Polar Form Question 1

What is the slope of the line tangent to the polar curve $r=2\theta$ at the point $\theta=\dfrac{\pi}{2}$ ?

-\dfrac{\pi}{2}

-\dfrac{2}{\pi}

0

\dfrac{\pi}{2}

Approach

\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{dy}{dx}

y=r\sin{\theta}

y=2\theta\sin{\theta}

\frac{dy}{d\theta}=2\theta\cos{\theta}+2\sin{\theta}

\frac{dy}{d\theta}\Big|_{\theta=\pi/2} = 2\left(\frac{\pi}{2}\right)\cos{\frac{\pi}{2}}+2\sin{\frac{\pi}{2}}

\frac{dy}{d\theta}= 2

x=r\cos{\theta}

x=2\theta\cos{\theta}

\frac{dx}{d\theta}=-2\theta\sin{\theta}+2\cos{\theta}

\frac{dx}{d\theta}\Big|_{\theta=\pi/2}=-2\left(\frac{\pi}{2}\right)\sin{\frac{\pi}{2}}+2\cos{\frac{\pi}{2}}

\frac{dx}{d\theta} = -\pi

\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{dy}{dx} =\boxed{-\frac{2}{\pi}}