Chapter 13 - Functions of Several Variables - 13.7 Exercises - Page 933: 28

$$\eqalign{ & 8x + y - z = 0 \cr & {\text{Line: }}\frac{x}{{ - 8}} = \frac{{y - 2}}{{ - 1}} = \frac{{z - 2}}{1} \cr} $$

$$\eqalign{ & z = y{e^{2xy}},{\text{ }}\left( {0,2,2} \right) \cr & z - y{e^{2xy}} = 0 \cr & {\text{Consider}} \cr & F\left( {x,y,z} \right) = z - y{e^{2xy}} \cr & {\text{Calculating the partial derivatives}} \cr & {F_x}\left( {x,y,z} \right) = 0 - y\left( {2y} \right){e^{2xy}} = - 2{y^2}{e^{2xy}} \cr & {F_y}\left( {x,y,z} \right) = - {e^{2xy}} - y\left( {2x{e^{2xy}}} \right){\text{ = }} - {e^{2xy}} - 2xy{e^{2xy}} \cr & {F_z}\left( {x,y,z} \right) = 1,{\text{ }} \cr & {\text{At the point }}\left( {0,2,2} \right) \cr & {F_x}\left( {0,2,2} \right) = - 2{\left( 2 \right)^2}{e^{2\left( 0 \right)\left( 2 \right)}} = - 8{\text{ }} \cr & {F_y}\left( {0,2,2} \right) = - {e^{2\left( 0 \right)\left( 2 \right)}} - 2\left( 0 \right)\left( 2 \right){e^{2\left( 0 \right)\left( 2 \right)}} = - 1{\text{ }} \cr & {F_z}\left( {0,2,2} \right) = 1,{\text{ }} \cr & {\text{Direction numbers: }} - 8, - 1,1 \cr & {\text{An equation of the tangent plane at }}\left( {0,2,2} \right){\text{ is}} \cr & {F_x}\left( {0,2,2} \right)\left( {x - 3} \right) + {F_y}\left( {0,2,2} \right)\left( {y - 2} \right) + {F_z}\left( {0,2,2} \right)\left( {z - 5} \right) = 0 \cr & - 8\left( {x - 0} \right) - \left( {y - 2} \right) + \left( {z - 2} \right) = 0 \cr & - 8x - y + 2 + z - 2 = 0 \cr & - 8x - y + z = 0 \cr & 8x + y - z = 0 \cr & \cr & {\text{The corresponding symmetric equations are:}} \cr & {\text{Line: }}\frac{{x - 0}}{{ - 8}} = \frac{{y - 2}}{{ - 1}} = \frac{{z - 2}}{1} \cr & {\text{Line: }}\frac{x}{{ - 8}} = \frac{{y - 2}}{{ - 1}} = \frac{{z - 2}}{1} \cr} $$