Answer
$$\eqalign{
& {\text{At }}t = \frac{\pi }{3},{\text{ the slope is }} - 4\sqrt 3 \cr
& \frac{{{d^2}y}}{{d{x^2}}} > 0{\text{ we can conclude that the graph is concave upward}} \cr} $$
Work Step by Step
$$\eqalign{
& x = {\cos ^3}\theta ,{\text{ }}y = 4{\sin ^3}\theta ,{\text{ }}\theta = \frac{\pi }{3} \cr
& {\text{By theorem 10}}{\text{.7}}{\text{ the slope is}} \cr
& \frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}},{\text{ }}dx/dt \ne 0,{\text{ for this exercise consider }}t = \theta \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{d}{{d\theta }}\left[ {4{{\sin }^3}\theta } \right]}}{{\frac{d}{{d\theta }}\left[ {{{\cos }^3}\theta } \right]}} \cr
& \frac{{dy}}{{dx}} = \frac{{12{{\sin }^2}\theta \cos \theta }}{{ - 3{{\cos }^2}\theta \sin \theta }} \cr
& \frac{{dy}}{{dx}} = - 4\tan \theta \cr
& {\text{Evaluate at }}\theta = \frac{\pi }{3} \cr
& {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \pi /3}} = - 4\tan \left( {\frac{\pi }{3}} \right) \cr
& {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \pi /3}} = - 4\sqrt 3 \cr
& {\text{Calculate the second derivative }}\frac{{{d^2}y}}{{d{x^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{d\theta }}\left[ {\frac{{dy}}{{dx}}} \right]}}{{dx/d\theta }} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{d\theta }}\left[ { - 4\tan \theta } \right]}}{{ - 3{{\cos }^2}\theta \sin \theta }} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - 4{{\sec }^2}\theta }}{{ - 3{{\cos }^2}\theta \sin \theta }} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{4}{3}{\sec ^4}\theta \csc \theta \cr
& {\text{Evaluate at }}\theta = \frac{\pi }{3} \cr
& {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\theta = \frac{\pi }{3}}} = \frac{4}{3}{\sec ^4}\left( {\frac{\pi }{3}} \right)\csc \left( {\frac{\pi }{3}} \right) \cr
& {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\theta = \frac{\pi }{3}}} = \frac{4}{3}\left( {16} \right)\left( {\frac{{2\sqrt 3 }}{3}} \right) = \frac{{128\sqrt 3 }}{9} \cr
& {\text{At }}t = \frac{\pi }{3},{\text{ the slope is }} - 4\sqrt 3 \cr
& \frac{{{d^2}y}}{{d{x^2}}} > 0{\text{ we can conclude that the graph is concave upward}} \cr} $$
