Chapter 10 - Conics, Parametric Equations, and Polar Coordinates - 10.3 Exercises - Page 711: 16

$$\eqalign{ & {\text{At }}\left( { - 1,3} \right):{\text{ }}x = - 1 \cr & {\text{At }}\left( {2,5} \right):{\text{ }}y = 5 \cr & {\text{At }}\left( {\frac{{4 + 3\sqrt 3 }}{2},2} \right):{\text{ }}2\sqrt 3 x - 3y - 4\sqrt 3 - 3 = 0 \cr} $$

$$\eqalign{ & x = 2 - 3\cos \theta ,{\text{ }}y = 3 + 2\sin \theta \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 + 2\sin \theta } \right]}}{{\frac{d}{{d\theta }}\left[ {2 - 3\cos \theta } \right]}} \cr & \frac{{dy}}{{dx}} = \frac{{2\cos \theta }}{{3\sin \theta }} \cr & \frac{{dy}}{{dx}} = \frac{2}{3}\cot \theta \cr & {\text{*For }}\left( { - 1,3} \right) \to \theta = 0 \cr & {\text{Calculate the tangent line at the point }}\left( { - 1,3} \right) \cr & {\text{Evaluate at }}\frac{{dy}}{{dx}}{\text{ }}\theta = 0 \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = 0}} = \frac{2}{3}\cot \left( 0 \right) = {\text{ Undefined}} \cr & {\text{Vertical line: }}x = {x_0} \cr & x = - 1 \cr & \cr & {\text{*For }}\left( {2,5} \right) \to \theta = \frac{\pi }{2} \cr & {\text{Evaluate at }}\theta = \frac{\pi }{2} \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \frac{\pi }{2}}} = \frac{2}{3}\cot \left( {\frac{\pi }{2}} \right) \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \frac{\pi }{2}}} = 0 \cr & {\text{Calculate the tangent line at the point }}\left( {2,5} \right) \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 5 = 0\left( {x - 2} \right) \cr & y = 5 \cr & \cr & {\text{*For }}\left( {\frac{{4 + 3\sqrt 3 }}{2},2} \right) \to \theta = \frac{{7\pi }}{6} \cr & {\text{Calculate the tangent line at the point }}\left( {\frac{{4 + 3\sqrt 3 }}{2},2} \right) \cr & {\text{Evaluate at }}\theta = \frac{{7\pi }}{6} \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \frac{{7\pi }}{6}}} = \frac{2}{3}\cot \left( {\frac{{7\pi }}{6}} \right) \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \frac{{7\pi }}{6}}} = \frac{{2\sqrt 3 }}{3} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 2 = \frac{{2\sqrt 3 }}{3}\left( {x - \frac{{4 + 3\sqrt 3 }}{2}} \right) \cr & y - 2 = \frac{{2\sqrt 3 }}{3}x - \frac{{\sqrt 3 \left( {4 + 3\sqrt 3 } \right)}}{3} \cr & 3y - 6 = 2\sqrt 3 x - 4\sqrt 3 - 9 \cr & 2\sqrt 3 x - 3y - 4\sqrt 3 - 3 = 0 \cr} $$