Parametric Equations

PARAMETRIC EQUATIONS

Erica RottenbornMary Sha

Table of Contents

Topanga’s Real-World Applicability……………………………. Pg.3

Hyde’s Helpful Mathematician’s Contributions………….. Pg.4

Angela’s Analytical Examples…………………………............ Pg.5-8

Midge’s Multiple Choice Example…………………………….. Pg.9-10

Cory’s Conceptual Example………………………………………. Pg.11

Foreman’s Arc Length with Graphing Calculator………. Pg.12-13

Mr. Feeny’s Free Response Example………………………... Pg.14-16

Sabrina’s Works Cited………………………………………….…….Pg.17

Why Parametric?

How many times have you seen a cat walk in the path of x²? Almost never. Cats (like Salem) pounce, jump, and walk in circles. However, there are no functions to describe Salem’s movement. By incorporating time into a

function, parametric equations allow us to describe these paths of movements that we

could not describe before. One function relies on time, t, to describe the movement with

respect to x, and the other function relies on time, as well, to describe the movement with

respect to y. Together, they give you graphs that can describe movement we could not describe before like the movement of the eight ball as

Cory and Shawn play pool or the path of Forman’s smoke.

Mustapha Amrani is a professor at the University of Cambridge and is affiliated with the Isaac Newton Institute. He studies parametric equations in detail so as to find real-world applications for the concept and he gives talks on his findings. Amrani has discovered a use of parametric equations that was previously unknown-- he has begun to incorporate parametric equations and semi-parametric equations into electricity consumption forecasting. For utilities it is necessary to have a good idea of the future's electricity consumption and Armani has found that through the use of parametric equations in his work he is able to better predict the consumption trends that will occur thereby aiding the utilities. Since Armani uses primarily semi-parametric structures he is able to create an even more accurate prediction as his models create a complex but interesting trade-off between previous accurate forecasts and new adaptations to different data sets. Although these contributions are still gaining credibility and usage in the real-world, Armani has proven how parametric equations can be utilized with a multitude of different geographical climates and horizons in order to try to accurately predict the electricity consumption of the future.

Mustapha Amrani

First, let’s learn how to eliminate the parameter.

3)(

3

t

ttx 2

2

1)( 2 tty

ty

ty

ty

24

24

2

12

2

2

324

243

y

yx

Eliminating the parameter allows for you to go from parametric equations to polynomial equations. Ultimately we are

eliminating time, t, from the equation.

So, when told to eliminate the parameter, look at the x -and-y functions and decide in which equation would it be easier to solve for time, t. Although both of the equations above are quite messy, the y-equation seems to be easier

to solve for time, t.

The last step is to solve for time, t, in the y-equation and plug it in for time, t, in the x-equation.

You just eliminated the parameter! Wasn’t that easy?

Basic Analyti cal

Graph and find the equation of the tangent line to the graph at t=1 if x(t)=2t²-4 and y(t)=t²-3t+1

First, let’s graph the parametric equation!

t x(t) y(t)0 -4 1.5 -3.5 -.251 -2 -1

1.5 .5 -1.252 4 -1

Graphing parametric equations are quite different from regular equations. Because the equations are given with respect to time, we are actually calculating positions at certain times. So, you would want to create a table of values like the one to the bottom-left and graph

the points by plugging time into the x- and y-functions. For example, at t=0, the x-value is -4 and the y-value is 1. So, the first point will be (-4,1). Continue this with the rest of the points. Also, one thing greatly different from regular graphs is direction. So add an arrow to specify

what direction you are going.

Analyti cal with a Pinch of Graphing

ttx 4)('

Now, it’s time to find the tangent line!

32)(' tty

When finding tangent lines, it’s not too different from regular equations. First, find the derivatives of both x- and y- functions.

Next, it’s time to find the slope of the tangent line. But, how do we do this when we have two functions!?!? You simply divide the derivative of the y-function by the derivative of the x-function.

4

1

)1(4

3)1(2

)1('

)1('

/

/

x

y

dtdx

dtdy

Now that we have the slope of the tangent line, it’s time to find the x- and y-coordinates. This step is a bit more straight forward.

Just plug in t=1 into the functions!

11)1(3)1()1(

24)1(2)1(2

2

y

x

Finally, it’s time to put it all together and graph the tangent line!

)2(4

11

xy

)2(4

11

xy

Now you know how to graph parametric equations and find their tangent lines!

To find a vertical tangent line, we must solve for when the derivative of the x-function is equal to zero. This will give us

the time at which there is no movement in the x-direction and thus a vertical tangent line!

2sin4)(' tttx 2sin40 tt

So what will make this equation equal to zero? Obviously when t is equal to zero, the whole function will be zero so that is one

option. When sine is zero, the whole function will be zero as well. At 0 and , sine is equal to zero but because we already

have 0, we can consider .

0t

t

t

t2

2 0sin

At which time, t, does the graph of and have a vertical tangent line on the interval ?

a. 0 b. c. 2 d. 0, e. 0, , 2

AP Level Multi ple Choice

0t tWe now have the two times at which there is no movement in the x-direction. So we’re done now right!? Well not quite. We have to test these times in the derivative of the y-function and

make sure it does not equal zero. Why may you ask? Well, what’s happening to the graph when both the derivative of x and y are

zero? That’s right, nothing. There’s no movement, and therefore, there wouldn’t be a vertical tangent line at that moment.

tty

tty

2cos2)('

2sin)(

))0(2cos(2)0(' y

2 2cos2)(' y

840.1

Now that we have established that there is still movement at those times,

we can officially say we have gotten our answer! The times at which a

vertical tangent line occurs are at 0 and Letter A doesn’t work because it only has 0, Letter B only has , Letter C doesn’t have either, and Letter E has too many answers. The only answer that matches our choice is Letter D!

What is the concavity of the graph of and at time, t=-3?

Once you hear the word concavity, immediately think the second derivative! Wait, but how do we take the second derivative of a

parametric function? First we find the first derivative which we already learned how to do. Differentiate the x- and y-functions then set dy over dx. This is the first derivative. Now differentiate the first derivative to

get the second derivative, but wait a second! The second derivative is in terms of dt! It should be in terms of dx²! To solve this problem just

divide the second derivative we just found by dx. Then evaluate at t=-3!

0))3(22(

)22(3

22

1

)22(

223

)22(

223

)22(

223022

22

3

/

/

33

3

32

2

22

2

22

e

e

dx

yd

etet

e

dx

yd

dtdxdt

dxdy

d

et

e

et

eet

dtdxdy

d

etdx

dy

dtdx

dtdy

t

tt

t

t

t

t

tt

t

tetdt

dx 22 3dt

dy

Because the second derivative evaluated at t=-3 is greater than zero than the graph is concave up!

AP Level Conceptual

If the eight ball Eric hit is described in inches by the functions and , what is the distance the eight ball traveled

between time t=1 and t=3?

To find the distance the eight ball traveled, we will want to use the arc length equation.

𝐴𝑟𝑐 h𝑙𝑒𝑛𝑔𝑡 =∫𝑎

𝑏

√𝑥 ′ (𝑡)2+𝑦 ′ (𝑡)2𝑑𝑡

So first, we need to find the derivatives of the x- and y-functions.

18

)('3

t

tx 221)(' tty Now just plug in the derivatives into the arc length equation

and solve it by using your graphing calculator!

∫

3

1

222

3211

8dtt

t

Arc Length with Graphing Calculator

The eight ball that Eric hit traveled a total distance of 182.138 inches!

Sabrina is practicing spells with her aunts Zelda and Hilda when-OH NO!--she accidentally hits Salem with a jumping jinx! Salem begins to jump uncontrollably with an initial velocity of 40 ft/sec at angle

of 75to the ground. Despite his efforts to acquiesce to the -32ft/sec acceleration of gravity he is unable to stop jumping. If Salem was

initially resting 3ft above the ground on the arm of the couch, create X(t) and Y(t).

To begin to find the position functions you must utilize the general equations:

cosVsin

First focus on finding X(t). Substitute the information from the problem into the general equation to get the velocity function for X.

V

Next, integrate the equation to get the position function for X.

AP Level Free Response

Now that you have found X(t), focus on finding Y(t). Integrate the acceleration function to get the velocity of Y.

V

Add the constant of integration that you get from the general equation for the velocity of Y.

Integrate the velocity function to get the position function of Y.

Add the initial condition.

The final answer is:

EXERCISESEliminate the parameter:

1)

3)

1

3)(

4

2

t

tty

2)

1)(

1)(

3

tty

t

ttx

Graph the following parametric equations:

4)

12)(

)(

tty

ttx 5)22)(

3)(23

2

ttty

tttx

3

4)(

13

3)(

34

3

2

ttty

t

ttx

6)

Find :

7)

32)(

3)(3

2

tty

tttx

)2ln()(

2

1)(

2

24

tty

tttx8)

Find the tangent line:

9) At t=2:

54

1)(

64)(

2

2

tty

tttx

10) At t=1:

75)(

)532ln()(3

2

ttty

tttx

11) At t=3:

2107)(

534)(2

3

ttty

tttx

12) If and , then at t= equals

(A) (B) (C) (D) (E)

13) What is the length of the path described by the parametric equations and , where ?

(A)309.122 (B)36.993 (C)325.687 (D)39.046 (E) 369.826

14) The velocity vector of a particle moving in the xy-plane is given by At t=0, the particle is at the point (0,-1). What is the position vector at t=3?

(A) (C) (B) (D) (E)

15)At what time(s) t does the graph of and have a vertical tangent line?

(A) 0 & 2/5 (B) None of the above (C) 0, 2/5, 1/5 (D) 1/5 (E)-1/3

16) Find the speed of a particle traveling along the path of and at t=2.

(A) 7.348 (B)27.295 (C)45.100 (D)2601 (E)93.134

Free Response Time!Cory and Topanga’s relationship has had up and downs and everything in between over the years. The path of

their love is modeled by the velocity vector

for and at t=0 the position of their love is (3,0).

a) Find the speed of the particle and the acceleration vector at t=2.

b) Find the x-and y-values of their love at t=2.c) Find the equation for the line tangent at t=2.d) Find the length of their love from t=0 to t=3.

CALCULATOR

PERMITTED

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Multiple-Choice & Free-Response Questions in Preparation for the AP Calculus (BC) Examination by David Lederman