Chapter 10.1, Problem 20E

$\text{(a)}$

To determine

To sketch: The graph of the given $t$ interval.

Answer to Problem 20E

The answer:

  

Explanation of Solution

Given information:

The equations are:

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

Calculation:

The follow graph of the curve based on the information provided in this exercise.

  

$\text{(b)}$

To determine

To find: The requested points.

Answer to Problem 20E

The answer: $(0,2)$

Explanation of Solution

Given information:

The equations are:

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

Calculation:

From part $(a)$ conclude that the lowest point of the curve is equal to $(0,2)$ .

Therefore, the required requested point $(0,2)$ .

$(c)$

To determine

To prove: The requested points by analyze a appropriate derivative.

Explanation of Solution

Given information:

The equations are:

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

Calculation:

To find the lowest point of the curve, to find the minimum of the function $y=2\sec t$ where,

  $-1\le t\le 1$

Know that, $y$ has a relative minimum at the point $t=t_0$ if

  $\frac{dy}{dt}\left(t_0\right)=0$

To use the previous result,

  $\begin{array}{c}\frac{dy}{dt}=\frac{0\cdot \cos t-1\cdot (-\sin t)}{{\cos }^{2}t}\\ =\frac{1}{\cos t}\cdot \frac{\sin t}{\cos t}\\ =\sec t\tan t\\ \Rightarrow \frac{dy}{dt}(0)=0\end{array}$

To use the previous result, then

  $\begin{array}{c}x(0)=\tan (0)\\ =0\\ y(0)=2\sec (0)\\ =2\cdot 1\\ =2\end{array}$

which means that the lowest point of the curve is equal to $(0,2)$ .