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

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Answer 1 · 2015-10-25T09:09:26

#8#

#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#