# How do you find the slope of the polar curve r=3sec(2theta) at theta=pi/6 ?

Sep 21, 2014

The slope of the curve $r = 3 \sec \left(2 \theta\right)$ at $\theta = \frac{\pi}{6}$ is $\frac{3 \sqrt{3}}{5}$.

Let us look at some details.

We can write

$\left\{\begin{matrix}x = r \cos \theta = 3 \sec \left(2 \theta\right) \cos \theta \\ y = r \sin \theta = 3 \sec \left(2 \theta\right) \sin \theta\end{matrix}\right.$

$\frac{\mathrm{dx}}{d \theta} = 3 \left[2 \sec \left(2 \theta\right) \tan \left(2 \theta\right) \cdot \cos \theta + \sec \left(2 \theta\right) \cdot \left(- \sin \theta\right)\right]$

$= 3 \sec \left(2 \theta\right) \left[2 \tan \left(2 \theta\right) \cos \theta - \sin \theta\right]$

$\frac{\mathrm{dy}}{d \theta} = 3 \left[2 \sec \left(2 \theta\right) \tan \left(2 \theta\right) \cdot \sin \theta + \sec \left(2 \theta\right) \cdot \cos \theta\right]$

$= 3 \sec \left(2 \theta\right) \left[2 \tan \left(2 \theta\right) \sin \theta + \cos \theta\right]$

So,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{d \theta}}{\frac{\mathrm{dx}}{d \theta}} = \frac{3 \sec \left(2 \theta\right) \left[2 \tan \left(2 \theta\right) \sin \theta + \cos \theta\right]}{3 \sec \left(2 \theta\right) \left[2 \tan \left(2 \theta\right) \cos \theta - \sin \theta\right]}$

by cancelling out $3 \sec \left(2 \theta\right)$,

$= \frac{2 \tan \left(2 \theta\right) \sin \theta + \cos \theta}{2 \tan \left(2 \theta\right) \cos \theta - \sin \theta}$

Now, we can find the slope $m$ by pluggin in $\theta = \frac{\pi}{6}$.

$m = \frac{\mathrm{dy}}{\mathrm{dx}} {|}_{\theta = \frac{\pi}{6}} = \frac{2 \left(\sqrt{3}\right) \left(\frac{1}{2}\right) + \frac{\sqrt{3}}{2}}{2 \left(\sqrt{3}\right) \frac{\sqrt{3}}{2} - \frac{1}{2}} = \frac{3 \sqrt{3}}{5}$