# How do you find the exact length of the polar curve r=e^theta ?

Sep 7, 2014

If $\theta$ goes from ${\theta}_{1}$ to ${\theta}_{2}$, then the arc length is $\sqrt{2} \left({e}^{{\theta}_{2}} - {e}^{{\theta}_{1}}\right)$.

Let us look at some details.
$L = {\int}_{{\theta}_{1}}^{{\theta}_{2}} \sqrt{{r}^{2} + {\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2}} d \theta$
since $r = {e}^{\theta}$ and $\frac{\mathrm{dr}}{d \theta} = {e}^{\theta}$,
$= {\int}_{{\theta}_{1}}^{{\theta}_{2}} \sqrt{{\left({e}^{\theta}\right)}^{2} + {\left({e}^{\theta}\right)}^{2}} d \theta$
by pulling ${e}^{\theta}$ out of the square-root,
$= {\int}_{{\theta}_{1}}^{{\theta}_{2}} {e}^{\theta} \sqrt{2} d \theta = \sqrt{2} {\int}_{{\theta}_{1}}^{{\theta}_{2}} {e}^{\theta} d \theta$
by evaluating the integral,
$= \sqrt{2} {\left[{e}^{\theta}\right]}_{{\theta}_{1}}^{{\theta}_{2}} = \sqrt{2} \left({e}^{{\theta}_{2}} - {e}^{{\theta}_{1}}\right)$