add maths folio 2014 blah

ADDITIONAL

MATHEMATICS

PROJECT

2014

STATE: SELANGOR

NAME:

SCHOOL: SMK SUBANG UTAMA

CLASS:

TEACHER:

TITLE:

INTRODUCTION

The idea of calculus had been developed earlier in Egypt, Greece, China, India, Iraq, Persia and Japan. The use of calculus began in Europe, during the 17th century, when Isaac Newton and Gottfried Wilheim Leibniz built on the work of earlier mathematics to introduce the basic principles. The development of calculus was built on earlier concepts of instantaneous motion and area under the curve.

Application of differential calculus include computations involving velocity and acceleration, the slope of curve and optimization. Applications of integral calculus include computations involving area, volume, arc length, centre of mass, work and pressure. Calculus is also used to gain a more precise understanding of the nature of space, time and motion.

(Extend all of this so that it reaches at least 1 page)

INDEX

PART 1: INTRODUCING NEWTON

Finished

Method of Fluxions

In 1671, but it remained unpublished until after his death (in1736)

Fluxions=differentiation

Fluents=integration

Begin rigorous mathematics at around 22 training under Isaac Barrow, where he learnt analysis and infinity series.

Later he wrote De analysi per aequationes numero terminorum infinitas.

(unpublished until his death)

Involved and manipulated the calculus controversy (Priorittsstreit) against Gottfried W. Leibniz

Sir Isaac Newton

(1642-1727)

Infinitesimal calculus was not well received at that time because of contradictions and there were attempts to disprove it. Hence, Newton did not publish Method of Fluxions at that time.

Defined the derivative as the ratio of change and the integral as the inverse of the derivative.

Newtons Notation of calculus in his work

Differentiation:

Integration:

(Not widely use because it might cause confusion because of similarities in representation.)

Applies calculus in his written work

Philosophi Naturalis Principia Mathematica

(Mathematical Principles of Natural Philosophy) published in 1687. It was largely about science and geometry, but he gave convictions on calculus in it.

PART 2

D

C

B

A

Before we begin, let us count.

Line AB

It is given that the linear line between point A and B is:

v=60t+20

Calculation for point A : Calculation for point B:

When t=0 hour, v=60(0)+20 When t=1 hour, v=60(1)+20

=20km/h =80km/h

This means the car was travelling at a velocity This means the car was travelling

of 20km/h when time started to be measured. at a velocity of 80km/h during the first hour.

(We would like to assume there was a person

that was a timekeeper for the car)

Point A is (0.0,20) Point B is (1.0,80)

_______________________________________________________________________________

Line BP

It is shown that this line BP is a line parallel to the time axis. So, it can be said that the function of this line is:

v=80

Point P is (1.5, 80)

_______________________________________________________________________________

Line PQ

It is given that the linear line between point P and Q is:

v=-160t+320

So Q is:

v=-160(2)+320

= -320+320

= 0 (As seen on graph)

Point Q is (2.0, 0)

_______________________________________________________________________________

Line QR

It is shown that this line QR is a line at time axis. So, it can be said that the function of this line is:

v=0

Point R is (2.5, 0)

_______________________________________________________________________________

Line RS

It is stated in the question that PQ is parallel to RS, so it can be said that RS has the same gradient with PQ, which is -160. So, by using

v=mt+c

(a.k.a y=mx+c)

We put m=-160. Not only that, we know that this line intersects the point R, so we put ine equations of point R into the basic linear equation:

(0)=-160(2.5)+c

c=160(2.5)

Hence,

c=400

Therefore, the equation of the linear line RS is:

v=-160t+400

The t value for point S has already been given, now we only have to find the f(t) value, velocity.

v=-160(3)+400

= -480+400

=-80

Point S is (3.0, -80)

_______________________________________________________________________________

Line SC

Since the velocity shall be constantly -80 at this period, hence:

v=-80

Point C is (3.5, 80)

_______________________________________________________________________________

Line CD

Point D is intersects the time axis, therefore:

Point D is (4.0, 0)

To find the linear equation of line CD, we use this:

v=mt+c

But first we need to find the gradient, hence, Let C= () & D= ()

160

m=160

So, we insert point D,

0= 160(4)+c

c= -640

Therefore, the equation of the linear line CD is:

v=160t-640

v=160t-640, Point D is (4.0,0)

_______________________________________________________________________________

Overview

Line

Function

Point

Coordinate

AB

v=60t+20

A

(0.0,20)

BP

v=80

B

(1.0,80)

PQ

v=-160t+320

P

(1.5,80)

QR

v=0

Q

(2.0,0)

RS

v=-160t+400

R

(2.5,0)

SC

v=-80

S

(3.0,-80)

CD

v=160t-640

C

(3.5,-80)

D

(4.0,0)

Q1:

Ai) Find acceleration of the car in the first hour.

So, the acceleration at the first hour (Period AB) is:

Method 1 Method 2

Point A= (0.0,20) At period AB, f(t)=60t+20

Point B= (1.0,80) 60t+20]= 60km/h2

So we can just find the gradient: (apply power rule)

= = 60km/h2

The acceleration is 60km/h2 .The rate of change of velocity is constant.

Aii) Average speed of the car in the first 2 hours

Average speed= , time (h) =2

Method 1

To obtaining the distance, you can just count them by using:

Area of quadrilateral formula: Area of triangle formula: Area of trapezium formula:

(Length)(Width) (Length) (Width) (Width)

The graph of the first 2 hours can be easily separated into 3 shapes:

(0.0-1.0): A trapezium (1.0-1.5): a rectangular quadrilateral (1.5-2.0): a triangle

Area of Trapezium

Points needed: A (0.0,20) and B (1.0,80) Area= (Width)

= = Width=x= = (20+80) (1.0)

= 20 = 80 = 1.0-0.0 = 50(1.0)

(Velocity) = 1.0 (time) = 50unit2

Area of trapezium (below line AB)= 50unit2

Distance the car has travelled=50km

Area of Quadrilateral (rectangle)

Points needed: B (1.0, 80) and P (1.5, 80) Area= (Length) (Width)

Length= Width=x= = 80 0.5

=80 =1.5-1.0 = 40unit2

=0.5

Area of quadrilateral (below line BP)= 40unit2

Distance travelled=40km

Area of triangle

Points needed: P (1.5, 80) and Q (2.0, 0) Area= (Length) (Width)

Length=y= Width=x= = 80 0.5

= 0-80 = 2.0-1.5 = 400.5

=|-80| =0.5 = 20unit2

=80

Area of triangle (below line PQ)= 20unit2


_______________________________________________________________________________

Total distance in 2 hours = (50+40+20)km= 110km

Average speed of the car in the first 2 hours

= = = 55km/h

Average speed of the car in the first 2 hours is 55km/h

Method 2

We can use integration to find the distance travelled by the car of the car.

The displacement of the car in the first 2 hours consists of the area below 3 lines that are:

[Line AB] =60t+20 [Line BP] =80 [Line PQ] =-160t+320

Area under the line AB

Interval [0.0, 1.0]

=

=

= [30(1.0)2+20(1.0)] - [30(0.0)2+20(0.0)]

= (30+20) 0

= 50unit2

Area = 50unit2


Area under the line BP

Interval [1.0, 1.5]

=

=

= [80(1.5)] - [80(1.0)]

= (120) (80)

= 40unit2

Area = 40unit2



Intervals [1.5 and 2.0]

=

=

= [-80(2.0)2+320(2.0)] - [80(1.5)2+320(1.5)]

= (320) (300)

= 20unit2

Area = 20unit2


_______________________________________________________________________________

Total distance in 2 hours = (50+40+20)km= 110km

Average speed of the car in the first 2 hours

= = = 55km/h

Average speed of the car in the first 2 hours is 55km/h

B) What is the significance of the position of the graph:

i) Above the t-axis?

The car is moving towards a perceived positive direction. It can be said that the car is moving away from its original point (0.0, 0). Positive directions are usually said to be right (East) and up (North).

ii)Below the t-axis?

The car is moving towards a perceived negative direction. It can be said that the car is moving towards its original point (0.0, 0). Negative directions are usually said to be left (West) and down (South).

C) Using 2 different methods, find the total distance travelled by the car.

All value calculated are modules since distance is not a vector.

Method 1

To obtaining the distance, you can just count them by using:

Area of quadrilateral formula: Area of triangle formula: Area of trapezium formula:

(Length)(Width) (Length) (Width) (Width)

The graph of the first 2 hours can be easily separated into 3 shapes:

(0.0-1.0): A trapezium (1.0-1.5): a rectangular quadrilateral (1.5-2.0): a triangle

Area of Trapezium

Points needed: A (0.0,20) and B (1.0,80) Area= (Width)

= = Width=x= = (20+80) (1.0)

= 20 = 80 = 1.0-0.0 = 50(1.0)

(Velocity) = 1.0 (time) = 50unit2

Area of trapezium (below line AB)= 50unit2

Distance the car has travelled=50km

Area of Quadrilateral (rectangle)

Points needed: B (1.0, 80) and P (1.5, 80) Area= (Length) (Width)

Length= Width=x= = 80 0.5

=80 =1.5-1.0 = 40unit2

=0.5

Area of quadrilateral (below line BP)= 40unit2


Area of triangle

Points needed: P (1.5, 80) and Q (2.0, 0) Area= (Length) (Width)


= 0-80 = 2.0-1.5 = 400.5

=|-80| =0.5 = 20unit2

=80

Area of triangle (below line PQ)= 20unit2


_______________________________________________________________________________

Area below the line QR

Points needed: Q (2.0,0) and R (2.5,0)

Length= Width=x= Area = (Length) (Width)

=0 =2.5-2.0 = 0 0.5

(since both ys are constant) =0.5 = 0

Area of below line QR= 0unit2


______________________________________________________________________________

And now, we are going to find how much the car travelled in the last 1.5 hours. the last segment of the velocity-time graph that is located below the time axis can be separated into 3 shapes:

(2.5-3.0): A triangle (3.0-3.5): A rectangular quadrilateral (3.5-4.0): A triangle

Area of triangle (Above line RS)

Points needed: R (2.5,0) and S (3.0,-80) Area = (Length) (Width)


= -80-0 = 3.0-2.5 = 40 0.5

=|-80| = 0.5 = 20unit2

= 80

Area of triangle= 20unit2


Area of quadrilateral [rectangle] (above line SC)

Points needed: S (3.0, -80) and C (3.5,-80) Area= (Length) (Width)

Length = Width=x= = 80 0.5

=|-80| = 3.5-3.0 = 40unit2

= 80 = 0.5

Area of quadrilateral [rectangle] = 40unit2


Area of triangle (Above line CD)

Points needed: C (3.5,-80) and S (4.0,0) Area = (Length) (Width)


= 0-(-80) = 3.5-4.0 = 40 0.5

= 80 = 0.5 = 20unit2

Area of triangle = 20unit2


_______________________________________________________________________________

Total distance travelled by the car

= Area of ([AB]+[BP]+[PQ]+[QR[o]]+[RS]+[SC]+[CD])

= (50+40+20+20+40+20) unit2

= (50+40+20+20+40+20) km

= 190km


Method 2

Distance travelled by car can be found using integration.

The displacement (or speed) of the car in consists of the area below 3 lines that are:

[Line AB] =60t+20 [Line BP] =80 [Line PQ] =-160t+320

The one with nothing is:

[Line QR] =0

The displacement (or speed) of the car in consists of the area above 3 lines that are:

[Line RS] = -160t+400 [Line SC] =-80 [Line SC] = 160t-640

Area under the line AB

Interval [0.0, 1.0]

=

=

= [30(1.0)2+20(1.0)] - [30(0.0)2+20(0.0)]

= (30+20) 0

= 50unit2

Area = 50unit2



Interval [1.0, 1.5]

=

=

= [80(1.5)] - [80(1.0)]

= (120) (80)

= 40unit2

Area = 40unit2



Interval [1.5 , 2.0]

=

=

= [-80(2.0)2+320(2.0)] - [80(1.5)2+320(1.5)]

= (320) (300)

= 20unit2

Area = 20unit2


_______________________________________________________________________________

Area under the line QR

Interval [2.0, 2.5]

=

=

= [0] - [0]

= 0unit2

Area = 0 unit2


_______________________________________________________________________________

Area under the line RS

Intervals [2.5, 3.0]

=

=

= [-80(3.0)2+400(3.0)] - [-80(2.5)2+400(2.5)]

= [(-720+1200)-(-500+1000)]

= 480-500

= -20unit2

Area = |-20 unit2|, |Displacement=-20km|

= Distance travelled=20km

Area under the line SC

Interval [3.0, 3.5]

=

=

= [-80(3.5)] - [-80(3.0)]

= -280-(-240)

= -280+240

= -40unit2



Area under the line CD

Interval [3.5, 4.0]

=

=

= [80(4.0)2-640(4.0)] - [80(3.5)2-640(3.5)]

= [(1280-2560)-(980-2240)]

= -1280-(-1260)

= -1280+1260

= -20unit2



_______________________________________________________________________________

Total distance travelled

=Total area obtained from integration between 2 intervals

= (AB+BP+PQ+QR+RS+SC+CD) unit2

= (50+40+20+20+40+20) unit2

= 190 unit2

= 190km


D) Based on the above graph, write an interesting story of the journey in not more

than 100 words.

The car in the story must begin moving at 20km/h and accelerating constantly at 60km/h2 from 0hours-1hour

Then the car moves at a constant velocity of80km/h at 1.0hours-1.5hours

The car begins to decelerate at 160km/h2 at 1.5 hours-2.0hours

The car completely stops during 2.0 hours . The distance of the car form its starting point is 110km

The car begins to move 30 minutes later and accelerate at 160km/h2 towards the opposite direction at 2.5hours-3.0hours

The car travels at a constant velocity of 80km/h at an opposite direction at 3.0hours-3.5hours

The car then begins to decelerate by 160km/h2 at 3.5hours-4.0hours

At 4 hours the car has travelled 80km at the opposite direction and is now stopped. It is 30km from its starting point.

The behavior of the car in the story must be something like that.

PART 3

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

Method 1

Use the square method of quadratic equation:

y=ax2+bx+c

=a(x2 x)+c

=a(x+)2-()2+c

Sub in minimum point of graph, (0,4)

y = a(x+)2-()2+c

= a(x+0)2-(0)2+4

= ax2+4

y = ax2+4 implies b=0

Take a positive point (4,5) for convenience.

5=a(4)2+4

1=16a Sub a, and c into

a= y=ax2+bx+c

Hence, equation of curve y=f(x) is f(x) = x2+4

Method 2

We know there are 3 points, (-4,5),(0,4) and (4,5).

Use

y=ax2+bx+c

c=4 since when x=0,y=4=c. Hence sub all 3 known values to make 3 equations.

Point (-4, 5) Point (0,4) Point (4,5)

a(-4)2+b(-4)+4=5 a(0)2+b(0)+4=5 a(4)2+b(4)+4=5

16a-4b=1 1=0?! 16a+4b=1

(This is impossible hence shall not be used)

Do simultaneous equation.

16a-4b=1 Since b=0, therefore:

16a+4b=1 16a=1 Sub a, b and c into

-4b-1=1+4b a= y=ax2+bx+c

8b=0

b=0

Hence, equation of curve y=f(x) is f(x) = x2+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. Before that, use a scientific calculator or any suitable computer software, to the area bounded by the curve y = f(x) at (a), the x-axis, x = 0 and x = 4.

Using a scientific calculator of the CASIO-570 series, the answer obtained from the integration of y=f(x) is= 17unit2

On the topic of counting strips, we find that.

Rectangle

x (Interval)

Input [x value]

Output

[f(x) =x2+4]

Area of rectangle (unit2)[xOutput]

1

(0.5-0.0)=0.5

0.0

4.000000

2.0000000

2

(1.0-0.5)=0.5

0.5

4.015625

2.0078125

3

(1.5-1.0)=0.5

1.0

4.062500

2.0312500

4

(2.0-1.5)=0.5

1.5

4.140625

2.0703125

5

(2.5-2.0)=0.5

2.0

4.250000

2.1250000

6

(3.0-2.5)=0.5

2.5

4.390625

2.1953125

7

(3.5-3.0)=0.5

3.0

4.562500

2.2812500

8

(4.0-3.5)=0.5

3.5

4.765625

2.3828125

Total Area

17.0937500

Rectangle

x (Interval)

Input [x value]

Output

[f(x) =x2+4]


1

(0.5-0.0)=0.5

0.5

4.015625

2.0078125

2

(1.0-0.5)=0.5

1.0

4.062500

2.0312500

3

(1.5-1.0)=0.5

1.5

4.140625

2.0703125

4

(2.0-1.5)=0.5

2.0

4.250000

2.1250000

5

(2.5-2.0)=0.5

2.5

4.390625

2.1953125

6

(3.0-2.5)=0.5

3.0

4.562500

2.2812500

7

(3.5-3.0)=0.5

3.5

4.765625

2.3828125

8

(4.0-3.5)=0.5

4.0

5.000000

2.5000000

Total Area

17.5937500

Rectangle

x (Interval)

Input [x value]

Output

[f(x) =x2+4]


1

(1.0-0.0)=1.0

0.5

4.015625

4.015625

2

(2.0-1.0)=1.0

1.5

4.140625

4.140625

3

(3.0-2.0)=1.0

2.5

4.390625

4.390625

4

(4.0-3.0)=1.0

3.5

4.765625

4.765625

Total Area

17.312500

C i) Calculate the area under the curve using integration.

=

=

= [ + 4(4)] - [ + 4(0)]

= [( +16)-(0+0)]

= 17 - 0

= 17unit2

(which is surprisingly the same value displayed by the calculator)

ii) Compare your answer in c (i) with the values obtained in (b). Hence, discuss which

diagram gives the best approximate area.

Graph

Approx. area under the graph

Approx. deviation from accurate answer

100%]

(i)

17.09375

0.013822115

(ii)

17.59375

+0.015024058

(iii)

17.31250

0.001201923

Diagram (iii) from question B gives the best approximate area as its deviation value is the smallest compared with the other two. (this sentence is not that complete, ask ying sheng)

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

By decreasing the value of x into an infinitesimal value (0), the number of triangles under the graph can be increased . Thus, this makes the rectangles become thinner and thinner but it shall make the area of graph within an interval become more and more accurate.

D) Calculate the volume of the satellite disc

=

=

= {[ - 64(5)] - [ - 64(4)]}

= [(200-320)-(128-256)]

= (-120)-(-128)

= 8 unit3

= 8 m3

25.133 m3

Volume of satellite dish is approx.. 25.133 m3

(Even though is strange)

PART 4

A) Volume of gold needed is.

Step 1: Find the 360 o volume revolution of f(x) =1.2-5x2 at the interval [-0.2,0.2] with respect to x.

=

=

=

= {[1.44(0.2) - +] - [1.44(-0.2) - +]}

= [(0.288-0.032+0.0016)-(-0.288+0.032-0.0016)]

= (2)(0.2576)

= 0.5152 unit3

= 0.5152 cm3

Step 2: Find the 360 o volume revolution of f(x) =1 at the interval [-0.2,0.2] with respect to x. (Since a

Second line exists there and it is a constant f(x) value. When x= -0.2/0.2, = 1).

=

=

=

= [(0.2)-(-0.2)]

= (0.4)

= 0.4 unit3

Step 3: To get the volume of gold needed to make the gold ring, we need to:

- = 0.5152 cm3 - 0.4 unit3

= 0.1152 cm3

0.3619 cm3

Volume needed to make gold ring

=0.1152 cm3 0.3619 cm3

B) The cost of gold needed to make the ring

The price is based on International price and not retail price since gold sold in gold shops are more expensive. Let us assume that this gold ring is a 24 carat gold ring, since 24 carat gold has the highest percentage of gold (99.9%)

(as of 10th June 2014, 01:34 amMalaysian time)

1 gram of 24k gold = RM129.05 per gram

Density=

Volume=0.1152 cm3

Given gold density=19.3g/cm-3

Mass= Density Volume

= (19.3g/cm-3)(0.1152 cm3)

= 2.22336 g

Price of gold = RM129.05/gram (x) gram

= RM129.05/gram 2.22336 gram

= RM 286.924608

= RM 901.4002406

RM 901.40

Price of gold needed to make the gold ring

RM 901.40

Might have some errors.

Volume needed to make gold ring

REFLECTION

TALK SHIT HERE

To get my A+ in Add Maths I had done everything,But had done nothing but to be wrong and scolded of,The sound of scribbling became louder every day,Then I noticed the fact there was no timeI was a believer in maths to be maths always,And was asking whether I would be excel againGive me a reason why not to use these equations,Or mark me to be wrong of so many incurable errors,Tell me why, or why not complaining way too much,Maybe I overlooked something in add mathsThe whole world was at a complete standstill,And I was going crazy, at the mercy of add maths,The equations became harder every day,I knew but I did not bother give a f*c*Its method had been in the eyes of beholder all along,It had grown dark before I finally got to solveAmong the nonsense solutions, what on earth you are looking for?You have to work hard yourself and do it your own way.Then tell me why, or why not complaining way too much,Maybe I overlooked something and failed the examThere is nobody who knows this shit .

There will be nobody !!!!Including me, and all the world gone bat shit crazy !!!!So how many fcks are you are willing to give?How much hard work you are willing to do?Now what? So what? Dont you come interrupt me, oh please,While I am god damn doing add maths.

add maths folio 2014 blah

Documents