Chapter 11, Problem 12RE

(a)

To determine

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

Answer to Problem 12RE

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

Explanation of Solution

Given information:

The equations: $x=-2\cos t$

  $y=2\sin t$

Formula used:

Formulas of derivatives:

  $\frac{d}{dx}\sin x=\cos x$

  $\frac{d}{dx}\cos x=-\sin x$

Calculation:

The given equations are $x=-2\cos t$  ..... (1)

And, $y=2\sin t$  ..... (2)

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

  $\left(x,y\right)=\left(-2\cos t,2\sin t\right)$

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

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left(-2\cos t\right)\\ =-2\left(-\sin t\right)\\ =2\sin t\end{array}$

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

Now,

  $\begin{array}{c}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ \frac{dy}{dx}=\frac{2\cos t}{2\sin t}\\ \frac{dy}{dx}=\cot t\end{array}$

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 $\frac{dy}{dt}=2\cos t$ .

  $\begin{array}{c}2\cos t=0\\ \cos t=0\end{array}$

Simplify the above expression for t .

  $\begin{array}{c}\cos t=0\\ t=\frac{\pi }{2},\frac{3\pi }{2}\end{array}$

Substitute $\frac{\pi }{2}$ for t in the point $\left(x,y\right)=\left(-2\cos t,2\sin t\right)$ .

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

Also, substitute $\frac{3\pi }{2}$ for t in the point $\left(x,y\right)=\left(-2\cos t,2\sin t\right)$ .

  $\begin{array}{c}\left(x,y\right)=\left(-2\cos t,2\sin t\right)\\ =\left(-2\cos \frac{3\pi }{2},2\sin \frac{3\pi }{2}\right)\\ =\left(-2\left(0\right),2\left(-1\right)\right)\\ =\left(0,-2\right)\end{array}$

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

(b)

To determine

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

Answer to Problem 12RE

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

Explanation of Solution

Given information:

The equations: $x=-2\cos t$

  $y=2\sin t$

Formula used:

Formulas of derivatives:

  $\frac{d}{dx}\sin x=\cos x$

  $\frac{d}{dx}\cos x=-\sin x$

Calculation:

The given equations are $x=-2\cos t$  ..... (1)

And, $y=2\sin t$  ..... (2)

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

  $\left(x,y\right)=\left(-2\cos t,2\sin t\right)$

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

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left(-2\cos t\right)\\ =-2\left(-\sin t\right)\\ =2\sin t\end{array}$

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

Now,

  $\begin{array}{c}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ \frac{dy}{dx}=\frac{2\cos t}{2\sin t}\\ \frac{dy}{dx}=\cot t\end{array}$

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 $\frac{dx}{dt}=2\sin t$ .

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

Substitute 0 for t in the point $\left(x,y\right)=\left(-2\cos t,2\sin t\right)$ .

  $\begin{array}{c}\left(x,y\right)=\left(-2\cos \left(0\right),2\sin \left(0\right)\right)\\ =\left(-2\left(1\right),2\left(0\right)\right)\\ =\left(-2,0\right)\end{array}$

Also, substitute $\pi$ for t in the point $\left(x,y\right)=\left(-2\cos t,2\sin t\right)$ .

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

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