Answer
The slope of the tangent line to the given polar curve at the point where: $\theta = \pi$ is equal to:
$$-\pi$$
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:
$\frac{dr}{d\theta} = \frac{d(1/\theta)}{d\theta} = - \frac 1 {\theta^2}$
$\frac{dy}{dx} = \frac{-(\frac{1}{\theta^2})sin(\theta) + (1/\theta)cos(\theta)}{-(\frac{1}{\theta^2})cos(\theta) - (1/\theta)sin(\theta)} \div \frac{1/\theta}{1/\theta} = \frac{-(\frac{1}{\theta})sin(\theta) + cos(\theta)}{-(\frac{1}{\theta})cos(\theta) - sin(\theta)}$
3. Calculate the slope when $\theta = \pi$
$\frac{dy}{dx} = \frac{-(1/\pi)sin(\pi) + cos(\pi)}{-(1/\pi)cos(\pi) - sin(\pi)} = \frac{-(1/\pi)(0) + (-1)}{-(1/\pi)(-1) - 0} = \frac{-1}{1/\pi} = -\pi$
