Answer
$\frac{1}{3\sqrt{3}}$
Work Step by Step
Find $dy/d\theta$:
$\frac{dy}{d\theta}=\frac{d}{d\theta}(r\sin\theta)=\frac{d}{d\theta}((1+2\cos\theta)\sin\theta)=\frac{d}{d\theta}(\sin\theta+2\sin\theta\cos\theta)=\frac{d}{d\theta}(\sin\theta+\sin2\theta)=\cos\theta+2\cos2\theta$
Find $dx/d\theta$:
$\frac{dx}{d\theta}=\frac{d}{d\theta}(r\cos\theta)=\frac{d}{d\theta}((1+2\cos\theta)\cos\theta)=\frac{d}{d\theta}(\cos\theta+2\cos^2\theta)=\frac{d}{d\theta}(\cos\theta+\cos2\theta+1)=-\sin\theta-2\sin2\theta+0=-\sin\theta-2\sin2\theta$
Find the slope of the tangent line at $\theta=\pi/3$:
$m=\frac{dy}{dx}|_{\theta=\pi/3}=\frac{dy/d\theta}{dx/d\theta}|_{\theta=\pi/3}=\frac{\cos(\pi/3)+2\cos(2\pi/3)}{-\sin(\pi/3)-2\sin(2\pi/3)}$
$=\frac{\frac{1}{2}+2(-\frac{1}{2})}{-\frac{\sqrt{3}}{2}-2\cdot\frac{\sqrt{3}}{2}}=\frac{\frac{1}{2}-\frac{2}{2}}{-\frac{\sqrt{3}}{2}-\frac{2\sqrt{3}}{2}}=\frac{-\frac{1}{2}}{-\frac{3\sqrt{3}}{2}}=\frac{1}{3\sqrt{3}}$
Thus, the slope is $\frac{1}{3\sqrt{3}}$.
