Chapter 12 - Functions of Several Veriables - 12.7 Tangent Planes and Linear Approximation - 12.7 Exercises - Page 936: 16

$$\eqalign{ & x + y - z = 0 \cr & x + y + 3z + 4 = 0 \cr & \cr} $$

$$\eqalign{ & 2x + {y^2} - {z^2} = 0;\,\,\,\left( {0,1,1} \right){\text{ and }}\left( {4,1, - 3} \right) \cr & {\text{Let }}F\left( {x,y,z} \right) = {x^2} + y + z - 3 \cr & {\text{Calculate the partial derivatives }}{F_x}\left( {x,y,z} \right){\text{, }}{F_y}\left( {x,y,z} \right){\text{ }} \cr & {\text{and }}{F_z}\left( {x,y,z} \right){\text{ }} \cr & {F_x}\left( {x,y,z} \right) = 2 \cr & {F_y}\left( {x,y,z} \right) = 2y \cr & {F_z}\left( {x,y,z} \right) = - 2z \cr & {\text{Evaluate at the point }}\left( {0,1,1} \right){\text{ and }}\left( {4,1, - 3} \right) \cr & {\text{at }}{F_x}\left( {x,y,z} \right),\,{F_y}\left( {x,y,z} \right){\text{ and }}{F_z}\left( {x,y,z} \right) \cr & {F_x}\left( {0,1,1} \right) = 2 \cr & {F_y}\left( {0,1,1} \right) = 2 \cr & {F_z}\left( {0,1,1} \right) = - 2 \cr & and \cr & {F_x}\left( {4,1, - 3} \right) = 2 \cr & {F_y}\left( {4,1, - 3} \right) = 2 \cr & {F_z}\left( {4,1, - 3} \right) = 6 \cr & \cr & {\text{An equation of the plane tangent to the surface is}} \cr & {F_x}\left( {a,b,c} \right)\left( {x - a} \right) + {F_y}\left( {a,b,c} \right)\left( {y - b} \right) + {F_z}\left( {a,b,c} \right)\left( {z - c} \right) = 0 \cr & {\text{For the point }}\left( {0,1,1} \right) \cr & 2\left( {x - 0} \right) + 2\left( {y - 1} \right) - 2\left( {z - 1} \right) = 0 \cr & 2x + 2y - 2 - 2z + 2 = 0 \cr & 2x + 2y - 2z = 0 \cr & x + y - z = 0 \cr & {\text{For the point }}\left( {4,1, - 3} \right) \cr & 2\left( {x - 4} \right) + 2\left( {y - 1} \right) + 6\left( {z + 3} \right) = 0 \cr & 2x - 8 + 2y - 2 + 6z + 18 = 0 \cr & 2x + 2y + 6z + 8 = 0 \cr & x + y + 3z + 4 = 0 \cr} $$