which it is the integrated form. The solution often seems as diﬀerent from the original expression as a butterﬂy does from the caterpillar that it was. Who would have supposed that such an innocent thing as 1 dy = 2 dx a − x2 could blossom out into y= 1 a+x log + C? 2a a−x

yet the latter is the solution of the former. As a last example, let us work out the above together. By partial fractions, a2 1 1 1 = + , 2 −x 2a(a + x) 2a(a − x) dx dx dy = + , 2a(a + x) 2a(a − x) 1 dx dx y= + 2a a+x a−x

∗

This means that the actual result of solving it is called its “solution.” But many

mathematicians would say, with Professor Forsyth, “every diﬀerential equation is considered as solved when the value of the dependent variable is expressed as a function of the independent variable by means either of known functions, or of integrals, whether the integrations in the latter can or cannot be expressed in terms of functions already known.”

DODGES, PITFALLS, AND TRIUMPHS

231

=

1 (log (a + x) − log (a − x)) 2a 1 a+x = log + C. 2a a−x

Not a very diﬃcult metamorphosis! There are whole treatises, such as Boole’s Diﬀerential Equations, devoted to the subject of thus ﬁnding the “solutions” for diﬀerent original forms. Exercises XIX. (1) Find (3) Find (5) Find (7) Find (9) Find √ (See p. 266 for Answers.) (2) Find (4) Find (6) Find (8) Find x log x dx. x a2 − x2 dx.

xa log x dx. 1 cos(log x) dx. x (log x)a dx. x 5x + 1 dx. 2+x−2 x x2 b dx . − a2

cos x x

dx.

x2

dx.

dx . x log x (x2 − 3) dx . x3 − 7x + 6 4x dx . x4 − 1 dx . x a − bx2 √

(10) Find (12) Find (14) Find

(11) Find (13) Find

dx . 1 − x4

CHAPTER XXI.
FINDING SOME SOLUTIONS.
In this chapter we go to work ﬁnding solutions to some important diﬀerential equations, using for this purpose the processes shown in the preceding chapters. The beginner, who now knows how easy most of those processes are in themselves, will here begin to realize that integration is an art. As in all arts, so in this, facility can be acquired only by diligent and regular practice. He who would attain that facility must work out examples, and more examples, and yet more examples, such as are found abundantly in all the regular treatises on the Calculus. Our purpose here must be to aﬀord the briefest introduction to serious work.

Example 1. Find the solution of the diﬀerential equation ay + b Transposing we have b dy = 0. dx

dy = −ay. dx

FINDING SOLUTIONS

233

Now the mere inspection of this relation tells us that we have got dy is proportional to y. If we think of the to do with a case in which dx curve which will represent y as a function of x, it will be such that its slope at any point will be proportional to the ordinate at that point, and will be a negative slope if y is positive. So obviously the curve will be a die-away curve (p. 153), and the solution will contain work. As both y and dy occur in the equation and on opposite sides, we can do nothing until we get both y and dy to one side, and dx to the other. To do this, we must split our usually inseparable companions dy and dx from one another. a dy = − dx. y b Having done the deed, we now can see that both sides have got into 1 dy a shape that is integrable, because we recognize , or dy, as a diﬀery y ential that we have met with (p. 143) when diﬀerentiating logarithms. So we may at once write down the instructions to integrate, dy a = − dx; y b and doing the two integrations, we have: a log y = − x + log C, b where log C is the yet undetermined constant∗ of integration. Then,
∗

−x

as

a factor. But, without presuming on this bit of sagacity, let us go to

We may write down any form of constant as the “constant of integration,” and

the form log C is adopted here by preference, because the other terms in this line of equation are, or are treated as logarithms; and it saves complications afterward if the added constant be of the same kind.

CALCULUS MADE EASY

234

delogarizing, we get: y=C
−ax b

,

which is the solution required. Now, this solution looks quite unlike the original diﬀerential equation from which it was constructed: yet to an expert mathematician they both convey the same information as to the way in which y depends on x. Now, as to the C, its meaning depends on the initial value of y. For if we put x = 0 in order to see what value y then has, we ﬁnd that this makes y = C solution as y = y0
−ax b −0

; and as

−0

= 1 we see that C is nothing else than the

particular value∗ of y at starting. This we may call y0 , and so write the .

Example 2. Let us take as an example to solve ay + b dy = g, dx

where g is a constant. Again, inspecting the equation will suggest, (1) that somehow or other x will come into the solution, and (2) that

if at any part of the curve y becomes either a maximum or a minimum, dy g so that = 0, then y will have the value = . But let us go to work dx a as before, separating the diﬀerentials and trying to transform the thing
∗

Compare what was said about the “constant of integration,” with reference to

Fig. 48 on p. 184, and Fig. 51 on p. 187.

FINDING SOLUTIONS

235

into some integrable shape. dy = g − ay; dx dy a g = −y ; dx b a dy a g = − b dx. y− a b Now we have done our best to get nothing but y and dy on one side, and nothing but dx on the other. But is the result on the left side integrable? It is of the same form as the result on p. 145; so, writing the instructions to integrate, we have: g =− y− a dy a dx; b

and, doing the integration, and adding the appropriate constant, log whence and ﬁnally, which is the solution. If the condition is laid down that y = 0 when x = 0 we can ﬁnd C; for then the exponential becomes = 1; and we have g + C, a g C=− . a 0= y− g a = − x + log C; a b a g y − = C − b x; a a g y = + C − b x, a

or

CALCULUS MADE EASY

236

Putting in this value, the solution becomes g y = (1 − a
−ax b

).

But further, if x grows indeﬁnitely, y will grow to a maximum; for g when x = ∞, the exponential = 0, giving ymax. = . Substituting this, a we get ﬁnally y = ymax. (1 −
−ax b

).

This result is also of importance in physical science.

Example 3. Let ay + b dy = g · sin 2πnt. dt

We shall ﬁnd this much less tractable than the preceding. First divide through by b. dy a g + y = sin 2πnt. dt b b Now, as it stands, the left side is not integrable. But it can be made so by the artiﬁce—and this is where skill and practice suggest a plan—of multiplying all the terms by dy dt which is the same as dy dt at b at b at b

, giving us: · sin 2πnt,

a + y b

at b

=

g b

at b

d( b t ) g +y = dt b

a

at b

· sin 2πnt;

FINDING SOLUTIONS

237

and this being a perfect diﬀerential may be integrated thus:—since, if a d( b t ) dy a t a t du b , b +y = , u=y dt dt dt y or at b

g b g y= b =

at b

· sin 2πnt · dt + C, at b

−at b

· sin 2πnt · dt + C

−at b

.

[a]

The last term is obviously a term which will die out as t increases, and may be omitted. The trouble now comes in to ﬁnd the integral that appears as a factor. To tackle this we resort to the device (see p. 224) of integration by parts, the general formula for which is uv − vdu. For this purpose write   u= at b

udv =

;

 dv = sin 2πnt · dt. We shall then have   du =  at b

 v = − 1 cos 2πnt.  2πn Inserting these, the integral in question becomes: at b

×

a dt; b

· sin 2πnt · dt =− a a 1 1 a · b t · cos 2πnt − − cos 2πnt · b t · dt 2πn 2πn b at 1 at a b cos 2πnt + b · cos 2πnt · dt. =− 2πn 2πnb

[b]

CALCULUS MADE EASY

238

The last integral is still irreducible. To evade the diﬃculty, repeat the integration by parts of the left side, but treating it in the reverse way by writing:   u = sin 2πnt;  dv = a t · dt; b   du = 2πn · cos 2πnt · dt;   v= b  a at b

whence

Inserting these, we get at b

· sin 2πnt · dt = b · a at b

· sin 2πnt −

2πnb a

at b

· cos 2πnt · dt.

[c]

Noting that the ﬁnal intractable integral in [c] is the same as that 2πnb in [b], we may eliminate it, by multiplying [b] by , and multiplya a ing [c] by , and adding them. 2πnb The result, when cleared down, is: at b

· sin 2πnt · dt =

at b

ab · sin 2πnt − 2πnb2 · cos 2πnt a2 + 4π 2 n2 b2

[d]

Inserting this value in [a], we get y=g a · sin 2πnt − 2πnb · cos 2πnt a2 + 4π 2 n2 b2 .

To simplify still further, let us imagine an angle φ such that tan φ =

FINDING SOLUTIONS

239

2πnb . a Then and 2πnb , + 4π 2 n2 b2 a . cos φ = √ 2 + 4π 2 n2 b2 a sin φ = √ a2

Substituting these, we get: y=g which may be written sin(2πnt − φ) y = g√ , a2 + 4π 2 n2 b2 which is the solution desired. This is indeed none other than the equation of an alternating electric current, where g represents the amplitude of the electromotive force, n the frequency, a the resistance, b the coeﬃcient of self-induction of the circuit, and φ is an angle of lag. cos φ · sin 2πnt − sin φ · cos 2πnt √ , a2 + 4π 2 n2 b2

Example 4. Suppose that M dx + N dy = 0.

We could integrate this expression directly, if M were a function of x only, and N a function of y only; but, if both M and N are functions that depend on both x and y, how are we to integrate it? Is it itself an exact diﬀerential? That is: have M and N each been formed by

CALCULUS MADE EASY

240

partial diﬀerentiation from some common function U , or not? If they   ∂U  = M,  ∂x  ∂U  = N.  ∂y And if such a common function exists, then ∂U ∂U dx + dy ∂x ∂y is an exact diﬀerential (compare p. 172). Now the test of the matter is this. If the expression is an exact diﬀerential, it must be true that dM dN = ; dy dx d(dU ) d(dU ) = , dx dy dy dx have, then

for then which is necessarily true.

Take as an illustration the equation (1 + 3xy) dx + x2 dy = 0.

Is this an exact diﬀerential or not? Apply the test.   d(1 + 3xy)  = 3x,  dy  d(x2 )   = 2x, dx which do not agree. Therefore, it is not an exact diﬀerential, and the two functions 1 + 3xy and x2 have not come from a common original function.

FINDING SOLUTIONS

241

It is possible in such cases to discover, however, an integrating factor, that is to say, a factor such that if both are multiplied by this factor, the expression will become an exact diﬀerential. There is no one rule for discovering such an integrating factor; but experience will usually suggest one. In the present instance 2x will act as such. Multiplying by 2x, we get (2x + 6x2 y) dx + 2x3 dy = 0. Now apply the test to this.   d(2x + 6x2 y)  = 6x2 ,  dy  d(2x3 )   = 6x2 , dx which agrees. Hence this is an exact diﬀerential, and may be integrated. Now, if w = 2x3 y, dw = 6x2 y dx + 2x3 dy. Hence so that we get 6x2 y dx + 2x3 dy = w = 2x3 y;

U = x2 + 2x3 y + C.

d2 y + n2 y = 0. dt2 In this case we have a diﬀerential equation of the second degree, in Example 5. Let which y appears in the form of a second diﬀerential coeﬃcient, as well as in person. Transposing, we have d2 y = −n2 y. dt2

CALCULUS MADE EASY

242

It appears from this that we have to do with a function such that its second diﬀerential coeﬃcient is proportional to itself, but with reversed sign. In Chapter XV. we found that there was such a function—namely, the sine (or the cosine also) which possessed this property. So, without further ado, we may infer that the solution will be of the form y = A sin(pt + q). However, let us go to work. Multiply both sides of the original equation by 2 giving us 2 d2 y dy dy + 2x2 y = 0, and, as 2 dt dt dt d2 y dy = dt2 dt d dy dt dt
2

dy and integrate, dt

2

,

dy dt

2

+ n2 (y 2 − C 2 ) = 0,

C being a constant. Then, taking the square roots, dy = −n y 2 − C 2 dt and dy C 2 − y2 y ) C ; = n · dt.

But it can be shown that (see p. 168) 1 C 2 − y2 y = nt + C1 C d(arc sin = dy

whence, passing from angles to sines, arc sin and y = C sin(nt + C1 ),

where C1 is a constant angle that comes in by integration. Or, preferably, this may be written y = A sin nt + B cos nt, which is the solution.

FINDING SOLUTIONS

243

d2 y − n2 y = 0. dt2 Here we have obviously to deal with a function y which is such Example 6. that its second diﬀerential coeﬃcient is proportional to itself. The only function we know that has this property is the exponential function (see p. 139), and we may be certain therefore that the solution of the equation will be of that form. Proceeding as before, by multiplying through by 2 ing, we get 2 dy d2 y dy − 2x2 y = 0, 2 dx dx dx
2

dy , and integratdx

dy 2 2 dy d y dy dx and, as 2 2 = , − n2 (y 2 + c2 ) = 0, dx dx dx dx dy − n y 2 + c2 = 0, dx dy where c is a constant, and = n dx. y 2 + c2 Now, if w = log (y + y 2 + c2 ) = log u, d dw 1 = , du u and du =1+ dy dw = dy y y 2 + c2 1 y 2 + c2 = . y+ y 2 + c2 y 2 + c2

Hence, integrating, this gives us log (y + y+ Now whence (y + y 2 + c2 ) = nx + log C, y 2 + c2 = C nx .

(1)

y 2 + c2 ) × (−y + −y + y 2 + c2 =

y 2 + c2 ) = c2 ; c2 C
−nx

.

(2)

CALCULUS MADE EASY

244

Subtracting (2) from (1) and dividing by 2, we then have 1 1 c2 y = C nx − 2 2C which is more conveniently written y=A nx −nx

,

+B

−nx

.

Or, the solution, which at ﬁrst sight does not look as if it had anything to do with the original equation, shows that y consists of two terms, one of which grows logarithmically as x increases, and of a second term which dies away as x increases.

Example 7. dy d2 y + a + gy = 0. 2 dt dt Examination of this expression will show that, if b = 0, it has the Let b form of Example 1, the solution of which was a negative exponential. On the other hand, if a = 0, its form becomes the same as that of Example 6, the solution of which is the sum of a positive and a negative exponential. It is therefore not very surprising to ﬁnd that the solution of the present example is y=( where m=
−mt

)(A

nt

+B

−nt

),

a a2 g and n = − . 2 2b 4b b The steps by which this solution is reached are not given here; they

may be found in advanced treatises.

FINDING SOLUTIONS

245

d2 y d2 y = a2 2 . dt2 dx It was seen (p. 174) that this equation was derived from the original y = F (x + at) + f (x − at), where F and f were any arbitrary functions of t. Another way of dealing with it is to transform it by a change of variables into d2 y = 0, du · dv where u = x + at, and v = x − at, leading to the same general solution. If we consider a case in which F vanishes, then we have simply y = f (x − at); and this merely states that, at the time t = 0, y is a particular function of x, and may be looked upon as denoting that the curve of the relation of y to x has a particular shape. Then any change in the value of t is equivalent simply to an alteration in the origin from which x is reckoned. That is to say, it indicates that, the form of the function being conserved, it is propagated along the x direction with a uniform velocity a; so that whatever the value of the ordinate y at any particular time t0 at any particular point x0 , the same value of y will appear at the subsequent time t1 at a point further along, the abscissa of which is x0 +a(t1 −t0 ). In this case the simpliﬁed equation represents x direction. the propagation of a wave (of any form) at a uniform speed along the

Example 8.

CALCULUS MADE EASY

246

If the diﬀerential equation had been written m d2 y d2 y = k 2, dt2 dx

the solution would have been the same, but the velocity of propagation would have had the value a= k . m

You have now been personally conducted over the frontiers into the enchanted land. And in order that you may have a handy reference to the principal results, the author, in bidding you farewell, begs to present you with a passport in the shape of a convenient collection of standard forms (see pp. 249–251). In the middle column are set down a number of the functions which most commonly occur. The results of diﬀerentiating them are set down on the left; the results of integrating them are set down on the right. May you ﬁnd them useful!

EPILOGUE AND APOLOGUE.
It may be conﬁdently assumed that when this tractate “Calculus made Easy” falls into the hands of the professional mathematicians, they will (if not too lazy) rise up as one man, and damn it as being a thoroughly bad book. Of that there can be, from their point of view, no possible manner of doubt whatever. It commits several most grievous and deplorable errors. First, it shows how ridiculously easy most of the operations of the calculus really are. Secondly, it gives away so many trade secrets. By showing you that what one fool can do, other fools can do also, it lets you see that these mathematical swells, who pride themselves on having mastered such an awfully diﬃcult subject as the calculus, have no such great reason to be puﬀed up. They like you to think how terribly diﬃcult it is, and don’t want that superstition to be rudely dissipated. Thirdly, among the dreadful things they will say about “So Easy” is this: that there is an utter failure on the part of the author to demonstrate with rigid and satisfactory completeness the validity of sundry methods which he has presented in simple fashion, and has even dared to use in solving problems! But why should he not? You don’t forbid the use of a watch to every person who does not know how

CALCULUS MADE EASY

248

to make one? You don’t object to the musician playing on a violin that he has not himself constructed. You don’t teach the rules of syntax to children until they have already become ﬂuent in the use of speech. It would be equally absurd to require general rigid demonstrations to be expounded to beginners in the calculus. One other thing will the professed mathematicians say about this thoroughly bad and vicious book: that the reason why it is so easy is because the author has left out all the things that are really diﬃcult. And the ghastly fact about this accusation is that—it is true! That is, indeed, why the book has been written—written for the legion of innocents who have hitherto been deterred from acquiring the elements of the calculus by the stupid way in which its teaching is almost always presented. Any subject can be made repulsive by presenting it bristling with diﬃculties. The aim of this book is to enable beginners to learn its language, to acquire familiarity with its endearing simplicities, and to grasp its powerful methods of solving problems, without being compelled to toil through the intricate out-of-the-way (and mostly irrelevant) mathematical gymnastics so dear to the unpractical mathematician. There are amongst young engineers a number on whose ears the adage that what one fool can do, another can, may fall with a familiar sound. They are earnestly requested not to give the author away, nor to tell the mathematicians what a fool he really is.

TABLE OF STANDARD FORMS. dy dx ←− −→

y

y dx

Algebraic. 1 0 1 a 2x nxn−1 −x−2 du dv dw ± ± dx dx dx dv du u +v dx dx du dv v −u dx dx v2 du dx x a x±a ax x2 xn x−1 u±v±w uv u v u
1 2 x 2

+C

ax + C
1 2 x 2

± ax + C +C +C

1 ax2 2 1 3 x 3

1 xn+1 + C n+1 log x + C u dx ± v dx ± w dx

No general form known No general form known ux − x du + C

CALCULUS MADE EASY

250

dy dx x ←−

y

−→ x y dx

Exponential and Logarithmic. x +C

x−1 0.4343 × x−1 ax log a

log x log10 x

0.4343x(log x − 1) + C ax ax +C log a Trigonometrical. − cos x + C sin x + C − log cos x + C √ 1 − x2 + C √ 1 − x2 + C

x(log x − 1) + C

cos x − sin x sec2 x 1 − (1 − x2 ) 1 (1 − x2 ) 1 1 + x2 cosh x sinh x sech2 x

sin x cos x tan x arc sin x arc cos x arc tan x

Circular (Inverse).

x · arc sin x + x · arc cos x −

x · arc tan x − 1 log (1 + x2 ) + C 2 cosh x + C sinh x + C log cosh x + C

Hyperbolic. sinh x cosh x tanh x

TABLE OF STANDARD FORMS

251

dy dx 1 (x + a)2 x
3 2

←−

y

−→ Miscellaneous.

y dx

− −

−

(a2 + x2 ) b (a ± bx)2 3a2 x

1 x+a 1 √ 2 + x2 a 1 a ± bx a2 (a2 + x2 ) 2 sin ax cos ax tan ax sin2 x cos2 x sinn x 1 sin x 1 sin2 x 1 sin x · cos x sin mx · sin nx sin2 ax cos2 ax
1 2
3

log (x + a) + C log (x + √ a2 + x2 ) + C

(a2 + x2 ) 2 a · cos ax

5

1 ± log (a ± bx) + C b x √ +C 2 + x2 a 1 − cos ax + C a 1 sin ax + C a 1 − log cos ax + C a x sin 2x − +C 2 4 x sin 2x + +C 2 4
− n−1 cos x sinn−1 x + n n sinn−2 x dx + C

−a · sin ax a · sec2 ax sin 2x − sin 2x n · sinn−1 x · cos x cos x sin2 x sin 2x − 4 sin x sin2 x − cos2 x sin2 x · cos2 x −

log tan

x +C 2

− cotan x + C log tan x + C cos(m − n)x − 1 cos(m + n)x + C 2 x sin 2ax − +C 2 4a x sin 2ax + +C 2 4a

n · sin mx · cos nx + m · sin nx · cos mx

2a · sin 2ax −2a · sin 2ax

ANSWERS.
Exercises I. (1) (4) (7) (9) dy = 13x12 . dx du = 2.4t1.4 . dt 8 13 du = − x− 5 . dx 5 dy 3 3−q = x q . dx q (2) (5) (p. 24.) (3) (6) dy = 2ax(2a−1) . dx dy 5 8 = − x− 3 . dx 3

dy 3 5 = − x− 2 . dx 2 dz 1 2 = u− 3 . du 3 (8) (10)

dy = 2axa−1 . dx dy m m+n = − x− n . dx n

Exercises II. (1) (4) (7) dy = 3ax2 . dx dy 1 1 1 = c 2 x− 2 . dx 2 (2) (5)

(p. 31.) (3) (6) dy 1 = 6x− 2 . dx dy = 2.36t. dt

dy 1 = 13 × 3 x 2 . 2 dx du an n−1 = z . dz c

dlt = 0.000012 × l0 . dt dC (8) = abV b−1 , 0.98, 3.00 and 7.47 candle power per volt respectively. dV (9) dn 1 =− dD LD2 dn 1 =− dσ 2DL gT dn 1 gT , =− , 2 πσ dL DL πσ gT dn 1 g , = . 3 πσ dT 2DL πσT

ANSWERS TO EXERCISES

253

D Rate of change of P when t varies =− . Rate of change of P when D varies t dD 0.000012lt (11) 2π, 2πr, πl, 2 πrh, 8πr, 4πr2 . (12) = . 3 dT π (10)

Exercises III. x2 x3 x4 (1) (a) 1 + x + + + + ... 2 6 24 (d ) 3x2 + 6ax + 3a2 . (2) dw = a − bt. dt (3)

(p. 45.) (b) 2ax + b. (c) 2x + 2a.

dy = 2x. dx

(4) 14110x4 − 65404x3 − 2244x2 + 8192x + 1379. (5) (7) (9) dx = 2y + 8. dy −5 . (3x + 2)2 ad − bc . (cx + d)2 (6) 185.9022654x2 + 154.36334. 6x4 + 6x3 + 9x2 . (1 + x + 2x2 )2 anx−n−1 + bnxn−1 + 2nx−1 . (x−n + b)2

(8) (10)

(11) b + 2ct. b R0 (a + 2bt) (12) R0 (a+2bt), R0 a + √ , − (1 + at + bt2 )2 2 t dE k =b+ , dl i dE c + kl =− 2 . di i or R2 (a + 2bt) . R0

(13) 1.4340(0.000014t − 0.001024), −0.00117, −0.00107, −0.00097. (14)

CALCULUS MADE EASY

254

Exercises IV. (1) 17 + 24x; 24. (2)

(p. 50.) x2 + 2ax − a ; (x + a)2 2a(a + 1) . (x + a)3

x2 x3 x2 + ; 1+x+ . 1×2 1×2×3 1×2 (4) (Exercises III.): (3) 1 + x + (1) (a) d3 y d2 y = 3 = 1 + x + 1 x2 + 1 x3 + . . .. 2 6 dx2 dx (b) 2a, 0. (c) 2, 0. (d ) 6x + 6a, 6. (3) 2, 0.

(2) −b, 0.

(4) 56440x3 − 196212x2 − 4488x + 8192. (5) 2, 0. (6) 371.80453x, 371.80453. 30 270 (7) , − . 3 (3x + 2) (3x + 2)4 (Examples, p. 40): 6a 6a (1) 2 x, . b b2 2 1.056 (3) √ − √ , 3 8 5 11 θ θ √ √ 3a b 6b 3 a (2) √ − , x3 2 x 2.3232 16 √ − √ . 5 16 3 11 θ 3 θ √ √ 18b 3 a 3a b − √ . x4 4 x3 169320x2 − 392424x − 4488.

(4) 810t4 − 648t3 + 479.52t2 − 139.968t + 26.64. (5) 12x + 2, 12. (6) 6x2 − 9x, 12x − 9. 3 1 1 1 15 1 √ +√ √ −√ (7) + . 5 7 4 4 θ θ θ θ3 3 1 1 15 7 1 √ −√ √ +√ − . 8 8 θ5 θ3 θ9 θ7 3240t3 − 1944t2 + 959.04t − 139.968.

ANSWERS TO EXERCISES

255

Exercises V. (2) 64; 147.2; and 0.32 feet per second. (3) x = a − gt; x = −g. ¨

(p. 63.)

(4) 45.1 feet per second.

(5) 12.4 feet per second per second. Yes. (6) Angular velocity = 11.2 radians per second; angular acceleration = 9.6 radians per second per second. (7) v = 20.4t2 − 10.8. a = 40.8t. 172.8 in./sec., 122.4 in./sec2 . (8) v = 1 30
3

(t −

125)2

, a=−

1

45

3

(9) v = 0.8 −

8t 24t2 − 32 , a= , 0.7926 and 0.00211. (4 + t2 )2 (4 + t2 )3

(t − 125)5

.

(10) n = 2, n = 11.

Exercises VI. (1) √ (4) x . +1 ax (a − x2 )3 . (2) √ x . + a2

(p. 72.) (3) − 1 2 (a + x)3 .

x2

x2

2a2 − x2 √ (5) . x 3 x 2 − a2

CALCULUS MADE EASY
3 2 x 2 8 x (x3 9

256

(6) (8)

(x4 + a) 3 (x3 + a) 2
5 3 y . 2

+ a) − (x4 + a)
2 3

(7)

2a (x − a) . (x + a)3 1 √ . (1 − θ) 1 − θ2

(9)

Exercises VII. (1) (2) 3x2 (3 + 3x3 ) dw = 3. dx 27 1 x3 + 1 x6 2 4 dv =− dx du =− dx 1+ √ 2+ √ 3 + x3 3x2 12x √ 3+4

(p. 74.)

√ 1 + 2 + 3x2

2.

(3)

x2

x3 1+ 1+ √ 3

2 3

Exercises VIII. (2) 1.44.

(p. 88.)

dy = 3x2 + 3; and the numerical values are: 3, 3 3 , 6, and 15. 4 dx √ (5) ± 2. (4) (6) dy 4x = − . Slope is zero where x = 0; and is dx 9y

1 √ where x = 1. 3 2

ANSWERS TO EXERCISES

257

(7) m = 4, n = −3. (8) Intersections at x = 1, x = −3. Angles 153◦ 26 , 2◦ 28 . (9) Intersection at x = 3.57, y = 3.50. Angle 16◦ 16 .
5 (10) x = 1 , y = 2 1 , b = − 3 . 3 3

Exercises IX.

(p. 107.)

(1) Min.: x = 0, y = 0; max.: x = −2, y = −4. (2) x = a. (5) √ (4) 25 3 square inches.

(6) Max. for x = −1; min. for x = 1. R , no max. 2

dy 10 10 =− 2 + ; x = 4; y = 5. dx x (8 − x)2

(7) Join the middle points of the four sides. (8) r = 2 R, r = 3 (9) r = R

2 R , r = √ , r = 0.8506R. 3 2 8 (10) At the rate of square feet per second. r √ R 8 NR (11) r = . (12) n = . 3 r

CALCULUS MADE EASY

258

Exercises X.

(p. 115.)

(1) Max.: x = −2.19, y = 24.19; min.:, x = 1.52, y = −1.38. (2) b d2 y b dy = − 2cx; 2 = −2c; x = (a maximum). dx a dx 2ac

(3) (a) One maximum and two minima. (b) One maximum. (x = 0; other points unreal.) (4) Min.: x = 1.71, y = 6.14. (6) Max.: x = 1.414, y = 1.7675. Min.: x = −1.414, y = 1.7675. (7) Max.: x = −3.565, y = 2.12. Min.: x = +3.565, y = 7.88. a . c (5) Max: x = −.5, y = 4.

(8) 0.4N , 0.6N .

(9) x =

(10) Speed 8.66 nautical miles per hour. Time taken 115.47 hours. Minimum cost £112. 12s. (11) Max. and min. for x = 7.5, y = ±5.414. (See example no. 10, p. 71.)
1 (12) Min.: x = 1 , y = 0.25; max.: x = − 3 , y = 1.408. 2

ANSWERS TO EXERCISES

259

Exercises XI. (1) (4) (6) 1 2 + . x−3 x+4 (2)

(p. 127.) (3) 2 1 + . x−3 x+4

5 4 − . x−4 x−3

1 2 + . x−1 x−2 (5)

19 22 − . 13(2x + 3) 13(3x − 2)

2 4 5 + − . x−2 x−3 x−4 1 11 1 (7) + + . 6(x − 1) 15(x + 2) 10(x − 3) (8)

7 71 5 + − . 9(3x + 1) 63(3x − 2) 7(2x + 1) 1 2x + 1 2 1 − 2x (9) + . (10) x + + . 3(x − 1) 3(x2 + x + 1) 3(x + 1) 3(x2 − x + 1) (11) (13) (14) (15) (16) (17) (18) 2x + 1 3 + 2 . (x + 1) x + x + 1 (12) 1 1 2 − + . x − 1 x − 2 (x − 2)2

1 1 1 − + . 4(x − 1) 4(x + 1) 2(x + 1)2

4 4 1 . − − 9(x − 1) 9(x + 2) 3(x + 2)2

1 x−1 1 − 2 − 2 . x + 2 x + x + 1 (x + x + 1)2 5 32 36 − + . 2 x + 4 (x + 4) (x + 4)3 7 55 73 + + . 9(3x − 2)2 9(3x − 2)3 9(3x − 2)4

x 1 1 + − . 2 2 + 2x + 4) 6(x − 2) 3(x − 2) 6(x

CALCULUS MADE EASY

260

Exercises XII. (1) ab( ax (p. 150.) (3) log n. 3 − x−1 (7) . (x − 1)2 x +

−ax

).

2 (2) 2at + . t (6) n . x . (9) axa−1 . xa + a

(5) npv n−1 .

(8) 6x

−5x

− 5(3x2 + 1)

−5x

(10) (11)

3x2

1 − log (x + 3) . (x + 3)2

1 6x + √ − 1 2 ( x + x)

√ (3x2 − 1) ( x + 1).

(12) ax (axa−1 + xa log a). (15) 1+x . x

(14) Min.: y = 0.7 for x = 0.694. (16) 3 (log ax)2 . x

Exercises XIII. (1) Let

(p. 160.)

t = x (∴ t = 8x), and use the Table on page 157. T

(2) T = 34.627; 159.46 minutes. (3) Take 2t = x; and use the Table on page 157. (5) (a) xx (1 + log x); (b) 2x( x )x ; (c) xx × xx (1 + log x).

ANSWERS TO EXERCISES

261

(6) 0.14 second. (8) µ = 0.00037, 31m 1 . 4

(7) (a) 1.642; (b) 15.58.

(9) i is 63.4% of i0 , 220 kilometres. (10) 0.133, 0.145, 0.155, mean 0.144; −10.2%, −0.9%, +77.2%. 1 (11) Min. for x = . (13) Min. for x = log a. (12) Max. for x = .

Exercises XIV. (1) (i)

(p. 170.)

dy π ; = A cos θ − dθ 2 dy dy (ii) = 2 sin θ cos θ = sin 2θ and = 2 cos 2θ; dθ dθ dy dy (iii) = 3 sin2 θ cos θ and = 3 cos 3θ. dθ dθ π dy (2) θ = 45◦ or radians. (3) = −n sin 2πnt. 4 dt (5) cos x = cotan x sin x

(4) ax log a cos ax . (6) 18.2 cos (x + 26◦ ).

CALCULUS MADE EASY

262

(7) The slope is

When θ = 75◦ the slope is 100 cos(75◦ − 15◦ ) = 100 cos 60◦ = 100 ×
1 2

dy = 100 cos (θ − 15◦ ), which is a maximum when dθ (θ −15◦ ) = 0, or θ = 15◦ ; the value of the slope being then = 100. = 50.

(8) cos θ sin 2θ + 2 cos 2θ sin θ = 2 sin θ cos2 θ + cos 2θ = 2 sin θ 3 cos2 θ − 1 . (9) amnθn−1 tanm−1 (θn ) sec2 θn . (10) x sin2 x + sin 2x ;

x

sin2 x + 2 sin 2x + 2 cos 2x .
−x b

(11) (i) (12) (i)

dy ab a = 2 ; (ii) dx b (x + b)

; (iii)

1◦ ab . × 2 90 (b + x2 )

dy = sec x tan x; dx 1 dy = −√ (ii) ; dx 1 − x2 dy 1 = ; (iii) dx 1 + x2 dy 1 (iv) = √ ; dx x x2 − 1 √ 3 sec x (3 sec2 x − 1) dy (v) = . dx 2

dy = 4.6 (2θ + 3)1.3 cos (2θ + 3)2.3 . dθ dy (14) = 3θ2 + 3 cos (θ + 3) − log 3 cos θ × 3sin θ + 3θ . dθ (13) (15) θ = cot θ; θ = ±0.86; is max. for +θ, min. for −θ.

ANSWERS TO EXERCISES

263

Exercises XV. (1) x3 − 6x2 y − 2y 2 ;
1 3

(p. 177.)

− 2x3 − 4xy.

(2) 2xyz + y 2 z + z 2 y + 2xy 2 z 2 ; 2xyz + x2 z + xz 2 + 2x2 yz 2 ; 2xyz + x2 y + xy 2 + 2x2 y 2 z. (3) (x + y + z) − (a + b + c) 3 1 {(x − a) + (y − b) + (z − c)} = ; . r r r

(4) dy = vuv−1 du + uv log u dv. (5) dy = 3 sin vu2 du + u3 cos v dv, dy = u sin xu−1 cos x dx + (sin x)u log sin xdu, 1 1 1 du − log u 2 dv. dy = v u v (7) Minimum for x = y = − 1 . 2 (8) (a) Length 2 feet, width = depth = 1 foot, vol. = 2 cubic feet. 2 (b) Radius = feet = 7.46 in., length = 2 feet, vol. = 2.54. π (9) All three parts equal; the product is maximum. (10) Minimum for x = y = 1. (11) Min.: x =
1 2

and y = 2. √ 3 2V .

(12) Angle at apex = 90◦ ; equal sides = length =

CALCULUS MADE EASY

264

Exercises XVI.
1 (1) 1 3 .

(p. 187.) (3) 0.2624.

(2) 0.6344.

(4) (a) y = 1 x2 + C; (b) y = sin x + C. 8 (5) y = x2 + 3x + C.

Exercises XVII. √ 3 4 ax 2 + C. (1) 3 (4)
1 3 x 3

(p. 202.) (3)
5

(2) −

1 + C. x3

x4 + C. 4a

+ ax + C.

(5) −2x− 2 + C. ax2 bx3 cx4 (7) + + + C. 4 9 16

(6) x4 + x3 + x2 + x + C.

(8)

x2 + a a2 + a = x−a+ by division. Therefore the answer is x+a x+a x2 − ax + (a2 + a) log (x + a) + C. (See pages 196 and 198.) 2 x4 27 x3 2 − a 2 + 3x3 + x2 + 27x + C. (10) + x − 2ax + C. 4 2 3 2
3 4

(9)

9 (11) a2 (2x 2 + 4 x 3 ) + C.

(12) − 1 cos θ − 1 θ + C. 3 6 (14) θ sin 2θ − + C. 2 4

(13)

θ sin 2aθ + + C. 2 4a

ANSWERS TO EXERCISES

265

(15)

θ sin 2aθ − + C. 2 4a

(16)

1 3x . 3

(17) log(1 + x) + C.

(18) − log (1 − x) + C.

Exercises XVIII. (1) Area = 60; mean ordinate = 10. √ (2) Area = 2 of a × 2a a. 3 (3) Area = 2; mean ordinate =

(p. 221.)

2 = 0.637. π (4) Area = 1.57; mean ordinate = 0.5. (5) 0.572, 0.0476. (7) 1.25. h (6) Volume = πr2 . 3 (8) 79.4.

(9) Volume = 4.9348; area of surface = 12.57 (from 0 to π). a (10) a log a, log a. a−1 (12) Arithmetical mean = 9.5; quadratic mean = 10.85. 1 (13) Quadratic mean = √ A2 + A2 ; arithmetical mean = 0. 1 3 2 The ﬁrst involves a somewhat diﬃcult integral, and may be stated thus: By deﬁnition the quadratic mean will be 1 2π
2π

(A1 sin x + A3 sin 3x)2 dx.
0

CALCULUS MADE EASY

266

Now the integration indicated by (A2 sin2 x + 2A1 A3 sin x sin 3x + A2 sin2 3x) dx 1 3 is more readily obtained if for sin2 x we write 1 − cos 2x . 2 For 2 sin x sin 3x we write cos 2x − cos 4x; and, for sin2 3x, 1 − cos 6x . 2 Making these substitutions, and integrating, we get (see p. 198) A2 1 2 x− sin 2x 2 + A1 A3 sin 2x sin 4x − 2 4 + A2 3 2 x− sin 6x 6 .

At the lower limit the substitution of 0 for x causes all this to vanish, whilst at the upper limit the substitution of 2π for x gives A2 π + A2 π. And hence the answer follows. 1 3 (14) Area is 62.6 square units. Mean ordinate is 10.42. (16) 436.3. (This solid is pear shaped.)

Exercises XIX.

(p. 231.)

√ x2 x a2 − x 2 a2 x (log x − 1 ) + C. (1) + sin−1 + C. (2) 2 2 2 2 a

ANSWERS TO EXERCISES

267

(3)

xa+1 a+1

log x −

1 a+1

+ C. (4) sin (6) x x

+ C.

(5) sin(log x) + C. (7) 1 (log x)a+1 + C. a+1

(x2 − 2x + 2) + C.

(8) log (log x) + C.

(9) 2 log (x − 1) + 3 log (x + 2) + C. (10) (11) (13)
1 2

log (x − 1) + 1 log (x − 2) + 5

3 10

log (x + 3) + C. x2 − 1 + C. x2 + 1

b x−a log + C. 2a x+a
1 4

(12) log

1+x 1 + arc tan x + C. 1−x 2 √ √ 1 a − a − bx2 1 √ (14) √ log . (Let = v; then, in the result, let x a x a b v 2 − = v − u.) a You had better diﬀerentiate now the answer and work back to log the given expression as a check. Every earnest student is exhorted to manufacture more examples for himself at every stage, so as to test his powers. When integrating he can always test his answer by diﬀerentiating it, to see whether he gets back the expression from which he started. There are lots of books which give examples for practice. It will sufﬁce here to name two: R. G. Blaine’s The Calculus and its Applications, and F. M. Saxelby’s A Course in Practical Mathematics.
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transcriber’s note
The diagrams have been re-created, using accompanying formulas or descriptions from the text where possible. In Chapter XIV, pages 132–159, numerical values of 1 + x, 1 n n ,

and related quantities of British currency have been veriﬁed

and rounded to the nearest digit. On page 142 (page 146 in the original), the graphs of the natural logarithm and exponential functions, Figures 38 and 39, have been interchanged to match the surrounding text. The vertical dashed lines in the natural logarithm graph, Figure 39 (Figure 38 in the original), have been moved to match the data in the corresponding table. On page 164 (page 167 in the original), the graphs of the sine and cosine functions, Figures 44 and 45, have been interchanged to match the surrounding text.

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