Chapter 10.1, Problem 24E

$\text{(a)}$

To determine

To find: The points of horizontal.

Answer to Problem 24E

The answer: does not exist.

Explanation of Solution

Given information:

The equations are:

  $\begin{array}{c}x=\sec t,\\ y=\tan t\end{array}$

Calculation:

Since need to find the points at which the tangent line to the curve is horizontal, need to find the relative extrema of the function $y=-1+\sin t$ .

Know that function $y$ has a relative extremum at point $t=t_0$ .

So,

  $\frac{dy}{dt}\left(t_0\right)=0$

Know that,

  $\begin{array}{c}t\tan t=\sin t/\cos t\\ \sec t=1/\cos t\end{array}$

Then,

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

Because,

  ${\sin }^{2}t+{\cos }^{2}t=1$

To use the previous result,

  $dy/dt\ne 0$

According to the previous result,

  $y=\tan t$ doesn't have the relative extremum, which means that the point at which the tangent line to the curve is horizontal doesn't exist.

Therefore, the required horizontal points does not exists.

$\text{(b)}$

To determine

To find: The points of vertical.

Answer to Problem 24E

The answer: $(1,0)$ and $(-1,0)$

Explanation of Solution

Given information:

The equations are:

  $\begin{array}{c}x=\sec t,\\ y=\tan t\end{array}$

Calculation:

Since need to find the points at which the tangent line to the curve is vertical, need to find the relative extrema of the function $x=\sec t$

Know that function $y$ has a relative extremum at point $t=t_0$ .

So,

  $\frac{dy}{dt}\left(t_0\right)=0$

Hence, know that,

  $\sec t=1/\cos t$

Then,

  $\begin{array}{c}\frac{dx}{dt}=-\frac{1}{{\cos }^{2}t}\cdot (-\sin t)\\ =\frac{1}{\cos t}\cdot \frac{\sin t}{\cos t}\\ =\frac{\sin t}{{\cos }^{2}t}\end{array}$

To use the previous result,

  $\begin{array}{c}\frac{dx}{dt}=\frac{\sin t}{{\cos }^{2}t}\\ \Rightarrow \frac{dx}{dt}(k\pi )=0\end{array}$

For

   , know that $\sin (k\pi )=0$ for then the function

  $x=\sec t$

has the relative extrema at points

  $\begin{array}{c}t=0\\ t=\pi \end{array}$

In this exercise,

  $y=\tan t$

Then,

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

which means that the points at which the tangent line to the curve is vertical are equal to $(1,0)$ and $(-1,0)$ .

Therefore, the required vertical points are $(1,0)$ and $(-1,0)$ .