What is the exact length of the spiraling polar curve r=5e^(2theta) from 0 to 2pi?

May 19, 2018

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

Explanation:

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

Arc length:

$s = {\int}_{C} \dot{s} \setminus \mathrm{dt} q \quad = {\int}_{C} \sqrt{\boldsymbol{v} \cdot \boldsymbol{v}} \setminus \mathrm{dt} q \quad \triangle$

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

$= 10 {e}^{2 \theta} \dot{\theta} \setminus \boldsymbol{\hat{r}} + 5 {e}^{2 \theta} \frac{d}{\mathrm{dt}} \left(\setminus \boldsymbol{\hat{r}}\right) q \quad \square$

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

$\left(\begin{matrix}- \sin \theta \\ \cos \theta\end{matrix}\right) \dot{\theta} = \boldsymbol{\hat{\theta}} \dot{\theta}$

So $\square$ is:

$\boldsymbol{v} \left(\theta\right) = 10 {e}^{2 \theta} \dot{\theta} \setminus \boldsymbol{\hat{r}} + 5 {e}^{2 \theta} \dot{\theta} \setminus \boldsymbol{\hat{\theta}}$

And $\triangle$ becomes:

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

$= {\int}_{C} \setminus {e}^{2 \theta} \sqrt{{10}^{2} + {5}^{2}} q \quad \setminus \dot{\theta} \setminus \mathrm{dt}$

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

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