additional mathematics project work 2013 wpkl
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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
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Additional Mathematics Project Work W.P.K.L. 2013
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
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Additional Mathematics Project Work W.P.K.L. 2013
Page 3 of 19
Part 1
1.1 Photos related to parabola
McDonald’s Sign Board Satellite Antenna
Bellagio's Fountains
Leeds Grand Mosque
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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.
Additional Mathematics Project Work W.P.K.L. 2013
P
age 4 o
f 19
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Additional Mathematics Project Work W.P.K.L. 2013
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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 = ±
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Additional Mathematics Project Work W.P.K.L. 2013
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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
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Additional Mathematics Project Work W.P.K.L. 2013
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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
From question, the information we have are:
The bird can keep it balance up to a maximum
gradient with magnitude of 2
−
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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
From question, the information we have are:
The construction cost for the required partition
is RM 100 per meter square
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1.6 Determination of capacity of the air conditioning
From question, the information we have are:
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
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Additional Mathematics Project Work W.P.K.L. 2013
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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
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Additional Mathematics Project Work W.P.K.L. 2013
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2.2 Determine
From question, the information we have are:
( ℎ 0ℎ)
ℎ
ℎ
( ℎ 0ℎ)
ℎ ( ℎ 0ℎ)
( ℎ 0)
[ ( ) 0]
ℎ ℎ
( ) ( )
The rate of change of volume,
is
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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.
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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
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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.
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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.
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Additional Mathematics Project Work W.P.K.L. 2013
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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
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Additional Mathematics Project Work W.P.K.L. 2013
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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.
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Additional Mathematics Project Work W.P.K.L. 2013
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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.
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Additional Mathematics Project Work W.P.K.L. 2013
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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!
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