Graphing and Representing Parametric Equations

Learning Outcomes

Plot a curve described by parametric equations
Convert the parametric equations of a curve into the form [latex]y=f\left(x\right)[/latex]
Recognize the parametric equations of basic curves, such as a line and a circle

Parametric Equations and Their Graphs

Consider the orbit of Earth around the Sun. Our year lasts approximately 365.25 days, but for this discussion we will use 365 days. On January 1 of each year, the physical location of Earth with respect to the Sun is nearly the same, except for leap years, when the lag introduced by the extra [latex]\frac{1}{4}[/latex] day of orbiting time is built into the calendar. We call January 1 “day 1” of the year. Then, for example, day 31 is January 31, day 59 is February 28, and so on.

The number of the day in a year can be considered a variable that determines Earth’s position in its orbit. As Earth revolves around the Sun, its physical location changes relative to the Sun. After one full year, we are back where we started, and a new year begins. According to Kepler’s laws of planetary motion, the shape of the orbit is elliptical, with the Sun at one focus of the ellipse. We study this idea in more detail in Conic Sections.

An ellipse with January 1 (t = 1) at the top, April 2 (t = 92) on the left, July 1 (t = 182) on the bottom, and October 1 (t = 274) on the right. The focal points of the ellipse have F2 on the left and the Sun on the right.

Figure 1 depicts Earth’s orbit around the Sun during one year. The point labeled [latex]{F}_{2}[/latex] is one of the foci of the ellipse; the other focus is occupied by the Sun. If we superimpose coordinate axes over this graph, then we can assign ordered pairs to each point on the ellipse (Figure 2). Then each x value on the graph is a value of position as a function of time, and each y value is also a value of position as a function of time. Therefore, each point on the graph corresponds to a value of Earth’s position as a function of time.

We can determine the functions for [latex]x\left(t\right)[/latex] and [latex]y\left(t\right)[/latex], thereby parameterizing the orbit of Earth around the Sun. The variable [latex]t[/latex] is called an independent parameter and, in this context, represents time relative to the beginning of each year.

A curve in the [latex]\left(x,y\right)[/latex] plane can be represented parametrically. The equations that are used to define the curve are called parametric equations.

Definition

If [latex]x[/latex] and [latex]y[/latex] are continuous functions of [latex]t[/latex] on an interval [latex]I[/latex], then the equations

[latex]x=x\left(t\right)\text{and}y=y\left(t\right)[/latex]

are called parametric equations and [latex]t[/latex] is called the parameter. The set of points [latex]\left(x,y\right)[/latex] obtained as [latex]t[/latex] varies over the interval [latex]I[/latex] is called the graph of the parametric equations. The graph of parametric equations is called a parametric curve or plane curve, and is denoted by [latex]C[/latex].

Notice in this definition that [latex]x[/latex] and [latex]y[/latex] are used in two ways. The first is as functions of the independent variable [latex]t[/latex]. As [latex]t[/latex] varies over the interval [latex]I[/latex], the functions [latex]x\left(t\right)[/latex] and [latex]y\left(t\right)[/latex] generate a set of ordered pairs [latex]\left(x,y\right)[/latex]. This set of ordered pairs generates the graph of the parametric equations. In this second usage, to designate the ordered pairs, [latex]x[/latex] and [latex]y[/latex] are variables. It is important to distinguish the variables [latex]x[/latex] and [latex]y[/latex] from the functions [latex]x\left(t\right)[/latex] and [latex]y\left(t\right)[/latex].

Example: Graphing a Parametrically Defined Curve

Sketch the curves described by the following parametric equations:

[latex]x\left(t\right)=t - 1,y\left(t\right)=2t+4,-3\le t\le 2[/latex]
[latex]x\left(t\right)={t}^{2}-3,y\left(t\right)=2t+1,-2\le t\le 3[/latex]
[latex]x\left(t\right)=4\cos{t},y\left(t\right)=4\sin{t},0\le t\le 2\pi [/latex]

Show Solution

To create a graph of this curve, first set up a table of values. Since the independent variable in both [latex]x\left(t\right)[/latex] and [latex]y\left(t\right)[/latex] is t, let t appear in the first column. Then [latex]x\left(t\right)[/latex] and [latex]y\left(t\right)[/latex] will appear in the second and third columns of the table.

t [latex]x\left(t\right)[/latex] [latex]y\left(t\right)[/latex]

−3 −4 −2

−2 −3 0

−1 −2 2

0 −1 4

1 0 6

2 1 8

The second and third columns in this table provide a set of points to be plotted. The graph of these points appears in Figure 3. The arrows on the graph indicate the orientation of the graph, that is, the direction that a point moves on the graph as t varies from −3 to 2.

Figure 3. Graph of the plane curve described by the parametric equations in part a.
To create a graph of this curve, again set up a table of values.

t [latex]x\left(t\right)[/latex] [latex]y\left(t\right)[/latex]

−2 1 −3

−1 −2 −1

0 −3 1

1 −2 3

2 1 5

3 6 7

The second and third columns in this table give a set of points to be plotted (Figure 4). The first point on the graph (corresponding to [latex]t=-2[/latex]) has coordinates [latex]\left(1,-3\right)[/latex], and the last point (corresponding to [latex]t=3[/latex]) has coordinates [latex]\left(6,7\right)[/latex]. As [latex]t[/latex] progresses from −2 to 3, the point on the curve travels along a parabola. The direction the point moves is again called the orientation and is indicated on the graph.

Figure 4. Graph of the plane curve described by the parametric equations in part b.

t	[latex]x\left(t\right)[/latex]	[latex]y\left(t\right)[/latex]
−3	−4	−2
−2	−3	0
−1	−2	2
0	−1	4
1	0	6
2	1	8

t	[latex]x\left(t\right)[/latex]	[latex]y\left(t\right)[/latex]
−2	1	−3
−1	−2	−1
0	−3	1
1	−2	3
2	1	5
3	6	7

In this case, use multiples of [latex]\frac{\pi}{6}[/latex] for t and create another table of values:

t	[latex]x\left(t\right)[/latex]	[latex]y\left(t\right)[/latex]	t	[latex]x\left(t\right)[/latex]	[latex]y\left(t\right)[/latex]
0	4	0	[latex]\frac{7\pi }{6}[/latex]	[latex]-2\sqrt{3}\approx -3.5[/latex]	2
[latex]\frac{\pi }{6}[/latex]	[latex]2\sqrt{3}\approx 3.5[/latex]	[latex]2[/latex]	[latex]\frac{4\pi }{3}[/latex]	−2	[latex]-2\sqrt{3}\approx -3.5[/latex]
[latex]\frac{\pi }{3}[/latex]	[latex]2[/latex]	[latex]2\sqrt{3}\approx 3.5[/latex]	[latex]\frac{3\pi }{2}[/latex]	0	−4
[latex]\frac{\pi }{2}[/latex]	0	4	[latex]\frac{5\pi }{3}[/latex]	2	[latex]-2\sqrt{3}\approx -3.5[/latex]
[latex]\frac{2\pi }{3}[/latex]	−2	[latex]2\sqrt{3}\approx 3.5[/latex]	[latex]\frac{11\pi }{6}[/latex]	[latex]2\sqrt{3}\approx 3.5[/latex]	2
[latex]\frac{5\pi }{6}[/latex]	[latex]-2\sqrt{3}\approx -3.5[/latex]	2	[latex]2\pi [/latex]	4	0
[latex]\pi [/latex]	−4	0

The graph of this plane curve appears in the following graph.

A circle with radius 4 centered at the origin is graphed with arrow going counterclockwise. The point (4, 0) is marked t = 0, the point (0, 4) is marked t = π/2, the point (−4, 0) is marked t = π, and the point (0, −4) is marked t = 3π/2. On the graph there are also written three equations: x(t) = 4 cos(t), y(t) = 4 sin(t), and 0 ≤ t ≤ 2π.

This is the graph of a circle with radius 4 centered at the origin, with a counterclockwise orientation. The starting point and ending points of the curve both have coordinates [latex]\left(4,0\right)[/latex].

Watch the following video to see the worked solution to Example: Graphing a Parametrically Defined Curve.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “7.1 Parametric Equations” here (opens in new window).

try it

Sketch the curve described by the parametric equations

[latex]x\left(t\right)=3t+2,y\left(t\right)={t}^{2}-1,-3\le t\le 2[/latex].

Hint

Show Solution

Try It

Eliminating the Parameter

To better understand the graph of a curve represented parametrically, it is useful to rewrite the two equations as a single equation relating the variables x and y. Then we can apply any previous knowledge of equations of curves in the plane to identify the curve. For example, the equations describing the plane curve in part (b) of the previous example are

[latex]x\left(t\right)={t}^{2}-3,y\left(t\right)=2t+1,-2\le t\le 3[/latex].

Solving the second equation for t gives

[latex]t=\frac{y - 1}{2}[/latex].

This can be substituted into the first equation:

[latex]x={\left(\frac{y - 1}{2}\right)}^{2}-3=\frac{{y}^{2}-2y+1}{4}-3=\frac{{y}^{2}-2y - 11}{4}[/latex].

This equation describes x as a function of y. These steps give an example of eliminating the parameter. The graph of this function is a parabola opening to the right. Recall that the plane curve started at [latex]\left(1,-3\right)[/latex] and ended at [latex]\left(6,7\right)[/latex]. These terminations were due to the restriction on the parameter t.

Before working through an example on how to eliminate the parameter, it is useful to recall the Pythagorean Identity as well as the equations of circles.

Recall: Pythagorean identity and equation of circles

For any angle [latex] t, \sin^2 t + \cos^2 t = 1 [/latex]
A circle of radius [latex] a [/latex] centered at the origin is given by [latex] x^2 + y^2 = a^2 [/latex]

Example: Eliminating the Parameter

Eliminate the parameter for each of the plane curves described by the following parametric equations and describe the resulting graph.

[latex]x\left(t\right)=\sqrt{2t+4},y\left(t\right)=2t+1,-2\le t\le 6[/latex]
[latex]x\left(t\right)=4\cos{t},y\left(t\right)=3\sin{t},0\le t\le 2\pi [/latex]

Show Solution

To eliminate the parameter, we can solve either of the equations for t. For example, solving the first equation for t gives

[latex]\begin{array}{ccc}\hfill x& =\hfill & \sqrt{2t+4}\hfill \\ \hfill {x}^{2}& =\hfill & 2t+4\hfill \\ \hfill {x}^{2}-4& =\hfill & 2t\hfill \\ \hfill t& =\hfill & \frac{{x}^{2}-4}{2}.\hfill \end{array}[/latex]

Note that when we square both sides it is important to observe that [latex]x\ge 0[/latex]. Substituting [latex]t=\frac{{x}^{2}-4}{2}[/latex] this into [latex]y\left(t\right)[/latex] yields

[latex]\begin{array}{ccc}\hfill y\left(t\right)& =\hfill & 2t+1\hfill \\ \hfill y& =\hfill & 2\left(\frac{{x}^{2}-4}{2}\right)+1\hfill \\ \hfill y& =\hfill & {x}^{2}-4+1\hfill \\ \hfill y& =\hfill & {x}^{2}-3.\hfill \end{array}[/latex]

This is the equation of a parabola opening upward. There is, however, a domain restriction because of the limits on the parameter t. When [latex]t=-2[/latex], [latex]x=\sqrt{2\left(-2\right)+4}=0[/latex], and when [latex]t=6[/latex], [latex]x=\sqrt{2\left(6\right)+4}=4[/latex]. The graph of this plane curve follows.

Figure 7. Graph of the plane curve described by the parametric equations in part a.
Sometimes it is necessary to be a bit creative in eliminating the parameter. The parametric equations for this example are

[latex]x\left(t\right)=4\cos{t}\text{and}y\left(t\right)=3\sin{t}[/latex].

Solving either equation for t directly is not advisable because sine and cosine are not one-to-one functions. However, dividing the first equation by 4 and the second equation by 3 (and suppressing the t) gives us

[latex]\cos{t}=\frac{x}{4}\text{and}\sin{t}=\frac{y}{3}[/latex].

Now use the Pythagorean identity [latex]{\cos}^{2}t+{\sin}^{2}t=1[/latex] and replace the expressions for [latex]\sin{t}[/latex] and [latex]\cos{t}[/latex] with the equivalent expressions in terms of x and y. This gives

[latex]\begin{array}{ccc}\hfill {\left(\frac{x}{4}\right)}^{2}+{\left(\frac{y}{3}\right)}^{2}& =\hfill & 1\hfill \\ \hfill \frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}& =\hfill & 1.\hfill \end{array}[/latex]

This is the equation of a horizontal ellipse centered at the origin, with semimajor axis 4 and semiminor axis 3 as shown in the following graph.

Figure 8. Graph of the plane curve described by the parametric equations in part b.

As t progresses from [latex]0[/latex] to [latex]2\pi [/latex], a point on the curve traverses the ellipse once, in a counterclockwise direction. Recall from the section opener that the orbit of Earth around the Sun is also elliptical. This is a perfect example of using parameterized curves to model a real-world phenomenon.

Watch the following video to see the worked solution to Example: Eliminating the Parameter.

You can view the transcript for this segmented clip of “7.1 Parametric Equations” here (opens in new window).

try it

Eliminate the parameter for the plane curve defined by the following parametric equations and describe the resulting graph.

[latex]x\left(t\right)=2+\frac{3}{t},y\left(t\right)=t - 1,2\le t\le 6[/latex]

Hint

Show Solution

Try It

So far we have seen the method of eliminating the parameter, assuming we know a set of parametric equations that describe a plane curve. What if we would like to start with the equation of a curve and determine a pair of parametric equations for that curve? This is certainly possible, and in fact it is possible to do so in many different ways for a given curve. The process is known as parameterization of a curve.

Example: Parameterizing a Curve

Find two different pairs of parametric equations to represent the graph of [latex]y=2{x}^{2}-3[/latex].

Show Solution

Watch the following video to see the worked solution to Example: Parameterizing a Curve.

You can view the transcript for this segmented clip of “7.1 Parametric Equations” here (opens in new window).

try it

Find two different sets of parametric equations to represent the graph of [latex]y={x}^{2}+2x[/latex].

Hint

Show Solution

Module 7: Parametric Equations and Polar Coordinates