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

Jun 8, 2017

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

We obtained:

$s = {\int}_{{\theta}_{1}}^{{\theta}_{2}} \sqrt{{r}^{2} + {\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2}} d \theta$

and used it to get about $2.734$, using technology.

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 $\sin x = \cos \left(x - \frac{\pi}{2}\right)$.

Then, we should first square $r$ to get:

${r}^{2} = 9 {\sin}^{2} \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) - 6 \theta \sin \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) \cos \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) + {\theta}^{2} {\cos}^{2} \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right)$

Alright, and now, the derivative with respect to $\theta$...

$\frac{\mathrm{dr}}{d \theta} = 3 \cos \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) \cdot - \frac{1}{2} - \left[\theta \cdot - \sin \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) \cdot - \frac{1}{2} + \cos \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right)\right]$

$= - \frac{3}{2} \cos \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) - \frac{\theta}{2} \cdot \sin \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) - \cos \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right)$

$= - \frac{5}{2} \cos \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) - \frac{\theta}{2} \sin \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right)$

This squared gives:

${\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2} = {\left[- \frac{5}{2} \cos \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) - \frac{\theta}{2} \sin \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right)\right]}^{2}$

$= \frac{25}{4} {\cos}^{2} \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) + \frac{5}{2} \theta \sin \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) \cos \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) + {\theta}^{2} / 4 {\sin}^{2} \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right)$

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

$s = {\int}_{11 \pi / 8}^{13 \pi / 8} \sqrt{9 {\sin}^{2} \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) - 6 \theta \sin \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) \cos \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) + {\theta}^{2} {\cos}^{2} \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) + \frac{25}{4} {\cos}^{2} \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) + \frac{5}{2} \theta \sin \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) \cos \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) + {\theta}^{2} / 4 {\sin}^{2} \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right)} d \theta$

$= {\int}_{11 \pi / 8}^{13 \pi / 8} \sqrt{\left(9 + {\theta}^{2} / 4\right) {\sin}^{2} \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) - \frac{7}{2} \theta \sin \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) \cos \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right) + \left({\theta}^{2} + \frac{25}{4}\right) {\cos}^{2} \left(\frac{3 \pi}{16} - \frac{\theta}{2}\right)} d \theta$

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!

$\textcolor{b l u e}{s \approx 2.734}$