Answer
$TLS\vert_{\frac{\pi}{3}}=\frac{1}{\sqrt{3}}$
Work Step by Step
Given:
$r=2cos(\theta)$
$\theta=\frac{\pi}{3}$
Use the equation for tangent line slope for polar coordinates:
$TLS=\frac{\frac{dr}{d\theta}sin\theta+rcos\theta}{\frac{dr}{d\theta}cos\theta-rsin\theta}$
Find $\frac{dr}{d\theta}$:
$\frac{dr}{d\theta}=(2cos\theta)^\prime=-2sin\theta$
Plug in for $r$, $\theta$, and $\frac{dr}{d\theta}$ and evaluate:
$TLS\vert_{\frac{\pi}{3}}=\frac{(-2sin\frac{\pi}{3})sin\frac{\pi}{3}+(2cos\frac{\pi}{3})cos\frac{\pi}{3}}{(-2sin\frac{\pi}{3})cos\frac{\pi}{3}-(2cos\frac{\pi}{3})sin\frac{\pi}{3}}=\frac{-2(\frac{\sqrt{3}}{2})^2+2(\frac{1}{2})^2}{(-4(\frac{\sqrt{3}}{2}))(\frac{1}{2})}$
Simplify:
$TLS\vert_{\frac{\pi}{3}}=\frac{1}{\sqrt{3}}$
