8.7: Parametric Equations
- Page ID
- 114058
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- Parameterize a curve.
- Eliminate the parameter.
- Find a rectangular equation for a curve defined parametrically.
- Find parametric equations for curves defined by rectangular equations.
Consider the path a moon follows as it orbits a planet, which simultaneously rotates around the sun, as seen in Figure 1. At any moment, the moon is located at a particular spot relative to the planet. But how do we write and solve the equation for the position of the moon when the distance from the planet, the speed of the moon’s orbit around the planet, and the speed of rotation around the sun are all unknowns? We can solve only for one variable at a time.
Figure 1
In this section, we will consider sets of equations given by x(t)x(t) and y(t)y(t) where tt is the independent variable of time. We can use these parametric equations in a number of applications when we are looking for not only a particular position but also the direction of the movement. As we trace out successive values of t,t, the orientation of the curve becomes clear. This is one of the primary advantages of using parametric equations: we are able to trace the movement of an object along a path according to time. We begin this section with a look at the basic components of parametric equations and what it means to parameterize a curve. Then we will learn how to eliminate the parameter, translate the equations of a curve defined parametrically into rectangular equations, and find the parametric equations for curves defined by rectangular equations.
Parameterizing a Curve
When an object moves along a curve—or curvilinear path—in a given direction and in a given amount of time, the position of the object in the plane is given by the x-coordinate and the y-coordinate. However, both xx and yy vary over time and so are functions of time. For this reason, we add another variable, the parameter, upon which both xx and yy are dependent functions. In the example in the section opener, the parameter is time, t.t. The xx position of the moon at time, t,t, is represented as the function x(t),x(t), and the yy position of the moon at time, t,t, is represented as the function y(t).y(t). Together, x(t)x(t) and y(t)y(t) are called parametric equations, and generate an ordered pair (x(t),y(t)).(x(t),y(t)). Parametric equations primarily describe motion and direction.
When we parameterize a curve, we are translating a single equation in two variables, such as xx and y ,y , into an equivalent pair of equations in three variables, x,y,x,y, and t.t. One of the reasons we parameterize a curve is because the parametric equations yield more information: specifically, the direction of the object’s motion over time.
When we graph parametric equations, we can observe the individual behaviors of xx and of y.y. There are a number of shapes that cannot be represented in the form y=f(x),y=f(x), meaning that they are not functions. For example, consider the graph of a circle, given as r2=x2+y2.r2=x2+y2. Solving for yy gives y=±r2−x2−−−−−−√,y=±r2−x2, or two equations: y1=r2−x2−−−−−−√y1=r2−x2 and y2=−r2−x2−−−−−−√.y2=−r2−x2. If we graph y1y1 and y2y2 together, the graph will not pass the vertical line test, as shown in Figure 2. Thus, the equation for the graph of a circle is not a function.
Figure 2
However, if we were to graph each equation on its own, each one would pass the vertical line test and therefore would represent a function. In some instances, the concept of breaking up the equation for a circle into two functions is similar to the concept of creating parametric equations, as we use two functions to produce a non-function. This will become clearer as we move forward.
Suppose tt is a number on an interval, I.I. The set of ordered pairs, (x(t),(x(t), y(t)),y(t)), where x=f(t)x=f(t) and y=g(t),y=g(t), forms a plane curve based on the parameter t.t. The equations x=f(t)x=f(t) and y=g(t)y=g(t) are the parametric equations.
EXAMPLE 1
Parameterizing a Curve
Parameterize the curve y=x2−1y=x2−1 letting x(t)=t.x(t)=t. Graph both equations.
- Answer
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Analysis
The arrows indicate the direction in which the curve is generated. Notice the curve is identical to the curve of y=x2−1.y=x2−1.
Construct a table of values and plot the parametric equations: x(t)=t−3,x(t)=t−3, y(t)=2t+4;−1≤t≤2.y(t)=2t+4;−1≤t≤2.
EXAMPLE 2
Finding a Pair of Parametric Equations
Find a pair of parametric equations that models the graph of y=1−x2,y=1−x2, using the parameter x(t)=t.x(t)=t. Plot some points and sketch the graph.
- Answer
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Parameterize the curve given by x=y3−2y.x=y3−2y.
EXAMPLE 3
Finding Parametric Equations That Model Given Criteria
An object travels at a steady rate along a straight path (−5,3)(−5,3) to (3,−1)(3,−1) in the same plane in four seconds. The coordinates are measured in meters. Find parametric equations for the position of the object.
- Answer
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Analysis
Again, we see that, in Figure 5(c), when the parameter represents time, we can indicate the movement of the object along the path with arrows.
Eliminating the Parameter
In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as xx and y.y. Eliminating the parameter is a method that may make graphing some curves easier. However, if we are concerned with the mapping of the equation according to time, then it will be necessary to indicate the orientation of the curve as well. There are various methods for eliminating the parameter tt from a set of parametric equations; not every method works for every type of equation. Here we will review the methods for the most common types of equations.
Eliminating the Parameter from Polynomial, Exponential, and Logarithmic Equations
For polynomial, exponential, or logarithmic equations expressed as two parametric equations, we choose the equation that is most easily manipulated and solve for t.t. We substitute the resulting expression for tt into the second equation. This gives one equation in xx and y.y.
EXAMPLE 4
Eliminating the Parameter in Polynomials
Given x(t)=t2+1x(t)=t2+1 and y(t)=2+t,y(t)=2+t, eliminate the parameter, and write the parametric equations as a Cartesian equation.
- Answer
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Analysis
This is an equation for a parabola in which, in rectangular terms, xx is dependent on y.y. From the curve’s vertex at (1,2),(1,2), the graph sweeps out to the right. See Figure 6. In this section, we consider sets of equations given by the functions x(t)x(t) and y(t),y(t), where tt is the independent variable of time. Notice, both xx and yy are functions of time; so in general yy is not a function of x.x.
Figure 6
Given the equations below, eliminate the parameter and write as a rectangular equation for yy as a function of x.x.
x(t)=2t2+6y(t)=5−tx(t)=2t2+6y(t)=5−t
EXAMPLE 5
Eliminating the Parameter in Exponential Equations
Eliminate the parameter and write as a Cartesian equation: x(t)=e−tx(t)=e−t and y(t)=3et,y(t)=3et, t>0.t>0.
- Answer
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Analysis
The graph of the parametric equation is shown in Figure 7(a). The domain is restricted to t>0.t>0. The Cartesian equation, y=3xy=3x is shown in Figure 7(b) and has only one restriction on the domain, x≠0.x≠0.
Figure 7
EXAMPLE 6
Eliminating the Parameter in Logarithmic Equations
Eliminate the parameter and write as a Cartesian equation: x(t)=t√+2x(t)=t+2 and y(t)=log(t).y(t)=log(t).
- Answer
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Analysis
To be sure that the parametric equations are equivalent to the Cartesian equation, check the domains. The parametric equations restrict the domain on x=t√+2x=t+2 to t>0;t>0; we restrict the domain on xx to x>2.x>2. The domain for the parametric equation y=log(t)y=log(t) is restricted to t>0;t>0; we limit the domain on y=log(x−2)2y=log(x−2)2 to x>2.x>2.
Eliminate the parameter and write as a rectangular equation.
x(t)=t2y(t)=lntt>0x(t)=t2y(t)=lntt>0
Eliminating the Parameter from Trigonometric Equations
Eliminating the parameter from trigonometric equations is a straightforward substitution. We can use a few of the familiar trigonometric identities and the Pythagorean Theorem.
First, we use the identities:
x(t)=acosty(t)=bsintx(t)=acosty(t)=bsint
Solving for costcost and sint,sint, we have
xa=costyb=sintxa=costyb=sint
Then, use the Pythagorean Theorem:
cos2t+sin2t=1cos2t+sin2t=1
Substituting gives
cos2t+sin2t=(xa)2+(yb)2=1cos2t+sin2t=(xa)2+(yb)2=1
EXAMPLE 7
Eliminating the Parameter from a Pair of Trigonometric Parametric Equations
Eliminate the parameter from the given pair of trigonometric equations where 0≤t≤2π0≤t≤2π and sketch the graph.
x(t)=4costy(t)=3sintx(t)=4costy(t)=3sint
- Answer
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Analysis
Applying the general equations for conic sections (introduced in Analytic Geometry, we can identify x216+y29=1x216+y29=1 as an ellipse centered at (0,0).(0,0). Notice that when t=0t=0 the coordinates are (4,0),(4,0), and when t=π2t=π2 the coordinates are (0,3).(0,3). This shows the orientation of the curve with increasing values of t.t.
Eliminate the parameter from the given pair of parametric equations and write as a Cartesian equation: x(t)=2costx(t)=2cost and y(t)=3sint.y(t)=3sint.
Finding Cartesian Equations from Curves Defined Parametrically
When we are given a set of parametric equations and need to find an equivalent Cartesian equation, we are essentially “eliminating the parameter.” However, there are various methods we can use to rewrite a set of parametric equations as a Cartesian equation. The simplest method is to set one equation equal to the parameter, such as x(t)=t.x(t)=t. In this case, y(t)y(t) can be any expression. For example, consider the following pair of equations.
x(t)=ty(t)=t2−3x(t)=ty(t)=t2−3
Rewriting this set of parametric equations is a matter of substituting xx for t.t. Thus, the Cartesian equation is y=x2−3.y=x2−3.
EXAMPLE 8
Finding a Cartesian Equation Using Alternate Methods
Use two different methods to find the Cartesian equation equivalent to the given set of parametric equations.
x(t)=3t−2y(t)=t+1x(t)=3t−2y(t)=t+1
- Answer
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Write the given parametric equations as a Cartesian equation: x(t)=t3x(t)=t3 and y(t)=t6.y(t)=t6.
Finding Parametric Equations for Curves Defined by Rectangular Equations
Although we have just shown that there is only one way to interpret a set of parametric equations as a rectangular equation, there are multiple ways to interpret a rectangular equation as a set of parametric equations. Any strategy we may use to find the parametric equations is valid if it produces equivalency. In other words, if we choose an expression to represent x,x, and then substitute it into the yy equation, and it produces the same graph over the same domain as the rectangular equation, then the set of parametric equations is valid. If the domain becomes restricted in the set of parametric equations, and the function does not allow the same values for xx as the domain of the rectangular equation, then the graphs will be different.
EXAMPLE 9
Finding a Set of Parametric Equations for Curves Defined by Rectangular Equations
Find a set of equivalent parametric equations for y=(x+3)2+1.y=(x+3)2+1.
- Answer
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Access these online resources for additional instruction and practice with parametric equations.
8.6 Section Exercises
Verbal
1.
What is a system of parametric equations?
2.
Some examples of a third parameter are time, length, speed, and scale. Explain when time is used as a parameter.
3.
Explain how to eliminate a parameter given a set of parametric equations.
4.
What is a benefit of writing a system of parametric equations as a Cartesian equation?
5.
What is a benefit of using parametric equations?
6.
Why are there many sets of parametric equations to represent on Cartesian function?
Algebraic
For the following exercises, eliminate the parameter tt to rewrite the parametric equation as a Cartesian equation.
7.
{x(t)=5−ty(t)=8−2t{ x(t)=5−ty(t)=8−2t
8.
{x(t)=6−3ty(t)=10−t{ x(t)=6−3ty(t)=10−t
9.
{x(t)=2t+1y(t)=3t√{ x(t)=2t+1y(t)=3t
10.
{x(t)=3t−1y(t)=2t2{ x(t)=3t−1y(t)=2t2
11.
{x(t)=2ety(t)=1−5t{ x(t)=2ety(t)=1−5t
12.
{x(t)=e−2ty(t)=2e−t{ x(t)=e−2ty(t)=2e−t
13.
{x(t)=4log(t)y(t)=3+2t{x(t)=4log(t)y(t)=3+2t
14.
{x(t)=log(2t)y(t)=t−1−−−−√{x(t)=log(2t)y(t)=t−1
15.
{x(t)=t3−ty(t)=2t{ x(t)=t3−ty(t)=2t
16.
{x(t)=t−t4y(t)=t+2{ x(t)=t−t4y(t)=t+2
17.
{x(t)=e2ty(t)=e6t{ x(t)=e2ty(t)=e6t
18.
{x(t)=t5y(t)=t10{ x(t)=t5y(t)=t10
19.
{x(t)=4costy(t)=5sint{x(t)=4costy(t)=5sint
20.
{x(t)=3sinty(t)=6cost{ x(t)=3sinty(t)=6cost
21.
{x(t)=2cos2ty(t)=−sint{x(t)=2cos2ty(t)=−sint
22.
{x(t)=cost+4y(t)=2sin2t{ x(t)=cost+4y(t)=2sin2t
23.
{x(t)=t−1y(t)=t2{ x(t)=t−1y(t)=t2
24.
{x(t)=−ty(t)=t3+1{ x(t)=−ty(t)=t3+1
25.
{x(t)=2t−1y(t)=t3−2{ x(t)=2t−1y(t)=t3−2
For the following exercises, rewrite the parametric equation as a Cartesian equation by building an x-yx-y table.
26.
{x(t)=2t−1y(t)=t+4{ x(t)=2t−1y(t)=t+4
27.
{x(t)=4−ty(t)=3t+2{ x(t)=4−ty(t)=3t+2
28.
{x(t)=2t−1y(t)=5t{ x(t)=2t−1y(t)=5t
29.
{x(t)=4t−1y(t)=4t+2{ x(t)=4t−1y(t)=4t+2
For the following exercises, parameterize (write parametric equations for) each Cartesian equation by setting x(t)=tx(t)=t or by setting y(t)=t.y(t)=t.
30.
y(x)=3x2+3y(x)=3x2+3
31.
y(x)=2sinx+1y(x)=2sinx+1
32.
x(y)=3log(y)+yx(y)=3log(y)+y
33.
x(y)=y√+2yx(y)=y+2y
For the following exercises, parameterize (write parametric equations for) each Cartesian equation by using x(t)=acostx(t)=acost and y(t)=bsint.y(t)=bsint. Identify the curve.
34.
x24+y29=1x24+y29=1
35.
x216+y236=1x216+y236=1
36.
x2+y2=16x2+y2=16
37.
x2+y2=10x2+y2=10
38.
Parameterize the line from (3,0)(3,0) to (−2,−5)(−2,−5) so that the line is at (3,0)(3,0) at t=0,t=0, and at (−2,−5)(−2,−5) at t=1.t=1.
39.
Parameterize the line from (−1,0)(−1,0) to (3,−2)(3,−2) so that the line is at (−1,0)(−1,0) at t=0,t=0, and at (3,−2)(3,−2) at t=1.t=1.
40.
Parameterize the line from (−1,5)(−1,5) to (2,3)(2,3) so that the line is at (−1,5)(−1,5) at t=0,t=0, and at (2,3)(2,3) at t=1.t=1.
41.
Parameterize the line from (4,1)(4,1) to (6,−2)(6,−2) so that the line is at (4,1)(4,1) at t=0,t=0, and at (6,−2)(6,−2) at t=1.t=1.
Technology
For the following exercises, use the table feature in the graphing calculator to determine whether the graphs intersect.
42.
{x1(t)=3ty1(t)=2t−1 and {x2(t)=t+3y2(t)=4t−4{ x1(t)=3ty1(t)=2t−1 and { x2(t)=t+3y2(t)=4t−4
43.
{x1(t)=t2y1(t)=2t−1 and {x2(t)=−t+6y2(t)=t+1{ x1(t)=t2y1(t)=2t−1 and { x2(t)=−t+6y2(t)=t+1
For the following exercises, use a graphing calculator to complete the table of values for each set of parametric equations.
44.
{x1(t)=3t2−3t+7y1(t)=2t+3{ x1(t)=3t2−3t+7y1(t)=2t+3
| tt | xx | yy |
|---|---|---|
| –1 | ||
| 0 | ||
| 1 |
45.
{x1(t)=t2−4y1(t)=2t2−1{ x1(t)=t2−4y1(t)=2t2−1
| tt | xx | yy |
|---|---|---|
| 1 | ||
| 2 | ||
| 3 |
46.
{x1(t)=t4y1(t)=t3+4{ x1(t)=t4y1(t)=t3+4
| tt | xx | yy |
|---|---|---|
| -1 | ||
| 0 | ||
| 1 | ||
| 2 |