Parametric Equations

Copyright © 2007 Pearson Education, Inc. Slide 10-1

Parametric Equations

• Here are some examples of trigonometric functions used in parametric equations.


INTRODUCTION

• Imagine that a particle moves along the curve C shown here.

– It is impossible to describe C by an equation of the form y = f(x).

– This is because C fails the Vertical Line Test.


• However, the x- and y-coordinates of the particle are functions of time.So, we can write x = f(t) and y = g(t).

INTRODUCTION


• Such a pair of equations is often a convenient way of describing a curve and gives rise to the following definition.

INTRODUCTION


• Suppose x and y are both given as functions of a third variable t (called a parameter) by the equations

x = f(t) and y = g(t)

– These are called parametric equations.

PARAMETRIC EQUATIONS


• Each value of t determines a point (x, y),which we can plot in a coordinate plane.

• As t varies, the point (x, y) = (f(t), g(t)) varies and traces out a curve C.

– This is called a parametric curve.

PARAMETRIC CURVE


• The parameter t does not necessarily represent time.

PARAMETER t


PARAMETER t• However, in many applications of

parametric curves, t does denote time.

– Thus, we can interpret (x, y) = (f(t), g(t)) as the position of a particle at time t.


Graphing a Circle with Parametric Equations

Example Graph x = 2 cos t and y = 2 sin t for 0 2. Find an equivalent equation using rectangular coordinates.

Solution Let X1T = 2 cos (T) and Y1T = 2 sin (T), and graph these parametric equations as shown.

Technology Note Be sure the calculator is set in parametric mode. A square window is necessary for the curve to appear circular.


Graphing a Circle with Parametric Equations

To verify that this is a circle, consider the following.

The parametric equations are equivalent to x2 + y2 = 4, which is a circle with center (0, 0) and radius 2.

4

)sin(cos4

sin4cos4

)sin2()cos2(

22

22

2222

tt

tt

ttyx x = 2 cos t, y = 2sin t

cos2 t + sin2 t = 1


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.

42

yx 2

2 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!


Sketch the curve described below

2 4

2

x t

ty


Sketch the curve described below

24 4x t

y t


Eliminate the parameter:

2 4

2

x t

ty


Your turn to practice. Do Example 4


Sketch the curve

3cos

3sin

x

y


Finding parametric equations

2 2 3y x x


Finding parametric equations

2 4

3 1

xy

x



Graphing an Ellipse with Parametric Equations

Example Graph the plane curve defined by x = 2 sin t and y = 3 cos t for t in [0, 2].

Solution

Now add both sides of the equation.

tx

tx

tx

22

22

sin4

sin4

sin2

ty

ty

ty

22

22

cos9

cos9

cos3

194

cossin94

22

2222

yx

ttyx


Applications of Parametric Equations

• Parametric equations are used frequently in computer graphics to design a variety of figures and letters.

Example Graph a “smiley” face using parametric equations.

SolutionHead Use the circle centered at the origin. If the radius is 2, then let x = 2 cos t and y = 2 sin t for 0 t 2.


Applications of Parametric Equations

Eyes Use two small circles. The eye in the first quadrant can be modeled by x = 1 + .3 cos t and y = 1 + .3 sin t. This represents a circle centered at (1, 1) with radius .3. The eye in quadrant II can be modeled by x = –1 + .3 cos t and y = 1 + .3 sin t for 0 t 2, which is a circle centered at (–1, 1) with radius 0.3.

Mouth Use the lower half of a circle. Try x = .5 cos ½t and y = –.5 –.5 sin ½t. This is a semicircle centered at (0, –.5) with radius .5. Since t is in [0, 2], the term ½t ensures that only half

the circle will be drawn.


Example Graph the plane curve defined by x = 2cos t + 2 cos (4t) y = sin t + sin(4t) and for t in [0, 5].


Example Graph the plane curve defined by x = t - 2sin t and y = 2 - 2cos t for t in [0, 10].


Simulating Motion with Parametric Equations

• If a ball is thrown with a velocity v feet per second at an angle with the horizontal, its flight can be modeled by the parametric equations

where t is in seconds and h is the ball’s initial height above the ground. The term –16t2 occurs because gravity pulls the ball downward.

,16)sin(and)cos( 2 httvytvx

Figure 80 pg 10-128



Example Three golf balls are hit simultaneously into the air at

132 feet per second making angles of 30º, 50º, and 70º with the horizontal.(a) Assuming the ground is level, determine graphically which

ball travels the farthest. Estimate this distance.(b) Which ball reaches the greatest height? Estimate this height.

Solution(a) The three sets of parametric equations with h = 0 are

as follows.

X1T = 132 cos (30º) T, Y1T = 132 sin (30º) T – 16T2





With 0 t 9, a graphing calculator in simultaneous mode shows all three balls in flight at the same time.

The ball hit at 50º goes the farthest at an approximate distance of 540 feet.

(b) The ball hit at 70º reaches the greatest height of about 240 feet.


Examining Parametric Equations of Flight

Example For each problems, use your calculator to graph two parametric equations for a projectile fired at the given angle at the given initial speed at ground level. Then estimate the max height and the range of the object.

30 , 90 / sec

30 , 120 / sec

oo

oo

v ft

v ft

30 , 90 / sec

70 , 90 / sec

oo

oo

v ft

v ft


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


A baseball player is at bat and makes contact with the ball at a height of 3 ft. The ball leaves the bat at 110 miles per hour towards the center field fence, 425 feet away which is 12 feet high. If the ball leaves the bat at the following angles of elevation, determine whether or not the ball will be a home run.

17o 18o

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Parametric Equations