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: 45

$$4,{\text{ 4}}$$

$$\eqalign{ & {\text{Let }}x = \sqrt t ,{\text{ }}y = 2t + 4,{\text{ }}t = 1 \cr & {\text{From the equation }}\frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}}{\text{ we have,}} \cr & \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {2t + 4} \right] = 2 \cr & \frac{{{d^2}y}}{{d{t^2}}} = 0 \cr & \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {\sqrt t } \right] = \frac{1}{{2\sqrt t }} \cr & \frac{{{d^2}y}}{{d{x^2}}} = - \frac{1}{{4{t^{3/2}}}} \cr & \cr & \frac{{dy}}{{dx}} = \frac{2}{{\frac{1}{{2\sqrt t }}}} \cr & \frac{{dy}}{{dx}} = 4\sqrt t \cr & {\text{Evaluate at }}t = 1 \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{t = 1}} = 4\sqrt 1 \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{t = 1}} = 4 \cr & \cr & {\text{Where }}\frac{{{d^2}y}}{{d{x^2}}} = \frac{{dy'/dt}}{{dx/dt}},{\text{ and }}y' = 4\sqrt t \cr & \frac{{dy'}}{{dt}} = \frac{d}{{dt}}\left[ {4\sqrt t } \right] \cr & \frac{{dy'}}{{dt}} = \frac{2}{{\sqrt t }} \cr & {\text{Therefore}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{2}{{\sqrt t }}}}{{\frac{1}{{2\sqrt t }}}} = 4 \cr & {\text{Evaluate at }}t = 1 \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{t = 1}} = 4 \cr & \cr & 4,{\text{ 4}} \cr} $$