Chapter 11.3, Problem 40E

To determine

To calculate: The slope of the curve at each indicated point.

Answer to Problem 40E

The required slope is 0 at $\theta =\pm \frac{\pi }{2}$ and undefined at $\theta =0,\pi$ .

Explanation of Solution

Given information:

The curve: $r=\cos 2\theta$

The points: $\theta =0,\pm \frac{\pi }{2},\pi$

Calculation:

The given curve is $r=\cos 2\theta$ .

It is known that in a polar coordinate system, if $r=f\left(\theta \right)$ is a curve, then in rectangular coordinate system, $x=r\cos \theta =f\left(\theta \right)\cos \theta$ and $y=r\sin \theta =f\left(\theta \right)\sin \theta$ .

Rewrite x and y as:

  $\begin{array}{c}x=r\cos \theta \\ =\cos 2\theta \cos \theta \end{array}$

And, $\begin{array}{c}y=r\sin \theta \\ =\cos 2\theta \sin \theta \end{array}$

Since, polar curves lie in xy -plane; so, the slope of the polar curve and the slope of the tangent are same that is, $\frac{dy}{dx}$ .

The slope can be written as:

  $\begin{array}{c}\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}\\ \frac{dy}{dx}=\frac{\frac{d}{d\theta }\left[\cos 2\theta \sin \theta \right]}{\frac{d}{d\theta }\left[\cos 2\theta \cos \theta \right]}\\ \frac{dy}{dx}=\frac{\cos 2\theta \cos \theta -2\sin \theta \sin 2\theta }{-\sin \theta \cos 2\theta -2\cos \theta \sin 2\theta }\end{array}$

Now, substitute 0 for $\theta$ in the above derivative.

  $\begin{array}{c}\frac{dy}{dx}=\frac{\cos 2\left(0\right)\cos \left(0\right)-2\sin \left(0\right)\sin 2\left(0\right)}{-\sin \left(0\right)\cos 2\left(0\right)-2\cos \left(0\right)\sin 2\left(0\right)}\\ =\frac{1-0}{0-0}\\ =\text{Undefined}\end{array}$

Substitute $-\frac{\pi }{2}$ for $\theta$ in the above derivative.

  $\begin{array}{c}\frac{dy}{dx}=\frac{\cos 2\left(-\frac{\pi }{2}\right)\cos \left(-\frac{\pi }{2}\right)-2\sin \left(-\frac{\pi }{2}\right)\sin 2\left(-\frac{\pi }{2}\right)}{-\sin \left(-\frac{\pi }{2}\right)\cos 2\left(-\frac{\pi }{2}\right)-2\cos \left(-\frac{\pi }{2}\right)\sin 2\left(-\frac{\pi }{2}\right)}\\ =\frac{0-0}{-1-0}\\ =0\end{array}$

Substitute $\frac{\pi }{2}$ for $\theta$ in the above derivative.

  $\begin{array}{c}\frac{dy}{dx}=\frac{\cos 2\left(\frac{\pi }{2}\right)\cos \left(\frac{\pi }{2}\right)-2\sin \left(\frac{\pi }{2}\right)\sin 2\left(\frac{\pi }{2}\right)}{-\sin \left(\frac{\pi }{2}\right)\cos 2\left(\frac{\pi }{2}\right)-2\cos \left(\frac{\pi }{2}\right)\sin 2\left(\frac{\pi }{2}\right)}\\ =\frac{0-0}{1-0}\\ =0\end{array}$

Substitute $\pi$ for $\theta$ in the above derivative.

  $\begin{array}{c}\frac{dy}{dx}=\frac{\cos 2\left(\pi \right)\cos \left(\pi \right)-2\sin \left(\pi \right)\sin 2\left(\pi \right)}{-\sin \left(\pi \right)\cos 2\left(\pi \right)-2\cos \left(\pi \right)\sin 2\left(\pi \right)}\\ =\frac{-1-0}{0-0}\\ =\text{Undefined}\end{array}$

Hence, the required slope is 0 at $\theta =\pm \frac{\pi }{2}$ and undefined at $\theta =0,\pi$ .