ADDITIONAL MATHEMATICS PROJECT WORK 2014

-----THE FUNDAMENTALS OF CALCULUS-----

SCHOOL: HIGH SHOOL KLANG

NAME: HARIHARAN A/L THAMILARASU

INDEX NUMBER:

IC NO: 970219-10-6617

CLASS: 5 SN 1

TEACHER: EN.NASRUDDIN

CONTENTS

Content

Page

Acknowledgement

Introduction

Part 1

Part 2

Part 3

Further Exploration

Reflection

Attachments

ACKNOWLEDGEMENT

First and foremost, I would like to thank my Additional Mathematics teacher, En.Nasruddin as he gave me important guidance and commitment during this project work. He has been a very supportive figure throughout thewhole project.

Not to be forgotten, I would also like to express my sincere gratitude to my parents, Mr.Thamilarasu and Pn.Uma, for supplying the equipment and money needed for the resources to complete this project and for their continuous moral support.

Finally, I also would like to acknowledge to all my friends for helping me directly and indirectly to help complete this project work.

Thank you.

INTRODUCTION

Calculus is the broad area of mathematics dealing with such topics as instantaneous rates of change, areas under curves, and sequences and series. Underlying all of these topics is the concept of a limit, which consists of analyzing the behavior of a function at points ever closer to a particular point, but without ever actually reaching that point. As a typical application of the methods of calculus, consider a moving car. It is possible to create a function describing the displacement of the car (where it is located in relation to a reference point) at any point in time as well as a function describing the velocity (speed and direction of movement) of the car at any point in time. If the car were traveling at a constant velocity, then algebra would be sufficient to determine the position of the car at any time; if the velocity is unknown but still constant, the position of the car could be used (along with the time) to find the velocity.However, the velocity of a car cannot jump from zero to 35 miles per hour at the beginning of a trip, stay constant throughout, and then jump back to zero at the end. As the accelerator is pressed down, the velocity rises gradually, and usually not at a constant rate (i.e., the driver may push on the gas pedal harder at the beginning, in order to speed up). Describing such motion and finding velocities and distances at particular times cannot be done using methods taught in pre-calculus, whereas it is not only possible but straightforward with calculus.Calculus has two basic applications: differential calculus and integral calculus. The simplest introduction to differential calculus involves an explicit series of numbers. Given the series (42, 43, 3, 18, 34), the differential of this series would be (1, -40, 15, 16). The new series is derived from the difference of successive numbers which gives rise to its name "differential". Rarely, if ever, are differentials used on an explicit series of numbers as done here. Instead, they are derived from a continuous function in a manner which is described later.Integral calculus, like differential calculus, can also be introduced via series of numbers.

PART 1

Gottfried Wilhelm LeibnizBorn July 1st, 1646 in Leipzig, Germany, Gottfried Wilhelm Leibniz soon became known as a great philosopher, mathematician and physicist of his time. At the young age of six, his father died. Despite the tragedy, he continued the advancement of his education. With only twelve years of age, he managed to teach himself to read Latin, and by age thirteen, he took upon the challenge of improving Aristotles theory of categories. By age fourteen, he entered the Leipzig University where he focused his studies on law. Due to controversy, he did not receive his doctorate in law from Leipzig, but instead from the University of Altdorf. Following this achievement, he moved to Nuremberg and became involved in the German political spectrum. As a result of the patronage of his dear friend Baron Johann Christian von Boineburg, Leibniz served as the legal adviser to the Elector of Mainz. When he was not fulfilling his duties as an adviser, Leibniz took interest in Latin poetry and frequently studied Vergil and Platos works.

In 1672, Leibniz developed a strategic plan to distract Louis XIV away from Northern Europe with an appealing scheme for the conquest of Egypt. Subsequently, he was invited by Boineburg to travel to Paris and proposed it to the French government. There in Paris, nothing came of his plan, but he did receive the opportunity to study geometry with Huygens. He described this experience as one that opened a new world to him. Although he had already written various tracts on mathematics, this moment sparked a deeper concentration in the mathematics world.

In 1675 Leibniz invented infinitesimal calculus, which is differential calculus and integral calculus combined. He first published his system in 1684, while Newton invented his system for calculus in 1666 and published it in 1687. Although Leibniz developed his system independently of Newton, there was an extensive dispute about whether Leibniz truly discovered calculus independently of Newton or whether he had derived his ideas from Newton and simply invented a different notation. Today Leibniz along with Newton is acknowledged for discovering infinitesimal calculus, and the product rule in differential calculus is referred to as Leibnizs Rule. The notation Leibniz used for calculus is still used today. Those symbols are dx for differentiation and for integration. Leibniz was also the first person to use a period for multiplication and ~ in geometry for similarity.

Although a large portion of his life was spent in controversy and dishonor due to the continuous argument over the true creator of calculus, Leibniz still contributed significantly to the world of philosophy, physics, and more importantly, to mathematics. He died in Hanover, Germany in 1716 with seventy years of age.

-BUBBLE MAP-

-BRACE MAP-

PART 2

A 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, findi) The acceleration of the car in the first hourWhen t = 0,v = 60(0) + 20 v = 20Whent = 1,v = 60(1) + 20v = 80Acceleration of the car in the first hour,= 80km/h 20km/h 1 hour= 60km/

ii) The average speed of the car in the first two hoursAverage speed =total distancetotal time = 110km 2 hours = 55km/h b) What is the significance of the position of the graphi) Above the t-axis?The position of the graph above the t-axis shows that the car travels to its destination.

ii) Below the t-axis?The position of the graph below the t-axis shows that the car travels in the opposite direction, travelling back where it came from.

c) Using two different methods, find the total distance travelled by the carMethod 1: Area under graph

Area of A= x (20 +80) x 1= 50km Area of B= Rectangle B Right angle Q= (80 x 1) ( x 80 x 0.5)= 80 20= 60kmArea of C = x (1.5 + 0.5) x 80= x 2 hours x 80= 80km

Hence, the total distance travelled by the car is:= 50 + 60 + 80= 190kmMethod 2: Integration

Since PQ is parallel to RS, the gradient of RS is -160 too. Instead of using y=mx+c, v=mt + c is used. Substitute R (2.5, 0) into v= mt + c0 = -160 (2.5) + cc = 400

Hence, the equation of RS is v = -160t + 400

The gradient of TU is 160 v = mt + cv =160t + cSubstitute U (4,0) into v = 160t + c0 = 160 (4) + cc = -640Hence, the equation of TU IS v = 160t 640

Area of A = == = = - = 50kmArea of B= == = 80(1.5) 80(1)= 40km

Area of C = == = = - = 20km

Area of D = == = - 20= = 20km

Area of E= == - 40= = 40km

Area of F= == = = - = -1280 (-1260)= - 20= =20km

So, the total distance travelled by the car is= 50 + 40 + 20 + 20 + 40 + 20= 190km d) Based on the above graph, write an interesting story of the journey.

I bought a new Mercedes and proudly drove it off the showroom floor to take home. I attempted to change radio stations and saw that theres only one station. I immediately turned and accelerated back to the dealer.

Once at the dealer, I found the salesman and he told me that the car radio was voice-activated, and that I would only need to state aloud the type of music that I wanted and the car would find it.

I got into the car and started the engine and then said the word "country," and the radio changed to a station playing a George Strait song. I was satisfied and started home. After a while I decided to try out the radio and said "rock 'n' roll;" the radio station changed and a song by the Rolling Stones came from the speakers. Quite pleased, I continued driving.

A few blocks from my house, another driver ran a light causing me to slam on my brakes to avoid a collision. I angrily exclaimed, "Ludicrous!"

The radio cut over to George Bush's press conference. I was hopeless over my sensitive voice-activated radio and drove back home slowly.

PART 3

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

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

The parabolic satellite disc is symmetrical at the y-axis, the curve y, f(x) can be written as y = ax + c. The curve y = f(x) cuts the y-axis at the point (0,4). Substitute (0,4) into y = + c y = + 4

Substitute point (4,5) into y = + 4 5 = a() + 4 a =

Thus, y = f(x) is now written as y = + 4So, f(x) = + 4b) To find the approximate area under a curve, we can divide the region into several vertical strips, 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 = f(x)y = + 4f(x) = + 4

First strip:when x = 0, f(0) = + 4f(0) = 4

Second strip:when x = 0.5, f(0.5) = + 4f(0.5) = 4.0156

Third strip:when x = 1, f(1) = + 4f(1) = 4.0625

Fourth strip:when x = 1.5, f(1.5) = + 4f(1.5) = 4.1406

Fifth strip:when x = 2, f(2) = + 4f(2) = 4.25

Sixth strip:when x = 2.5, f(2.5) = + 4f(2.5) = 4.3906

Seventh strip:when x = 3, f(3) = + 4f(3) = 4.5625

Eight strip:when x = 3.5, f(3.5) = + 4f(3.5) = 4.7656

i)

Area of the first strip = 4m x 0.5m = 2

Area of the second strip = 4.0156m x 0.5m = 2

Area of the third strip = 4.0625m x 0.5m = 2

Area of the fourth strip = 4.1406m x 0.5m= 2

Area of the fifth strip = 4.25m x 0.5m = 2

Area of the sixth strip = 4.3906m x 0.5m = 2

Area of the seventh strip= 4.5625m x 0.5m = 2

Area of the eight strip = 4.7656m x 0.5m = 2.3828

Total area= 2 + 2.0078 + 2.0313 + 2.0703 + 2.125 + 2.1953 + 2.2813 + 2.3828=

ii)

Area of the first strip = 4.0156m x 0.5m = 2.0078

Area of the second strip = 4.0625m x 0.5m = 2

Area of the third strip = 4.1406m x 0.5m = 2

Area of the fourth strip = 4.25m x 0.5m= 2

Area of the fifth strip = 4.3906m x 0.5m = 2

Area of the sixth strip = 4.5625m x 0.5m = 2

Area of the seventh strip= 4.7656m x 0.5m = 2

Area of the eight strip = 5m x 0.5m = 2.5

Total area=2.0078 + 2.0313 + 2.0703 + 2.125 + 2.1953 + 2.2813 + 2.3828 + 5=

iii)

Area of the first and second strips = 4.0156m x 1m = 4.0156

Area of the third and fourth strips = 4.1406m x 1m= 4.1046

Area of the fifth and sixth strips = 4.3906m x 1m = 4.3906

Area of the seventh and eight strips = 4.7656m x 1m= 4.7656

Total area= 4.0156 + 4.1046 + 4.3906 + 4.7656=

c) i) Calculate the area under the curve using integrationArea = dx= dx= = (0) = 17

ii) Compare the area in c (i) with the values obtained in (b). Hence, discuss which diagram gives the best approximate area. The diagram 3 (iii) gives the best approximate area, which is 17.3124

iii) Explain how you can improve the value in c(ii)The value in c(ii) can be improved by having more strips from x = 0 to x = 4, and obeying the Trapezium Rule.

d) Calculate the volume of the satellite disc.y = = 16 (y 4) = 16y 64 Volume = dy= dy= = = - = 8

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.

Finda) The volume of gold needed

y = 1.2 y = 1 = = = 1.44 + - = 1

Volume of gold needed= dx - dx= - = - = 0.5152 0.4= 0.36191

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

Gold density is 19.3. Let r be the weight of the gold ring.

1 = 19.3g0.36191 = r gSo, r = 6.9849 gPrice of 1g of gold = RM 58.35

Finally, the price of 1g of gold x 6.9849g= RM 58.35 x 6.9849g= RM 407. 57

REFLECTION

Derivatives we find- The power rule is our tool. We never ever whine Cause math is really cool. Related rates galore And differentials too, Looking for the tangent, Solving the graph,Finding the acceleration of a vehicle,How could you say that add maths a bore? When its taught by you-know-who?Calculus, Calculus You make our lives complete Without learning calculus Wed end up on the street. To study for the test, Wild parties are the key, Then get a good nights rest, And success youll surely see. But if you do not pass, Corrections are your friend. So hit the books and go to class- The fun will never end! Thank you, Add Maths for introducing Calculus!

Sung to the tune of Jingle Bells

ATTACHMENTS