Chapter 13 - Functions of Several Variables - 13.7 Exercises - Page 933: 30

$$\eqalign{ & \frac{2}{e}x + y + 4z = 8 \cr & {\text{Line: }}\frac{{x - e}}{{2/e}} = \frac{{y - 2}}{1} = \frac{{z - 1}}{4} \cr} $$

$$\eqalign{ & y\ln x{z^2} = 2,{\text{ }}\left( {e,2,1} \right) \cr & y\ln x{z^2} - 2 = 0 \cr & {\text{Consider}} \cr & F\left( {x,y,z} \right) = y\ln x{z^2} - 2 \cr & {\text{Calculating the partial derivatives}} \cr & {F_x}\left( {x,y,z} \right) = y\left( {\frac{{{z^2}}}{{x{z^2}}}} \right) - 0 = \frac{y}{x} \cr & {F_y}\left( {x,y,z} \right) = \left( 1 \right)\ln x{z^2} - 0 = \ln x{z^2} \cr & {F_z}\left( {x,y,z} \right) = y\left( {\frac{{2xz}}{{x{z^2}}}} \right) - 0 = \frac{{2y}}{z}{\text{ }} \cr & {\text{At the point }}\left( {e,2,1} \right) \cr & {F_x}\left( {e,2,1} \right) = \frac{y}{x} = \frac{2}{e} \cr & {F_y}\left( {e,2,1} \right) = \ln x{z^2} = \ln \left( e \right){\left( 1 \right)^2} = 1 \cr & {F_z}\left( {e,2,1} \right) = \frac{{2y}}{z}{\text{ = }}\frac{{2\left( 2 \right)}}{1}{\text{ = 4 }} \cr & {\text{Direction numbers: }}\frac{2}{e},1,4 \cr & {\text{An equation of the tangent plane at }}\left( {e,2,1} \right){\text{ is}} \cr & {F_x}\left( {e,2,1} \right)\left( {x - e} \right) + {F_y}\left( {e,2,1} \right)\left( {y - 2} \right) + {F_z}\left( {e,2,1} \right)\left( {z - 1} \right) = 0 \cr & \frac{2}{e}\left( {x - e} \right) + \left( {y - 2} \right) + 4\left( {z - 1} \right) = 0 \cr & \frac{2}{e}x - 2 + y - 2 + 4z - 4 = 0 \cr & \frac{2}{e}x + y + 4z = 8 \cr & \cr & {\text{The corresponding symmetric equations are:}} \cr & {\text{Line: }}\frac{{x - e}}{{2/e}} = \frac{{y - 2}}{1} = \frac{{z - 1}}{4} \cr} $$