Chapter H.2, Problem 40E

To determine

To calculate:

The length of the following curve:

  $r=4\sin 3\theta$

Answer to Problem 40E

The length of the curve is $26.73$

Explanation of Solution

Given information:

  $r=4\sin 3\theta$

Formula Used:

The length of the curve is given by the following expression:

  $L={\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}}d\theta$

Calculation:

The equation of the curve is:

  $r=4\sin 3\theta$

So, the curve is plotted as below:

  

The full length of the $r=4\sin 3\theta$ is in the interval of $0\le \theta \le 2\pi$

Know that:

  $\begin{array}{l}r=4\sin 3\theta \\ \frac{dr}{d\theta }=\frac{d}{d\theta }\left(4\sin 3\theta \right)\\ \frac{dr}{d\theta }=12\cos 3\theta \end{array}$

Now, the length of the given curve calculated as:

  $\begin{array}{l}L={\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}}d\theta \\ L={\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{{\left(4\sin 3\theta \right)}^2+{\left(12\cos 3\theta \right)}^2}}d\theta \\ L={\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{16{\sin }^{2}3\theta +144{\cos }^{2}3\theta }}d\theta \\ L={\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{16\left(\frac{1-\cos 6\theta }{2}\right)+144\left(\frac{1+\cos 6\theta }{2}\right)}}d\theta \\ L={\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{8+72-8\cos 6\theta +72\cos 6\theta }}d\theta \\ L={\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{80-64\cos 6\theta }}d\theta \end{array}$

This is difficult to solve so, need to use the Simpson’s method:

  $\begin{array}{l}n=10\\ \Delta \theta =\frac{2\pi }{10}=\frac{\pi }{5}\end{array}$

So, the subintervals are $\left[0,\frac{\pi }{5}\right]\left[\frac{\pi }{5},\frac{2\pi }{5}\right],\left[\frac{2\pi }{5},\frac{23\pi }{5}\right]\mathrm{...}\left[\frac{9\pi }{5},2\pi \right]$

Use the Simpson’s rule as:

  $\begin{array}{l}f\left(\theta \right)=\sqrt{80-64\cos 6\theta }\\ L=\frac{\Delta \theta }{3}\left[f\left(0\right)+4f\left(\frac{\pi }{5}\right)+2f\left(2\frac{\pi }{5}\right)+\mathrm{...}+4f\left(\frac{9\pi }{5}\right)+f\left(2\pi \right)\right]\\ L=\frac{\pi }{15}\left[f\left(0\right)+4f\left(\frac{\pi }{5}\right)+2f\left(2\frac{\pi }{5}\right)+\mathrm{...}+4f\left(\frac{9\pi }{5}\right)+f\left(2\pi \right)\right]\\ L=26.73\end{array}$

Hence, the length of the curve is $L=26.73$