# What is the arclength of the polar curve f(theta) = csc^2theta-cot^2theta  over theta in [pi/3,pi/2] ?

Hi there!

Let's start off with the formula for the arc length of a polar curve whereby:

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

#### Explanation:

First of all, it's important to note that $r = f \left(\Theta\right)$ and thus, $r = {\csc}^{2} \left(\Theta\right) - {\cot}^{2} \left(\Theta\right)$.

Well, $1 + {\cot}^{2} \left(\Theta\right) = {\csc}^{2} \left(\Theta\right)$, and rearranged:

$1 = {\csc}^{2} \left(\Theta\right) - {\cot}^{2} \left(\Theta\right)$

Therefore, this function would simplify down to r = 1! This makes everything much easier!

Substituting everything into the formula we get:

$L = {\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sqrt{{\left(1\right)}^{2} + {\left(\frac{d}{\mathrm{dT} h \eta} \left(1\right)\right)}^{2}} \mathrm{dT} h \eta$

Simplifying everything:

$L = {\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sqrt{1} \mathrm{dT} h \eta$

$L = {\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} \left(1\right) \mathrm{dT} h \eta$

Integrating a constant with respect to theta we get:

$L = \left(\Theta\right) {|}_{\frac{\pi}{3}}^{\frac{\pi}{2}}$ ------ Note that this bar means "evaluated from"

Now substitute the bounds in:

$L = \left(\frac{\pi}{2}\right) - \left(\frac{\pi}{3}\right)$

$L = \frac{\pi}{6}$

Therefore, the arclength of the polar curve is $\frac{\pi}{6}$ rad.

Hopefully everything was clear and concise! If you have any questions, feel free to ask! :)