Chapter 10 - Conics, Parametric Equations, and Polar Coordinates - Review Exercises - Page 743: 43

$$\eqalign{ & {\text{At }}t = \frac{\pi }{6},{\text{ the slope is }} - 4\sqrt 3 \cr & \frac{{{d^2}y}}{{d{x^2}}} < 0{\text{ we can conclude that the graph is concave downward}} \cr} $$

$$\eqalign{ & x = 5 + \cos \theta ,{\text{ }}y = 3 + 4\sin \theta ,{\text{ }}\theta = \frac{\pi }{6} \cr & {\text{By theorem 10}}{\text{.7}}{\text{ the slope is}} \cr & \frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}},{\text{ }}dx/dt \ne 0,{\text{ for this exercise consider }}t = \theta \cr & \frac{{dy}}{{dx}} = \frac{{\frac{d}{{d\theta }}\left[ {3 + 4\sin \theta } \right]}}{{\frac{d}{{d\theta }}\left[ {5 + \cos \theta } \right]}} \cr & \frac{{dy}}{{dx}} = \frac{{4\cos \theta }}{{ - \sin \theta }} \cr & \frac{{dy}}{{dx}} = - 4\cot \theta \cr & {\text{Evaluate at }}\theta = \frac{\pi }{6} \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \pi /6}} = - 4\cot \left( {\frac{\pi }{6}} \right) \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \pi /6}} = - 4\sqrt 3 \cr & {\text{Calculate the second derivative }}\frac{{{d^2}y}}{{d{x^2}}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{d\theta }}\left[ {\frac{{dy}}{{dx}}} \right]}}{{dx/d\theta }} \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{d\theta }}\left[ { - 4\cot \theta } \right]}}{{ - \sin \theta }} \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{{4{{\csc }^2}\theta }}{{ - \sin \theta }} \cr & \frac{{{d^2}y}}{{d{x^2}}} = - 4{\csc ^3}\theta \cr & {\text{Evaluate at }}\theta = \frac{\pi }{6} \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\theta = \frac{\pi }{6}}} = - 4{\csc ^3}\left( {\frac{\pi }{6}} \right) \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\theta = \frac{\pi }{6}}} = - 32 \cr & {\text{At }}t = \frac{\pi }{6},{\text{ the slope is }} - 4\sqrt 3 \cr & \frac{{{d^2}y}}{{d{x^2}}} < 0{\text{ we can conclude that the graph is concave downward}} \cr} $$