Chapter 1.7, Problem 19E

To determine

To find: The motion of the particle with the position $\left(x,y\right)$ as it varies in the in the given interval.

Answer to Problem 19E

The curve represents the parametric equations traced by the particle that moves three times clockwise around the eclipse and start at the point $\left(0,-2\right)$ .

Explanation of Solution

Given:

The value of $x$ and $y$ is given by,

  $\begin{array}{l}x=5\sin t\\ y=\cos t\end{array}$

The value of the range is,

  $-\pi \le t\le 5\pi$

Calculation:

Consider the given interval,

  $-\pi \le t\le 5\pi$

Consider the given expressions,

  $\begin{array}{l}x=5\sin t\\ y=\cos t\end{array}$

The variation in the variable $x$ and $y$ is shown in Table 1

Table 1

$t$	$x$	$y$
$-\pi$	$0$	$-2$
$\frac{-\pi }{2}$	$-5$	$0$
$0$	$0$	$2$
$\frac{\pi }{2}$	$5$	$0$
$\pi$	$0$	$2$
$\frac{3\pi }{2}$	$5$	$0$
$2\pi$	$0$	$2$
$\frac{5\pi }{2}$	$5$	$0$
$3\pi$	$0$	$-2$
$\frac{7\pi }{2}$	$-5$	$0$
$4\pi$	$0$	$2$
$\frac{9\pi }{2}$	$5$	$0$
$5\pi$	$0$	$-2$

The required graph is shown in Figure 1

  

Figure 1

From the given equation $x=5\sin t$ evaluate as,

  $\begin{array}{l}x=5\sin t\\ {\sin }^{2}t=\frac{x^2}{5^2}\end{array}$

From the given equation $y=2\cos t$ evaluate as,

  $\begin{array}{l}y=2\cos t\\ \frac{y}{2}={\cos }^{2}t\\ \frac{y^2}{2^2}={\cos }^{2}t\end{array}$

Then,

  $\begin{array}{l}{\cos }^{2}t+{\sin }^{2}t=\frac{x^2}{5^2}+\frac{y^2}{2^2}\\ \frac{x^2}{5^2}+\frac{y^2}{2^2}=1\end{array}$

The above equation is for the eclipse that is centredat $\left(0,0\right)$ .

The above equation ranges from $\left(3,3\right)$ to $\left(3,-1\right)$ .

Thus, the curve represents the parametric equations traced by the particle that moves three times clockwise around the eclipse and start at the point $\left(0,-2\right)$ .