Parametric Equations: Graphs

By the end of this section, you will be able to:

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 $45^\circ$ 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.

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.

Construct a table with three columns: $t$ , $x(t)$ , and $y(t)$ .
Evaluate $x$ and $y$ for values of $t$ over the interval for which the functions are defined.
Plot the resulting pairs $(x,y)$ .

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

Solution. Construct a table of values for $t$ , $x(t)$ , and $y(t)$ , and plot the points in a plane.

$t$	$x(t)=t^2+1$	$y(t)=2+t$
$-5$	$26$	$-3$
$-4$	$17$	$-2$
$-3$	$10$	$-1$
$-2$	$5$	$0$
$-1$	$2$	$1$
$0$	$1$	$2$
$1$	$2$	$3$
$2$	$5$	$4$
$3$	$10$	$5$
$4$	$17$	$6$
$5$	$26$	$7$

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

Figure: The plane curve x(t)=t squared plus 1, y(t)=2+t traced for t from -3 to 3: a parabola opening to the right with vertex at (1,2), a lower branch through (10,-1), (5,0), (2,1), and an upper branch through (2,3), (5,4), (10,5), with an arrow on the lower branch pointing toward the vertex and an arrow on the upper branch pointing away from it, both in the direction of increasing t.

Analysis. As values for $t$ progress in a positive direction from $0$ to $5$ , the plotted points trace out the top half of the parabola. As values of $t$ 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 $t$ . 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 $t$ , and the negative values for $t$ .

Sketch the graph of the parametric equations $x=\sqrt t$ , $y=2t+3$ , $0\le t\le3$ .

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

\begin{array}{l} x=2\cos t \\[4pt] y=4\sin t \end{array}

Solution. Construct a table like the one below using angle measure in radians as inputs for $t$ , and evaluating $x$ and $y$ . Using angles with known sine and cosine values for $t$ makes calculations easier.

$t$	$x=2\cos t$	$y=4\sin t$
$0$	$2$	$0$
$\tfrac{\pi}{6}$	$\sqrt3$	$2$
$\tfrac{\pi}{3}$	$1$	$2\sqrt3$
$\tfrac{\pi}{2}$	$0$	$4$
$\tfrac{2\pi}{3}$	$-1$	$2\sqrt3$
$\tfrac{5\pi}{6}$	$-\sqrt3$	$2$
$\pi$	$-2$	$0$
$\tfrac{7\pi}{6}$	$-\sqrt3$	$-2$
$\tfrac{4\pi}{3}$	$-1$	$-2\sqrt3$
$\tfrac{3\pi}{2}$	$0$	$-4$
$\tfrac{5\pi}{3}$	$1$	$-2\sqrt3$
$\tfrac{11\pi}{6}$	$\sqrt3$	$-2$
$2\pi$	$2$	$0$

The figure below shows the graph.

Figure: An ellipse taller than it is wide, traced by x=2cos t, y=4sin t, marked at (2,0) for t=0, (0,4) for t=pi/2, (-1.73,2) for t=5pi/6, and (1.73,-2) for t=11pi/6, with an arrow indicating a counterclockwise direction of travel.

By the symmetry shown in the values of $x$ and $y$ , we see that the parametric equations represent an ellipse. The ellipse is mapped in a counterclockwise direction as shown by the arrows indicating increasing $t$ 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 $Y=$ window should show

\begin{array}{l} X_{1T}= \\[4pt] Y_{1T}= \end{array}

instead of $Y_1=$ .

Graph the parametric equations: $x=5\cos t$ , $y=3\sin t$ .

Example. Graph the parametric equations $x=5\cos t$ and $y=2\sin t$ . 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 the one below.

$t$	$x=5\cos t$	$y=2\sin t$
$0$	$x=5\cos(0)=5$	$y=2\sin(0)=0$
$1$	$x=5\cos(1)\approx2.7$	$y=2\sin(1)\approx1.7$
$2$	$x=5\cos(2)\approx-2.1$	$y=2\sin(2)\approx1.8$
$3$	$x=5\cos(3)\approx-4.95$	$y=2\sin(3)\approx0.28$
$4$	$x=5\cos(4)\approx-3.3$	$y=2\sin(4)\approx-1.5$
$5$	$x=5\cos(5)\approx1.4$	$y=2\sin(5)\approx-1.9$
$-1$	$x=5\cos(-1)\approx2.7$	$y=2\sin(-1)\approx-1.7$
$-2$	$x=5\cos(-2)\approx-2.1$	$y=2\sin(-2)\approx-1.8$
$-3$	$x=5\cos(-3)\approx-4.95$	$y=2\sin(-3)\approx-0.28$
$-4$	$x=5\cos(-4)\approx-3.3$	$y=2\sin(-4)\approx1.5$
$-5$	$x=5\cos(-5)\approx1.4$	$y=2\sin(-5)\approx1.9$

Plot the $(x,y)$ values from the table. See the figure below.

Figure: An ellipse wider than it is tall, plotted from the parametric table, marked at (5,0) for t=0, (-2.1,1.8) for t=2, and (-2.1,-1.8) for t=-2, with an arrow indicating a counterclockwise direction of travel.

(a) Plotted from the parametric table of values.

(b) Plotted from the rectangular equation $\tfrac{x^2}{25}+\tfrac{y^2}{4}=1$ .

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

\begin{array}{lrcl} & x &=& 5\cos t \\[4pt] \text{Solve for}\ \cos t. & \tfrac{x}{5} &=& \cos t \\[4pt] & y &=& 2\sin t \\[4pt] \text{Solve for}\ \sin t. & \tfrac{y}{2} &=& \sin t \end{array}

Then, use the Pythagorean Theorem.

\begin{array}{lrcl} & \cos^2t+\sin^2t &=& 1 \\[4pt] & \left(\tfrac{x}{5}\right)^2+\left(\tfrac{y}{2}\right)^2 &=& 1 \\[4pt] & \tfrac{x^2}{25}+\tfrac{y^2}{4} &=& 1 \end{array}

Analysis. In the figure below, the data from the parametric equations and the rectangular equation are plotted together. The parametric equations are plotted as a solid curve; the graph for the rectangular equation is drawn on top of the parametric plot in a dashed style. Clearly, both forms produce the same graph.

Figure: The two overlaid ellipses coincide exactly: a solid ellipse traced from the parametric equations lies directly under a dashed ellipse traced from the rectangular equation, wider than it is tall.

(c) The parametric plot (solid) and the rectangular plot (dashed) coincide exactly.

Example. Graph the parametric equations $x=t+1$ and $y=\sqrt t$ , $t\ge0$ , and the rectangular equivalent $y=\sqrt{x-1}$ on the same coordinate system.

Solution. Construct a table of values for the parametric equations, as we did in the previous example, and graph $y=\sqrt t$ , $t\ge0$ on the same grid, as in the figure below.

Figure: A curve rising from (1,0) to about (9,2.83), plotted twice: a solid curve from the parametric equations and a dashed curve from the rectangular equation y equals the square root of x minus 1, exactly overlapping, with an arrow pointing in the direction of increasing t.

Analysis. With the domain on $t$ restricted, we only plot positive values of $t$ . The parametric data is graphed as a solid curve and the graph of the rectangular equation is dashed on top of it. Once again, we see that the two forms overlap.

Sketch the graph of the parametric equations $x=2\cos\theta$ and $y=4\sin\theta$ , 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 $\theta$ to the horizontal, with an initial speed of $v_0$ , and at a height $h$ above the horizontal.

The path of an object propelled at an inclination of $\theta$ to the horizontal, with initial speed $v_0$ , and at a height $h$ above the horizontal, is given by

\begin{array}{l} x=(v_0\cos\theta)t \\[4pt] y=-\tfrac12gt^2+(v_0\sin\theta)t+h \end{array}

where $g$ accounts for the effects of gravity and $h$ is the initial height of the object. Depending on the units involved in the problem, use $g=32\text{ ft/s}^2$ or $g=9.8\text{ m/s}^2$ . The equation for $x$ gives horizontal distance, and the equation for $y$ gives the vertical distance.

How to: given a projectile motion problem, use parametric equations to solve.

The horizontal distance is given by $x=(v_0\cos\theta)t$ . Substitute the initial speed of the object for $v_0$ .
The expression $\cos\theta$ indicates the angle at which the object is propelled. Substitute that angle in degrees for $\cos\theta$ .
The vertical distance is given by the formula $y=-\tfrac12gt^2+(v_0\sin\theta)t+h$ . The term $-\tfrac12gt^2$ represents the effect of gravity. Depending on units involved, use $g=32\text{ ft/s}^2$ or $g=9.8\text{ m/s}^2$ . Again, substitute the initial speed for $v_0$ , and the height at which the object was propelled for $h$ .
Proceed by calculating each term to solve for $t$ .

Example. 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 $45^\circ$ to the horizontal, making contact $3$ feet above the ground.

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

Solution. ⓐ Use the formulas to set up the equations. The horizontal position is found using the parametric equation for $x$ . Thus,

\begin{array}{lrcl} & x &=& (v_0\cos\theta)t \\[4pt] & x &=& (140\cos(45^\circ))t \end{array}

The vertical position is found using the parametric equation for $y$ . Thus,

\begin{array}{lrcl} & y &=& -16t^2+(v_0\sin\theta)t+h \\[4pt] & y &=& -16t^2+(140\sin(45^\circ))t+3 \end{array}

ⓑ Substitute $2$ into the equations to find the horizontal and vertical positions of the ball.

\begin{array}{lrcl} & x &=& (140\cos(45^\circ))(2) \\[4pt] & x &\approx& 198\ \text{feet} \end{array}

\begin{array}{lrcl} & y &=& -16(2)^2+(140\sin(45^\circ))(2)+3 \\[4pt] & y &\approx& 137\ \text{feet} \end{array}

After $2$ seconds, the ball is $198$ feet away from the batter’s box and $137$ feet above the ground.

\begin{array}{lrcl} & y &=& -16t^2+(140\sin(45^\circ))t+3 \\[4pt] \text{Set}\ y(t)=0\ \text{and solve the quadratic.} & 0 &=& -16t^2+(140\sin(45^\circ))t+3 \\[4pt] & t &\approx& 6.2173 \end{array}

When $t\approx6.2173$ 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.)

ⓓ 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 $t$ , and input $t$ into $y$ .

\begin{array}{lrcl} & x &=& (140\cos(45^\circ))t \\[4pt] & 400 &=& (140\cos(45^\circ))t \\[4pt] & t &\approx& 4.04 \end{array}

\begin{array}{lrcl} & y &=& -16(4.04)^2+(140\sin(45^\circ))(4.04)+3 \\[4pt] & y &\approx& 141.8 \end{array}

The ball is $141.8$ feet in the air when it soars out of the ballpark. It was indeed a home run. See the figure below.

Figure: The trajectory of the baseball, a parabolic arc from (0,3) rising to a peak of about 156 feet near 306 feet out, then falling back to the ground at about 615 feet; a short vertical mark at 400 feet rises to 10 feet, showing the outfield wall, which the trajectory clears well above.

Key concepts

When there is a third variable, a third parameter on which $x$ and $y$ depend, parametric equations can be used.
To graph parametric equations by plotting points, make a table with three columns labeled $t$ , $x(t)$ , and $y(t)$ . Choose values for $t$ in increasing order. Plot the last two columns for $x$ and $y$ .
When graphing a parametric curve by plotting points, note the associated $t$ -values and show arrows on the graph indicating the orientation of the curve.
Parametric equations allow the direction or the orientation of the curve to be shown on the graph. Equations that are not functions can be graphed and used in many applications involving motion.
Projectile motion depends on two parametric equations: $x=(v_0\cos\theta)t$ and $y=-16t^2+(v_0\sin\theta)t+h$ . Initial velocity is symbolized as $v_0$ . $\theta$ represents the initial angle of the object when thrown, and $h$ represents the height at which the object is propelled.

Practice

Graph plane curves described by parametric equations by plotting points

Graph the set of parametric equations by making a table of values for $t=-3,-2,-1,0,1,2$ , and include the orientation: $x(t)=t-1$ , $y(t)=t^2$ .

Graph the set of parametric equations by making a table of values, and include the orientation: $x(t)=2\sin t$ , $y(t)=4\cos t$ .

Sketch the curve and include the orientation: $x(t)=-\sqrt t$ , $y(t)=t$ .

Graph parametric equations

Eliminate the parameter to write the Cartesian equation of $x(t)=-2-2t$ , $y(t)=3+t$ in slope-intercept form.

Eliminate the parameter to write the Cartesian equation of $x=2\cos t$ , $y=-\sin t$ .

Write the parametric equations of a circle centered at the origin with radius $5$ and a counterclockwise orientation, using the form $x(t)=5\cos t$ . Enter $x(t)$ first, then $y(t)$ , separated by a comma.

A dart is thrown upward from the ground with an initial velocity of $64$ ft/s at an angle of elevation of $52^\circ$ . Neglecting air resistance, when will the dart hit the ground? Round to the nearest tenth of a second.

A dart is thrown upward from the ground with an initial velocity of $64$ ft/s at an angle of elevation of $52^\circ$ . Neglecting air resistance, at what time will the dart reach its maximum height? Round to the nearest tenth of a second.

This section is adapted from Precalculus 2e, Section 8.7: Parametric Equations: Graphs by Jay Abramson and OpenStax, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: omitted the decorative photo of a baseball batter (Figure 1), which carries no mathematics beyond the credit line, and the “Graphing Parametric Equations on the TI-84” media resource box, an external video link with no mathematical content of its own. Recreated every instructional figure as an accessible spec-first SVG: the two-branch parabola of Example 1 (drawn from the exact sideways parabola $x=(y-2)^2+1$ , with the source’s own $t$ -labels and two tangent arrows placed on the exact table segments); the two ellipses of Examples 2 and 3 (drawn from the exact rx/ry primitive rather than sampled points, with the source’s own labeled points and orientation arrows computed from the parametric derivative); the parametric/rectangular overlay pairs of Examples 3 and 4 (a solid curve for the parametric plot and a dashed curve of the identical primitive for the rectangular plot, reproducing the source’s solid-blue/dashed-red overlay convention in one color); and the baseball trajectory of Example 5 (sampled from the exact solved formula $x(t)=(140\cos45^\circ)t$ , $y(t)=-16t^2+(140\sin45^\circ)t+3$ over $t\in[0,6.2173]$ , with the outfield wall and hitter/wall/trajectory labels from the source figure; the source’s leader-line callouts were simplified to plain text labels). Every retained Try It became a real interactive component. Because the grader has no primitive that judges a drawn curve, each “sketch/graph the curve” Try It became a multiplechoice in graph mode, with distractors that vary the drawn shape, semi-axis lengths, domain branch, or overlay agreement (never only a marked point) and options ordered so the page does not use one answer position throughout. Adapted eight selected end-of-section exercises into a closing Practice block, one group per objective: three “graph by making a table”/“sketch the curve” items became graph-mode multiplechoice questions for the same reason the Try Its did; a fourth graph-by-table item ( $x(t)=-2-2t$ , $y(t)=3+t$ , whose printed answer is the graph of a negative-slope line) was recast as an eliminate-the-parameter fillin asking for that line’s slope-intercept equation, since the drawn answer is exactly the line the equation names; and four items whose skill survives as a value — eliminating the parameter to a Cartesian equation, constructing a pinned circle parametrization, and two applied timing questions about a thrown dart — became fillin questions, each independently re-derived (including by running the arithmetic in Node) rather than read off the source key. The two dart questions restate the shared scenario from the source’s introductory sentence, since each interactive component must stand alone. Every polar/degree angle in an applied formula (e.g. $140\cos45^\circ$ ) is graded by the engine’s exact treatment of $^\circ$ , so no answerForm token was needed on the numeric applied answers beyond decimal.

Parametric Equations Vectors