What is the arclength of the polar curve $f (θ) = {csc}^{2} θ - {cot}^{2} θ$ $f(theta) = csc^2theta-cot^2theta$ over $θ \in [\frac{π}{3}, \frac{π}{2}]$ $theta in [pi/3,pi/2]$ ?

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Answer 1 · 2016-03-24T02:31:36

Hi there!

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

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

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! :)

What is the arclength of the polar curve f(theta) = csc^2theta-cot^2theta f(θ)=csc2θ−cot2θf(theta) = csc^2theta-cot^2theta over theta in [pi/3,pi/2] θ∈[π3,π2]theta in [pi/3,pi/2] ?