Chapter 10 - Conics, Parametric Equations, and Polar Coordinates - 10.3 Exercises - Page 711: 18

$$\eqalign{ & {\text{At }}\left( {2,0} \right):{\text{ }}x = 2 \cr & {\text{At }}\left( {3, - 2} \right):{\text{ }}y = - x + 1 \cr & {\text{At }}\left( {18,10} \right):{\text{ }}y = \frac{{13}}{{32}}x + \frac{{43}}{{16}} \cr} $$

$$\eqalign{ & x = {t^4} + 2,{\text{ }}y = {t^3} + t \cr & {\text{By theorem 10}}{\text{.7}}{\text{ the slope is}} \cr & \frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}},{\text{ }}dx/dt \ne 0,{\text{ for this exercise consider }}t = \theta \cr & \frac{{dy}}{{dx}} = \frac{{\frac{d}{{dt}}\left[ {{t^3} + t} \right]}}{{\frac{d}{{dt}}\left[ {{t^4} + 2} \right]}} \cr & \frac{{dy}}{{dx}} = \frac{{3{t^2} + 1}}{{4{t^3}}} \cr & \cr & {\text{*For }}\left( {2,0} \right) \to t = 0 \cr & {\text{Calculate the tangent line at the point }}\left( {2,0} \right) \cr & {\text{Evaluate at }}\frac{{dy}}{{dt}}{\text{ }}t = 0 \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{t = 0}} = \frac{{3{{\left( 0 \right)}^2} + 1}}{{4{{\left( 0 \right)}^3}}} = {\text{ Undefined}} \cr & {\text{Vertical line: }}x = {x_0} \cr & x = 2 \cr & \cr & {\text{*For }}\left( {3, - 2} \right) \to t = - 1 \cr & {\text{Evaluate }}\left. {\frac{{dy}}{{dx}}} \right|{\text{ at }}t = - 1 \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{t = - 1}} = \frac{{3{{\left( { - 1} \right)}^2} + 1}}{{4{{\left( { - 1} \right)}^3}}} \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{t = - 1}} = - 1 \cr & {\text{Calculate the tangent line at the point }}\left( {3, - 2} \right) \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y + 2 = - \left( {x - 3} \right) \cr & y + 2 = - x + 3 \cr & y = - x + 1 \cr & \cr & {\text{*For }}\left( {18,10} \right) \to t = 2 \cr & {\text{Calculate the tangent line at the point }}\left( {18,10} \right) \cr & {\text{Evaluate }}\frac{{dy}}{{dx}}{\text{ at }}t = 2 \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{t = 2}} = \frac{{3{{\left( 2 \right)}^2} + 1}}{{4{{\left( 2 \right)}^3}}} \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{t = 2}} = \frac{{13}}{{32}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 10 = \frac{{13}}{{32}}\left( {x - 18} \right) \cr & y - 10 = \frac{{13}}{{32}}x - \frac{{117}}{{16}} \cr & y = \frac{{13}}{{32}}x + \frac{{43}}{{16}} \cr} $$