Chapter 11.3, Problem 42E

To determine

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

Answer to Problem 42E

The required slope of the curve at $\left(1.5,\frac{\pi }{3}\right)$ , $\left(4.5,\frac{2\pi }{3}\right)$ , $\left(6,\pi \right)$ and $\left(3,\frac{3\pi }{2}\right)$ are undefined, 0, 1 and undefined respectively.

Explanation of Solution

Given information:

The curve: $r=3\left(1-\cos \theta \right)$

The graph of the curve:

  

Calculation:

The given curve is $r=3\left(1-\cos \theta \right)$ .

The points are $\left(1.5,\frac{\pi }{3}\right)$ , $\left(4.5,\frac{2\pi }{3}\right)$ , $\left(6,\pi \right)$ , $\left(3,\frac{3\pi }{2}\right)$ .

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:

  $\begin{array}{c}x=r\cos \theta \\ =3\left(1-\cos \theta \right)\cos \theta \end{array}$

And, $\begin{array}{c}y=r\sin \theta \\ =3\left(1-\cos \theta \right)\sin \theta \end{array}$

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:

  $\begin{array}{c}\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}\\ \frac{dy}{dx}=\frac{\frac{d}{d\theta }\left[3\left(1-\cos \theta \right)\sin \theta \right]}{\frac{d}{d\theta }\left[3\left(1-\cos \theta \right)\cos \theta \right]}\\ \frac{dy}{dx}=\frac{3\left[\left(1-\cos \theta \right)\cos \theta +\left(\sin \theta \right)\sin \theta \right]}{3\left[\left(1-\cos \theta \right)\left(-\sin \theta \right)+\left(\sin \theta \right)\cos \theta \right]}\\ \frac{dy}{dx}=\frac{\cos \theta -\cos 2\theta }{-\sin \theta +\sin 2\theta }\end{array}$

Now, write the slope at $\left(1.5,\frac{\pi }{3}\right)$ as:

  $\begin{array}{c}\frac{dy}{dx}=\frac{\cos \left(\frac{\pi }{3}\right)-\cos 2\left(\frac{\pi }{3}\right)}{-\sin \left(\frac{\pi }{3}\right)+\sin 2\left(\frac{\pi }{3}\right)}\\ =\text{Undefined}\end{array}$

Write the slope at $\left(4.5,\frac{2\pi }{3}\right)$ as:

  $\begin{array}{c}\frac{dy}{dx}=\frac{\cos \left(\frac{2\pi }{3}\right)-\cos 2\left(\frac{2\pi }{3}\right)}{-\sin \left(\frac{2\pi }{3}\right)+\sin 2\left(\frac{2\pi }{3}\right)}\\ =0\end{array}$

Write the slope at $\left(6,\pi \right)$ as:

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

Write the slope at $\left(3,\frac{3\pi }{2}\right)$ as:

  $\begin{array}{c}\frac{dy}{dx}=\frac{\cos \left(\frac{3\pi }{2}\right)-\cos 2\left(\frac{3\pi }{2}\right)}{-\sin \left(\frac{3\pi }{2}\right)+\sin 2\left(\frac{3\pi }{2}\right)}\\ =\frac{\cos \left(\frac{3\pi }{2}\right)-\cos \left(3\pi \right)}{-\sin \left(\frac{3\pi }{2}\right)+\sin \left(3\pi \right)}\\ =1\end{array}$

Hence, the required slope of the curve at $\left(1.5,\frac{\pi }{3}\right)$ , $\left(4.5,\frac{2\pi }{3}\right)$ , $\left(6,\pi \right)$ and $\left(3,\frac{3\pi }{2}\right)$ are undefined, 0, 1 and undefined respectively.