Question

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

Answer 1

Hi there!

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

`L=int_a^bsqrt(r^2+((dr)/(dTheta))^2)dTheta`

Explanation:

First of all, it's important to note that `r = f(Theta)` and thus, `r=csc^2(Theta)-cot^2(Theta)`.

Notice anything about this function? A trig identity perhaps?

Well, `1 + cot^2(Theta) = csc^2(Theta)`, and rearranged:

`1 = csc^2(Theta) - cot^2(Theta)`

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

Substituting everything into the formula we get:

`L=int_(pi/3)^(pi/2)sqrt((1)^2+((d)/(dTheta)(1))^2)dTheta`

Simplifying everything:

`L=int_(pi/3)^(pi/2)sqrt(1)dTheta`

`L=int_(pi/3)^(pi/2)(1)dTheta`

Integrating a constant with respect to theta we get:

`L=(Theta)|_(pi/3)^(pi/2)` ------ Note that this bar means "evaluated from"

Now substitute the bounds in:

`L = (pi/2) - (pi/3)`

`L = pi/6`

Therefore, the arclength of the polar curve is `pi/6` rad.

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