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Math Internal Assessment: Investigating Parametric Equations in Motion Problems

Subject: Mathematics SL Name of Candidate: Martin Lai Candidate Number: 2934-19 Teacher: Mr. Curic Session: May 2014 School Name: Monarch Park Collegiate School IB Number: 2394 Word Count: 3,484 Date: Wednesday, March 26th, 2014 Page 1 of 16

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Parametric equations are a set of equations that describe the co-ordinates of a function or curve at a specific point in terms of a parameter, usually t to represent time, or another variable. This math exploration will be focusing on parametric equations of curves in the two-dimensional xy plane. Hence, these parametric equations take the form: 𝑥 = 𝑥(𝑡) 𝑦 = 𝑦(𝑡) In this way, we can describe the positions of both the x and y co-ordinates as functions of a third variable, known as the parameter. By using these parametric equations, the position of an object can be written as (x,y) = (x(t), y(t)). This is particularly useful for describing the specific location of a moving object at any time t that is moving along a vector or curved path. However, parametric equations also have many other applications that will be explained later in this exploration. First, we will investigate the method of parameterizing equations. Then, we will investigate the properties of parametric equations to model circular motion and vectors. Last, calculus concepts will be implemented through the differentiation and integration of the parametric equations to develop a better understanding of how parametric equations can reveal valuable information about motion and curves. Above all, real life situations will be integrated throughout this exploration in order to give a sense to the practical purposes and connections we can make to the real world and our own lives. The aim of this exploration will be to explore parametric equations and how they can model and reveal properties about motion problems. My rationale for choosing parametric equations as the topic for this investigation is due to my interest with physics. After completing two years of physics classes I have started to apply physics concepts and connections to other areas of my life and this has been an enlightening experience that gave me a new perspective of the world. The area of physics that deals with the motion of objects is known as Kinematics. After some research, I found that we can also study motion of an object through the use of mathematical modeling and equations, such as parametric equations. Being able to know the location of a specific object at any time is an immense asset to understanding the world around us, such as the movement of boats, airplanes and animals, but also on a subatomic level with particles. It is the sheer practicality of parametric equations that made this an appealing choice for an investigation. Moreover, parametric equations also integrate many concepts that I have been learning throughout high school such as curve sketching, functions, calculus and vectors, allowing me to apply those concepts to a new area of study and make connections to them in a meaningful way.

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First, we will investigate how to put basic functions into parametric form, which is known as parameterizing1 an equation. Example: 𝑦 = 𝑓(𝑥) = −0.5𝑥 2 + 3𝑥 + 2. This is a quadratic function, as it is in the form y = ax2 + bx + c. To parameterize this equation in terms of the variable t, let 𝑥 = 𝑡, such that: 𝑦 = 𝑓(𝑡) 𝑎𝑛𝑑 𝑓(𝑡) = −0.5𝑡^2 + 3𝑡 + 2. We are left with the parametric equations: 𝑥 = 𝑡

𝑡≥0

𝑦 = −0.5𝑡 2 + 3𝑡 + 2 With a simple substitution we can now determine the position of an object at time t on the curve given by the function f(x) in terms of both x and y. For example, suppose that these parametric equations represent a tennis ball’s position as it moves through the air:

This may seem like a regular parabolic curve that models the motion of an object; however, while the xaxis usually represents time, here we can describe the x-axis as its horizontal distance. Being able to

1

"Parametric Equations and Curves." Calculus II. N.p., n.d. Web. .

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keep track of both horizontal and vertical distances separately allows for a very specific understanding of the location of the ball at a certain time. If I wanted to find out how far I could hit a tennis ball in 2 seconds, I could simply substitute 𝑡 = 2 to the parametric equations: 𝑥=2 𝑦 = −0.5(2)2 + 3(2) + 2 = 6 The position of the tennis ball would be at (3,6), and if we were to define distance in this situation in metres, the tennis ball would be 3 meters away from me horizontally and 6 meters above the ground after 2 seconds of flight. Another major addition with the use of parametric equations is the use of orientation arrows2. Notice that the arrows drawn on the parameterized curve point towards a specific direction. These arrows tell us the direction of motion that the object is experiencing at any t value. Using the example above, at 𝑡 = 2𝑠 the tennis ball is moving up and to the right. The orientation tells us the direction of a moving object as the value of 𝑡 increases. Parametric equations can also be used to model circular motion. We can recall that the equation of a circle is given by: 𝑥2 + 𝑦2 = 𝑟2 Working with this equation, it is very difficult to solve for the co-ordinates of a specific point without having the value of the radius and either the x or y co-ordinate. The solution to this problem is parametric equations. First, we consider a right triangle, whose hypotenuse starts at the center of the circle and extends the length of the radius:

2

http://www.mcs.sdsmt.edu/tkowalsk/notes/Basic-facts-about-parametric-equations.pdf

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To find the co-ordinates of point A on the outside of the circle, I revisited trigonometry that I learned in previous grades, which states: 𝒔𝒊𝒏𝜭 =

𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆(𝒐) 𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕(𝒂) 𝒂𝒏𝒅 𝒄𝒐𝒔𝜭 = 𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆(𝒉) 𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆(𝒉)

Looking at the diagram, side 𝑜 is the y co-ordinate of point A, and h is the radius of the circle. Therefore, we can say: 𝒚

𝒔𝒊𝒏𝜭 = 𝒓

𝒚 = 𝒓𝒔𝒊𝒏𝜭. Since the 𝑎 can be described as the x co-ordinate of point A, we can likewise say 𝒙 = 𝒓𝒄𝒐𝒔𝜭. Therefore, the parametric equations for a circle can be described as: 𝒚 = 𝒓𝒔𝒊𝒏(𝜭) 𝒂𝒏𝒅 𝒙 = 𝒓𝒄𝒐𝒔(𝜭) By manipulating these parametric equations, we can model many different types of circular motion. We can also incorporate the angular speed (how fast an object rotates around a circle) into parametric equations3. Noting that: 𝒂𝒏𝒈𝒍𝒆 𝒐𝒇 𝒓𝒐𝒕𝒂𝒕𝒊𝒐𝒏 (𝜭) = 𝒂𝒏𝒈𝒖𝒍𝒂𝒓 𝒔𝒑𝒆𝒆𝒅 (𝝎)𝒙 𝒕𝒊𝒎𝒆 (𝒕), We can say that: 𝒚 = 𝒓𝒔𝒊𝒏(𝝎𝒕) 𝒂𝒏𝒅 𝒙 = 𝒓𝒄𝒐𝒔(𝝎𝒕) The parameter is now in terms of time, 𝑡. For example, we can model the movement of point A moving along a bicycle wheel with the center on the origin, shown below:

3

"Parametric Equations." University of Washington. N.p., n.d. Web. .

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The radius of the bicycle 2 units, so we substitute 2 for r in our parametric equations: 𝑦 = 2𝑠𝑖𝑛(𝛳) 𝑎𝑛𝑑 𝑥 = 2cos(𝛳) If we assume that I am riding the bicycle and pedaling at a constant rate such that the angular speed is 𝜋 4

radians/sec, we have the parametric equations 𝜋

𝑦 = 2𝑠𝑖𝑛( 4 𝑡)

{ 𝜋 𝑥 = 2𝑐𝑜𝑠( 4 𝑡) Let us graph these equations to observe the motion of Point A moving along the circle:

t

𝒙 = 𝟐𝐜𝐨𝐬(

𝝅 𝒕) 𝟒

𝒚 = 𝟐𝒔𝒊𝒏(

1s

𝑥 = 2 cos (4 (1)) = 2 cos ( 4 ) = 2( 2 ) =√2 ≈ 1.4

2s 3s 4s 5s 6s 7s 8s

0 -1.4 -2 -1.4 0 1.4 2

𝜋

𝜋

√2

𝜋

𝜋

𝝅 𝒕) 𝟒 √2

𝑦 = 2𝑠𝑖𝑛 ( 4 (1)) = 2 sin ( 4 ) =2( 2 ) = √2 ≈ 1.4 2 1.4 0 -1.4 -2 -1.4 0

Note how at t=8s, Point A has moved back to its original position. If we wanted to only look at one rotation of the point on the wheel, we could set the restriction as: 0 ≤ 𝑡 ≤ 8𝑠

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Immediately, I notice that according to the orientation arrows on the parameterized curve, the direction of rotation is counterclockwise. Obviously, this does not fit the real life situation since the wheel would be rotating backwards. How can we change the direction of the orientation arrows? Notice 𝝅

that the positive angular speed of 𝟒 𝒓𝒂𝒅/𝒔𝒆𝒄 always corresponds to a counterclockwise direction; to reverse this, we simply make the angular speed negative. From learning about transformations of functions, I know that this is the same as a flip across the y-axis. Referring to the graph above, it is obvious that this would make the orientation arrows move in a clockwise direction. 𝝅

𝝅

𝒚 = 𝟐𝒔𝒊𝒏(− 𝟒 𝒕) 𝒂𝒏𝒅 𝒙 = 𝟐𝐜𝐨𝐬(− 𝟒 𝒕). This also helps us conclude that if angular speed, 𝜔 > 0, then the direction of motion will be counterclockwise, and clockwise when 𝜔 < 0. This type of modelling can be very interesting and useful for investigating the circular motion of objects, and can be applied to rotating engines, or even knowing the location of a planet in its orbit at a certain time by building off of this concept of parametric equations. In further study, these parametric equations can also be modified using vertical and horizontal transformations, or using different trigonometric operators aside from sine and cosine. Parametric equations can also be applied to vector problems. Recall that the general form of a vector equation for the position object moving at constant velocity is: ⃗ 𝒇𝒐𝒓 𝒕 ≥ 𝟎 ⃗⃗𝒓 = 𝒂 ⃗ + 𝒕𝒃 𝑎 is the initial position vector,⃗⃗𝑏 is the velocity vector, and⃗⃗𝑟 is the object’s position relative to the origin at time 𝑡. In essence, a position vector is the vector that represents the position of Point P relative to the origin at (0,0), the velocity vector tells us the direction of motion as well as its speed, and vector ⃗⃗𝑟 represents my displacement, or change in position, as I walk. For example, suppose I wanted to model my motion on a morning walk relative to my house at point O (0,0). At 9:00 AM, I start at Point P, which is the start of the hiking trail, located at the point (1,2). Using an online graphing tool, I was able to model this situation using vectors:

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The position vector 𝑎 is therefore [1,2] as point P is 1 metre East and 2 metres North of the origin (0,0), which in this case is my house. I walk at a constant velocity in a straight line down the path, which can be modelled by the velocity vector 𝑏⃗, [2,-1]. ⃗: ⃗ + 𝒕𝒃 Now I can create a vector equation in the form ⃗⃗𝒓 = 𝒂 𝒙 𝟏 𝟐 𝟏 𝟐 ⃗ = [ ] + 𝒕 [ ] , 𝒐𝒓 [𝒚] = [ ] + 𝒕 [ ] 𝒓 𝟐 −𝟏 𝟐 −𝟏

To change this vector equation into parametric equations, we deal with the x and y components within the vector equation separately: 𝒙 𝟏 𝟐 [𝒚] = [ ] + 𝒕 [ ] 𝟐 −𝟏 Parametric equations: 𝒙 = 𝟏 + 𝟐𝒕 𝒚=𝟐−𝒕 Therefore my position at time 𝑡, in hours after 9:00 AM, is given by: (x,y) → (1+2t, 2-t) If I wanted to find my position relative to my house at the end of a 2 hour walk, I would simply substitute t=2 into the parametric equations:f 𝒙 = 𝟏 + 𝟐𝒕 = 𝟏 + 𝟐(𝟐) = 𝟓 𝒚 = 𝟐 − 𝒕 = 𝟐 − (𝟐) = 𝟎 Hence I would be at the position (5,0) or directly East of my house. But how would we model this situation with a single equation? Since parametric equations set x and y in terms of a third variable (the parameter), a conjecture can be made that we can simply solve for the parameter, in this case t, in order to end up with a single equation in terms of x and y only. 𝒙 = 𝟏 + 𝟐𝒕 𝒚=𝟐−𝒕

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First, solve for the parameter in one of the equations: 𝒙 = 𝟏 + 𝟐𝒕 → 𝒕 =

𝒙−𝟏 𝟐

Next, substitute this for 𝑡 in the second equation: 𝒚=𝟐−𝒕 𝒙−𝟏 ) 𝟐

𝒚=𝟐−(

Multiplying both sides by 2 to eliminate denominator: 𝟐𝒚 = 𝟒 − (𝒙 − 𝟏)

→

𝟐𝒚 = 𝟒 − 𝒙 − 𝟐 →

𝟐𝒚 = 𝟒 − 𝒙

Hence we are left with the equation: 𝒙 + 𝟐𝒚 = 𝟒 Since this equation is in terms of x and y only, which are the variables used in the Cartesian coordinate system, it is said to be put into Cartesian form. This can be a useful form because it allows for y to be defined in terms of x for a clear relationship to be shown between the two variables. In the next section we will attempt to incorporate calculus concepts into our understanding of parametric equations. First, we will look at the differentiation of parametric equations: 𝑑𝑥 = 𝑥 ′ (𝑡) 𝑑𝑡

𝑑𝑦 = 𝑦 ′ (𝑡) 𝑑𝑡

The derivatives, x’(t) and y’(t) are known as velocity equations as they reveal information about the motion of an object. x’(t) and y’(t) are respectively called the horizontal and vertical components of velocity. Derivatives are functions that measure the instantaneous rate of change, or slope of the tangent at one point on the graph. A tangent is simply a line that touches only one point on a curve. Like any line, we could create an equation that tells us about the motion of the object at that specific point. These equations can tell us the direction of travel of an object on the curve. We know that if the slope of a tangent is positive, then function is increasing at the time and moving right and upwards. Oppositely, a negative tangent slope means the object is moving left or downwards: 𝑥 ′ (𝑡) > 0 → 𝑀𝑜𝑣𝑖𝑛𝑔 𝑟𝑖𝑔ℎ𝑡 { ′ 𝑥 (𝑡) < 0 → 𝑀𝑜𝑣𝑖𝑛𝑔 𝑙𝑒𝑓𝑡

𝑦 ′ (𝑡) > 0 → 𝑀𝑜𝑣𝑖𝑛𝑔 𝑈𝑝𝑤𝑎𝑟𝑑𝑠 { ′ 𝑦 (𝑡) < 0 → 𝑀𝑜𝑣𝑖𝑛𝑔 𝐷𝑜𝑤𝑛𝑤𝑎𝑟𝑑𝑠

This is a simple way to confirm the direction of motion at any time 𝑡 without graphing.

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Let us revisit the parametric equations for the tennis ball example earlier in this exploration:

𝑥(𝑡) = 𝑡

𝑡≥0

𝑦(𝑡) = −0.5𝑡 2 + 3𝑡 + 2 𝒙’(𝒕) = 𝟏 𝒚’(𝒕) = −𝒕 + 𝟑 From this alone we see that x’(t) will always be positive, and that the object will always be moving right. This makes sense if we look at the graph as the value of x is always increasing as t increases. Now, let us find the vertical direction of the tennis ball when 𝑡 = 2𝑠 and 𝑡 = 5𝑠. t = 2s: 𝒚’(𝟐) = −(𝟐) + 𝟑 = 𝟏, which confirms that the ball is moving upwards at 2s. 𝑡 = 5𝑠: 𝒚’(𝟓) = −(𝟓) + 𝟑 = −𝟐, which confirms that the ball is moving downwards at 5s. We also know from calculus that 𝒙’(𝒕) = 𝟎 𝒂𝒏𝒅 𝒚’(𝒕) = 𝟎 are turning points, since they correspond to where a local maximum or minimum point is found, also meaning that there is a change in direction. This allows us to find the intervals of upwards and downwards motion. 𝒚’(𝒕) = 𝟎 → 𝟎 = −𝒕 + 𝟑 →

𝒕=𝟑

∴ 𝒐𝒃𝒋𝒆𝒄𝒕 𝒔𝒕𝒐𝒑𝒔 𝒐𝒓 𝒄𝒉𝒂𝒏𝒈𝒆𝒔 𝒅𝒊𝒓𝒆𝒄𝒕𝒊𝒐𝒏 𝒗𝒆𝒓𝒕𝒊𝒄𝒂𝒍𝒍𝒚 𝒂𝒕 𝒕 = 𝟑𝒔,

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We can observe the intervals at which the tennis ball moves upwards and downwards without graphing: before 3s, the ball is moving upwards, while after 3s it is moving downwards. We know this because we used the test values of 𝑡 = 2𝑠 𝑎𝑛𝑑 𝑡 = 5𝑠 to check the direction of motion before and after 3s. The second application of parametric equation derivatives is to find the speed of an object at time 𝑡.4 Recall that to find the magnitude of the sum of orthogonal, or right-angled components (x’(t) and y’(t) in this case) we need to use the Pythagorean Theorem. Since x’(t) and y’(t) are components of velocity, we can combine them to find the resultant vector, or the speed. Pythagorean Theorem states that the sum of the squares of the two shorter sides in a right-angled triangle equals the square of the hypotenuse, or the long side opposite of the right angle. This rule can be represented by the following diagram: 𝟐

𝟐

(𝑺𝒑𝒆𝒆𝒅)𝟐 = (𝒙′ (𝒕)) + (𝒚′ (𝒕))

𝟐

𝟐

𝑺𝒑𝒆𝒆𝒅 = √(𝒙′ (𝒕)) + (𝒚′ (𝒕))

Let us find the speed of the tennis ball at 𝑡 = 5𝑠: 2

2

2

2

𝑆𝑝𝑒𝑒𝑑 = √(𝑥 ′ (𝑡)) + (𝑦 ′ (𝑡)) = √(𝑥 ′ (5)) + (𝑦 ′ (5)) = √(1)2 + (−5)2 = √26 ≈ 5.1 ∴ 𝟓 𝒔𝒆𝒄𝒐𝒏𝒅𝒔 𝒂𝒇𝒕𝒆𝒓 𝒊𝒕 𝒊𝒔 𝒓𝒆𝒍𝒆𝒂𝒔𝒆𝒅, 𝒕𝒉𝒆 𝒕𝒆𝒏𝒏𝒊𝒔 𝒃𝒂𝒍𝒍 𝒊𝒔 𝒎𝒐𝒗𝒊𝒏𝒈 𝒂𝒕 𝟓. 𝟏𝒎/𝒔. Integration can also tell us valuable information about the motion of an object. Integration is the inverse process of differentiation – if differentiation tells us the equation of the tangent through the equation of the curve, then integration does the opposite by providing us with information about the curve using the equation of the tangent. One of the many uses is to find the arc length of an object’s trajectory, which is simply the total distance the object has travelled within a time interval. To find the total distance, we integrate the speed equation.5 Let 𝐿 be total distance: 𝐿 = ∫ 𝑠𝑝𝑒𝑒𝑑 𝑑𝑡

4

http://www.mcs.sdsmt.edu/tkowalsk/notes/precalc-trig/Basic-facts-about-parametric-equations.pdf

5

http://www.mcs.sdsmt.edu/tkowalsk/notes/precalc-trig/Basic-facts-about-parametric-equations.pdf

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Now substituting our equation for speed as introduced earlier: 𝐿 = ∫ √(𝑥′(𝑡)2 + (𝑦 ′ (𝑡))2 𝑑𝑡 Since we want to find the value of L within an interval of time, we introduce definite integrals, which is simply an integral where we define boundaries, known as the upper bound, t1 and the lower bound, t0 , in order to calculate a finite value for the distance within that time interval: 𝑡1

𝐿 = ∫ √(𝑥′(𝑡))2 + (𝑦 ′ (𝑡))2 𝑑𝑡 𝑡0

Where 𝑡0 is the beginning of the time interval and 𝑡0 is the end of the interval. Again revisiting the tennis ball example encountered earlier, we will find the total distance it covered while in flight: 𝑡1

2

2

𝐿 = ∫ √(𝑥 ′ (𝑡)) + (𝑦 ′ (𝑡)) 𝑑𝑡 𝑡0 𝑡1

= ∫ ((1)2 + (−𝑡 + 3)2 )0.5 𝑑𝑡 𝑡0 𝑡1

= ∫ ((1)2 + (−𝑡 + 3)2 )0.5 𝑑𝑡 𝑡0 𝑡1

= ∫ (1 + 𝑡 2 − 6𝑡 + 9)0.5 𝑑𝑡 𝑡0 𝑡1

= ∫ (𝑡 2 − 6𝑡 + 10)0.5 𝑑𝑡 𝑡0

Now, all that is missing is our upper and lower bounds – or our time interval in which the ball was in flight. To account for the whole time interval that it was in flight, the lower bound, t0 , can be set to be 0 since it was the initial time of release. But how can we find the upper bound, t1 , when the ball hits the ground? A conjecture can be made that when the ball hits the ground, the vertical distance is y(t) = 0, so we can solve the y parametric equation to find the t-value at which the ball hits the ground. The y-parametric equation of the tennis ball’s flight was: 𝑦(𝑡) = −0.5𝑡 2 + 3𝑡 + 2 0 = −0.5𝑡 2 + 3𝑡 + 2

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We can solve this using a formula called the quadratic equation, by substituting in the values of our equation. 𝑡=

−𝑏 ± √𝑏 2 − 4𝑎𝑐 2𝑎

Note that the coefficient in front of the t2 term from our y-parametric equation is a, the coefficient of the t term is b, and the c term is the constant; Hence, 𝑡=

−(3) ± √(3)2 − 4(−0.5)(2) 2(−0.5)

∴𝑡=

−3 ± √13 −1

𝑡 ≈ −0.60555 s or 𝑡 ≈ 6.60555 s Being wary of the real life situation, 𝑡 ≥ 0 since time cannot be negative. Therefore, the ball will land at 𝑡 ≈ 6.60555 s. Now that we know the full time interval for which the ball was in flight, which is from 0 seconds to 6.60555 seconds. Hence, 6.60555 will be our upper bound. Now we can calculate the value of the definite integral: 𝑡1

𝐿 = ∫ (𝑡 2 − 6𝑡 + 10)0.5 𝑑𝑡 𝑡0 6.60555

=∫

(𝑡 2 − 6𝑡 + 10)0.5 𝑑𝑡

0

Using a GDC, we find that:

𝐿 ≈ 13.395 𝑚𝑒𝑡𝑟𝑒𝑠 Hence the arc length, or the total distance covered by the ball while in flight was 13.395 metres.

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Parametric equations can have many fascinating and practical applications, and through this mathematical exploration I have been able to develop a greater sense of how the motion of objects can be modelled, as well as how to draw on math concepts I learned in class such as integration, finding derivatives, trigonometry, vector equations and the properties of curves in order to manipulate these equations in a variety of ways. I also learned that parametric curves are different from ordinary Cartesian curves since they have orientation arrows, which are arrows that point in the direction the object is moving as time elapses. This property of parametric equations was completely new to me, but it further developed my engagement with the topic as modelling and investigating the motion of objects was initially what made me choose this topic. Before, I believed that math was simply a tool used by physicists; after the exploration, I now believe that physics itself is a subset of math where we are simply more aware of the real life implications. For example, when I looked at the circular motion of a bicycle tire, the parametric equations initially showed that the wheels were spinning counterclockwise with the orientation arrows, but I knew to manipulate the equations to spin clockwise since that is how the wheels move on a bicycle when pedaling forward. Additionally, the projectile motion question had a part of the curve start before 𝑡 = 0, and it was obvious that time cannot be negative, so I had to state the restriction that 𝑡 ≥ 0 accordingly. The applications of parametric equations were so numerous that I could not cover all of them in this exploration – I chose motion problems because that was the area that appealed to me. For further exploration, there are a number of other uses for parametric equations that I came across in my early stages of research, such as modelling complex yet artistic spirals and curves; I even came across a parametric curve in the shape of a butterfly6 that showed me how vastly different areas of knowledge such as arts and mathematics can intersect. Moreover, there was an article that pointed to the use of parametric equations in the field of biological research to model the shape and abnormalities of spinal disks.7 Parametric equations can also go further than modelling the motion of a tennis ball. For example, it occurred to me that even a tsunami or tidal wave can be represented by parametric equations. Information such as its vertical height, velocity, total distance travelled, horizontal position and other properties that I applied to my examples can similarly be used for natural phenomena, and can help scientists or the government predict the motion of a large tidal wave and take measures to protect against it. Moreover, further explorations of parametric equation can be done involving the third dimension, so instead of just modelling motion in the x and y planes, the z plane can also be defined in terms for a parameter, and the width or depth or objects can also be modelled, for example how wide a tidal wave is. Parametric equations can precisely measure and reveal information about the motion of an object and how phenomena can change over time, and a wealth of knowledge can be acquired from studying how they can be applied to our everyday lives, in nature as well as on a theoretical level.

6

http://cims.nyu.edu/~kiryl/Precalculus/Section_8.4Plane%20Curves%20and%20Parametric%20Equations/Plane%20Curves%20and%20Parametric%20Equations.pdf 7 http://eprints.qut.edu.au/11142/

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Works Cited "Basic Facts About Parametric Equations." Southern Dakota Institute of Mines and Technology. N.p., n.d. Web. . "Calculus with Parametric Equations." University of Portland. N.p., n.d. Web. . "Calculus/Parametric Introduction." Wikibooks. N.p., n.d. Web. . "Definite Integral." Wolfram MathWorld. N.p., n.d. Web. . "Graphmatica by KSoft." Graphmatica by KSoft. N.p., n.d. Web. . "Parametric Curves." MIT Open Courseware. N.p., n.d. Web. . "Parametric Equations - HMC Calculus Tutorial." Parametric Equations - HMC Calculus Tutorial. N.p., n.d. Web. . "Parametric Equations 1 | Parametric Equations |Khan Academy." Khan Academy. N.p., n.d. Web. . "Parametric Equations and Curves." Calculus II. N.p., n.d. Web. . "Parametric Equations For Curves In Space." Docbenton.com. N.p., n.d. Web. . "Parametric Equations." Parametric Equations. N.p., n.d. Web. . 15

"Parametric Equations to Represent the Profile of the Human Intervertebral Disc in the Transverse Plane." QUT EPrints. N.p., n.d. Web. . "Parametric Equations." University of Washington. N.p., n.d. Web. . "Plane Curves and Parametric Equations." Courant Institute of Mathematical Sciences. N.p., n.d. Web. .

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