Answer
$-1$
Work Step by Step
Given: $r=(3+cos3\theta)$ and $\theta =\pi/2$
In Cartesian Coordinates, $x=rcos\theta$
Then $x=(3+cos3\theta)cos(\theta)$
$\frac{dx}{d\theta}=\frac{d[(3+cos3\theta)cos(\theta)]}{d\theta}$
Use Product rule of differentiation.
$\frac{dx}{d\theta}=(-3sin3\theta)cos(\theta)-(3+cos3\theta)sin\theta$
At $\theta=\pi/2$
$=(-3sin3\pi/2)cos(\pi/2)-(3+cos3\pi/2)sin(\pi/2)$
Thus,
$\frac{dx}{d\theta}=((-3)(-1))(0)-(3+0)(1)=-3$
In Cartesian Coordinates, $y=rcos\theta$
Then $y=(3+cos3\theta)cos(\theta)$
$\frac{dy}{d\theta}=\frac{d[(3+cos3\theta)cos(\theta)]}{d\theta}$
Use Product rule of differentiation.
$\frac{dy}{d\theta}=(-3sin3\theta)sin(\theta)+(3+cos3\theta)cos\theta$
At $\theta=\pi/2$
$=(-3sin3\pi/2)sin(\pi/2)+(3+cos3\pi/2)cos(\pi/2)$
Thus,
$\frac{dy}{d\theta}=((-3)(-1))(1)+(3+0)(0)=3$
Let $m$ be the slope of the tangent line to the given curve.
$m=\frac{dy}{dx}|_{\theta=\pi/2}$
$=\frac{3}{-3}$
$=-1$
Hence, $m=-1$
