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: 50

$$3\sqrt 3 ,{\text{ }} - {\text{24}}$$

$$\eqalign{ & {\text{Let }}x = \cos \phi ,{\text{ }}y = 3\sin \phi ,{\text{ }}\phi = 5\pi /6 \cr & {\text{Calculate the derivatives }}\frac{{dy}}{{d\theta }}{\text{ and }}\frac{{dx}}{{d\theta }} \cr & \frac{{dy}}{{d\phi }} = \frac{d}{{d\phi }}\left[ {3\sin \phi } \right] = 3\cos \phi \cr & \frac{{dx}}{{d\phi }} = \frac{d}{{d\phi }}\left[ {\cos \phi } \right] = - \sin \phi \cr & {\text{From the equation }}\frac{{dy}}{{dx}} = \frac{{dy/d\phi }}{{dx/d\phi }}{\text{ we have,}} \cr & \frac{{dy}}{{dx}} = \frac{{dy/d\phi }}{{dx/d\phi }} = \frac{{3\cos \phi }}{{ - \sin \phi }} \cr & \frac{{dy}}{{dx}} = - 3\cot \phi \cr & {\text{Evaluate at }}\phi = \frac{{5\pi }}{6} \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{\phi = \frac{{5\pi }}{6}}} = - 3\cot \left( {\frac{{5\pi }}{6}} \right) \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{\phi = \frac{{5\pi }}{6}}} = 3\sqrt 3 \cr & \cr & {\text{Where }}\frac{{{d^2}y}}{{d{x^2}}} = \frac{{dy'/d\phi }}{{dx/d\phi }},{\text{ and }}y' = - 3\cot \phi \cr & \frac{{dy'}}{{d\phi }} = \frac{d}{{dt}}\left[ { - 3\cot \phi } \right] \cr & \frac{{dy'}}{{d\phi }} = - 3\left( { - {{\csc }^2}\phi } \right)\left( 1 \right) \cr & \frac{{dy'}}{{d\phi }} = 3{\csc ^2}\phi \cr & {\text{Therefore}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{{dy'/d\phi }}{{dx/d\phi }} = \frac{{3{{\csc }^2}\phi }}{{ - \sin \phi }} = - 3{\csc ^3}\phi \cr & {\text{Evaluate at }}\phi = \frac{{5\pi }}{6} \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\phi = \frac{{5\pi }}{6}}} = - 3{\csc ^3}\left( {\frac{{5\pi }}{6}} \right) \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\phi = \frac{{5\pi }}{6}}} = - 3{\left( 2 \right)^3} \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\phi = \frac{{5\pi }}{6}}} = - 24 \cr} $$