Chapter 11.3, Problem 41E

To determine

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

Answer to Problem 41E

The required slope of the curve at $\left(2,0\right)$ , $\left(-1,\frac{\pi }{2}\right)$ , $\left(2,\pi \right)$ and $\left(5,\frac{3\pi }{2}\right)$ are $-\frac{2}{3}$ , 0, $\frac{2}{3}$ and 0 respectively.

Explanation of Solution

Given information:

The curve: $r=2-3\sin \theta$

The graph of the curve:

  

Calculation:

The given curve is $r=2-3\sin \theta$ .

The points are $\left(2,0\right)$ , $\left(-1,\frac{\pi }{2}\right)$ , $\left(2,\pi \right)$ , $\left(5,\frac{3\pi }{2}\right)$ .

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:

  $\begin{array}{c}x=r\cos \theta \\ =\left(2-3\sin \theta \right)\cos \theta \end{array}$

And, $\begin{array}{c}y=r\sin \theta \\ =\left(2-3\sin \theta \right)\sin \theta \end{array}$

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:

  $\begin{array}{c}\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}\\ \frac{dy}{dx}=\frac{\frac{d}{d\theta }\left[\left(2-3\sin \theta \right)\cos \theta \right]}{\frac{d}{d\theta }\left[\left(2-3\sin \theta \right)\sin \theta \right]}\\ \frac{dy}{dx}=\frac{\left(2-6\sin \theta \right)\cos \theta }{\sin \theta \left(3\sin \theta -2\right)-3{\cos }^2\theta }\end{array}$

Now, write the slope at $\left(2,0\right)$ as:

  $\begin{array}{c}\frac{dy}{dx}=\frac{\left(2-6\sin \left(0\right)\right)\cos \left(0\right)}{\sin \left(0\right)\left(3\sin \left(0\right)-2\right)-3{\cos }^2\left(0\right)}\\ =\frac{\left(2-0\right)\left(1\right)}{0-3}\\ =-\frac{2}{3}\end{array}$

Write the slope at $\left(-1,\frac{\pi }{2}\right)$ as:

  $\begin{array}{c}\frac{dy}{dx}=\frac{\left(2-6\sin \left(\frac{\pi }{2}\right)\right)\cos \left(\frac{\pi }{2}\right)}{\sin \left(\frac{\pi }{2}\right)\left(3\sin \left(\frac{\pi }{2}\right)-2\right)-3{\cos }^2\left(\frac{\pi }{2}\right)}\\ =\frac{\left(2-6\right)\left(0\right)}{1\left(3-2\right)-0}\\ =\frac{0}{1-0}\\ =0\end{array}$

Write the slope at $\left(2,\pi \right)$ as:

  $\begin{array}{c}\frac{dy}{dx}=\frac{\left(2-6\sin \left(\pi \right)\right)\cos \left(\pi \right)}{\sin \left(\pi \right)\left(3\sin \left(\pi \right)-2\right)-3{\cos }^2\left(\pi \right)}\\ =\frac{\left(2-0\right)\left(-1\right)}{0-3}\\ =\frac{-2}{-3}\\ =\frac{2}{3}\end{array}$

Write the slope at $\left(5,\frac{3\pi }{2}\right)$ as:

  $\begin{array}{c}\frac{dy}{dx}=\frac{\left(2-6\sin \left(\frac{3\pi }{2}\right)\right)\cos \left(\frac{3\pi }{2}\right)}{\sin \left(\frac{3\pi }{2}\right)\left(3\sin \left(\frac{3\pi }{2}\right)-2\right)-3{\cos }^2\left(\frac{3\pi }{2}\right)}\\ =\frac{\left(2+6\right)\left(0\right)}{\left(-1\right)\left(-3-2\right)-0}\\ =\frac{0}{5-0}\\ =0\end{array}$

Hence, the required slope of the curve at $\left(2,0\right)$ , $\left(-1,\frac{\pi }{2}\right)$ , $\left(2,\pi \right)$ and $\left(5,\frac{3\pi }{2}\right)$ are $-\frac{2}{3}$ , 0, $\frac{2}{3}$ and 0 respectively.