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Parametric Equations

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

- 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).

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

- 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.

- 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.

- The parameter t does not necessarily represent time.

- 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.

ExampleGraph x = 2 cos t and y = 2 sin t for 0 2. Find

an equivalent equation using rectangular coordinates.

SolutionLet X1T = 2 cos (T) and Y1T = 2 sin (T), and graph

these parametric equations as shown.

Technology NoteBe sure the calculator is set in parametric mode.

A square window is necessary for the curve to appear circular.

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.

x = 2 cos t, y = 2sin t

cos2t + sin2t = 1

Graph the plane curve represented by the parametric equations

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

t

0

1

2

3

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

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

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).

Substitute for t in the other equation.

2

2

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

Graph the plane curve represented by the parametric equations

The t values we pick must be from 0 to 2

t

0

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

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

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

2

2

4

4

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:

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

Sketch the curve described below

Sketch the curve described below

Eliminate the parameter:

Your turn to practice. Do Example 4

Sketch the curve

Finding parametric equations

Finding 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.

- Parametric equations are used frequently in computer graphics to design a variety of figures and letters.
ExampleGraph a “smiley” face using parametric

equations.

Solution

HeadUse 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.

EyesUse 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.

MouthUse 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].

- 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.

Figure 80 pg 10-128

ExampleThree golf balls are hit simultaneously into the air at

132 feet per second making angles of 30º, 50º, and 70º with the

horizontal.

- Assuming the ground is level, determine graphically which ball travels the farthest. Estimate this distance.
- 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

X2T = 132 cos (50º) T, Y2T = 132 sin (50º) T – 16T2

X3T = 132 cos (70º) T, Y3T = 132 sin (70º) 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.

- The ball hit at 70º reaches the greatest height of about 240 feet.

ExampleFor 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.

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:

angle measured from horizontal

horizontal position

initial velocity

gravitational constant which is 9.8 m/s2

time

initial height

vertical position

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.

How long is the ball in the air?

When the ball hits the ground, the vertical position y will be 0.

use the quadratic formula

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.

What is the maximum height?

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.

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.