Chapter 10 - Parametric And Polar Curves; Conic Sections - 10.1 Parametric Equations; Tangent Lines And Arc Length For Parametric Curves - Exercises Set 10.1 - Page 703: 46

$$\frac{3}{2},{\text{ }}\frac{5}{8}$$

$$\eqalign{ & {\text{Let }}x = \frac{1}{2}{t^2} + 1,{\text{ }}y = \frac{1}{3}{t^3} - t,{\text{ }}t = 2 \cr & {\text{From the equation }}\frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}}{\text{ we have,}} \cr & \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {\frac{1}{3}{t^3} - t} \right] = {t^2} - 1 \cr & \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {\frac{1}{2}{t^2} + 1} \right] = t \cr & \cr & \frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}} = \frac{{{t^2} - 1}}{t} \cr & {\text{Evaluate at }}t = 2 \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{t = 2}} = \frac{{{{\left( 2 \right)}^2} - 1}}{2} \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{t = 2}} = \frac{3}{2} \cr & \cr & {\text{Where }}\frac{{{d^2}y}}{{d{x^2}}} = \frac{{dy'/dt}}{{dx/dt}},{\text{ and }}y' = \frac{{{t^2} - 1}}{t} = t - \frac{1}{t} \cr & \frac{{dy'}}{{dt}} = \frac{d}{{dt}}\left[ {t - \frac{1}{t}} \right] \cr & \frac{{dy'}}{{dt}} = 1 + \frac{1}{{{t^2}}} \cr & {\text{Therefore}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{{1 + \frac{1}{{{t^2}}}}}{t} = \frac{1}{t} + \frac{1}{{{t^3}}} \cr & {\text{Evaluate at }}t = 2 \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{t = 2}} = \frac{1}{2} + \frac{1}{{{{\left( 2 \right)}^3}}} \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{t = 2}} = \frac{5}{8} \cr} $$