Polar Form Question 2

What is the slope of the line tangent to the polar curve $r=1+2\sin{\theta}$ at $\theta=0$ ?

2

\dfrac{1}{2}

0

-\dfrac{1}{2}

Approach

We could attempt to convert the polar equation into a rectangular equation. However, this is a bit difficult. Instead, we can find:

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

y=r\sin{\theta} \hskip{2em} x=r\cos{\theta}

y=(1+2\sin{\theta})\sin{\theta}

\frac{dy}{d\theta} = (1+2\sin{\theta})\cos{\theta}+\sin{\theta}(2\cos{\theta})

\frac{dy}{d\theta}\Big|_{\theta=0} = [1+2(0)](1)+(0)[2(1)]

\frac{dy}{d\theta}\Big|_{\theta=0} =1

x=(1+2\sin{\theta})\cos{\theta}

\frac{dx}{d\theta} = (1+2\sin{\theta})(-\sin{\theta}) + \cos{\theta}(2\cos{\theta})

\frac{dx}{d\theta}\Big|_{\theta=0}=[1+2(0)](-0)+(1)[2(1)]

\frac{dx}{d\theta}\Big|_{\theta=0} = 2

\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \boxed{\frac{1}{2}}