Question

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

Answer 1

`L=sqrt(13/3)(pi/3)+sqrt(52/3)sum_(n=1)^oosum_(m=0)^(2n)((1/2),(n))((2n),(m))6^m/(-52)^nint_(pi/3)^(pi/2)cos^m2thetad theta` units.

Explanation:

`f(theta)=cos^2theta-3sin^2theta=2cos2theta-1`

`f'(theta)=4sin2theta`

Arclength is given by:

`L=int_(pi/3)^(pi/2)sqrt((2cos2theta-1)^2+16sin^2 2theta)d theta`

Expand the square and combine terms:

`L=int_(pi/3)^(pi/2)sqrt(17-4cos2theta-12cos^2 2theta)d theta`

Complete the square in the square root:

`L=1/sqrt3int_(pi/3)^(pi/2)sqrt(52-(6cos2theta+1)^2)d theta`

Factor out the larger piece:

`L=sqrt(52/3)int_(pi/3)^(pi/2)sqrt(1-1/52(6cos2theta+1)^2)d theta`

For `theta in [pi/3, pi/2]`, `1/52(6cos2theta+1)^2<1`. Take the series expansion of the square root:

`L=sqrt(52/3)int_(pi/3)^(pi/2)sum_(n=0)^oo((1/2),(n))(-1/52(6cos2theta+1)^2)^nd theta`

Isolate the `n=0` term:

`L=sqrt(52/3)int_(pi/3)^(pi/2)d theta+sqrt(52/3)sum_(n=1)^oo((1/2),(n))(-1/52)^nint_(pi/3)^(pi/2)(6cos2theta+1)^(2n)d theta`

Apply binominal expansion:

`L=sqrt(13/3)(pi/3)+sqrt(52/3)sum_(n=1)^oo((1/2),(n))(-1/52)^nint_(pi/3)^(pi/2)sum_(m=0)^(2n)((2n),(m))(6cos2theta)^md theta`

Simplify:

`L=sqrt(13/3)(pi/3)+sqrt(52/3)sum_(n=1)^oosum_(m=0)^(2n)((1/2),(n))((2n),(m))6^m/(-52)^nint_(pi/3)^(pi/2)cos^m2thetad theta`