somi5fjt

2023-03-20

What is the exact length of the spiraling polar curve $r=5{e}^{2\theta}$ from 0 to $2\pi$?

dalematealreypq3a

Beginner2023-03-21Added 10 answers

$r\left(\theta \right)=5{e}^{2\theta}\hat{r}$

Arc length:

$s={\int}_{C}\stackrel{.}{s}dt={\int}_{C}\sqrt{v\cdot v}dt\u25b3$

$v\left(\theta \right)=\frac{d}{dt}(5{e}^{2\theta}\hat{r})$

Product rule:

$=10{e}^{2\theta}\stackrel{.}{\theta}\hat{r}+5{e}^{2\theta}\frac{d}{dt}(\hat{r})\square$

$\frac{d}{dt}(\hat{r})=\frac{d}{dt}\left(\begin{array}{c}\mathrm{cos}\theta \\ \mathrm{sin}\theta \end{array}\right)$

$\left(\begin{array}{c}-\mathrm{sin}\theta \\ \mathrm{cos}\theta \end{array}\right)\stackrel{.}{\theta}=\hat{\theta}\stackrel{.}{\theta}$

Therefore, $\square$ is:

$v\left(\theta \right)=10{e}^{2\theta}\stackrel{.}{\theta}\hat{r}+5{e}^{2\theta}\stackrel{.}{\theta}\hat{\theta}$

And $\u25b3$ becomes:

$={\int}_{C}\sqrt{{\left(10{e}^{2\theta}\stackrel{.}{\theta}\right)}^{2}+{(5{e}^{2\theta}\stackrel{.}{\theta})}^{2}}dt$

$={\int}_{C}{e}^{2\theta}\sqrt{{10}^{2}+{5}^{2}}\stackrel{.}{\theta}dt$

$=5\sqrt{5}{\int}_{0}^{2\pi}{e}^{2\theta}d\theta$

$=\frac{5\sqrt{5}}{2}({e}^{4\pi}-1)$

Arc length:

$s={\int}_{C}\stackrel{.}{s}dt={\int}_{C}\sqrt{v\cdot v}dt\u25b3$

$v\left(\theta \right)=\frac{d}{dt}(5{e}^{2\theta}\hat{r})$

Product rule:

$=10{e}^{2\theta}\stackrel{.}{\theta}\hat{r}+5{e}^{2\theta}\frac{d}{dt}(\hat{r})\square$

$\frac{d}{dt}(\hat{r})=\frac{d}{dt}\left(\begin{array}{c}\mathrm{cos}\theta \\ \mathrm{sin}\theta \end{array}\right)$

$\left(\begin{array}{c}-\mathrm{sin}\theta \\ \mathrm{cos}\theta \end{array}\right)\stackrel{.}{\theta}=\hat{\theta}\stackrel{.}{\theta}$

Therefore, $\square$ is:

$v\left(\theta \right)=10{e}^{2\theta}\stackrel{.}{\theta}\hat{r}+5{e}^{2\theta}\stackrel{.}{\theta}\hat{\theta}$

And $\u25b3$ becomes:

$={\int}_{C}\sqrt{{\left(10{e}^{2\theta}\stackrel{.}{\theta}\right)}^{2}+{(5{e}^{2\theta}\stackrel{.}{\theta})}^{2}}dt$

$={\int}_{C}{e}^{2\theta}\sqrt{{10}^{2}+{5}^{2}}\stackrel{.}{\theta}dt$

$=5\sqrt{5}{\int}_{0}^{2\pi}{e}^{2\theta}d\theta$

$=\frac{5\sqrt{5}}{2}({e}^{4\pi}-1)$

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