Chapter 11.3, Problem 59E

To determine

To graph: The given polar curve. Also, find the area enclosed the curve and find the slope of the curve at the point where $\theta =\frac{\pi }{4}$ .

Answer to Problem 59E

The required the area enclosed the curve and the slope of the curve at the point where $\theta =\frac{\pi }{4}$ are $\pi$ and $\frac{1}{2}$ respectively.

Explanation of Solution

Given information:

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

Calculation:

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

It is known that in a polar coordinate system, if $r=f\left(\theta \right)$ is a curve; then the area of the region within the interval $\alpha \le \theta \le \beta$ is $A={\displaystyle {\int }_{\alpha }^{\beta }\frac{1}{2}{r}^{2}d\theta }={\displaystyle {\int }_{\alpha }^{\beta }\frac{1}{2}{\left(f\left(\theta \right)\right)}^{2}d\theta }$ .

The graph of the given curve is shown below:

  

It is seen that the shaded region between the curves lie within the interval $0\le \theta \le \pi$ .

Write the area of the given region as:

  $\begin{array}{c}A={\displaystyle {\int }_{\alpha }^{\beta }\frac{1}{2}{\left(f\left(\theta \right)\right)}^{2}d\theta }\\ ={\displaystyle {\int }_0^{\pi }{\left(2\sin 3\theta \right)}^{2}d\theta }\\ ={\displaystyle {\int }_0^{\pi }2\left(\frac{1-\cos 6\theta }{2}\right)d\theta }\\ ={\displaystyle {\int }_0^{\pi }\left(1-\cos 6\theta \right)d\theta }\end{array}$

Simplify the above expression further.

  $\begin{array}{c}{\displaystyle {\int }_0^{\pi }\left(1-\cos 6\theta \right)d\theta }={\left[\theta -\frac{1}{6}\sin 6\theta \right]}_0^{\pi }\\ =\left[\left(\pi -\frac{1}{6}\sin 6\pi \right)-\left(0-\frac{1}{6}\sin \left(0\right)\right)\right]\\ =\left[\left(\pi -0\right)-0\right]\\ =\pi \end{array}$

Thus, the area of the region is $\pi$ .

Now, 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:

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

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

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:

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

Now, write the slope at $\theta =\frac{\pi }{4}$ as:

  $\begin{array}{c}\frac{dy}{dx}=\frac{6\sin \left(\frac{\pi }{4}\right)\cos 3\left(\frac{\pi }{4}\right)+2\cos \left(\frac{\pi }{4}\right)\sin 3\left(\frac{\pi }{4}\right)}{6\cos \left(\frac{\pi }{4}\right)\cos 3\left(\frac{\pi }{4}\right)-2\sin \left(\frac{\pi }{4}\right)\sin 3\left(\frac{\pi }{4}\right)}\\ =\left(\frac{6\left(\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right)+2\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)}{6\left(\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right)-2\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)}\right)\\ =\left(\frac{-3+1}{-3-1}\right)\\ =\frac{1}{2}\end{array}$

Hence, the required the area enclosed the curve and the slope of the curve at the point where $\theta =\frac{\pi }{4}$ are $\pi$ and $\frac{1}{2}$ respectively.