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