# How do you find the slope of the polar curve r=1+sin(theta) at theta=pi/4 ?

Oct 10, 2014

By converting into parametric equations,

$\left\{\begin{matrix}x \left(\theta\right) = r \left(\theta\right) \cos \theta = \left(1 + \sin \theta\right) \cos \theta \\ y \left(\theta\right) = r \left(\theta\right) \sin \theta = \left(1 + \sin \theta\right) \sin \theta\end{matrix}\right.$

By finding the derivatives using Product Rule,

$x ' \left(\theta\right) = \cos \theta \cdot \cos \theta + \left(1 + \sin \theta\right) \cdot \left(- \sin \theta\right)$

$= \left({\cos}^{2} \theta - {\sin}^{2} \theta\right) - \sin \theta$

$= \cos 2 \theta - \sin \theta$

$R i g h t a r r o w x ' \left(\frac{\pi}{4}\right) = \cos \left(\frac{\pi}{2}\right) - \sin \left(\frac{\pi}{4}\right) = - \frac{1}{\sqrt{2}}$

$y ' \left(\theta\right) = \cos \theta \cdot \sin \theta + \left(1 + \sin \theta\right) \cdot \cos \theta$

$= 2 \sin \theta \cos \theta + \cos \theta$

$= \sin 2 \theta + \cos \theta$

$R i g h t a r r o w y ' \left(\frac{\pi}{4}\right) = \sin \left(\frac{\pi}{2}\right) + \cos \left(\frac{\pi}{4}\right) = 1 + \frac{1}{\sqrt{2}}$

The slope $m$ we are looking for is:

m={dy}/{dx}|_{theta=pi/4}={y'(pi/4)}/{x'(pi/4)}={1+1/sqrt{2}}/{-1/sqrt{2}} =-(sqrt{2}+1)

I hope that this was helpful.