Chapter 12 - Functions of Several Veriables - 12.7 Tangent Planes and Linear Approximation - 12.7 Exercises - Page 936: 14

$$\eqalign{ & 32x + 4y + 2z - 16 = 0 \cr & - 8x - y - 8z -16= 0 \cr} $$

$$\eqalign{ & yz{e^{xz}} - 8 = 0,{\text{ }}\left( {0,2,4} \right){\text{ and }}\left( {0, - 8, - 1} \right) \cr & {\text{Let }}F\left( {x,y,z} \right) = yz{e^{xz}} - 8 \cr & {\text{Calculate the partial derivatives }}{F_x}\left( {x,y,z} \right),{\text{ }}{F_y}\left( {x,y,z} \right) \cr & {\text{and }}{F_z}\left( {x,y,z} \right) \cr & {F_x}\left( {x,y,z} \right) = yz\left( {z{e^{xz}}} \right) = y{z^2}{e^{xz}} \cr & {F_y}\left( {x,y,z} \right) = z{e^{xz}} \cr & {F_z}\left( {x,y,z} \right) = y{e^{xz}} \cr & {\text{Evaluate the point at }}\left( {0,2,4} \right){\text{ and }}\left( {0, - 8, - 1} \right) \cr & {F_x}\left( {0,2,4} \right) = 32{\text{ ;}}{F_x}\left( {0, - 8, - 1} \right) = - 8 \cr & {F_y}\left( {0,2,4} \right) = 4{\text{ ;}}{F_y}\left( {0, - 8, - 1} \right) = - 1 \cr & {F_z}\left( {0,2,4} \right) = 2{\text{ ;}}{F_z}\left( {0, - 8, - 1} \right) = - 8 \cr & \cr & {\text{An equation of the tangent plane to the surface is }} \cr & {F_x}\left( {a,b,c} \right)\left( {x - a} \right) + {F_y}\left( {a,c,b} \right)\left( {y - b} \right) + {F_z}\left( {a,b,c} \right)\left( {z - c} \right) = 0 \cr & {\text{At the point }}\left( {0,2,4} \right){\text{ we obtain}} \cr & 32\left( {x - 0} \right) + 4\left( {y - 2} \right) + 2\left( {z - 4} \right) = 0 \cr & 32x + 4y - 8 + 2z - 8 = 0 \cr & 32x + 4y + 2z - 16 = 0 \cr & \cr & {\text{At the point }}\left( {0, - 8, - 1} \right){\text{ we obtain}} \cr & - 8\left( {x - 0} \right) - 1\left( {y +8} \right) - 8\left( {z + 1} \right) = 0 \cr & - 8x - y - 8 - 8z - 8 = 0 \cr & - 8x - y - 8z -16= 0 \cr} $$