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

1 Answer
Apr 14, 2017

Answer:

I used WolframAlpha #"Length" ~~ 12.6829#

Explanation:

From the reference Arc Length with Polar Coordinates

#L = int_alpha^ beta sqrt(r^2+((dr)/(d theta))^2)d theta#

Given: #r=2sin(theta) -3theta, alpha = (-3pi)/8, and beta=(7pi)/8#

Square the function:

#r^2=4sin^2(theta) -12thetasin(theta)+9theta^2#

Compute the derivative of the function:

#(dr)/(d theta) = 2cos(theta)-3#

Square the derivative:

#((dr)/(d theta))^2 = 4cos^2(theta)-12cos(theta) + 9#

Compute the argument under the radical:

#r^2+((dr)/(d theta))^2 = 4sin^2(theta) -12thetasin(theta)+9theta^2+4cos^2(theta)-12cos(theta) + 9 #

Use the identity #sin^2(theta)+cos^2(theta) = 1#:

#r^2+((dr)/(d theta))^2 = 4(1) -12thetasin(theta)+9theta^2-12cos(theta) + 9 #

Combine like terms:

#r^2+((dr)/(d theta))^2 = -12thetasin(theta)+9theta^2-12cos(theta) + 13#

Substitute into the integral:

#L = int_(-3pi/8)^(7pi/8) sqrt(-12thetasin(theta)+9theta^2-12cos(theta) + 13)d theta#

I used WolframAlpha to evaluate the integral:

#L = int_(-3pi/8)^(7pi/8) sqrt(-12thetasin(theta)+9theta^2-12cos(theta) + 13)d theta ~~ 12.6829#