ADDITIONAL MATHEMATICS

PROJECT WORK 2014

Name :xxxxxx .

I/C Number :xxxxxxxxx

Class :xxxx

Teacher’s Name :xxxx

School :xxxxxx

CONTENT

NO TITLE PAGE1 Acknowledge2 Objective3 Introduction4 Part 15 Part 26 Part 37 Further Exploration8 Conclusion9 Reflection

ACKNOWLEGMENT First of all, I would like to say Alhamdulillah, for giving me the strength and

health to do this project work.

Not forgotten my parents for providing everything, such as money, to buy anything that are related to this project work and their advise, which is the most

needed for this project. Internet, books, computers and all that. They also supported me and encouraged me to complete this task so that I will not

procrastinate in doing it.

Then I would like to thank my teacher, Puan Siti Rafidah Bt. Mohd Rejab for guiding me and my friends throughout this project. We had some difficulties in doing this task, but she taught us patiently until we knew what to do. She tried and tried to teach us until we understand what we supposed to do with the project work.

Last but not least, my friends who were doing this project with me and sharing our ideas. They were helpful that when we combined and discussed together, we had this task done.

OBJECTIVES

The aims of carrying out this project work are :

To apply and adapt a variety of problem-solving strategies to solve problems.

To improve thinking skills.

To promote effective mathematical communication.

To develop mathematical knowledge through problem solving in away that increases student’s interest and confidence.

To use the language of mathematics to express mathematical ideas precisely.

To provide learning environment that stimulates and enhances effective learning.

To develop positive attitude towards mathematics.

INTRODUCTIONCalculus

Calculus is the mathematical study of change, in the same way that geometry is the study of shape and algebra is the study of operations and their application to solving equations. It has two major branches, differential calculus (concerning rates of change and slopes of curves), and integral calculus (concerning accumulation of quantities and the areas under and between curves); these two branches are related to each other by the fundamental theorem of calculus. Both branches make use of the fundamental notions of convergence of infinite sequences and infinite series to a well-defined limit. Generally considered to have been founded in the 17th century by Isaac Newton and Gottfried Leibniz, today calculus has widespread uses in science, engineering and economics and can solve many problems that algebra alone cannot.

Calculus is a part of modern mathematics education. A course in calculus is a gateway to other, more advanced courses in mathematics devoted to the study of functions and limits, broadly called mathematical analysis. Calculus has historically been called "the calculus of infinitesimals", or "infinitesimal calculus". The word "calculus" comes from Latin (calculus) and refers to a small stone used for counting. More generally, calculus (plural calculi) refers to any method or system of calculation guided by the symbolic manipulation of expressions. Some examples of other well-known calculi are propositional calculus, calculus of variations, lambda calculus.

Newton kept his discovery to himself. However, enough was known of his abilities to effect his appointment in 1669 as Lucasian Professor of Mathematics at the University of Cambridge.

Newton published a detailed exposition of his fluxional method in 1704

He generalized the methods that were being used to draw tangents to curves and to calculate the area swept by curves, and he recognized that the two procedures were inverse operations. By joining them in what he called the fluxional method, Newton developed in the autumn of 1666 a kind of mathematics that is now known as Although

Newton was its inventor, he did not introduce calculus into European mathematics.

Achievement of Issac Newton in

calculus

PART 1 Choose one pioneer of modern calculus that you like and write about his background history. Hence, present your findings using one or more i-Think maps.

PART 2A car travels along a road and its velocity-time function is illustrated in Diagram 1. The straight line PQ is parallel to the straight line RS.

(a) From the graph, find

(i) the acceleration of the car in the first hour.

v = 60t + 20∴a =dvdt = 60t +20

When t = 1, = 60 ms -2

v = 60(1) + 20

= 80

When t = 0,

v = 60(0) + 20

= 20

v = 60t + 20

v = -160t+320

P

Q R

S

v (km/h)

t (h)1.0 1.5 2.0 2.5 3.0 3.5 4.0

Diagram 1

(ii) the average speed of the car in the first two hours.

Average speed = Totaldistance travelledTotal time

Total distance = Area under the graph

Area of A =12

×1.0 ×60 Area of BC = 12

× (0.5+1.0 )× 80

= 30 = 60

Area of D = 1.0 ×20

= 20

Total distance = 30 + 60 + 20

=110

∴ The average speed of the car in the first two hour = 1102

= 55 km/h.

A B

D

C

(1.0 , 80)

(1.5 , 80)

(2.0 , 0)(0 , 20)

0 1.0 1.5 2.0

(b) What is the significance of the position of the graph

(i) above the t-axis

The car move to a destination.

(ii) below the t-axis

The car move opposite ways the destination.

(c) Using two different methods, find the total distance travelled by the car.

Method 1

Based on calculation at 1(a)(ii), the total distance at region A is 110 km.

Total distance travelled at region B = Area of trapezium

= 12

׿1.5 + 0.5) × 80

= 80 km

∴ Total distance travelled by the car = 110 + 80

= 190 km.

A

B

(1.0, 80) (1.5 , 80)

(2.5 , 0) (4.0 , 0)

(3.0 , -80) (3.5 , -80)

(2.0 , 0)

(0 , 20)

Method 2

At region A, v = 60t + 20 At region D, equation PQ = v = -160t + 320

a = 60 mPQ= -160

s = ∫0

1

60 t +20 dt mRS = -160

= 50 y – y1= m(x – x1)Thus, area = 50 km v – 0 = -160(t –

2.5) v = -160t + 400

At region B, v = 80 s =

∫2.5

3.0

−160 t+400dt

s = ∫1.0

1.5

80 = -20

= 40 Thus, area = 20 kmThus, area = 40 km

At region C, v = -160t + 320 At region E, when t = 3.0,a = -160 v = -160t + 400

s = ∫1.5

2.0

−160 t+320dt = -160(3.0) + 400

= 20 = -80

v = 80

v = -80

0 1.0 1.5 2.0 2.5 3.0 3.5 4.0

v = 60t +20

v = -160t +320A B C

D E F

P

QR

S

Thus, area = 20 km s = ∫3.0

3.5

−80 dt

= -40 Thus, area = 40 km

At region F, gradient = m = −80−03.5−4.0

= 160 y – y1 = m(x – x1) v – 0 = 160(t – 4.0) v = 160t – 640

s = ∫3.5

4.0

160 t−640 dt

= -20 Thus, area = 20 km

∴Total distance travelled = Sum of all areas = 50 + 40 + 20 + 20 + 40 + 20

= 190 km

(d) Based on the above graph, write an interesting story of the journey in not more than 100 words.

Ramli was in his journey to join a convoy from Johor to Kelantan. On that day, Ramli was late and drove his car accelerating from 20 km/h to 80 km/h. After the first hour, Ramli found his convoy’s members that are moving together on the highway. He then followed them with a constant velocity, 80 km/h for half an hour. The group then decided to take a rest at any R&R, so they reduced their velocity for 30 minutes before they reached there. At that moment ,realize Ramli that he forgot to bring his

wallet so he headed home and decided to turn back home to take his wallet.

He took the opposite way and drove directly to his home with increasing acceleration from 0 km/h to 80 km/h. Unfortunately, there was a traffic jam that forces him to drive at a constant velocity, 80 km/h for 30 minutes. He arrived his homel half and hour later with a reduced velocity from 80 km/h to 0 km/h.

PART 3Diagram 2 shows a parabolic satellite disc which is symmetrical at the y – axis. Given that the diameter of the disc is 8 m and the depth is 1 m.

Diagram 2

(a) Find the equation of the curve y = f(x).

y = a(x - p)2 + q

Minimum point = (0 , 4)

y = a(x - 0)2 + 4

y = ax2 + 4

x – axis, 82= 4 y – axis, 4 + 1 = 5

At point = (4 ,5)

5 = a(4)2 + 4

5 – 4 = 16a

116 = a

y

8 m

1 m

4

0

y = f(x)

x

∴f(x) = 116x2 + 4

(b) To find the approximate area under a curve, we can divide the region into several vertical strips, and then we add up the areas of all the strips.Using a scientific calculator or any suitable computer software, estimate the area bounded by the curve y = f(x) at (a), the x– axis, x = 0 and x= 4.

y

0 0.5 1 1.5 2 2.5 3 3.5 4 x

4

A B C D E F G H

Area of region = (value of x× value of y) + [ 12

(value of x× value of y )]Area of region A, y = 1

16x2 + 4 Area of region E, y = 116

(2)2+ 4 = 1

16(0)2 + 4 = 4.25

= 4 Thus, area = (0.5)(4.25) + 12 (0.5)

(0.14)Thus, area = (0.5)(4) + 1

2(0.5)(0.02) = 2.17

= 2.00 Area of region F, y = 116 (2.5)2 +

4Area of region B, y = 1

16(0.5)2 + 4 = 4.39

= 4.02 Thus, area = (0.5)(4.39) + 12

(0.5)(0.17)Thus, area = (0.5)(4.02) + 1

2 (0.5)(0.04) = 2.24

= 2.02 Area of region G, y = 116 (3)2 + 4

Area of region C, y = 116(1)2 + 4 = 4.56

Diagram 3 (i)

y = f(x)

(i)

= 4.06 Thus, area = (0.5)(4.56) + 1

2 (0.5)(0.21)

Thus, area = (0.5)(4.06) + 12 (0.5)(0.08) = 2.33

= 2.05 Area of region H, y = 116(3.5)2 + 4

Area of region D, y = 116(1.5)2 + 4 = 4.77

= 4.14 Thus, area = (0.5)(4.77) + 12

(0.5)(0.23)Thus, area = (0.5)(4.14) + 1

2(0.5)(0.11) = 2.46 = 2.10

∴ Total area under the curve = Sum of all areas= 2.00 + 2.02 + 2.05 + 2.10 + 2.17 + 2.24

+ 2.33 + 2.46 = 17.37 m 2

(ii)

Area of region = (value of x× value of y) – [12

(value of x× value of y )]Based on the values of y obtained in the calculations at Diagram 3 (i),

Area of region A, = 0.5 × 4= 2

Area of region B, =(0.5 × 4.02 ) – [12(0.5)(0.04)]

= 2

Diagram 3 (ii)

y = f(x)

Area of region C, = (0.5 × 4.06 ) – [ 12

(0.5 ) (0.08 )]= 2.01

Area of region D, = (4.14 ×0.5 ) – [12(0.5)(0.11)]

= 2.04

Area of region E, = (0.5 × 4.25 ) – [12(0.5)(0.14)]

= 2.09

Area of region F, = (0.5× 4.39 ) – [ 12(0.5)(0.17)]

= 2.16

Area of region G, = (0.5 × 4.56 ) – [12(0.5)(0.21)]

= 2.23

Area of region H, = (0.5 × 4.77 ) – [ 12(0.5)(0.23)]

= 2.34

∴ Total area under the curve = Sum of all areas= 2 + 2 + 2.01 + 2.04 + 2.09 + 2.16 + 2.23 +

2.34 = 16.87 m 2

(iii)

x

0 0.5 1 1.5 2 2.5 3 3.5 4

A B C D E F G H

Based on the values of y obtained in the calculations at Diagram 3 (i),

y

Diagram 3(iii)

Area of region = ( value of x ×value of y ) + [ 12

(value of x× value of y )]

y = f(x)

Area of region A = 0.5 × 4= 2

Area of region B, = (0.5 × 4.02 ) +[12(0.5)(0.04)]

= 2.02

Area of region C, = (0.5 × 4.06 ) – [ 12

(0.5 ) (0.08 )]= 2.01

Area of region D, = (4.14 ×0.5 ) +[12(0.5)(0.11)]

= 2.10

Area of region E, = (0.5 × 4.25 ) – [12(0.5)(0.14)]

= 2.09

Area of region F, = (0.5× 4.39 ) +[12(0.5)(0.17)]

= 2.24

Area of region G, = (0.5 × 4.56 ) – [12(0.5)(0.21)]

= 2.23

Area of region H, = (0.5 × 4.77 ) +[12(0.5)(0.23)]

= 2.34

∴ Total area under the curve = Sum of all areas = 2 + 2.02 + 2.01 + 2.10 + 2.09 + 2.24 + 2.23 + 2.46 = 17.15 m 2

(c)(i) Calculate the area under the curve using integration.

Area = ∫0

4 110

x2+4 d x

= 1713x2

(c) (ii) Compare your answer in c (i) with the values obtained in (b). Hence, discuss which diagram gives the best approximate area.

Based on question (b) Diagram 3 (i), the values obtained is 17.37, in Diagram 3 (ii) is 16.87 while in Diagram 3 (iii) is 17.15. However, when we calculate using integration method, the values obtained is 17.33.

In a conclusion, the best approximate area among the three diagrams is Diagram 3 (i) with the values obtained 17.37 which are almost same with the calculation using integration method.

(iii) Explain how you can improve the value in (c) (ii).

The closer the stripes are located and the more the stripes are, the more approximate value will be obtained.

(d) Calculate the volume of the disc.

y = 116x2 + 4

16y = x2 + 64

16y – 64 = x2

∴ v = π∫ x2 dy

=π∫4

5

16 y−64 dy

=227 [ 16 y2

2−64 y ]54

= 227

[8 y2−64 y ]54

= 25 17 m3

FURTHER EXPLORATION

A gold ring in Diagram 4 (a) has the same volume as the solid of revolution obtained when the shaded region in Diagram 4 (b) is rotated 360° about the x-axis.

Find(a) the volume of gold needed.

y = 1.2 – 5x2

y2 = (1.2 – 5x2)2

= 1.44 + 25x4 – 12x2

∴ Volume =π ∫−0.2

0.2

y2 dx

= 227 ∫

−0.2

0.2

1.44+25 x4−12 x2

= 1.619

(b) the cost of gold needed for the ring. (Gold density is 19.3 gcm-3. The price of gold can be obtained from the goldsmith)

Density = mass

volume

19.3 =mass1.619

y

x0

Diagram 4 (a)

x

y

-0.2 0 0.2

Diagram 4 (b)

f(x)= 1.2 – 5x2

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mass = 19.3 ×1.1619 = 31.25g

On 29th May 2014, 1g of gold costs RM 155.00∴ The cost of gold needed for the ring = mass of gold × RM 155

= 31.25 × RM 155 = RM 4843.75