What is the arclength of the polar curve #f(theta) = 4sin(2theta)-2sec^2theta # over #theta in [0,pi/8] #?

1 Answer
Anjali G
May 2, 2017

#"Arclength " approx 2.509#

Explanation:

Arclength on the polar plane has the following formula:
#L=int_(theta_1)^(theta_2)sqrt(r^2+((dr)/(d theta))^2)" "d theta#

Given the equation #f(theta)=r=4sin(2theta)-2sec^2theta#, we can differentiate to find
#(dr)/(d theta)=4cos(2theta)(2)-4sectheta(sec theta tan theta)#

#(dr)/(d theta)=8cos(2theta)-4sec^2thetatantheta#

Plug in the expressions for #r# and #(dr)/(d theta)# into the polar arclength formula to get:
#"Arclength "= int_0^(pi/8)sqrt((4sin(2theta)-2sec^2theta)^2+(8cos(2theta)-4sec^2theta tantheta)^2)" "d theta#

Use a graphing calculator to evaluate:

#approx 2.509258859#

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