# What is the arclength of #r=-thetasin(theta/2-(27pi)/16) -3cos(theta/2 -(11pi)/16)# on #theta in [(11pi)/8,(13pi)/8]#?

##### 1 Answer

See here for a derivation of the arc length in polar coordinates.

We obtained:

#s = int_(theta_1)^(theta_2) sqrt(r^2 + ((dr)/(d theta))^2)d theta#

and used it to get about

First, let's rewrite

Using Wolfram Alpha, we got:

#r = 3 sin((3 π)/16 - θ/2) - θ cos((3 π)/16 - θ/2)# using identities like

#sinx = cos(x-pi/2)# .

Then, we should first square

#r^2 = 9 sin^2((3 pi)/16 - theta/2) - 6 theta sin((3 pi)/16 - theta/2)cos((3 pi)/16 - theta/2) + theta^2 cos^2((3 pi)/16 - theta/2)#

Alright, and now, the derivative with respect to

#(dr)/(d theta) = 3cos((3pi)/16 - theta/2)cdot-1/2 - [theta cdot -sin((3pi)/16 - theta/2)cdot-1/2 + cos((3pi)/16 - theta/2)]#

#= -3/2cos((3pi)/16 - theta/2) - theta/2 cdot sin((3pi)/16 - theta/2) - cos((3pi)/16 - theta/2)#

#= -5/2cos((3pi)/16 - theta/2) - theta/2 sin((3pi)/16 - theta/2)#

This squared gives:

#((dr)/(d theta))^2 = [-5/2cos((3pi)/16 - theta/2) - theta/2 sin((3pi)/16 - theta/2)]^2#

#= 25/4 cos^2((3pi)/16 - theta/2) + 5/2thetasin((3pi)/16 - theta/2)cos((3pi)/16 - theta/2) + theta^2/4sin^2((3pi)/16 - theta/2)#

Well... let's see where we are now.

#s = int_(11pi//8)^(13pi//8) sqrt(9 sin^2((3 pi)/16 - theta/2) - 6 theta sin((3 pi)/16 - theta/2)cos((3 pi)/16 - theta/2) + theta^2 cos^2((3 pi)/16 - theta/2) + 25/4 cos^2((3pi)/16 - theta/2) + 5/2thetasin((3pi)/16 - theta/2)cos((3pi)/16 - theta/2) + theta^2/4sin^2((3pi)/16 - theta/2))d theta#

#= int_(11pi//8)^(13pi//8) sqrt((9 + theta^2/4)sin^2((3 pi)/16 - theta/2) - 7/2 theta sin((3 pi)/16 - theta/2)cos((3 pi)/16 - theta/2) + (theta^2 + 25/4) cos^2((3pi)/16 - theta/2))d theta#

Welp, this is as simplified as I'm willing to make it. If this was a less painful expression to evaluate, I think you could do it.

But for this one, plug it into Wolfram Alpha and save yourself some brain. I only did two steps because the first step was too long for Wolfram Alpha to accept!

#color(blue)(s ~~ 2.734)#