Chapter 10 - Conics, Parametric Equations, and Polar Coordinates - 10.3 Exercises - Page 711: 25

$$\eqalign{ & {\text{At }}t = 0:{\text{ }}y = 1 \cr & {\text{At }}t = \pi :{\text{ }}y = 3x - 5 \cr} $$

$$\eqalign{ & x = {t^2} - t,{\text{ }}y = {t^3} - 3t - 1 \cr & {\text{The graph crosses itself at the point }}\left( {2,1} \right) \cr & {\text{At the point }}\left( {2,1} \right) \cr & {\text{For }}x = {t^2} - t \cr & 2 = {t^2} - t \cr & {t^2} - t - 2 = 0 \cr & \left( {t - 2} \right)\left( {t + 1} \right) \cr & t = 2,{\text{ }}t = - 1 \cr & {\text{For }}y = {t^3} - 3t - 1 \cr & 1 = {t^3} - 3t - 1 \cr & t = 2,{\text{ }}t = - 1 \cr & {\text{At the point }}\left( {2,1} \right){\text{ we have }}t = 2,{\text{ }}t = - 1 \cr & \cr & {\text{By theorem 10}}{\text{.7}}{\text{ the slope is}} \cr & \frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}},{\text{ }}dx/dt \ne 0,{\text{ for this exercise consider }}t = \theta \cr & \frac{{dy}}{{dx}} = \frac{{\frac{d}{{dt}}\left[ {{t^3} - 3t - 1} \right]}}{{\frac{d}{{dt}}\left[ {{t^2} - t} \right]}} \cr & \frac{{dy}}{{dx}} = \frac{{3{t^2} - 3}}{{2t - 1}} \cr & \cr & {\text{*For }}t = - 1 \cr & {\text{Calculate the tangent line at the point }}\left( {2,1} \right) \cr & {\text{Evaluate at }}\frac{{dy}}{{dt}}{\text{ }}t = - 1 \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{t = - 1}} = \frac{{3{{\left( { - 1} \right)}^2} - 3}}{{2\left( { - 1} \right)1}} = 0 \cr & y - 1 = 0\left( {x - 2} \right) \cr & y = 1 \cr & \cr & {\text{*For }}t = 2 \cr & {\text{Calculate the tangent line at the point }}\left( {2,1} \right) \cr & {\text{Evaluate at }}\frac{{dy}}{{dt}}{\text{ }}t = 2 \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{t = 2}} = \frac{{3{{\left( 2 \right)}^2} - 3}}{{2\left( 2 \right)1}} = 3 \cr & y - 1 = 3\left( {x - 2} \right) \cr & y = 3x - 6 + 1 \cr & y = 3x - 5 \cr & \cr & {\text{At }}t = 0:{\text{ }}y = 1 \cr & {\text{At }}t = \pi :{\text{ }}y = 3x - 5 \cr} $$