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: 51

$$\eqalign{ & \left( {\text{a}} \right){\text{ }}y = - \frac{1}{{{e^2}}}x + \frac{2}{e} \cr & \left( {\text{b}} \right){\text{ }}y = - \frac{1}{{{e^2}}}x + \frac{2}{e} \cr} $$

$$\eqalign{ & x = {e^t},{\text{ }}y = {e^{ - t}},{\text{ }}t = 1 \cr & {\text{Evaluate at }}t = 1 \cr & x = {e^1} = e \cr & y = {e^{ - 1}} = \frac{1}{e} \cr & {\text{We obtain the point }}\left( {e,\frac{1}{e}} \right) \cr & \cr & {\text{Calculate the derivatives }}\frac{{dy}}{{dt}}{\text{ and }}\frac{{dx}}{{dt}} \cr & \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {{e^t}} \right] = {e^t} \cr & \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {{e^{ - t}}} \right] = - {e^{ - t}} \cr & {\text{From the equation }}\frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}}{\text{ we have,}} \cr & \frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}} = \frac{{ - {e^{ - t}}}}{{{e^t}}} \cr & \frac{{dy}}{{dx}} = - {e^{ - 2t}} \cr & {\text{Evaluate at }}t = 1 \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{t = 0}} = - {e^{ - 2\left( 1 \right)}} \cr & m = - {e^{ - 2}} \cr & \cr & {\text{The equation of the tangent line is at }}\left( {e,\frac{1}{e}} \right){\text{ is}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - \frac{1}{e} = - {e^{ - 2}}\left( {x - e} \right) \cr & y - \frac{1}{e} = - \frac{1}{{{e^2}}}\left( {x - e} \right) \cr & y - \frac{1}{e} = - \frac{1}{{{e^2}}}x + \frac{1}{e} \cr & y = - \frac{1}{{{e^2}}}x + \frac{2}{e} \cr & \cr & \left( {\text{b}} \right){\text{By eliminating the parameter}} \cr & y = {e^{ - t}} \cr & y = \frac{1}{{{e^t}}} \cr & {\text{Substitute }}x{\text{ for }}{e^t} \cr & y = \frac{1}{x} \cr & {\text{Differentiate}} \cr & \frac{{dy}}{{dx}} = - \frac{1}{{{x^2}}} \cr & {\text{Calculate the slope at the point }}\left( {e,\frac{1}{e}} \right){\text{ }} \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{x = e}} = - \frac{1}{{{x^2}}} \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{x = e}} = - \frac{1}{{{e^2}}} \cr & {\text{The equation of the tangent line is at }}\left( {e,\frac{1}{e}} \right){\text{ is}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - \frac{1}{e} = - {e^{ - 2}}\left( {x - e} \right) \cr & y - \frac{1}{e} = - \frac{1}{{{e^2}}}\left( {x - e} \right) \cr & y - \frac{1}{e} = - \frac{1}{{{e^2}}}x + \frac{1}{e} \cr & y = - \frac{1}{{{e^2}}}x + \frac{2}{e} \cr} $$