Chapter 11, Problem 11RE

(a)

To determine

To determine: The points at which the tangent to the curve is horizontal.

Answer to Problem 11RE

The required points are $\left(0,\frac{1}{2}\right)$ and $\left(0,-\frac{1}{2}\right)$ .

Explanation of Solution

Given information:

The equations: $x=\left(\frac{1}{2}\right)\tan t$

  $y=\left(\frac{1}{2}\right)\sec t$

Formula used:

Formulas of derivatives:

  $\frac{d}{dx}\tan x={\sec }^{2}x$

  $\frac{d}{dx}\sec x=\sec x\tan x$

Calculation:

The given equations are $x=\left(\frac{1}{2}\right)\tan t$  ..... (1)

And, $y=\left(\frac{1}{2}\right)\sec t$  ..... (2)

Write the point at which the tangent to the curve is horizontal as:

  $\left(x,y\right)=\left(\frac{1}{2}\tan t,\frac{1}{2}\sec t\right)$

Find derivative of equation (1) and equation (2) with respect to t .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left(\frac{1}{2}\right)\tan t\\ =\frac{1}{2}{\sec }^{2}t\end{array}$

And, $\begin{array}{c}\frac{dy}{dt}=\frac{d}{dt}\left(\frac{1}{2}\right)\sec t\\ \frac{dy}{dt}=\frac{1}{2}\sec t\tan t\end{array}$

Now,

  $\begin{array}{c}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ \frac{dy}{dx}=\frac{\frac{1}{2}\sec t\tan t}{\frac{1}{2}{\sec }^{2}t}\\ \frac{dy}{dx}=\frac{\tan t}{\sec t}\\ \frac{dy}{dx}=\frac{\frac{\sin t}{\cos t}}{\frac{1}{\cos t}}\\ \frac{dy}{dx}=\sin t\end{array}$

It is known that a tangent to the curve is horizontal if $\frac{dy}{dt}=0$ .

Substitute 0 for $\frac{dy}{dt}$ in the above expression $\frac{dy}{dt}=\frac{1}{2}\sec t\tan t$ .

  $\begin{array}{c}\frac{1}{2}\sec t\tan t=0\\ \sec t\tan t=0\\ \frac{1}{\cos t}\frac{\sin t}{\cos t}=0\\ \frac{\sin t}{{\cos }^{2}t}=0\end{array}$

Simplify the above expression for t .

  $\begin{array}{c}\frac{\sin t}{{\cos }^{2}t}=0\\ \sin t=0\\ t=0,\pi \end{array}$

Substitute 0 for t in the point $\left(x,y\right)=\left(\frac{1}{2}\tan t,\frac{1}{2}\sec t\right)$ .

  $\begin{array}{c}\left(x,y\right)=\left(\frac{1}{2}\tan \left(0\right),\frac{1}{2}\sec \left(0\right)\right)\\ =\left(\frac{1}{2}\left(0\right),\frac{1}{2}\left(1\right)\right)\\ =\left(0,\frac{1}{2}\right)\end{array}$

Also, substitute $\pi$ for t in the point $\left(x,y\right)=\left(\frac{1}{2}\tan t,\frac{1}{2}\sec t\right)$ .

  $\begin{array}{c}\left(x,y\right)=\left(\frac{1}{2}\tan \left(\pi \right),\frac{1}{2}\sec \left(\pi \right)\right)\\ =\left(\frac{1}{2}\left(0\right),\frac{1}{2}\left(-1\right)\right)\\ =\left(0,-\frac{1}{2}\right)\end{array}$

Hence, the required points are $\left(0,\frac{1}{2}\right)$ and $\left(0,-\frac{1}{2}\right)$ .

(b)

To determine

To determine: The points at which the tangent to the curve is vertical.

Answer to Problem 11RE

The tangent to the curve is not vertical at any point.

Explanation of Solution

Given information:

The equations: $x=\left(\frac{1}{2}\right)\tan t$

  $y=\left(\frac{1}{2}\right)\sec t$

Formula used:

Formulas of derivatives:

  $\frac{d}{dx}\tan x={\sec }^{2}x$

  $\frac{d}{dx}\sec x=\sec x\tan x$

Calculation:

The given equations are $x=\left(\frac{1}{2}\right)\tan t$  ..... (1)

And, $y=\left(\frac{1}{2}\right)\sec t$  ..... (2)

Write the point at which the tangent to the curve is horizontal as:

  $\left(x,y\right)=\left(\frac{1}{2}\tan t,\frac{1}{2}\sec t\right)$

Find derivative of equation (1) and equation (2) with respect to t .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left(\frac{1}{2}\right)\tan t\\ \frac{dx}{dt}=\frac{1}{2}{\sec }^{2}t\end{array}$

And, $\begin{array}{c}\frac{dy}{dt}=\frac{d}{dt}\left(\frac{1}{2}\right)\sec t\\ \frac{dy}{dt}=\frac{1}{2}\sec t\tan t\end{array}$

Now,

  $\begin{array}{c}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ \frac{dy}{dx}=\frac{\frac{1}{2}\sec t\tan t}{\frac{1}{2}{\sec }^{2}t}\\ \frac{dy}{dx}=\frac{\tan t}{\sec t}\\ \frac{dy}{dx}=\frac{\frac{\sin t}{\cos t}}{\frac{1}{\cos t}}\\ \frac{dy}{dx}=\sin t\end{array}$

It is known that a tangent to the curve is vertical if $\frac{dx}{dt}=0$ .

Substitute 0 for $\frac{dx}{dt}$ in the above expression $\frac{dx}{dt}=\frac{1}{2}{\sec }^{2}t$ .

  $\begin{array}{c}\frac{1}{2}{\sec }^{2}t=0\\ \frac{1}{2{\cos }^{2}t}=0\\ t=\varphi \end{array}$

Since, t has no value for which $\frac{dx}{dt}=0$ ; so the tangent is never vertical.

Hence, the tangent to the curve is not vertical at any point.