additional mathematics project work 2013 wpkl

Name: JOSH, LRT

Class: SCIENCE STREAM PHYSICS CLASS

IC Number: 940901 – 14 – ****

SPM Registration Number: WP043 A000

Additional Mathematics Project Work

W.P.K.L. 2013

Author information

Josh, LRT

[email protected] // [email protected]

+6018 – 397 6808

Additional Mathematics Project Work W.P.K.L. 2013

Page 1 of 19

Introduction

A parabola is a two-dimensional, mirror-symmetrical curve, which is approximately U-

shaped when oriented as shown in the diagram, but which can be in any orientation in its

plane. It fits any of several superficially different mathematical descriptions which can all be

proved to define curves of exactly the same shape.

One description of a parabola involves a point (the focus) and a line

(the directrix). The focus does not lie on the directrix. The locus of

points in that plane that are equidistant from both the directrix and

the focus is the parabola. Another description of a parabola is as a

conic section, created from the intersection of a right circular conical

surface and a plane which is parallel to another plane which is

tangential to the conical surface. A third description is algebraic. A

parabola is a graph of a quadratic function, such as or

( ) , where a, b and c is real number.

The line perpendicular to the directrix and passing through the

focus (that is, the line that splits the parabola through the

middle) is called the "axis of symmetry". The point on the

axis of symmetry that intersects the parabola is called the

"vertex", and it is the point where the curvature is greatest.

The distance between the vertex and the focus, measured

along the axis of symmetry, is the "focal length". The "latus

rectum" is the chord of the parabola which is parallel to the

directrix and passes through the focus. Parabolas can open up,

down, left, right, or in some other arbitrary direction. Any parabola can be repositioned and

rescaled to fit exactly on any other parabola — that is, all parabolas are geometrically similar.

Parabolas have the property that, if they are made of material that reflects light, then light

which enters a parabola travelling parallel to its axis of symmetry is reflected to its focus,

regardless of where on the parabola the reflection occurs. Conversely, light that originates

from a point source at the focus is reflected ("collimated") into a parallel beam, leaving the

parabola parallel to the axis of symmetry. The same effects occur with sound and other forms

of energy. This reflective property is the basis of many practical uses of parabolas.

The parabola has many important applications, from a parabolic antenna or parabolic

microphone to automobile headlight reflectors to the design of ballistic missiles. They are

frequently used in physics, engineering, and many other areas.

Strictly, the adjective parabolic should be applied only to things that are shaped as a parabola,

which is a two-dimensional shape. However, as shown in the last paragraph, the same

adjective is commonly used for three-dimensional objects, such as parabolic reflectors, which

are really paraboloids. Sometimes, the noun parabola is also used to refer to these objects.

Though not perfectly correct, this usage is generally understood.

Conic Sections

Parabola Components

2013 © LRT DOCUMENTS Copyrighted. All rights reserved.


Page 2 of 19

The parabola was explored by Menaechmus (380 BC to 320 BC), who was a pupil of Plato

and Eudoxus. He was trying to dublicate the cube by finding the side of the cube that has an

area double the cube. Instead, Menaechmus solved it by finding the intersection of the two

parabolas and . Euclid (325 BC to 265 BC) wrote about the parabola.

Apollonius (262 BC to 190 BC) named the parabola. Pappus (290 to 350) considered the

focus and directrix of the parabola. Pascal (1623 to 1662) considered the parabola as a

projection of a circle. Galileo (1564 to 1642) showed that projectiles falling under uniform

gravity follow parabolic paths. Gregory (1638 to 1675) and Newton (1643 tp 1727)

considered the properties of a parabola.

The earliest known work on conic sections was by

Menaechmus in the fourth century BC. He discovered a

way to solve the problem of doubling the cube using

parabolae. (The solution, however, does not meet the

requirements imposed by compass and straightedge

construction.) The area enclosed by a parabola and a line

segment, the so-called "parabola segment", was computed

by Archimedes via the method of exhaustion in the third

century BC, in his The Quadrature of the Parabola. The

name "parabola" is due to Apollonius, who discovered

many properties of conic sections. It means "application",

referring to "application of areas" concept, that has a

connection with this curve, as Apollonius had proved. The

focus–directrix property of the parabola and other conics is due to Pappus.

Galileo showed that the path of a projectile follows a parabola, a consequence of uniform

acceleration due to gravity.

The idea that a parabolic reflector could produce an image was already well known before the

invention of the reflecting telescope. Designs were proposed in the early to mid-seventeenth

century by many mathematicians including René Descartes, Marin Mersenne and James

Gregory. When Isaac Newton built the first reflecting telescope in 1668 he skipped using a

parabolic mirror because of the difficulty of fabrication, opting for a spherical mirror.

Parabolic mirrors are used in most modern reflecting telescopes and in satellite dishes and

radar receivers.

Thus, parabola is important in our daily applications. Human should continue apply the

knowledge of parabolas to lead us into a more modern society.

Parabola Compass



Page 3 of 19

Part 1

1.1 Photos related to parabola

McDonald’s Sign Board Satellite Antenna

Bellagio's Fountains

Leeds Grand Mosque


1.2 Mind Map of Parabolas

Equations

𝑥 ±4 𝑎𝑦

About 𝑦-axis

𝑦 ±4 𝑎𝑥

About 𝑥-axis

Vertex (0,0) 𝑥 4𝑎𝑦 𝑦 4𝑎𝑥

𝑥 −4𝑎𝑦 𝑦 −4𝑎𝑥

(𝑦 − 𝑘) ±4𝑎(𝑥 − ℎ)

(𝑥 − ℎ) ±4𝑎(𝑦 − 𝑘)

When the vertex is not at origin,

Where (ℎ,𝑘) is the vertex.

𝑎 is the key of solution to the vertex, focus, directrix and latus

rectum.


P

age 4 o

f 19

2013 ©

LRT D

OCU

MEN

TS Copyrighted. All rights reserved.


Page 5 of 19

1.3 Determination equation of parabola

As the vertex are not at origin, equation of ( − ℎ) −4 ( − ) is chosen.

(ℎ, ) (0, 0) and ( , ) ( , 0)

( − 0) −4 (0 − 0)

0

0

−4.

0/ ( − 0)

−

4

−4

0

The equation of the parabolic dome is −

0.

𝑦

𝑥

From question, the information we have are:

Vertex = (0, 0) Roots = ±



Page 6 of 19

Generation of different parabolas

0

No Shape Information Equation

1.

original

Vertex: (0, 0) Roots: ± −

4

0

2.

inverted

Vertex: (0, − 0) Roots: ±

− −.−4

0/

4

− 0

3.

shifted

5 units right

Vertex: ( , 0) Roots: − 0, 0

( − ℎ) −4 ( − )

( − ) −4.

0/ ( − 0)

4( − ) − 00

−4

( − ) 0

4.

shifted + inverted

Vertex: ( , − 0) Roots: − 0, 0

− −(−4

( − ) 0)

4

( − ) − 0



Page 7 of 19

1.4 Determination the maximum distance / coordinates of point

−4

0

Gradient of curve

, So

−

Magnitude gradient | |

−

−

4

−4

(−

4)

0

−4

4

− −

4

−4

(

4)

0

−4

4

The maximum distance from horizontal and vertical axes that the bird can walk without

slipping downwards is up to the coordinates (−

, −

) or (

, −

).

𝑦

𝑥

𝑜


The bird can keep it balance up to a maximum

gradient with magnitude of 2

−



Page 8 of 19

1.5 Determination the budget of construction

To calculate the required area, the method involve are integration of curve to find bounded

area along x-axis and area of triangle.

0∫ .−4

0/

1 [

( )( )]

[ 0

] [ ]

4

Construction cost,

4

00

The budget required to construct the shaded partition is RM 16533.33

𝑦

𝑥

( , )

𝑜


The construction cost for the required partition

is RM 100 per meter square



Page 9 of 19

1.6 Determination of capacity of the air conditioning


Temperature to keep constant 4

000 ℎ ⁄ 0 ℎ

00

00

−4

0

−

4

∫

∫ ( −

4 )

0

( 0 ) ( 00) 00

0 ℎ ⁄

000 ℎ ⁄ 0 ℎ

0 ℎ ⁄ ℎ

0 0 000 4 ℎ

The capacity of air conditioner required is 4 ℎ



Page 10 of 19

Part 2

2.1 Observations

When a cylindrical beaker is filled with water until half full and circular motion of stirring

process begins, the observations are as below:

1. When stirring, the circular motion of glass rod will form a spiral depth, which will

increase in vertical downwards depth which reached the bottom of cylindrical beaker.

2. Through observations, height of spiral motion (ℎ ) increases toward bottom of

cylinder beaker while the water level at the center is displaced more results increase in

water level near the wall of the beaker.

3. So, as the spiral motion of water goes deeper, the higher the displace of water level near

the wall of beaker.

4. The vertical cross-sectional of water level forms a parabola as it goes deeper. At the same

time, the water level near to the wall of beaker is increases to maintain the volume of

water in the beaker which is constant. So, spiral movement of water forms various

parabolas with different heights.

Cylindrical beaker Spiral Movement 2D Spiral Water Movement 3D

ℎ𝑠𝑝𝑖𝑟𝑎𝑙

Direction of water

moving to the wall

of beaker



Page 11 of 19

2.2 Determine


( ℎ 0ℎ)

ℎ

ℎ

( ℎ 0ℎ)

ℎ ( ℎ 0ℎ)

( ℎ 0)

[ ( ) 0]

ℎ ℎ

( ) ( )

The rate of change of volume,

is



Page 12 of 19

Part 3

3.1 Find

and form a progression

Given

and distance between each cable is 0

4

When 0 ,

(0 )

When 0,

( 0)

When ,

( )

When 0,

( 0)

Progression of

are

,

,

,

Identify type of progression, AP (Arithmetic progression) or GP (Geometric progression)

GP should have same ( )

4

4

Since, there are having different r, it should not be an GP.

AP should have same ( )

4−

−

Since, there are having same , it should be an AP.



Page 13 of 19

Formula of AP, ( − )

,

(

) (

−

)

3.2 Find the cost of 19th

cable from a progression

4

4

(

)

Length of 1 meter cable is cost RM 100

Thus, length of 19th

cable length is

and cost 00



Page 14 of 19

3.3 Determine the term of an value

( )

Given length of cable is 4

is the value of y.

4

( )

√(

− )

±

Since, only the cable on the right side of the origin needs to be repaired.

So,

4

4( )

4

Where is the value of

4

4

The 34th

cable needed to be repaired.



Page 15 of 19

Further Exploration

Topics Chosen: Parabola in Physics

In physics the equations involving variety of sections including graphing, physical defining,

gravity, object falling and so on. This time, I gonna to explore about the object falling or

release in a parabolic motion which known as projectile motion.

Projectile, is when a particle is projected under gravity at a velocity u at an angle θ to the

horizontal (neglecting air resistance) it follows the curve of a parabola.

This motion – Projectile is a 2D motion due to exist of 2 components in the action of kinetic

projectiles. The components are vertical (y-axis) and horizontal (x-axis).

Diagram below is Oblique Projectile:

At Fy, the motion should be constant

acceleration (due to GRAVITY).

At Fx, the motion should be constant

velocity (due to linear).

What we can saw in the diagram is:

1. The 𝒗𝒙 is constant, because there is no

any force [horizontal] acting on it.

2. The 𝒗𝒚 is changing, because the height

of motion per second is different.

[Decreasing towards maximum point and

increasing towards same level of initial

point]

3. At highest point of the trajectory:

𝑣𝑦 0 but 𝑣𝑥 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.

4. Acceleration is constant and vertically

downwards. Therefore, 𝑎 −𝑔.



Page 16 of 19

Projectile Equations and Formula:

Symbol define:

1.

2.

3.

4.

5.

6.

Since, the gravity is equal to acceleration and it is always towards to the earth. Therefore the

equation is derived as: − .

To calculate Y component, information’s needed is:

a.

b. −

c. ℎ ℎ

Given that,

−

Since, maximum point is reached, final velocity of 0.

0 ( ) −

( )

. → used to calculate MAXIMUM height for component Y.

𝜃

𝑎 −𝑔

𝑢

𝐻



Page 17 of 19

d. Since, an object is projected from a starting time and end with a ending time.

Therefore,

Given that,

−

0 −

→ calculate time of HALF projectile for component Y.

e. Instantaneous can calculate at any time by using the formula below:

Given that,

−

( ) −

→ calculate any time of projectile for component Y.

f. To calculate FULL TIME of an object projectile, just simply take answer of

⟨

⟩ → calculate time of FULL projectile.

To calculate X component, information’s needed is:

a.

b. Instantaneous horizontal displacement at any time is

( )

c. To find Range, R which is the total distance from start point X to end point.

( )

( ) (

)

(

)

So, by knowing the method to gain the projectile equation, you can find out any range, angle,

height, speed required in order to project object from starting to its destination. It is important

to applied in scientific field and kept as knowledge towards future.



Page 18 of 19

Conclusion

Parabolas have widely used sports, physics and architecture field. Human design a parabolic

satellite station to transmit their data to stay connected with peoples around the world.

Sportsman uses the physics theory to determine what the best angle, speed and direction they

may create another record for the next competitions. Besides, it also applied in architecture

and arts, design of bridge, building and also drawing in the art block. Thus, learning

parabolas is actually fun and interesting.

So, would like to thanks to this great man which involve in my further explore of parabola in

physics - Galileo Galilei which found that all objects thrown form a parabolic path, no matter

what. He deduced this by the simple observation of watching objects being thrown. Galileo is

responsible for the modern concepts of velocity and acceleration to explain projectile motion

that is studied today:

A projectile which is carried by a uniform horizontal motion compounded with a naturally

accelerated vertical motion describes a path which is a semi-parabola.

Go deeply, the knowledge of parabolas is the key for everything especially in engineering

field, graphs of parabola show many information which tends to make our world getting more

modern one day.



Page 19 of 19

Reflection

∯ ∑ ∐ ⋂ − ⏞

±√ − *

+

⇒

∭

∫

∐

→

←

………… ………… ………… a complicated math equation combined with few of single

equation and finally had derived into this level.

I had use about

weeks of time to complete this project. On the ,

→ | -,

I had try all the best to answer the questions by finding the method, guide and the most

important – understanding. For this, I finally knew that why I am learning ∑

4 ℎ…, the important is to be improved and to maximize

the human brain, the power to solve all the questions that is naturally unsolved. With this

learning, we may able to find out the ℎ that is still in the ―Unsolved list‖ in

the log book written by Qian Xuesen, the Aerodynamics Scientist who had develop missiles,

rockets, and flying technology a great inventor, discover person, a historical future person

that can change the world.

∬∭∬ , , ± ∑∏∐⋃⋂ , , ‖√

√

‖,

Doing, deriving, solving, trying, continuous………

Overnight, seeking the solutions, finally ask for the pro,

Found out the answer; blow out with a feeling,

The nature of forest blossom with tears and cheers………

( ) ∑(

)

Understand the solutions, trying to applying on the questions,

And….finally you are the success one!


additional mathematics project work 2013 wpkl

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