Question

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

Answer 1

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 `2.734`, using technology.

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First, let's rewrite `r` so that it has the same arguments... might help us later if we need to use any trig identities.

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` to get:

`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 `theta`...

`(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)`