# What is the arclength of the polar curve f(theta) = 4sin(2theta)-2sec^2theta  over theta in [0,pi/8] ?

May 2, 2017

$\text{Arclength } \approx 2.509$

#### Explanation:

Arclength on the polar plane has the following formula:
$L = {\int}_{{\theta}_{1}}^{{\theta}_{2}} \sqrt{{r}^{2} + {\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2}} \text{ } d \theta$

Given the equation $f \left(\theta\right) = r = 4 \sin \left(2 \theta\right) - 2 {\sec}^{2} \theta$, we can differentiate to find
$\frac{\mathrm{dr}}{d \theta} = 4 \cos \left(2 \theta\right) \left(2\right) - 4 \sec \theta \left(\sec \theta \tan \theta\right)$

$\frac{\mathrm{dr}}{d \theta} = 8 \cos \left(2 \theta\right) - 4 {\sec}^{2} \theta \tan \theta$

Plug in the expressions for $r$ and $\frac{\mathrm{dr}}{d \theta}$ into the polar arclength formula to get:
$\text{Arclength "= int_0^(pi/8)sqrt((4sin(2theta)-2sec^2theta)^2+(8cos(2theta)-4sec^2theta tantheta)^2)" } d \theta$

Use a graphing calculator to evaluate:

$\approx 2.509258859$