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

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Answer 1 · 2018-03-14T08:32:02

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

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

$f'(theta)=4sin2theta$

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:

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

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