**10.5 Parametric Equations**

check out: http://colalg.math.csusb.edu/~devel/precalcdemo/param/src/param.html

Up to this point, you have been representing a graph by a single equation involving two variables such as

*x*and

*y*. In this section, you will study situations in which it is useful to introduce a third variable to represent a curve in the plane.

Let's consider the path followed by an object that is propelled into the air at an angle of 45 degrees. If the initial velocity of the object is 48 feet per second, it can be shown that the object follows the parabolic path

*y*= -

*x*

^{2}/72 +

*x*

However, this equation does not tell the whole story. Although it does tell us where the object has been, it doesn't tell us when the object was at a given point (

*x*,

*y*) on the path. To determine this time, you can introduce a third variable

*t*, which is called a

**parameter**. It is possible to write both

*x*and

*y*as functions of t to obtain the

**parametric equations**.

Example: x = 24Ö(2) t and y = -16t

^{2}+ 24Ö(2) t

Curvilinear motion: two variables for position, one for time.

A curve in the plane is said to be parameterized if the set of coordinates on the curve, (x,y), are represented as functions of a variable

*t*. Namely, x = f(t), y = g(t).

Conversely, given a pair of parametric equations with parameter

*t*, the set of points (f(

*t*), g(

*t*)) form a curve in the plane.

For this particular motion problem,

*x*and

*y*are continuous functions of

*t*, and the resulting path is a

**plane curve**. (Recall that a continuous function is one whose graph can be traced without lifting the pencil from the paper).

**Definition of a Plane Curve**:

If

*f*and

*g*are continuous functions of

*t*on an interval I, the set of ordered pairs (f(

*t*), g(

*t*)) is a plane curve C. The equations

*x*= f(

*t*) and

*y*= g(

*t*)

are parametric equations for C, and

*t*is the parameter.

One way to sketch a curve represented by a pair of parametric equations is to plot points in the xy-plane. Each set of coordinates (x,y) is determined from a value chosen for the parameter t. By plotting the resulting points in the order of increasing values of t, you trace the curve in a specific direction. This is called the

**orientation**of the curve.

The orientation of a parameterized curve is the direction determined by increasing values of the parameter. Sometimes arrows are drawn on the curve to denote the orientation.

**Example 1**: you are given a pair of parametric equations. Select 6 valid values of

*t*and substitute each into both equations to get the coordinates of a point in the plane. Plot all six points in the plane.

*x*=

*t*

^{2}- 4 and

*y*=

*t*/2, where -2 ≤

*t*≤ 3

t = -2, x = 0, y = -1

t = -1, x = -3, y = -.5

t = 0, x = -4, y = 0

t = 1, x = -3, y = .5

t = 2, x = 0, y = 1

t = 3, x = 5, y = 1.5

By plotting these points in the order of increasing t, you obtain the curve that does not define y as a function of x. The graph looks like a parabola that opens to the right with t = 0, x = -4, and y = 0 as its vertex.

to graph this in your TI-83 plus calculator, go to mode, arrow down to the line with func and arrow over to par (for parametric). Then press enter. Your calculator is now in parametric mode. Go to the y=. For X

_{1T}= T

^{2}-4, Y

_{1T}= T/2. Now go to window, Tmin = 0, Tmax = 10, Tstep = 1, Xmin = -10, Xmax = 10, Xscl = 1, Ymin = -10, Ymax = 10, and Yscl = 1. Press 2nd graph and you will see the table above. Now press graph and you can see this curve.

Sometimes you can have two different sets of parametric equations that have the same graph. For example, x = 4t

^{2}- 4 and y = t, -1 ≤ t ≤ 1.5

When you plot these, they have the same graph as the one above but the values of

*t*are different. This means that the second graph is traced out more rapidly than the first. Therefore, in applying this to real situations, different parametric representations can be used to represent various speeds at which objects travel along a given path.

**Eliminating the Parameter**:

A second technique to identifying the curve of the parametric equations is to try to eliminate the parameter from the equations. This will result in an equation involving only

*x*and

*y*which we may recognize. For example, let's look again at an example.

Given x = 3 - 2t and y = 2 + 3t

Sketching this graph, we have the following data:

t = -2, x = 7, y = -4

t = -1, x = 5, y = -1

t = 0, x = 3, y = 2

t = 1, x = 1, y = 5

t = 2, x = -1, y = 8

then solving for t in the first equation

x = 3 - 2t

x - 3 = -2t

t = (x-3)/(-2)

Substitute for t into the second equation to get

y = 2 + 3t

y = 2 + 3((x-3)/(-2)

y = 2 + (3x - 9)/(-2)

y = (3x)/(-2) + 13/2

Now sketch this graph

x = 7, y = -4

x = 5, y = -1

x = 3, y = 2

x = 1, y = 5

x = -1, y = 8

We can sketch its graph using function graphing techniques. Be careful, however, to take into account any restrictions on the value of the parameter.

Notice anything about the two separate parts?

Converting equations from parametric to rectangular form can change the ranges of x and y. In such cases, you should restrict x and y in the rectangular equation so that its graph matches the graph of the parametric equations.

**Example:**Given x =Ö t and y = 1 - t

1st solve for t in the first equation:

x =Ö t

x

^{2}= t with the constraint that x ³ 0

Therefore, y = 1 - t

y = 1 - x

^{2}

Parametric Equation:

t = 0, x = 0, y = 1

t = 1, x = 1, y = 0

t = 2, x = Ö2, y = -1

t = 3, x = Ö3, y = -2

t = 4, x = 2, y = -3

Now the rectangular Equation:

x = 0, y = 1

x = 1, y = 0

x = Ö2, y = -1

x = Ö3, y = -2

x = 2, y = -3

As you can see when you graph these, you have to restrict x ³

**Example**: Sketch the curve represented by

x = 4 + 2 cosq and y = 1 + 2 sinq

Begin by solving for cosq and sinq

x = 4 + 2 cosq

x - 4 = 2 cosq

(x - 4)/2 = cosq

y = 1 + 2 sinq

y - 1 = 2 sinq

(y - 1)/2 = sinq

Now make use of the identity sin

^{2}q + cos

^{2}q = 1

((y-1)/2)

^{2}+ ((x-4)/2)

^{2}= 1

(y-1)

^{2}/4 + (x-4)

^{2}/4 = 1

(x-4)

^{2}+ (y-1)

^{2}= 4

This is the graph of a circle with center point (4,1) and radius of 2

Now you can graph this putting your calculator in parametric mode

X

_{1}= 4 + 2 cosT

Y

_{1}= 1 + 2 sinT

Now your window Tmin = 0

Tmax = either 2p radians or 360 degrees depending whether you are in degrees or radians.

Tstep = play with this - try 90 degrees, 30 degrees, 1 unit and see what happens to your graph

Xmin = 0, Xmax = 7, Xscl = 1

Ymin = -3, Ymax = 4, Yscl = 1

**Example**: Find 2 different sets of parametric equations for the given rectangular equation:

y = x

^{2}

a). Let x = t so then using substitution y = t

^{2}

t = 0, x = 0, y = 0

t = 1, x = 1, y = 1

t = 2, x = 2, y = 4

t = 3, x = 3, y = 9

t = 4, x = 4, y = 16

b). Let x = t

^{2}then using substitution y = t

^{4}

t = 0, x = 0, y = 0

t = 1, x = 1, y = 1

t =2, x = 4, y = 16

The x-values and the y-values are the same, but the t-values are faster.

You can think of this as a track race. In theory, the runners are following the same path. But they all go at a different speed. Therefore their location placements are the same (the x-values and y-values) but the time is different. Some slower and some quicker.