Chapter 10 - Conics, Parametric Equations, and Polar Coordinates - 10.3 Exercises - Page 711: 10

$$\eqalign{ & {\text{Slope: }}\infty \cr & {\text{Vertical tangent line at}}\,{\text{ }}\left( {1,0} \right) \cr} $$

$$\eqalign{ & x = \cos \theta ,{\text{ }}y = 3\sin \theta ,{\text{ }}\theta = 0 \cr & {\text{By theorem 10}}{\text{.7}}{\text{. The slope is}} \cr & \frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}},{\text{ }}dx/dt \ne 0,{\text{ for this exercise consider }}t = \theta \cr & \frac{{dy}}{{dx}} = \frac{{\frac{d}{{d\theta }}\left[ {3\sin \theta } \right]}}{{\frac{d}{{d\theta }}\left[ {\cos \theta } \right]}} \cr & \frac{{dy}}{{dx}} = \frac{{3\cos \theta }}{{ - \sin \theta }} \cr & \frac{{dy}}{{dx}} = - 3\cot \theta \cr & {\text{Evaluate at }}\theta = 0 \cr & m = \frac{{dy}}{{dx}} = - 3\cot \left( 0 \right) \cr & m = - \infty \cr & m = {\text{undefined}} \cr & {\text{Vertical line}} \cr & {\text{Calculate the second derivative }}\frac{{{d^2}y}}{{d{x^2}}} \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{d\theta }}\left[ {\frac{{dy}}{{dx}}} \right]}}{{dx/d\theta }} \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{d\theta }}\left[ { - 3\cot \theta } \right]}}{{ - \sin \theta }} \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{{3\csc \theta \cot \theta }}{{ - \sin \theta }} \cr & \frac{{{d^2}y}}{{d{x^2}}} = \frac{{3{{\csc }^2}\theta }}{{ - \sin \theta }} \cr & \frac{{{d^2}y}}{{d{x^2}}} = - 3{\csc ^3}\theta \cr & {\text{Evaluate at }}\theta = 0 \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\theta = 0}} = - 3{\csc ^3}\left( 0 \right) \cr & {\left. {\frac{{{d^2}y}}{{d{x^2}}}} \right|_{\theta = 0}} = - \infty \cr & {\text{Vertical line}} \cr & {\text{Then,}} \cr & {\text{Vertical tangent at }}\left( {\cos 0,3\sin 0} \right) = \left( {1,0} \right) \cr & {\text{Slope: }}\infty \cr & {\text{Vertical tangent line at}}\,{\text{ }}\left( {1,0} \right) \cr} $$