Chapter 11, Problem 13RE

(a)

To determine

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

Answer to Problem 13RE

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

Explanation of Solution

Given information:

The equations: $x=-\cos t$

  $y={\cos }^{2}t$

Formula used:

Formulas of derivatives:

  $\frac{d}{dx}\sin x=\cos x$

  $\frac{d}{dx}\cos x=-\sin x$

Calculation:

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

And, $y={\cos }^{2}t$  ..... (2)

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

  $\left(x,y\right)=\left(-\cos t,{\cos }^{2}t\right)$

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

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

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

Now,

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

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

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

Here, at the points $t=0,\pi$ the curve has no derivatives that is, the curve has no cusps.

So, the value of t are $\frac{\pi }{2}$ and $\frac{3\pi }{2}$ .

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

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

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

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

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

(b)

To determine

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

Answer to Problem 13RE

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

Explanation of Solution

Given information:

The equations: $x=-\cos t$

  $y={\cos }^{2}t$

Formula used:

Formulas of derivatives:

  $\frac{d}{dx}\sin x=\cos x$

  $\frac{d}{dx}\cos x=-\sin x$

Calculation:

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

And, $y={\cos }^{2}t$  ..... (2)

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

  $\left(x,y\right)=\left(-\cos t,{\cos }^{2}t\right)$

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

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

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

Now,

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

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

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

But, at the points $t=0,\pi$ the curve has no derivative that is, the curve has no cusps. So, the tangent is never vertical

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