ADDITIONAL MATHEMATICS
PROJECT WORK 1/2010

Content

| | | | | | | | | |

Introduction

The diagram below shows the gate of an art gallery. A concrete structure is built at the upper part of the gate and the words ‘ART GALLERY’ is written on it. The top of the concrete structure is flat whereas at the bottom is parabolic in shape. The concrete structure is supported by two vertical pillars at both ends.
The distance between the two pillars is 4 metres and the height of the pillar is 5 metres. The height of the concrete structure is 1 metre. The shortest distance from point A of the concrete structure to point B, that is the highest point on the parabolic shape, is 0.5 metres.

Question: (a) The parabolic shape of the concrete structure can be represented by various functions depending on the point of reference. Based on different points of reference, obtain at least three different functions which can be used to represent the curve of this concrete structure. (b) The front surface of this concrete structure will be painted before the words ‘ART GALLERY’ is written on it. Find the area to be painted.

Solution:
(a)Function 1

Maximum point (0,4.5) and pass through point (2,4) y=a(x-b)2+ c b=0, c=4.5 y=a(x-0)2+ 4.5 y=ax2+ 4.5---(1)
Substitute 2,4 into (1)
4=a(2)2+ 4.5
4a= -0.5 a= -0.125
∴y= -0.125x2+ 4.5

Function 2
Maximum point (0, 0.5) and pass through point (2, 0) y=a(x-b)2+ c b=0, c=0.5 y=a(x-0)2+ 0.5 y=ax2+ 0.5---(2)
Substitute 2, 0 into (2)
0=a(2)2+ 0.5
4a= -0.5 a= -0.125
∴y= -0.125x2+ 0.5

Function 3

Maximum point (2, 4.5) and pass through point (0, 4) y=a(x-b)2+ c b=2, c=4.5 y=a(x-2)2+ 4.5---(3)
Substitute 0, 4 into (3)
4=a(0-2)2+ 4.5
4a= -0.5 a= -0.125
∴y= -0.125(x-2)2+ 4.5

(b)
Area to be painted
= Area of rectangle - Area under the curve
=4 x 1-202-0.125x2+0.5dx
=4-2-0.125x33+0.5x02
=4-223-0
=223m2

Further Exploration (a) You are given four different shapes of concrete structures as shown in the diagrams below. All the structures have the same thickness of 40cm and are symmetrical (i) Given that the cost to construct 1 cubic metre of concrete is RM840.00, determine which structure will cost the minimum to construct. (ii) As the president of Arts Club, you are given the opportunity to decide on the shape of the gate to be constructed. Which shape would you choose? Explain and elaborate on your reasons for choosing the shape.

Solution:
(i) Structure 1

Area= 223 m2

Volume =Area x Thickness =223m2 x 0.4 m = 1615m3

Cost=1615m3 x RM840 =RM896

Structure 2

Area=Area of Rectangle-Area of Triangle =1m x 4m-12 x 4m x 0.5m =4m2- 1m2 =3m2
Volume =Area x Thickness =3m2 x 0.4m =1.2m3

Cost=1.2m3 x RM840 =RM1008

Structure 3

Area=Area of Rectangle-Area of Trapezium =1m x 4m-(4m+1m)2 x 0.5m =4m2- 54m2 =2.75m2
Volume =Area x Thickness =2.75m2 x 0.4m =1.1m3

Cost=1.1m3 x RM840 =RM924

Structure 4

Area=Area of Rectangle-Area of Trapezium =1m x 4m-(2m+4m)2 x 0.5m =4m2- 1.5m2 =2.5m2
Volume =Area x Thickness =2.5m2 x 0.4m =1m3
Cost=1m3 x RM840 =RM840

∴Structure 4 will cost the minimum to construct, that is RM 840.
(ii) As the president of the Arts Club, I will decide Structure 4 as the shape of the gate to be constructed. It is because Structure 4 will cost the minimum and it is easier to be constructed compared to Structure 1 which is a curve.

(b) The following questions refer to the concrete structure in the diagram below.
If the value of k increases with a common difference of 0.25 m; (i) Complete Table 1 by finding the values of k and the corresponding areas of the concrete structure to be painted. (ii) Observe the values of the area to be painted from Table 1. Do you see any pattern? Discuss. k (m) | Area to be painted(m2) | 0.00 | 4 x 1- 0+42 x 0.5= 3 | 0.25 | 4 x 1- 0.25+42 x 0.5= 2.9375 | 0.50 | 4 x 1- 0.5+42 x 0.5= 2.875 | 0.75 | 4 x 1- 0.75+42 x 0.5= 2.8125 | 1.00 | 4 x 1- 1+42 x 0.5= 2.75 | 1.25 | 4 x 1- 1.25+42 x 0.5= 2.6875 | 1.50 | 4 x 1- 1.5+42 x 0.5= 2.625 | 1.75 | 4 x 1- 1.75+42 x 0.5= 2.5625 | 2.00 | 4 x 1- 2+42 x 0.5= 2.5 | (ii) There is a pattern in the area to be painted.

The area to be painted decreases as the k increases 0.25m and form a series of numbers:

3, 2.9375, 2.875, 2.8125, 2.75, 2.6875, 2.625, 2.5625, 2.5

We can see that the difference between each term and the next term is the same.

2.9375-3 = -0.0625
2.875- 2.9375 = -0.0625
2.8125- 2.875 = -0.0625
2.75- 2.8125 = -0.0625
2.6875- 2.75 = -0.0625
2.625- 2.6875 = -0.0625
2.5625- 2.625 = -0.0625
2.5- 2.5625 = -0.0625

∴ We can deduce that this series of numbers is an Arithmetic Progression (AP), with a common difference, d= -0.0625

In conclusion, when k increases 0.25m, the area to be painted decreases by -0.0625m2

(c) Express the area of the concrete structure to be painted in terms of k. Find the area when k approaches the value of 4 and predict the shape of the concrete structure.

The area of the concrete structure to be painted
=4 x 1- ( k+4 )2 x 0.5 =4- k4+ 1
=3- k4

∴k →4 k4 → 1
Area of concrete structure to be painted →3-1 →2m2

The shape of the concrete structure will be a rectangle with length 4m and breadth 0.5m, which may look like this:

Reflection