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

1 Answer
Mar 14, 2018

Answer:

#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#