Parametric Equations: Graphs

Learning Objectives

In this section you will:

• Graph plane curves described by parametric equations by plotting points.
• Graph parametric equations.

While not every fan (or team manager) appreciates it, baseball and many other sports have become dependent on analytics, which involve complex data recording and quantitative evaluation used to understand and predict behavior. The earliest influence of analytics was mostly statistical; more recently, physics and other sciences have come into play. Foremost among these is the focus on launch angle and exit velocity, which when at certain values can almost guarantee a home run. On the other hand, emphasis on launch angle and focusing on home runs rather than overall hitting results in far more outs. Consider the following situation: it is the bottom of the ninth inning, with two outs and two players on base. The home team is losing by two runs. The batter swings and hits the baseball at 140 feet per second and at an angle of approximately 45°45° to the horizontal. How far will the ball travel? Will it clear the fence for a game-winning home run? The outcome may depend partly on other factors (for example, the wind), but mathematicians can model the path of a projectile and predict approximately how far it will travel using parametric equations. In this section, we’ll discuss parametric equations and some common applications, such as projectile motion problems.

Figure 1 Parametric equations can model the path of a projectile. (credit: Paul Kreher, Flickr)

Graphing Parametric Equations by Plotting Points

In lieu of a graphing calculator or a computer graphing program, plotting points to represent the graph of an equation is the standard method. As long as we are careful in calculating the values, point-plotting is highly dependable.

HOW TO

Given a pair of parametric equations, sketch a graph by plotting points.

1. Construct a table with three columns: t,x(t),andy(t).𝑡,𝑥(𝑡),and𝑦(𝑡).
2. Evaluate x𝑥 and y𝑦 for values of t𝑡 over the interval for which the functions are defined.
3. Plot the resulting pairs (x,y).(𝑥,𝑦).

EXAMPLE 1

Sketching the Graph of a Pair of Parametric Equations by Plotting Points

Sketch the graph of the parametric equations x(t)=t2+1,𝑥(𝑡)=𝑡2+1, y(t)=2+t.𝑦(𝑡)=2+𝑡.

Solution

Construct a table of values for t,x(t),𝑡,𝑥(𝑡),  and y(t),𝑦(𝑡), as in Table 1, and plot the points in a plane.

t𝑡	x(t)=t2+1𝑥(𝑡)=𝑡2+1	y(t)=2+t𝑦(𝑡)=2+𝑡
−5−5	2626	−3−3
−4−4	1717	−2−2
−3−3	1010	−1−1
−2−2	55	00
−1−1	22	11
00	11	22
11	22	33
22	55	44
33	1010	55
44	1717	66
55	2626	77

The graph is a parabola with vertex at the point (1,2),(1,2), opening to the right. See Figure 2.

Figure 2

Analysis

As values for t𝑡 progress in a positive direction from 0 to 5, the plotted points trace out the top half of the parabola. As values of t𝑡 become negative, they trace out the lower half of the parabola. There are no restrictions on the domain. The arrows indicate direction according to increasing values of t.𝑡. The graph does not represent a function, as it will fail the vertical line test. The graph is drawn in two parts: the positive values for t,𝑡, and the negative values for t.𝑡.

TRY IT #1

Sketch the graph of the parametric equations x=t√,𝑥=𝑡, y=2t+3,𝑦=2𝑡+3, 0≤t≤3.0≤𝑡≤3.

EXAMPLE 2

Sketching the Graph of Trigonometric Parametric Equations

Construct a table of values for the given parametric equations and sketch the graph:

x=2costy=4sint𝑥=2cos𝑡𝑦=4sin𝑡

Solution

Construct a table like that in Table 2 using angle measure in radians as inputs for t,𝑡, and evaluating x𝑥 and y.𝑦. Using angles with known sine and cosine values for t𝑡 makes calculations easier.

t𝑡	x=2cost𝑥=2cos𝑡	y=4sint𝑦=4sin𝑡
0	x=2cos(0)=2𝑥=2cos(0)=2	y=4sin(0)=0𝑦=4sin(0)=0
π6𝜋6	x=2cos(π6)=3–√𝑥=2cos(𝜋6)=3	y=4sin(π6)=2𝑦=4sin(𝜋6)=2
π3𝜋3	x=2cos(π3)=1𝑥=2cos(𝜋3)=1	y=4sin(π3)=23–√𝑦=4sin(𝜋3)=23
π2𝜋2	x=2cos(π2)=0𝑥=2cos(𝜋2)=0	y=4sin(π2)=4𝑦=4sin(𝜋2)=4
2π32𝜋3	x=2cos(2π3)=−1𝑥=2cos(2𝜋3)=−1	y=4sin(2π3)=23–√𝑦=4sin(2𝜋3)=23
5π65𝜋6	x=2cos(5π6)=−3–√𝑥=2cos(5𝜋6)=−3	y=4sin(5π6)=2𝑦=4sin(5𝜋6)=2
π𝜋	x=2cos(π)=−2𝑥=2cos(𝜋)=−2	y=4sin(π)=0𝑦=4sin(𝜋)=0
7π67𝜋6	x=2cos(7π6)=−3–√𝑥=2cos(7𝜋6)=−3	y=4sin(7π6)=−2𝑦=4sin(7𝜋6)=−2
4π34𝜋3	x=2cos(4π3)=−1𝑥=2cos(4𝜋3)=−1	y=4sin(4π3)=−23–√𝑦=4sin(4𝜋3)=−23
3π23𝜋2	x=2cos(3π2)=0𝑥=2cos(3𝜋2)=0	y=4sin(3π2)=−4𝑦=4sin(3𝜋2)=−4
5π35𝜋3	x=2cos(5π3)=1𝑥=2cos(5𝜋3)=1	y=4sin(5π3)=−23–√𝑦=4sin(5𝜋3)=−23
11π611𝜋6	x=2cos(11π6)=3–√𝑥=2cos(11𝜋6)=3	y=4sin(11π6)=−2𝑦=4sin(11𝜋6)=−2
2π2𝜋	x=2cos(2π)=2𝑥=2cos(2𝜋)=2	y=4sin(2π)=0𝑦=4sin(2𝜋)=0

Figure 3 shows the graph.

Figure 3

By the symmetry shown in the values of x𝑥 and y,𝑦, we see that the parametric equations represent an ellipse. The ellipse is mapped in a counterclockwise direction as shown by the arrows indicating increasing t𝑡 values.

Analysis

We have seen that parametric equations can be graphed by plotting points. However, a graphing calculator will save some time and reveal nuances in a graph that may be too tedious to discover using only hand calculations.

Make sure to change the mode on the calculator to parametric (PAR). To confirm, the Y=𝑌= window should show

X1T=Y1T=𝑋1𝑇=𝑌1𝑇=

instead of Y1=.𝑌1=.

TRY IT #2

Graph the parametric equations: x=5cost,𝑥=5cos𝑡, y=3sint.𝑦=3sin𝑡.

EXAMPLE 3

Graphing Parametric Equations and Rectangular Form Together

Graph the parametric equations x=5cost𝑥=5cos𝑡 and y=2sint.𝑦=2sin𝑡. First, construct the graph using data points generated from the parametric form. Then graph the rectangular form of the equation. Compare the two graphs.

Solution

Construct a table of values like that in Table 3.

t𝑡	x=5cost𝑥=5cos𝑡	y=2sint𝑦=2sin𝑡
00	x=5cos(0)=5𝑥=5cos(0)=5	y=2sin(0)=0𝑦=2sin(0)=0
11	x=5cos(1)≈2.7𝑥=5cos(1)≈2.7	y=2sin(1)≈1.7𝑦=2sin(1)≈1.7
22	x=5cos(2)≈−2.1𝑥=5cos(2)≈−2.1	y=2sin(2)≈1.8𝑦=2sin(2)≈1.8
33	x=5cos(3)≈−4.95𝑥=5cos(3)≈−4.95	y=2sin(3)≈0.28𝑦=2sin(3)≈0.28
44	x=5cos(4)≈−3.3𝑥=5cos(4)≈−3.3	y=2sin(4)≈−1.5𝑦=2sin(4)≈−1.5
55	x=5cos(5)≈1.4𝑥=5cos(5)≈1.4	y=2sin(5)≈−1.9𝑦=2sin(5)≈−1.9
−1−1	x=5cos(−1)≈2.7𝑥=5cos(−1)≈2.7	y=2sin(−1)≈−1.7𝑦=2sin(−1)≈−1.7
−2−2	x=5cos(−2)≈−2.1𝑥=5cos(−2)≈−2.1	y=2sin(−2)≈−1.8𝑦=2sin(−2)≈−1.8
−3−3	x=5cos(−3)≈−4.95𝑥=5cos(−3)≈−4.95	y=2sin(−3)≈−0.28𝑦=2sin(−3)≈−0.28
−4−4	x=5cos(−4)≈−3.3𝑥=5cos(−4)≈−3.3	y=2sin(−4)≈1.5𝑦=2sin(−4)≈1.5
−5−5	x=5cos(−5)≈1.4𝑥=5cos(−5)≈1.4	y=2sin(−5)≈1.9𝑦=2sin(−5)≈1.9

Plot the (x,y)(𝑥,𝑦) values from the table. See Figure 4.

Figure 4

Next, translate the parametric equations to rectangular form. To do this, we solve for t𝑡 in either x(t)𝑥(𝑡) or y(t),𝑦(𝑡), and then substitute the expression for t𝑡 in the other equation. The result will be a function y(x)𝑦(𝑥) if solving for t𝑡 as a function of x,𝑥, or x(y)𝑥(𝑦) if solving for t𝑡 as a function of y.𝑦.

x=5costx5=costy=2sinty2=sintSolve for cost.Solve for sint.𝑥=5cos𝑡𝑥5=cos𝑡Solve for cos𝑡.𝑦=2sin𝑡Solve for sin𝑡.𝑦2=sin𝑡

Then, use the Pythagorean Theorem.

cos2t+sin2t=1(x5)2+(y2)2=1×225+y24=1cos2𝑡+sin2𝑡=1(𝑥5)2+(𝑦2)2=1𝑥225+𝑦24=1

Analysis

In Figure 5, the data from the parametric equations and the rectangular equation are plotted together. The parametric equations are plotted in blue; the graph for the rectangular equation is drawn on top of the parametric in a dashed style colored red. Clearly, both forms produce the same graph.

Figure 5

EXAMPLE 4

Graphing Parametric Equations and Rectangular Equations on the Coordinate System

Graph the parametric equations x=t+1𝑥=𝑡+1 and y=t√,𝑦=𝑡, t≥0,𝑡≥0, and the rectangular equivalent y=x−1−−−−√𝑦=𝑥−1 on the same coordinate system.

Solution

Construct a table of values for the parametric equations, as we did in the previous example, and graph y=t√,𝑦=𝑡, t≥0𝑡≥0 on the same grid, as in Figure 6.

Figure 6

Analysis

With the domain on t𝑡 restricted, we only plot positive values of t.𝑡. The parametric data is graphed in blue and the graph of the rectangular equation is dashed in red. Once again, we see that the two forms overlap.

TRY IT #3

Sketch the graph of the parametric equations x=2cosθandy=4sinθ,𝑥=2cos𝜃and𝑦=4sin𝜃, along with the rectangular equation on the same grid.

Applications of Parametric Equations

Many of the advantages of parametric equations become obvious when applied to solving real-world problems. Although rectangular equations in x and y give an overall picture of an object’s path, they do not reveal the position of an object at a specific time. Parametric equations, however, illustrate how the values of x and y change depending on t, as the location of a moving object at a particular time.

A common application of parametric equations is solving problems involving projectile motion. In this type of motion, an object is propelled forward in an upward direction forming an angle of θ𝜃 to the horizontal, with an initial speed of v0,𝑣0, and at a height hℎ above the horizontal.

The path of an object propelled at an inclination of θ𝜃 to the horizontal, with initial speed v0,𝑣0, and at a height hℎ above the horizontal, is given by

x=(v0cosθ)t  y=−12gt2+(v0sinθ)t+h𝑥=(𝑣0cos𝜃)𝑡  𝑦=−12𝑔𝑡2+(𝑣0sin𝜃)𝑡+ℎ

where g𝑔 accounts for the effects of gravity and hℎ is the initial height of the object. Depending on the units involved in the problem, use g=32ft/s2𝑔=32ft/s2 or g=9.8m/s2.𝑔=9.8m/s2. The equation for x𝑥 gives horizontal distance, and the equation for y𝑦 gives the vertical distance.

HOW TO

Given a projectile motion problem, use parametric equations to solve.

1. The horizontal distance is given by x=(v0cosθ)t.𝑥=(𝑣0cos𝜃)𝑡. Substitute the initial speed of the object for v0.𝑣0.
2. The expression cosθcos𝜃 indicates the angle at which the object is propelled. Substitute that angle in degrees for cosθ.cos𝜃.
3. The vertical distance is given by the formula y=−12gt2+(v0sinθ)t+h.𝑦=−12𝑔𝑡2+(𝑣0sin𝜃)𝑡+ℎ. The term −12gt2−12𝑔𝑡2 represents the effect of gravity. Depending on units involved, use g=32ft/s2𝑔=32ft/s2 or g=9.8m/s2.𝑔=9.8m/s2. Again, substitute the initial speed for v0,𝑣0, and the height at which the object was propelled for h.ℎ.
4. Proceed by calculating each term to solve for t.𝑡.

EXAMPLE 5

Finding the Parametric Equations to Describe the Motion of a Baseball

Solve the problem presented at the beginning of this section. Does the batter hit the game-winning home run? Assume that the ball is hit with an initial velocity of 140 feet per second at an angle of 45°45° to the horizontal, making contact 3 feet above the ground.

1. ⓐ Find the parametric equations to model the path of the baseball.
2. ⓑ Where is the ball after 2 seconds?
3. ⓒ How long is the ball in the air?
4. ⓓ Is it a home run?

Solution

1. ⓐUse the formulas to set up the equations. The horizontal position is found using the parametric equation for x.𝑥. Thus,x=(v0cosθ)tx=(140cos(45°))t𝑥=(𝑣0cos𝜃)𝑡𝑥=(140cos(45°))𝑡The vertical position is found using the parametric equation for y.𝑦. Thus,y=−16t2+(v0sinθ)t+hy=−16t2+(140sin(45°))t+3𝑦=−16𝑡2+(𝑣0sin𝜃)𝑡+ℎ𝑦=−16𝑡2+(140sin(45°))𝑡+3
2. ⓑSubstitute 2 into the equations to find the horizontal and vertical positions of the ball.x=(140cos(45°))(2)x=198 feety=−16(2)2+(140sin(45°))(2)+3y=137 feet𝑥=(140cos(45°))(2)𝑥=198 feet𝑦=−16(2)2+(140sin(45°))(2)+3𝑦=137 feetAfter 2 seconds, the ball is 198 feet away from the batter’s box and 137 feet above the ground.
3. ⓒTo calculate how long the ball is in the air, we have to find out when it will hit ground, or when y=0.𝑦=0. Thus,y=−16t2+(140sin(45∘))t+3y=0t=6.2173Set y(t)=0 and solve the quadratic.𝑦=−16𝑡2+(140sin(45∘))𝑡+3𝑦=0Set 𝑦(𝑡)=0 and solve the quadratic.𝑡=6.2173When t=6.2173𝑡=6.2173 seconds, the ball has hit the ground. (The quadratic equation can be solved in various ways, but this problem was solved using a computer math program.)
4. ⓓWe cannot confirm that the hit was a home run without considering the size of the outfield, which varies from field to field. However, for simplicity’s sake, let’s assume that the outfield wall is 400 feet from home plate in the deepest part of the park. Let’s also assume that the wall is 10 feet high. In order to determine whether the ball clears the wall, we need to calculate how high the ball is when x = 400 feet. So we will set x = 400, solve for t,𝑡, and input t𝑡 into y.𝑦.x=(140cos(45°))t400=(140cos(45°))tt=4.04y=−16(4.04)2+(140sin(45°))(4.04)+3y=141.8𝑥=(140cos(45°))𝑡400=(140cos(45°))𝑡𝑡=4.04𝑦=−16(4.04)2+(140sin(45°))(4.04)+3𝑦=141.8The ball is 141.8 feet in the air when it soars out of the ballpark. It was indeed a home run. See Figure 7.

Figure 7