Let C be the curve in the plane defined by $r=2+sin\frac{1}{2} \theta$. Calculate the slope of the tangent of C when $\theta = 0$ and when $\theta=2 \pi$. The formula $\frac {dy}{dx} = \frac {\frac {dy}{d \theta}}{\frac {dx}{d \theta}}$ may be helpful.
Ok so using that formula i get 4 and -4 for the slopes when $\theta = 0$ and when $\theta=2 \pi$ respectively. It just seems quite easy so I wanted to check if I didn't overlook something. Can anyone confirm that?