What is the arclength of #r=sin(theta^2-(7pi)/6) +2thetacos(theta -(3pi)/8)# on #theta in [(7pi)/8,pi]#?

Question

1677 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-13T07:51:23

The arc length #s=2.092001# units

The given polar equation:

#r=sin(θ^2−(7π)/6)+2θ*cos(θ−(3π)/8)#

The first derivative #(dr)/ (d theta)# is needed in the formula for arc length #s#:

#(dr)/(d theta)=cos(theta^2-(7 pi)/6)(2 theta)+2(1*cos(theta-(3 pi)/8)+ theta(-sin(theta-(3 pi)/8))*1)#

#(dr)/(d theta)=2 thetacos(theta^2-(7 pi)/6)+2 cos(theta-(3 pi)/8)-2 theta sin(theta-(3 pi)/8)#

The formula for arc length #s#:

#s=int_a^b sqrt((sin(θ^2−(7π)/6)+2θ*cos(θ−(3π)/8))^2+ (2 thetacos(theta^2-(7 pi)/6)+2 cos(theta-(3 pi)/8)-2 theta sin(theta-(3 pi)/8))^2 d theta#

#s=2.09201# units