Construct a table of values for [latex]t,x\left(t\right)[/latex], and [latex]y\left(t\right)[/latex], as in the table below, and plot the points in a plane.

The graph is a parabola with vertex at the point [latex]\left(1,2\right)[/latex], opening to the right. See Figure 2.

Figure 2

Analysis of the Solution

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

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

Figure 3 shows the graph.

Figure 3

By the symmetry shown in the values of [latex]x[/latex] and [latex]y[/latex], we see that the parametric equations represent an ellipse. The ellipse is mapped in a counterclockwise direction as shown by the arrows indicating increasing [latex]t[/latex] values.

Construct a table of values like the table below.

Plot the [latex]\left(x,y\right)[/latex] values from the table. See Figure 4.

Figure 4

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

[latex]\begin{align}&x=5\cos t \\ &\frac{x}{5}=\cos t&& \text{Solve for }\cos t. \\ &y=2\sin t&& \text{Solve for }\sin t. \\ &\frac{y}{2}=\sin t \end{align}[/latex]

Then, use the Pythagorean Theorem.

[latex]\begin{gathered} {\cos }^{2}t+{\sin }^{2}t=1\\ {\left(\frac{x}{5}\right)}^{2}+{\left(\frac{y}{2}\right)}^{2}=1\\ \frac{{x}^{2}}{25}+\frac{{y}^{2}}{4}=1\end{gathered}[/latex]

Analysis of the Solution

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

Construct a table of values for the parametric equations, as we did in the previous example, and graph [latex]y=\sqrt{t},t\ge 0[/latex] on the same grid, as in Figure 6.

Figure 6

Analysis of the Solution

With the domain on [latex]t[/latex] restricted, we only plot positive values of [latex]t[/latex]. 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.

1. Use the formulas to set up the equations. The horizontal position is found using the parametric equation for [latex]x[/latex]. Thus,
   [latex]\begin{align}&x=\left({v}_{0}\cos \theta \right)t\\ &x=\left(140\cos \left(45^\circ \right)\right)t \end{align}[/latex]

   The vertical position is found using the parametric equation for [latex]y[/latex]. Thus,

   [latex]\begin{align}&y=-16{t}^{2}+\left({v}_{0}\sin \theta \right)t+h \\ &y=-16{t}^{2}+\left(140\sin \left(45^\circ \right)\right)t+3 \end{align}[/latex]
2. Substitute 2 into the equations to find the horizontal and vertical positions of the ball.
   [latex]\begin{align} &x=\left(140\cos \left(45^\circ \right)\right)\left(2\right) \\ &x=198\text{ feet} \\ \text{ } \\ &y=-16{\left(2\right)}^{2}+\left(140\sin \left(45^\circ \right)\right)\left(2\right)+3 \\ &y=137\text{ feet} \end{align}[/latex]

   After 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 [latex]y=0[/latex]. Thus,
   [latex]\begin{align}&y=-16{t}^{2}+\left(140\sin \left({45}^{\circ }\right)\right)t+3 \\ &y=0&& \text{Set }y\left(t\right)=0\text{ and solve the quadratic}.\\ &t=6.2173 \end{align}[/latex]

   When [latex]t=6.2173[/latex] 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 [latex]t[/latex], and input [latex]t[/latex] into [latex]y[/latex].
   [latex]\begin{align}&x=\left(140\cos \left(45^\circ \right)\right)t \\ &400=\left(140\cos \left(45^\circ \right)\right)t \\ &t=4.04 \\ \text{ } \\ &y=-16{\left(4.04\right)}^{2}+\left(140\sin \left(45^\circ \right)\right)\left(4.04\right)+3 \\ &y=141.8\end{align}[/latex]

   The 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