Answer
The slope of the tangent line to the given polar curve when $\theta = \pi$ is equal to:
$$-\sqrt 3$$
Work Step by Step
1. Write the formula for the slope of the tangent line: $$\frac{dy}{dx} = \frac{\frac{dr}{d\theta}sin(\theta) + rcos(\theta)}{\frac{dr}{d\theta}cos(\theta) - rsin(\theta)} $$
2. Calculate $\frac{dr}{d\theta}$ and substitute the equation into the formula:
$\frac{dr}{d\theta} = \frac{d(cos(\theta/3))}{d\theta} = -\frac{1}{3}sin(\theta/3)$
$\frac{dy}{dx} = \frac{(-\frac{1}{3}sin(\theta/3))sin(\theta) + (cos(\theta/3))cos(\theta)}{(-\frac{1}{3}sin(\theta/3))cos(\theta) - (cos(\theta/3))sin(\theta)} $
3. Calculate the slope when $\theta = \pi$
$\frac{dy}{dx} = \frac{(-\frac{1}{3}sin(\pi/3))sin(\pi) + (cos(\pi/3))cos(\pi)}{(-\frac{1}{3}sin(\pi/3))cos(\pi) - (cos(\pi/3))sin(\pi)} = \frac{(-\frac{1}{3}\frac{\sqrt 3}{2})(0) + (\frac 1 2)(-1)}{(-\frac{1}{3}\frac{\sqrt 3}{2})(-1) - (\frac 12)(0)}$
$\frac {dy}{dx} = \frac{-\frac 12}{\frac {\sqrt 3}6} = - \frac {3}{\sqrt 3} = -\sqrt 3$
