Let C be the curve in the plane defined by $\displaystyle r=2+sin\frac{1}{2} \theta $. Calculate the slope of the tangent of C when $\displaystyle \theta = 0 $ and when $\displaystyle \theta=2 \pi $. The formula $\displaystyle \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 $\displaystyle \theta = 0 $ and when $\displaystyle \theta=2 \pi $ respectively. It just seems quite easy so I wanted to check if I didn't overlook something. Can anyone confirm that?