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

1 Answer

Answer:

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! :)