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: 52

$$\eqalign{ & \left( {\text{a}} \right){\text{ }}y = 7x - 32 \cr & \left( {\text{b}} \right){\text{ }}y = 7x - 32 \cr} $$

$$\eqalign{ & x = 2t + 4,{\text{ }}y = 8{t^2} - 2t + 4,{\text{ }}t = 1 \cr & {\text{Evaluate at }}t = 1 \cr & x = 2\left( 1 \right) + 4 = 6 \cr & y = 8{\left( 1 \right)^2} - 2\left( 1 \right) + 4 = 10 \cr & {\text{We obtain the point }}\left( {6,10} \right) \cr & {\text{Calculate the derivatives }}\frac{{dy}}{{dt}}{\text{ and }}\frac{{dx}}{{dt}} \cr & \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {2t + 4} \right] = 2 \cr & \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {8{t^2} - 2t + 4} \right] = 16t - 2 \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{{16t - 2}}{2} \cr & \frac{{dy}}{{dx}} = 8t - 1 \cr & {\text{Evaluate at }}t = 1 \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{t = 1}} = 8\left( 1 \right) - 1 \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{t = 1}} = 7 \cr & \cr & {\text{The equation of the tangent line is at }}\left( {6,10} \right){\text{ is}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 10 = 7\left( {x - 6} \right) \cr & y - 10 = 7x - 42 \cr & y = 7x - 32 \cr & \cr & \left( {\text{b}} \right){\text{By eliminating the parameter}} \cr & {\text{ }}y = 8{t^2} - 2t + 4 \cr & x = 2t + 4 \Rightarrow t = \frac{{x - 4}}{2} \cr & {\text{Substitute }}\frac{{x - 4}}{2}{\text{ for }}t{\text{ into }}y = 8{t^2} - 2t + 4 \cr & {\text{ }}y = 8{\left( {\frac{{x - 4}}{2}} \right)^2} - 2\left( {\frac{{x - 4}}{2}} \right) + 4 \cr & {\text{ }}y = 2\left( {{x^2} - 8x + 16} \right) - x + 4 + 4 \cr & {\text{ }}y = 2{x^2} - 16x + 32 - x + 8 \cr & {\text{ }}y = 2{x^2} - 17x + 40 \cr & {\text{Differentiate}} \cr & \frac{{dy}}{{dx}} = 4x - 17 \cr & {\text{Calculate the slope at the point }}\left( {6,10} \right){\text{ }} \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{x = 6}} = 4\left( 6 \right) - 17 \cr & {\left. {\frac{{dy}}{{dx}}} \right|_{x = 6}} = 7 \cr & {\text{The equation of the tangent line is at }}\left( {6,10} \right){\text{ is}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 10 = 7\left( {x - 6} \right) \cr & y - 10 = 7x - 42 \cr & y = 7x - 32 \cr} $$