Chapter 10 - Conics, Parametric Equations, and Polar Coordinates - Review Exercises - Page 743: 42

$$\eqalign{ & {\text{At }}t = - 2,{\text{ the slope is }}16 \cr & \frac{{{d^2}y}}{{d{x^2}}} > 0{\text{ we can conclude that the graph is concave upward}} \cr} $$

$$\eqalign{ & x = \frac{1}{t},{\text{ }}y = {t^2},{\text{ }}t = - 2 \cr & {\text{By theorem 10}}{\text{.7}}{\text{ the slope is}} \cr & \frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}},{\text{ }}dx/dt \ne 0 \cr & \frac{{dy}}{{dx}} = \frac{{\frac{d}{{dt}}\left[ {{t^2}} \right]}}{{\frac{d}{{dt}}\left[ {\frac{1}{t}} \right]}} \cr & \frac{{dy}}{{dx}} = \frac{{2t}}{{ - 1/{t^2}}} \cr & \frac{{dy}}{{dx}} = - 2{t^3} \cr & {\text{Evaluate at }}t = - 2 \cr & \frac{{dy}}{{dx}} = - 2{\left( { - 2} \right)^3} \cr & \frac{{dy}}{{dx}} = 16 \cr & {\text{Calculate the second derivative }}\frac{{{d^2}y}}{{d{x^2}}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{dt}}\left[ {\frac{{dy}}{{dx}}} \right]}}{{dx/dt}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{dt}}\left[ { - 2{t^3}} \right]}}{{ - 1/{t^2}}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - 6{t^2}}}{{ - 1/{t^2}}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = 6{t^4} \cr & {\text{Evaluate at }}t = - 2 \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{t = - 2}} = 6{\left( { - 2} \right)^4} = 96 \cr & \cr & {\text{At }}t = - 2,{\text{ the slope is }}16 \cr & \frac{{{d^2}y}}{{d{x^2}}} > 0{\text{ we can conclude that the graph is concave upward}} \cr} $$