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: 47

$$\frac{{2\sqrt 3 }}{3},{\text{ }} - \frac{{\sqrt 3 }}{9}$$

$$\eqalign{ & {\text{Let }}x = \sec t,{\text{ }}y = \tan t,{\text{ }}t = \pi /3 \cr & {\text{From the equation }}\frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}}{\text{ we have,}} \cr & \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {\tan t} \right] = {\sec ^2}t \cr & \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {\sec t} \right] = \sec t\tan t \cr & \cr & \frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}} = \frac{{{{\sec }^2}t}}{{\sec t\tan t}} \cr & \frac{{dy}}{{dx}} = \frac{{\sec t}}{{\tan t}} \cr & \frac{{dy}}{{dx}} = \csc t \cr & {\text{Evaluate at }}t = \pi /3 \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{t = \pi /3}} = \csc \left( {\frac{\pi }{3}} \right) \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{t = \pi /3}} = \frac{{2\sqrt 3 }}{3} \cr & \cr & {\text{Where }}\frac{{{d^2}y}}{{d{x^2}}} = \frac{{dy'/dt}}{{dx/dt}},{\text{ and }}y' = \frac{{{{\sec }^2}t}}{{\sec t}} = \csc t \cr & \frac{{dy'}}{{dt}} = \frac{d}{{dt}}\left[ {\csc t} \right] \cr & \frac{{dy'}}{{dt}} = - \csc t\cot t \cr & {\text{Therefore}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - \csc t\cot t}}{{\sec t\tan t}} = - \frac{{{{\cos }^3}t}}{{{{\sin }^3}t}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = - {\cot ^3}t \cr & {\text{Evaluate at }}t = \pi /3 \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{t = \pi /3}} = - {\cot ^3}\left( {\frac{\pi }{3}} \right) \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{t = \pi /3}} = {\left( { - \frac{{\sqrt 3 }}{3}} \right)^3} \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{t = \pi /3}} = - \frac{{\sqrt 3 }}{9} \cr} $$