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}$

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

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