Answer
$-1$
Work Step by Step
Given: $r=e^{-\theta}$ and $\theta =\pi$
In Cartesian Coordinates, $x=rcos\theta$
Then $x=e^{-\theta}cos(\theta)$
$\frac{dx}{d\theta}=\frac{d(e^{-\theta}cos(\theta)}{d\theta}$
Use Product rule of differentiation.
$\frac{dx}{d\theta}=-e^{-\theta}cos(\theta)-e^{-\theta}sin(\theta)$
At $\theta=\pi$
$=-e^{-\pi}(-1)-e^{-\pi}(0)$
Thus,
$\frac{dx}{d\theta}=e^{-\pi}$
In Cartesian Coordinates, $y=rsin\theta$
Then $y=e^{-\theta}sin(\theta)$
$\frac{dy}{d\theta}=\frac{d(e^{-\theta}sin(\theta)}{d\theta}$
Use Product rule of differentiation.
$\frac{dy}{d\theta}=-e^{-\theta}sin(\theta)+e^{-\theta}cos(\theta)$
At $\theta=\pi$
$=-e^{-\theta}sin(\pi)+e^{-\theta}cos(\pi)$
$=-e^{-\pi}(0)+e^{-\pi}(-1)$
Thus,
$\frac{dy}{d\theta}=-e^{-\pi}$
Let $m$ be the slope of the tangent line to the given curve.
$m=\frac{dy}{dx}|_{\theta=\pi}$
$=\frac{-e^{-\pi}}{e^{-\pi}}$
$=-1$
Hence, $m=-1$
