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: 48

$$0,1$$

$$\eqalign{ & {\text{Let }}x = \sinh t,{\text{ }}y = \cosh t,{\text{ }}t = 0 \cr & {\text{Calculate the derivatives }}\frac{{dy}}{{dt}}{\text{ and }}\frac{{dx}}{{dt}} \cr & \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {\cosh t} \right] = \sinh t \cr & \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {\sinh t} \right] = \cosh t \cr & {\text{From the equation }}\frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}}{\text{ we have,}} \cr & \frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}} = \frac{{\sinh t}}{{\cosh t}} \cr & \frac{{dy}}{{dx}} = \tanh t \cr & {\text{Evaluate at }}t = 0 \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{t = 0}} = \tanh \left( 0 \right) \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{t = 0}} = 0 \cr & \cr & {\text{Where }}\frac{{{d^2}y}}{{d{x^2}}} = \frac{{dy'/dt}}{{dx/dt}},{\text{ and }}y' = \tanh t \cr & \frac{{dy'}}{{dt}} = \frac{d}{{dt}}\left[ {\tanh t} \right] \cr & \frac{{dy'}}{{dt}} = {\text{sec}}{{\text{h}}^2}t \cr & {\text{Therefore}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{{{\text{sec}}{{\text{h}}^2}t}}{{\cosh t}} = {\left( {\frac{1}{{\cosh t}}} \right)^2}\left( {\frac{1}{{\cosh t}}} \right) = \frac{1}{{{{\cosh }^3}t}} \cr & {\text{Evaluate at }}t = 0 \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{t = 0}} = \frac{1}{{{{\cosh }^3}0}} \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{t = 0}} = \frac{1}{{{1^3}}} \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{t = 0}} = 1 \cr & \cr & 0,1 \cr} $$