Chapter 8.4, Problem 18E

(a)

To determine

To sketch: the curve of the given parametric equation

Answer to Problem 18E

  

Explanation of Solution

Given: $x=\cos 2t,y=\sin 2t$

Calculation:

Sketch the curve represented by the parametric equations

  $x=\cos 2t,y=\sin 2t$.

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

If $t=0$ , then

  $\begin{array}{l}x=\cos 2\left(0\right)\\ \text{ =1,}\end{array}$

and

  $\begin{array}{l}\text{y}\text{=}\sin 2\left(\text{0}\right)\\ \text{ =0,}\end{array}$

So the point corresponding to $t=0$is $\left(1,0\right)$ .

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

$t$	$x$	$y$
$0$	$1$	$0$
$\frac{\pi }{4}$	$0$	$1$
$\frac{\pi }{2}$	$-1$	$0$
$\frac{3\pi }{4}$	$0$	$-1$
$\pi$	$1$	$0$

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 18E

Here, the given curve has the rectangular coordinate equation $x^2+y^2=1$ .

Explanation of Solution

Calculation:

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

  $x=\cos 2t,y=\sin 2t,$

by eliminating the parameter.

To eliminate the parameter, use the trigonometric identity ${\sin }^{2}2t+{\cos }^{2}2t=1.$

From the given equations,

  $\sin 2t=y,$

and

  $\cos 2t=x.$

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

  $y^2+x^2=1$

Thus,

  $x^2+y^2=1$

Therefore, the given curve has the rectangular coordinate equation,

  $\boxed{x^2+y^2=1.}$

Conclusion:

Hence, the given curve has the rectangular coordinate equation $x^2+y^2=1$ .