Answer
$$\eqalign{
& {\text{At }}t = \frac{\pi }{4},{\text{ the slope is }} - 1 \cr
& \frac{{{d^2}y}}{{d{x^2}}} < 0{\text{ we can conclude that the graph is concave downward}} \cr} $$
Work Step by Step
$$\eqalign{
& x = 10\cos \theta ,{\text{ }}y = 10\sin \theta ,{\text{ }}\theta = \frac{\pi }{4} \cr
& {\text{By theorem 10}}{\text{.7}}{\text{ the slope is}} \cr
& \frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}},{\text{ }}dx/dt \ne 0,{\text{ for this exercise consider }}t = \theta \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{d}{{d\theta }}\left[ {10\sin \theta } \right]}}{{\frac{d}{{d\theta }}\left[ {10\cos \theta } \right]}} \cr
& \frac{{dy}}{{dx}} = \frac{{10\cos \theta }}{{ - 10\sin \theta }} \cr
& \frac{{dy}}{{dx}} = - \cot \theta \cr
& {\text{Evaluate at }}\theta = \frac{\pi }{4} \cr
& {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \pi /4}} = - \cot \left( {\frac{\pi }{4}} \right) \cr
& {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \pi /4}} = - 1 \cr
& {\text{Calculate the second derivative }}\frac{{{d^2}y}}{{d{x^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{d\theta }}\left[ {\frac{{dy}}{{dx}}} \right]}}{{dx/d\theta }} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{d\theta }}\left[ { - \cot \theta } \right]}}{{ - 10\sin \theta }} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{{{{\csc }^2}\theta }}{{ - 10\sin \theta }} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = - \frac{1}{{10}}{\csc ^3}\theta \cr
& {\text{Evaluate at }}\theta = \frac{\pi }{4} \cr
& {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\theta = \frac{\pi }{4}}} = - \frac{1}{{10}}{\csc ^3}\left( {\frac{\pi }{4}} \right) \cr
& {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\theta = \frac{\pi }{4}}} = - \frac{{\sqrt 2 }}{5} \cr
& {\text{At }}t = \frac{\pi }{4},{\text{ the slope is }} - 1 \cr
& \frac{{{d^2}y}}{{d{x^2}}} < 0{\text{ we can conclude that the graph is concave downward}} \cr} $$
