How do you find the arc length of the cardioid r = 1+cos(theta) from 0 to 2pi?

1 Answer
Oct 25, 2015

8

Explanation:

l=int sqrt(r^2+((dr)/(d theta))^2) d theta

r=1+costheta => (dr)/(d theta) = -sintheta

l = 2int_0^pi sqrt((1+costheta)^2+sin^2theta) d theta

l = 2int_0^pi sqrt(1+2costheta+cos^2theta+sin^2theta) d theta

l = 2int_0^pi sqrt(1+2costheta+1) d theta

l = 2int_0^pi sqrt(2+2costheta) d theta = 2 int_0^pi sqrt(2(1+costheta)) d theta

l = 2 int_0^pi sqrt(4cos^2(theta/2)) d theta = 2 int_0^pi 2cos(theta/2) d theta

l=8 sin(theta/2) |_0^pi = 8