Logan Glover

2023-03-02

How to find the length of the polar curve $r={\mathrm{cos}}^{3}\left(\frac{\theta }{3}\right)$?

Hunter Hendricks

We using the chain rule.
By Chain Rule,
$\frac{dr}{d\theta }=3{\mathrm{cos}}^{2}\left(\frac{\theta }{3}\right)\cdot \left[-\mathrm{sin}\left(\frac{\theta }{3}\right)\right]\cdot \frac{1}{3}$
by cleaning up a bit,
$=-{\mathrm{cos}}^{2}\left(\frac{\theta }{3}\right)\mathrm{sin}\left(\frac{\theta }{3}\right)$
Note that $\theta$ goes from 0 to $3\pi$ to complete the loop once.
Let us now calculate the curve's length L.
$L={\int }_{0}^{3\pi }\sqrt{{r}^{2}+{\left(\frac{dr}{d\theta }\right)}^{2}}d\theta$
$={\int }_{0}^{3\pi }\sqrt{{\mathrm{cos}}^{6}\left(\frac{\theta }{3}\right)+{\mathrm{cos}}^{4}\left(\frac{\theta }{3}\right){\mathrm{sin}}^{2}\left(\frac{\theta }{3}\right)}d\theta$
by pulling ${\mathrm{cos}}^{2}\left(\frac{\theta }{3}\right)$ out of the square-root,
$={\int }_{0}^{3\pi }{\mathrm{cos}}^{2}\left(\frac{\theta }{3}\right)\sqrt{{\mathrm{cos}}^{2}\left(\frac{\theta }{3}\right)+{\mathrm{sin}}^{2}\left(\frac{\theta }{3}\right)}d\theta$
by ${\mathrm{cos}}^{2}\theta =\frac{1}{2}\left(1+\mathrm{cos}2\theta \right)$ and ${\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta =1$,
$=\frac{1}{2}{\int }_{0}^{3\pi }\left[1+\mathrm{cos}\left(\frac{2\theta }{3}\right)\right]d\theta$
$=\frac{1}{2}{\left[\theta +\frac{3}{2}\mathrm{sin}\left(\frac{2\theta }{3}\right)\right]}_{0}^{3\pi }$
$=\frac{1}{2}\left[3\pi +0-\left(0+0\right)\right]=\frac{3\pi }{2}$

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