Chapter 13 - Functions of Several Variables - Review Exercises - Page 961: 53

$$2x + 6y - z = 8$$

$$\eqalign{ & z = {x^2} + {y^2} + 2,{\text{ }}\left( {1,3,12} \right) \cr & z - {x^2} - {y^2} - 2 = 0 \cr & {\text{Considering }} \cr & F\left( {x,y,z} \right) = z - {x^2} - {y^2} - 2 \cr & {\text{Calculate the partial derivatives }} \cr & {F_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {z - {x^2} - {y^2} - 2} \right] = - 2x \cr & {F_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {z - {x^2} - {y^2} - 2} \right] = - 2y \cr & {F_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {z - {x^2} - {y^2} - 2} \right] = 1 \cr & {\text{At the point }}\left( {1,3,12} \right){\text{ the partial derivatives are}} \cr & {F_x}\left( {1,3,12} \right) = - 2\left( 1 \right) = - 2 \cr & {F_y}\left( {1,3,12} \right) = - 2\left( 3 \right) = - 6 \cr & {F_z}\left( {1,3,12} \right) = 1 \cr & {\text{An equation of the tangent plane at }}\left( {{x_0},{y_0},{z_0}} \right){\text{ is}} \cr & {F_x}\left( {{x_0},{y_0},{z_0}} \right)\left( {x - {x_0}} \right) + {F_y}\left( {{x_0},{y_0},{z_0}} \right)\left( {y - {y_0}} \right) \cr & + {F_z}\left( {{x_0},{y_0},{z_0}} \right)\left( {z - {z_0}} \right) = 0 \cr & {\text{Substituting}} \cr & {F_x}\left( {1,3,12} \right)\left( {x - 1} \right) + {F_y}\left( {1,3,12} \right)\left( {y - 3} \right) + {F_z}\left( {1,3,12} \right)\left( {z - 12} \right) = 0 \cr & - 2\left( {x - 1} \right) - 6\left( {y - 3} \right) + \left( {z - 12} \right) = 0 \cr & {\text{Simplifying}} \cr & - 2x + 2 - 6y + 18 + z - 12 = 0 \cr & - 2x - 6y + z + 8 = 0 \cr & 2x + 6y - z - 8 = 0 \cr & 2x + 6y - z = 8 \cr} $$