Add Math Project

NUR FARRAH HANIS BT SALIHAN

CONTENT

No. Aspects Pages1 Acknowledgement 22 Objective 33 Moral values 44 Introduction 5

5 Part 11.1 Observations of parabola1.2 About parabola1.3 Equations of parabola1.4 Coordinates of points1.5 Construction budget1.6 Capacity of air conditioning required

7

6 Part 22.1 Observations of cylindrical beaker2.2 Rate of change of volume

17

7 Part 33.1 Gradients of point 3.2 Cost3.3 Progression

20

8 Further Exploration 249 Conclusion 28

ACKNOWLEDGEMENTADDITIONAL MATHEMATICS PROJECT 2013Page 1


First and foremost I would like to thank my Additional Mathematics teacher, Miss Azura, who has been guiding my whole class with the highest level of patience in this subject. I am forever grateful and in debt t o her. She has done her level best and if I failed it is I who have failed her but if I succeed, all the credits goes to her.

Next I would like to thank my helpful peers who have aided me throughout the process of completing this project. Among them are Teivardashni, Eugene Chee and Dhawina. These are the people I look for when I have problems regarding the project and without them I would not have been able to complete this project.

I would also like to express my gratitude to the school and the Ministry of Education of Malaysia for giving me a chance to relearn the topic parabola by making this project compulsory. This has been a very helpful experience for me as I have now delved further into this topic and came out with deeper understanding on how it works.

Last but not least, I would like to take this opportunity to thank my mother who have provided for me while I complete this folio. Without the internet, printer and paper, my work would have gone to waste.

~ THANK YOU ~

OBJECTIVE

ADDITIONAL MATHEMATICS PROJECT 2013Page 2


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

2. To improve thinking skills.

3. To promote effective mathematical communication.

4. To develop mathematical knowledge through problem solving in a way that increases students’ interest and confidence.

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

6. To provide learning environment that stimulates and enhances effective learning.

7. To develop positive attitude towards mathematics.

MORAL VALUESADDITIONAL MATHEMATICS PROJECT 2013Page 3


HARDWORK AND DETERMINATIONIn order to get things done, hard work and determination is the key. I have worked hard and burnt midnight oil to complete this project and my hard work and determination was worth it when my folio is done.

DILIGENCEI have put a huge amount of effort on this project. I spent a lot of time researching and solving the problems given and it goes to show that diligence is extremely important in getting to the finish line as I managed to finally complete this assignment.

TEAMWORKWithout the aid of my friends and teachers, I highly doubt I would have been able to complete this project. Therefore it is proven that teamwork is of utmost importance. My friends and I have sat in groups to discuss the solution to many of the questions given. With everyone’s different input and perspective we have a wider range of information in our folios and a better understanding of the topic being discussed.

INTRODUCTION



A parabola is a two-dimensional, mirror symmetrical curve which is approximately u-shaped when orientated, 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.

To form a parabola according to the ancient Greek definition, you would start with a line (the directrix) and a point (the focus) off to one side. The parabola is the curve formed from all the points ( x , y ) that are equidistant from the directrix and the focus. The line perpendicular to the directrix and passing through the focus (that is, the line that splits the parabola up the middle) is called the axis of symmetry. The point on this axis which is exactly midway between the focus and the directrix is the vertex. The vertex is the point where the parabola changes direction.

Regular or vertical parabola (blue), with the focus (green) inside the parabola, the directrix (purple) below the graph, the axis of symmetry (red) passing through the focus and perpendicular to the directrix, and the vertex (orange) on the graph.

Side ways or horizontal, the parabola (blue), with the focus (green) inside the parabola. The directrix (purple) to the left of the graph, the axis of symmetry (red) passing through the focus and the perpendicular through the directrix and the vertex (orange) on the graph.

Another way to describe 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. Lastly, a parabola can also be described as a graph of a quadratic function, such as y = x² or f(x) = ax² + bx + c, where a, b and c are real numbers.

The name parabola is derived from a New Latin term that means something similar to “compare” or “balance”, and refers to the fact that the distance from the parabola to the focus is always equal to the distance from the parabola to the directrix.Parabolas have a property that, if they are made of materials 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 into a parallel beam, leaving the parabola parallel to the axis of symmetry. The same effect occurs with sound and other forms of energy. This reflective property is the basis of many practical uses of parabolas.

Parabola has many important applications including a parabola antenna, parabolic microphone, automatic headlight reflectors, the designs of ballistic missiles, etc. they are frequently used in physics, engineering and many other areas.

The parabola was explored by Menaechmus (360 BC – 320 BC), who was a pupil of Plato and Eudoxus. He was trying to duplicate the cube by finding the side of the cube that has an area double the cube. Instead, he solved it by finding the intersection of two parabolas. Euclid (325 BC – 265 BC) wrote about the parabola. Apollonius (260 BC – 190 BC) named the parabola. Pascal (1638 – 1662) considered the parabola as a projection of a circle. Galileo (1564 – 1642) showed the projectiles falling under uniform gravity follows parabolic paths. Gregory (1638-1675) and Newton (1643 – 1727) considered the properties of a parabola.

The idea that a parabolic reflector could produce an image was already known well before the invention of the reflecting telescope. Designs were proposed in the early to mid-seventtenth century by many mathematicians including Rene 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 to apply the knowledge of parabolas to lead us to a better future of technology as our past mathematicians had done.



1.1

ADDITIONAL MATHEMATICS PROJECT 2013Page 7Sri Saujana Bridge

set of all points, P in a planeequidistancefixed point (focus)fixed line (directrix)

Definition

y² = ±4axabout x-axis x² = ±4ayabout y-axis

Equation


1.2

PARABOLA


Putrajaya Mosque

McDonald’s Sign Board

Satellite AntennaRoller coaster


1.3

Case 1Given maximum value, q = 20, axis of symmetry, p = 0 and roots of equation, x = 25, -25

y = a (x + p)² + q

sub p = 0 and q = 20

y = a (x + 0)² + 20

y = ax² + 20 ①

x = 25, -25

equation is:

y = a (x + 25) (x – 25)

y = a (x² - 625)


Sketching of Parabola

When the vertex is not at origin,

(y – k)² = ± 4a (x – h)(x – h)² = ± 4a (y – k)

Where (h , k) is the vertex


y = ax² - 625a ②

by comparing ① and ②,

20 = -625a

−2065 = a

a = 0.032

sub a = -0.032 into ①

y = -0.032 x² + 20

Case 2Switching the parabola to another position by moving x to the right by 10 units given value

q = 20

axis of symmetry, x = 0 + 10

x = 10

x – 10 = 0

hence, p = -10

new roots, x = 25 + 10, -25 + 10

x = 35, -15



y = a (x + p)² + q

sub p = -10 and q = 20

y = a (x – 10)² + 20

y = a (x² - 20x + 100) + 20

y = ax² - 20ax + 100a + 20 ①

x = 35, -15

equation is,

y = a (x – 35) (x + 15)

y = a (x² - 20x + 525)

y = ax² - 20ax + 525a ②

by comparing ① and ②,

100a + 20 = -525a

20 -625a

−20625 = a

a = -0.032

sub a = -0.032 into ①

y= -0.032x² - 20 (-0.032)x + 100 (-0.032) + 20

= -0.032x² + 0.64x + 16.8

Case 3Inverting the parabola in case 1

Maximum point changes to minimum point,

q = -20

p = 0

roots of equations does not change, x = 25, -25

y = a (x + p)² +

sub p = 0, q = -20

y = a (x + 0)² - 20

y = ax² - 20 ①



x = 25, -25

equation is,

y = a (x + 25) (x – 25)

y = ax² - 625a ②

by comparing ① with ②,

-20 = -625a

20625 = a

a = 0.032

sub a = 0.032 into ①

y = 0.032x² - 20

Case 4Inverting the parabola in case 2

Maximum point changes to minimum point,

q = -20

p = -10

roots of equation does not change, x = 35, -15

y = a (x + p)² + q

sub p = -10, q = -20

y = a (x – 10)² - 20

y = a (x² = 20x + 100) – 20

y = ax² - 20ax +100a – 20 ①



x = 35, -15

equation is,

y = a (x – 35) (x + 15)

y = a (x² + 20x – 525)

y = ax² + 20x – 525 a ②

by comparing ① with ②,

100a – 20 = - 525a

-20 = -62a

20625 = a

a = 0.032

sub a = 0.032 into ①,

y = 0.032x² - 20 (0.032x + 100 (0.032)) – 20

y = 0.032x² - 0.64x – 16.8

Conclusion:

From the equation of the parabolas, when the parabola are inverted, the values of a, b, and c is multiplied by -1. When the parabolas are inverted, the value of q is also multiplied by -1 whereas the value of p does not change.

1.4

From the question, the information obtained are:

The bird can keep it balanced up to a maximum gradient with the magnitude of 2.

y = - 4125x² + 20

gradient of curve = dydx , so



dydx = -

8125x

Magnitude gradient = | 2 |

2 = - 8125x

x = - 1254

y = - 4125 (-

1254 )² + 20

y = - 454

-2 = - 8125x

x = 1254

y = - 4125 (

1254 )² + 20

y = - 454

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

slipping downwards is up to the coordinates (- 1254 , -

454 ) or (

1254 , -

454 )

1.5


The construction cost for the required partition is RM 100 per metre square



To calculate the required area, the method involved are integration of curve to find bounded are along x-axis and area of triangle.

Area

= ¿

= [2083 ] + [96]

= 4963 unit²

Construction cost,

4963 × RM100 = RM 16 533.33

The budget required to construct the shaded partition is RM 16 533.33

1.6


- The temperature to keep constant = 24˚C



- 1000 Btu/hr = 0.393 hp- N = 100 people

Air conditioner capacity = 23 × volume + N × 500

y = - 4 x ²125 + 20

x² = 625 - 1254 y

volume of dome

= π∫a

b

x2dx

= π∫0

20

(625−1254 y)dy= 6250π unit³

Air conditioner capacity

= 23× (6250 π )+(100 )×500

= 63089.969 Btu/hr

1000 Btu/hr = 0.393 hp

63089.969 Btu/hr = Z hp

Z = 63089.969 × 0.393 ÷ 1000 = 24.79 hp

The capacity of air condidtioner required is 24.79 hp



2.1

When a cylindrical beaker is filled with water until half full and circular motion of stirring process begins, the observation 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, the height of the spiral motion (hspiral) increases toward bottom of cylinder beaker while the water level at the centre is displaced mre results increase in water level near the wall of the beaker.

3. So, as the spiral motion of water goes deeper, the higher the displacement of water level, near the wall of the beaker.

4. The vertical cross-section of water level forms a parabola as it goes deeper. At the same time, the water level near the wall of the beaker is increased to maintain the volume of water in the beaker that is constant. So, spiral movement of water forms various parabolas with different heights

2.2




V = 𝜋 ( 3h² + 10h )

dhdt = 2 mms ᶦˉ

H = 8 mm

V = 𝜋 (3h² + 10h)

dVdh = 𝜋 (3h² + 10h)

= 𝜋 (6h + 10)

= 𝜋 [6 (8) + 10]

= 58𝜋 mm

dVdt =

dVdh ×

dhdt

dVdt = (58𝜋) (2)

= 116𝜋 mm³ s ᶦˉ

The rate of change of volume, dVdt is 116𝜋 mm³ s ᶦˉ



3.1

Given y = 1xx ²+6 and distance between each cable is 0.5 m

y=18x2+6

dydx

=14x



When x = 0.5, dydx

=14

(0.5 )=18 = 0.125

When x = 1.0, dydx

=14

(1.0 )=14 = 0.25

When x = 1.5, dydx

=14

(1.5 )=38 = 0.375

When x = 2.0, dydx

=14

(2.0 )=12 = 0.5

Hence the gradient of the parabola form an arithmetic progression of

a = 0.125

d = 0.125

3.2

Cost of 19th cable from a progression



T 19=198

dydx

=14x

198

=14x

x=192

y=18 ( 192 )

2

+6

= 55332

1 metre of cable costs RM 100

Thus, length of 19th cable is

= 55332 m

And the cost is

= RM 100 × 55332

= RM 1728.13

3.3



Determining which cable needs to be repaired

4218

=18x2+6

x=17dydx

=14

(17 )=4 .25

Using Tn = a + (n − 1)d

4.25 = 0.125 + (n − 1)(0.125)4.26 n = 34

The cable that needs repairing is cable no. 34 on the right from the center of the bridge.

Further Exploration



Parabola in Physics (parabolic motions of projectiles)A projectile is an object upon which the only force is gravity. Gravity, being a downward force, causes a projectile to accelerate in the downward direction. The force of gravity could never alter the horizontal velocity of an object since perpendicular components of motion are independent of each other. A vertical force does not effect a horizontal motion. The result of a vertical force acting upon a horizontally moving object is to cause the object to deviate from its otherwise linear path. This is depicted in the animation below.

According to Newton's law of inertia, an object in motion in a horizontal direction would continue in its horizontal motion with the same horizontal speed and direction unless acted upon by an unbalanced horizontal force. The animation above shows a green sphere moving to the right at constant speed. The horizontal distance traveled in each second is a constant value. The red sphere undergoes a vertically accelerated motion which is typical of an object upon which only the force of gravity acts. If these two motions are combined - vertical free fall motion and constant horizontal motion - then the trajectory will be that of a parabola. An object which begins with an initial horizontal velocity and is acted upon only by the force of gravity will follow the path of the blue sphere. It will travel the same horizontal distance in each consecutive second but will fall vertically a greater distance in each consecutive second. The result is a parabolic path as shown in the animation above.



Based on the diagram:

1. the Vᵪ, is constant, because there is no horizontal force acting on it.

2. the Vᵧ, 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 trajectory Vᵧ = 0 but Vᵪ = constant

4. acceleration is constant and velocity is moving downwards. Therefore a = -g



The initial velocity

If the projectile is launched with an initial velocity v0, then it can be written as

.

The components v0x and v0y can be found if the angle, ϴ is known:

,

.

If the projectile's range, launch angle, and drop height are known, launch velocity can be found by

.

The launch angle is usually expressed by the symbol theta, but often the symbol alpha is used.

Kinematic quantities of projectile motion

In projectile motion, the horizontal motion and the vertical motion are independent of each other; that is, neither motion affects the other.

Acceleration

Since there is no acceleration in the horizontal direction velocity in horizontal direction is constant which is equal to ucosα. The vertical motion of the projectile is the motion of a particle during its free fall. Here the acceleration is constant, equal to g.[1] The components of the acceleration:

,

.Velocity

The horizontal component of the velocity remains unchanged throughout the motion. The vertical component of the velocity increases linearly, because the acceleration is constant. At any time t, the components of the velocity:

,

.

The magnitude of the velocity (under the Pythagorean theorem):

.Displacement

At any time t, the projectile's horizontal and vertical displacement:

,

.



Maximum height of a projectile

The highest height which the object will reach is known as the peak of the object's motion. The increase of the height will last, until , that is,

.

Time to reach the maximum height:

.

From the vertical displacement the maximum height of projectile:

.

Maximum distance of projectile

The horizontal range d of the projectile is the horizontal distance the projectile has travelled when it returns to its initial height (y = 0).

.

Time to reach ground: alue when

,

which necessarily corresponds to

,

or

.



CONCLUSION

After weeks of hard work and research, my journey in completing my Additional Mathematics Project 2013 has finally come to an end. Throughout the whole process I have learnt many things relating to the topic I have chosen; parabola and its importance in our daily life.

In nature, approximations of parabolae and paraboloids (such as catenary curves) are found in many diverse situations. The best-known instance of the parabola in the history of physics is thetrajectory of a particle or body in motion under the influence of a uniform gravitational field without air resistance (for instance, a baseball flying through the air, neglecting air friction).

The parabolic trajectory of projectiles was discovered experimentally by Galileo in the early 17th century, who performed experiments with balls rolling on inclined planes. He also later proved thismathematically in his book Dialogue Concerning Two New Sciences.[8][d] For objects extended in space, such as a diver jumping from a diving board, the object itself follows a complex motion as it rotates, but the center of mass of the object nevertheless forms a parabola. As in all cases in the physical world, the trajectory is always an approximation of a parabola. The presence of air resistance, for example, always distorts the shape, although at low speeds, the shape is a good approximation of a parabola. At higher speeds, such as in ballistics, the shape is highly distorted and does not resemble a parabola.

Another hypothetical situation in which parabolae might arise, according to the theories of physics described in the 17th and 18th Centuries by Sir Isaac Newton, is in two-body orbits; for example the path of a small planetoid or other object under the influence of the gravitation of the Sun. Parabolic orbits do not occur in nature; simple orbits most commonly resemble hyperbolas or ellipses. The parabolic orbit is the degenerate intermediate case between those two types of ideal orbit. An object following a parabolic orbit would travel at the exact escape velocity of the object it orbits; objects in elliptical or hyperbolic orbits travel at less or greater than escape velocity, respectively. Long-period comets travel close to the Sun's escape velocity while they are moving through the inner solar system, so their paths are close to being parabolic.

Approximations of parabolae are also found in the shape of the main cables on a simple suspension bridge. The curve of the chains of a suspension bridge is always an intermediate curve between a parabola and a catenary, but in practice the curve is generally nearer to a parabola, and in calculations the second degree parabola is used.[9][10] Under the influence of a uniform load (such as a horizontal suspended deck), the otherwise catenary-shaped cable is deformed toward a parabola. Unlike an inelastic chain, a freely hanging spring of zero unstressed length takes the shape of a parabola. Suspension-bridge cables are, ideally, purely in tension, without having to carry other, e.g. bending, forces. Similarly, the structures of parabolic arches are purely in compression.

Paraboloids arise in several physical situations as well. The best-known instance is the parabolic reflector, which is a mirror or similar reflective device that concentrates light or other forms ofelectromagnetic radiation to a common focal point, or conversely,


http://en.wikipedia.org/wiki/Focus_(optics)

http://en.wikipedia.org/wiki/Electromagnetic_radiation

http://en.wikipedia.org/wiki/Parabolic_reflector

http://en.wikipedia.org/wiki/Parabola#cite_note-14

http://en.wikipedia.org/wiki/Parabola#cite_note-Troyano-13

http://en.wikipedia.org/wiki/Catenary

http://en.wikipedia.org/wiki/Suspension_bridge

http://en.wikipedia.org/wiki/Comet

http://en.wikipedia.org/wiki/Hyperbolic_orbit

http://en.wikipedia.org/wiki/Elliptical_orbit

http://en.wikipedia.org/wiki/Escape_velocity

http://en.wikipedia.org/wiki/Degeneracy_(math)

http://en.wikipedia.org/wiki/Ellipse

http://en.wikipedia.org/wiki/Hyperbola

http://en.wikipedia.org/wiki/Parabolic_orbit

http://en.wikipedia.org/wiki/Sun

http://en.wikipedia.org/wiki/Two-body_orbit

http://en.wikipedia.org/wiki/Sir_Isaac_Newton

http://en.wikipedia.org/wiki/Hypothetical

http://en.wikipedia.org/wiki/Center_of_mass



http://en.wikipedia.org/wiki/Mathematical

http://en.wikipedia.org/wiki/Galileo

http://en.wikipedia.org/wiki/Friction

http://en.wikipedia.org/wiki/Air_resistance

http://en.wikipedia.org/wiki/Gravitational_field

http://en.wikipedia.org/wiki/Trajectory

http://en.wikipedia.org/wiki/Physics

http://en.wikipedia.org/wiki/Catenary


collimates light from a point source at the focus into a parallel beam. The principle of the parabolic reflector may have been discovered in the 3rd century BC by the geometer Archimedes, who, according to a legend of debatable veracity,[11] constructed parabolic mirrors to defend Syracuse against the Roman fleet, by concentrating the sun's rays to set fire to the decks of the Roman ships. The principle was applied to telescopes in the 17th century. Today, paraboloid reflectors can be commonly observed throughout much of the world in microwave and satellite-dish receiving and transmitting antennas.

In parabolic microphones, a parabolic reflector that reflects sound, but not necessarily electromagnetic radiation, is used to focus sound onto a microphone, giving it highly directional performance.

Paraboloids are also observed in the surface of a liquid confined to a container and rotated around the central axis. In this case, the centrifugal force causes the liquid to climb the walls of the container, forming a parabolic surface. This is the principle behind the liquid mirror telescope.

Thus it is clear, the importance of parabola in our daily life. With this I end my conclusion with a final note in hoping that mankind uses the knowledge and retrospect of parabola wisely into leading the world as we know into a modern fairytale.

THE END


http://en.wikipedia.org/wiki/Liquid_mirror_telescope

http://en.wikipedia.org/wiki/Liquid_mirror_telescope

http://en.wikipedia.org/wiki/Centrifugal_force

http://en.wikipedia.org/wiki/Sound

http://en.wikipedia.org/wiki/Parabolic_microphone

http://en.wikipedia.org/wiki/Microwave

http://en.wikipedia.org/wiki/Telescope

http://en.wikipedia.org/wiki/Roman_Empire

http://en.wikipedia.org/wiki/Syracuse,_Italy


http://en.wikipedia.org/wiki/Archimedes

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