# How do you find the slope of the polar curve r=3+8sin(theta) at theta=pi/6 ?

Oct 30, 2014

$r \left(\theta\right) = 3 + 8 \sin \theta$

by converting into parametric equations,

$R i g h t a r r o w \left\{\begin{matrix}x \left(\theta\right) = r \left(\theta\right) \cos \theta = \left(3 + 8 \sin \theta\right) \cos \theta \\ y \left(\theta\right) = r \left(\theta\right) \sin \theta = \left(3 + 8 \sin \theta\right) \sin \theta\end{matrix}\right.$

By differentiating with respect to $\theta$,

$x ' \left(\theta\right) = 8 \cos \theta \cdot \cos \theta + \left(3 + 8 \sin \theta\right) \cdot \left(- \sin \theta\right)$

$= 8 \left({\cos}^{2} \theta - {\sin}^{2} \theta\right) - 3 \sin \theta$

$= 8 \cos 2 \theta - 3 \sin \theta$

by evaluating at $\theta = \frac{\pi}{6}$,

$R i g h t a r r o w x ' \left(\frac{\pi}{6}\right) = 8 \cos \left(\frac{\pi}{3}\right) - 3 \sin \left(\frac{\pi}{6}\right) = 4 - \frac{3}{2} = \frac{5}{2}$

By differentiating with respect to $\theta$,

$y ' \left(\theta\right) = 8 \cos \theta \cdot \sin \theta + \left(3 + 8 \sin \theta\right) \cdot \cos \theta$

$= 16 \sin \theta \cos \theta + 3 \cos \theta$

$= 8 \sin 2 \theta + 3 \cos \theta$

by evaluating at $\theta = \frac{\pi}{6}$,

$R i g h t a r r o w y ' \left(\frac{\pi}{6}\right) = 8 \sin \left(\frac{\pi}{3}\right) + 3 \cos \left(\frac{\pi}{6}\right) = 4 \sqrt{3} + \frac{3 \sqrt{3}}{2} = \frac{11 \sqrt{3}}{2}$

So, the slope $m$ can be found by

$m = \frac{\mathrm{dy}}{\mathrm{dx}} {|}_{\theta = \frac{\pi}{6}} = \frac{\frac{\mathrm{dy}}{d \theta} {|}_{\theta = \frac{\pi}{6}}}{\frac{\mathrm{dx}}{d \theta} {|}_{\theta = \frac{\pi}{6}}} = \frac{y ' \left(\frac{\pi}{6}\right)}{x ' \left(\frac{\pi}{6}\right)} = \frac{\frac{11 \sqrt{3}}{2}}{\frac{5}{2}} = \frac{11 \sqrt{3}}{5}$

The graph along with its tangent line at $\theta = \frac{\pi}{6}$ looks like: I hope that this was helpful.