Chapter 8.4, Problem 19E

(a)

To determine

To sketch: the curve of the given parametric equation

Answer to Problem 19E

  

Explanation of Solution

Given: $x=\sec t,y=\tan t,0\le t<\frac{\pi }{2}$

Calculation:

Sketch the curve represented by the parametric equations

  $x=\sec t,y=\tan t,0\le t<\frac{\pi }{2}$.

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

If $t=0$ , then

  $\begin{array}{l}x=\sec 0\\ =\frac{1}{\cos 0}\\ \text{ =1,}\end{array}$

and

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

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.

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

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 19E

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

Explanation of Solution

Calculation:

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

  $x=\sec t,y=\tan t,0\le t<\frac{\pi }{2}$

by eliminating the parameter.

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

From the given equations,

  $\sec t=x,$

and

  $\tan t=y.$

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

  $x^2=1+y^2$

Therefore, the given curve has the rectangular coordinate equation,

  $\boxed{x^2=1+y^2.}$

Conclusion:

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