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The variable t (the parameter) often represents time. We can picture this like a particle moving along and we know its x position over time and its y position over time and we figure out each of these and plot them together to see the movement of the particle.

0;4,2 ttytx

Graph the plane curve represented by the parametric equations

We'll make a chart and choose some t values and find the corresponding x and y values.

t x y

0 002 004

The t values we pick must be greater than or equal to 0. Let's start with 0.

yx, 0,0

1 4.112 414 4,2

0,0

4,2

2 222 824 8,2

3 4.232 1234

8,2

12,6

12,6

We see the "path" of the particle. The orientation is the direction it would be moving over time (shown by the arrows)

0;4,2 ttytx

We could take these parametric equations and find an equivalent rectangular equation with substitution. This is called "eliminating the parameter."

Solve for the parameter t in one of equations (whichever one is easier).

0,0

4,2

8,2

12,6

4

yt Substitute for t in the other

equation.

42y

x 22 yx 2 2

yx 22

We recognize this as a parabola opening up. Since our domain for t started at 0, it is only the right half.

20;sin4,cos2 ttytxGraph the plane curve represented by the parametric equations

t x y

0 20cos2 00sin4

The t values we pick must be from 0 to 2

yx, 0,2

24

cos2

224

sin4 22,2

4,0

0,2

Make the orientation arrows based where the curve was as t increased.

4

2

02

cos2

42

sin4

2cos2 0sin4

2

30

2

3cos2

4

2

3sin4

4,0

4

52

4

5cos2

22

4

5sin4

22,2

You could fill in with more points to better see the curve.

20;sin4,cos2 ttytx

Let's eliminate the parameter. Based on our curve we'd expect to get the equation of an ellipse.

When you want to eliminate the parameter and you have trig functions, it is not easy to solve for t. Instead you solve for cos t and sin t and substitute them in the Pythagorean Identity:

1cossin 22 tt

22 4 4

tx

ty

cos2

sin4

:above From

124

22

xy 1

416

22

xy

Here is the rectangular version of our ellipse. You can see it matches!

When you then enter a graph, it will have t for the variable and you can enter more than one equation.

Your Casio graphic calculator can plot parametric equations. Select “Graph” mode and check that “Type” is set to Parm.

If you watch as it draws the graph, you will see the orientation (direction) of the curve.

If an object is dropped, thrown, launched etc. at a certain angle and has gravity acting upon it, the equations for its position at time t can be written as:

tvx o cos htvgty o sin2

1 2

horizontal position initial velocity angle measured from horizontal

time gravitational constant which is 9.8 m/s2

initial heightvertical position

tvx o cos htvgty o sin2

1 2

Adam throws a tennis ball off a cliff, 300 metres high with an initial speed of 40 metres per second at an angle of 45° to the horizontal. Find the parametric equations that describe the position of the ball at time t.

30045sin408.92

1 2 tty tx 45cos40

How long is the ball in the air? When the ball hits the ground, the vertical position y will be 0.

30028.289.40 2 tt

30028.289.4 2 ttytx 28.28

use the quadratic formula

sec 23.11or 45.5t

The negative time value doesn't make sense so we throw it out.

Adam throws a tennis ball off a cliff, 300 metres high with an initial speed of 40 metres per second at an angle of 45° to the horizontal. Find the parametric equations that describe the position of the ball at time t.

When is the ball at its maximum height?

The motion is parabolic (opening down) so maximum will be at the turning point.

30028.289.4 2 ttytx 28.28

a

bt

2TP of value

sec 89.29.42

28.28

What is the maximum height?

30089.228.2889.29.4 2 y metres 8.340

Adam throws a tennis ball off a cliff, 300 meters high with an initial speed of 40 meters per second at an angle of 45° to the horizontal. Find the parametric equations that describe the position of the ball at time t.

Determine the horizontal distance the ball traveled.Use time in air from first part of problem.

30028.289.4 2 ttytx 28.28

23.1128.28x metres 6.317

Acknowledgement

I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint.

www.slcc.edu

Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum.

Stephen CorcoranHead of MathematicsSt Stephen’s School – Carramarwww.ststephens.wa.edu.au