Chapter 8.4, Problem 14E

(a)

To determine

To sketch: the curve of the given parametric equation

Answer to Problem 14E

  

Explanation of Solution

Given: $x=2\cos t,y=3\sin t,0\le t\le 2\pi$

Calculation:

Sketch the curve represented by the parametric equations

  $x=2\cos t,y=3\sin t,0\le t\le 2\pi$.

For every value of $t$ , get a point on the curve.

If $t=0$ , then

  $\begin{array}{l}x=2\cos 0\\ \text{ =2,}\end{array}$

and

  $\begin{array}{l}\text{y}\text{=}3\sin \text{0}\\ \text{ =3,}\end{array}$

So the point corresponding to $t=0$is $\left(2,0\right)$ .

The points $\left(x,y\right)$determined for different values of $t$ are shown in the following table.

$t$	$x$	$y$
$0$	$2$	$0$
$\frac{\pi }{2}$	$0$	$3$
$\pi$	$-2$	$0$
$\frac{3\pi }{2}$	$0$	$-3$
$2\pi$	$2$	$0$

Plotting these points get the following graph of the curve.

  

Conclusion:

Hence, the curve represented by the given parametric equation is sketched.

(b)

To determine

To find: the rectangular coordinate equation for the curve.

Answer to Problem 14E

Here, the given curve has the rectangular coordinate equation $\frac{y^2}{9}+\frac{x^2}{4}=1$ .

Explanation of Solution

Calculation:

Find a rectangular-coordinate equation for the curve, represented by the parametric equations,

  $x=2\cos t,y=3\sin t,0\le t\le 2\pi$

by eliminating the parameter.

To eliminating the parameter, use the trigonometric identify ${\sin }^{2}t+{\cos }^{2}t=1$

From the given equations,

  $\sin t=\frac{y}{3},$

and

  $\cos t=\frac{x}{2}.$

Substituting the values of $\sin t$ and $\cos t$ , in ${\sin }^{2}t+{\cos }^{2}t=1$ ,

  ${\left(\frac{y}{3}\right)}^2+{\left(\frac{x}{2}\right)}^2=1$

Thus,

  $\frac{y^2}{9}+\frac{x^2}{4}=1$ .

Therefore, the given curve has the rectangular coordinate equation,

  $\boxed{\frac{y^2}{9}+\frac{x^2}{4}=1.}$

Conclusion:

Hence, the given curve has the rectangular coordinate equation $\frac{y^2}{9}+\frac{x^2}{4}=1$ .