# How do you find the slope of the polar curve r=cos(2theta) at theta=pi/2 ?

##### 1 Answer
Oct 16, 2014

By converting into parametric equations,

{(x(theta)=r(theta)cos theta=cos2theta cos theta), (y(theta)=r(theta)sin theta=cos2theta sin theta):}

By Product Rule,

$x ' \left(\theta\right) = - \sin 2 \theta \cos \theta - \cos 2 \theta \sin \theta$

$x ' \left(\frac{\pi}{2}\right) = - \sin \left(\pi\right) \cos \left(\frac{\pi}{2}\right) - \cos \left(\pi\right) \sin \left(\frac{\pi}{2}\right) = 1$

$y ' \left(\theta\right) = - \sin 2 \theta \sin \theta + \cos 2 \theta \cos \theta$

$y ' \left(\frac{\pi}{2}\right) = - \sin \left(\pi\right) \sin \left(\frac{\pi}{2}\right) + \cos \left(\pi\right) \cos \left(\frac{\pi}{2}\right) = 0$

So, the slope $m$ of the curve can be found by

$m = \frac{\mathrm{dy}}{\mathrm{dx}} {|}_{\theta = \frac{\pi}{2}} = \frac{y ' \left(\frac{\pi}{2}\right)}{x ' \left(\frac{\pi}{2}\right)} = \frac{0}{1} = 0$

I hope that this was helpful.