Chapter 11.1, Problem 20E

(a)

To determine

To find: The sketch for the curve over the given $t$ interval indicating the direction in which it is traced.

Answer to Problem 20E

The required sketch is shown in Figure 1

Explanation of Solution

Given Data:

The given equations are,

  $\begin{array}{l}x=\tan t\\ y=2\sec t\\ -1\le t\le 1\end{array}$

Calculation:

Consider the trigonometric equation as,

  ${\sec }^{2}t-{\tan }^{2}t=1$

Substitute the values in the given equation then,

  $\begin{array}{l}{\left(\frac{y}{2}\right)}^2-x^2=1\\ {\frac{y}{4}}^2-x^2=1\end{array}$

The sketch for the above function is shown in Figure 1

  

Figure 1

(b)

To determine

To find: The identification for the requested points.

Answer to Problem 20E

The lowest point on the graph is $\left(0,2\right)$ .

Explanation of Solution

Consider the graph is shown in Figure 1 the graph shows that the lowest point is $\left(0,2\right)$ .

(c)

To determine

To find: The justification that you have found the requested point by analysing an appropriate derivative.

Answer to Problem 20E

The lowest point on the graph is $\left(0,2\right)$ .

Explanation of Solution

To obtain the lowest point on the graph differentiate the given equation as,

  $\begin{array}{l}\frac{dy}{dt}=0\\ 2\sec t\tan t=0\\ \tan t=0\\ t=0\end{array}$

Then,

  $\begin{array}{c}x=\tan 0\\ =0\end{array}$

Then, the value of $y$ is,

  $\begin{array}{c}y=2\sec 0\\ =2\end{array}$

Thus, lowest point on the graph is $\left(0,2\right)$ .