Chapter 0.4, Problem 45E

a.

To determine

To graph: the curve and find the initial and terminal points. Also, indicate the direction in which the curve is traced.

Answer to Problem 45E

  

No initial and terminal point.

Explanation of Solution

Given information:

The parametric equations:

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

Formula Used:

  $\begin{array}{l}\sec x=\frac{1}{\cos x}\\ \tan x=\frac{\sin x}{\cos x}\end{array}$

Use the graphing calculator to graph the given curve.

First step is to set the mode to parametric.

For that press the MODE key.

Then Scroll down to FUNC.

Then move it to right till PAR

Press ENTER key.

  

Now, go back to main window by quitting (to do that press and then ).

Now, press the key

Now enter the given parametric equations as shown below:

  

Press the WINDOW key.

Enter values (Tmin = $-\frac{\pi }{2}=-\mathrm{1.570796...}$ , Tmax = $\frac{\pi }{2}=\mathrm{1.570796...}$ )

  

Press GRAPH (here observe the direction in which the graph is being traced.)

  

This is the required graph and the red arrow gives the direction in which it is traced.

From the graph, it can be clearly observed that it has no initial and terminal point.

b.

To determine

To find: Cartesian equation for the curve that contains the parameterized curve. Also explain what portion of the Cartesian equation is traced by the parameterized curve.

Answer to Problem 45E

  $x^2-y^2=1$

Only left half of the graph of the Cartesian equation is traced by the parameterized curve.

Explanation of Solution

Given information:

The parametric equations:

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

Formula Used:

  ${\sec }^{2}x=1+{\tan }^{2}x$

Squaring both $x=-\sec t$ and $y=\tan t$ ,

  $x^2={\sec }^{2}t$

  $y^2={\tan }^{2}t$

Substitute these values in the identity ${\sec }^{2}x=1+{\tan }^{2}x$,

  $\begin{array}{c}{\sec }^{2}t=1+{\tan }^{2}t\\ x^2=1+y^2\\ x^2-y^2=1\end{array}$

Thus, the required Cartesian equation is $x^2-y^2=1$ .

The Cartesian equation $x^2-y^2=1$ represents a horizontal hyperbola and from the graph in part (a), it is clear that the only left half of the graph of the Cartesian equation is traced by the parameterized curve.