# How do you find the length of the polar curve r=5^theta ?

Aug 31, 2014

You can find the length of this polar curve by applying the formula for Arc Length for Parametric Equations:

${L}_{=} {\int}_{a}^{b} \sqrt{{r}^{2} + {\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2}} d \theta$

$L = \frac{{5}^{\theta} \sqrt{1 + {\ln}^{2} \left(5\right)}}{\ln} 5 {|}_{a}^{b}$

Process:

The only extra component we need to find for this formula is $\frac{\mathrm{dr}}{d \theta}$, which we find by deriving our original function.

To derive an exponential function with a base other than $e$, we first rewrite the original function, multiply it by the $\ln$ of the base, then multiply by the derivative of the term in the exponent:

$\frac{\mathrm{dr}}{d \theta} = {5}^{\theta} \cdot \ln \left(5\right) \cdot \left(1\right) = {5}^{\theta} \ln 5$

Plugging this into our formula, we have:

$L = {\int}_{a}^{b} \sqrt{{\left({5}^{\theta}\right)}^{2} + {\left({5}^{\theta} \ln 5\right)}^{2}} d \theta$

Distribute the exponent:

L =int_a^b sqrt(5^(2theta) + 5^(2theta)ln^2(5) d theta

We can now pull out a ${5}^{\theta}$ from both terms in the radical:

L =int_a^b sqrt(5^(2theta)(1 + ln^2(5)) d theta

We can now take the square root of ${5}^{2 \theta}$ and pull it out of the radical:

$L = {\int}_{a}^{b} {5}^{\theta} \sqrt{1 + {\ln}^{2} \left(5\right)} d \theta$

The important thing to notice here is that $\sqrt{1 + {\ln}^{2} \left(5\right)}$

is actually a constant, which means it can be pulled out of the integral entirely:

$L = \sqrt{1 + {\ln}^{2} \left(5\right)} {\int}_{a}^{b} {5}^{\theta} d \theta$

Now to integrate this exponential function with a base other than $e$, we rewrite the original function and then divide by the $\ln$ of the base:

${\int}_{a}^{b} {5}^{\theta} d \theta = {5}^{\theta} / \ln 5$

You can derive this result to make sure it's correct, knowing that you can pull out $\frac{1}{\ln} 5$ since it's a constant.

We now have:

$L = \sqrt{1 + {\ln}^{2} \left(5\right)} {5}^{\theta} / \ln 5 {|}_{a}^{b}$

Simplifying, we arrive at our final answer:

$L = \frac{{5}^{\theta} \sqrt{1 + {\ln}^{2} \left(5\right)}}{\ln} 5 {|}_{a}^{b}$