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

Beginner2023-03-03Added 6 answers

We using the chain rule.

By Chain Rule,

$\frac{dr}{d\theta}=3{\mathrm{cos}}^{2}\left(\frac{\theta}{3}\right)\cdot [-\mathrm{sin}\left(\frac{\theta}{3}\right)]\cdot \frac{1}{3}$

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 ${\mathrm{cos}}^{2}\theta =\frac{1}{2}(1+\mathrm{cos}2\theta )$ and ${\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta =1$,

$=\frac{1}{2}{\int}_{0}^{3\pi}[1+\mathrm{cos}\left(\frac{2\theta}{3}\right)]d\theta$

$=\frac{1}{2}{[\theta +\frac{3}{2}\mathrm{sin}\left(\frac{2\theta}{3}\right)]}_{0}^{3\pi}$

$=\frac{1}{2}[3\pi +0-(0+0)]=\frac{3\pi}{2}$

By Chain Rule,

$\frac{dr}{d\theta}=3{\mathrm{cos}}^{2}\left(\frac{\theta}{3}\right)\cdot [-\mathrm{sin}\left(\frac{\theta}{3}\right)]\cdot \frac{1}{3}$

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 ${\mathrm{cos}}^{2}\theta =\frac{1}{2}(1+\mathrm{cos}2\theta )$ and ${\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta =1$,

$=\frac{1}{2}{\int}_{0}^{3\pi}[1+\mathrm{cos}\left(\frac{2\theta}{3}\right)]d\theta$

$=\frac{1}{2}{[\theta +\frac{3}{2}\mathrm{sin}\left(\frac{2\theta}{3}\right)]}_{0}^{3\pi}$

$=\frac{1}{2}[3\pi +0-(0+0)]=\frac{3\pi}{2}$

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