Chapter 13 - Functions of Several Variables - Review Exercises - Page 961: 54

$$3y - 8z + 25 = 0$$

$$\eqalign{ & 9{x^2} + {y^2} + 4{z^2} = 25,{\text{ }}\left( {0, - 3,2} \right) \cr & 9{x^2} + {y^2} + 4{z^2} - 25 = 0 \cr & {\text{Considering }} \cr & F\left( {x,y,z} \right) = 9{x^2} + {y^2} + 4{z^2} - 25 \cr & {\text{Calculate the partial derivatives }} \cr & {F_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {9{x^2} + {y^2} + 4{z^2} - 25} \right] = 18x \cr & {F_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {9{x^2} + {y^2} + 4{z^2} - 25} \right] = 2y \cr & {F_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {9{x^2} + {y^2} + 4{z^2} - 25} \right] = 8z \cr & {\text{At the point }}\left( {0, - 3,2} \right){\text{ the partial derivatives are}} \cr & {F_x}\left( {0, - 3,2} \right) = 18\left( 0 \right) = 0 \cr & {F_y}\left( {0, - 3,2} \right) = 2\left( { - 3} \right) = - 6 \cr & {F_z}\left( {0, - 3,2} \right) = 8\left( 2 \right) = 16 \cr & {\text{An equation of the tangent plane at }}\left( {{x_0},{y_0},{z_0}} \right){\text{ is}} \cr & {F_x}\left( {{x_0},{y_0},{z_0}} \right)\left( {x - {x_0}} \right) + {F_y}\left( {{x_0},{y_0},{z_0}} \right)\left( {y - {y_0}} \right) \cr & + {F_z}\left( {{x_0},{y_0},{z_0}} \right)\left( {z - {z_0}} \right) = 0 \cr & {\text{Substituting}} \cr & {F_x}\left( {0, - 3,2} \right)\left( {x - 0} \right) + {F_y}\left( {0, - 3,2} \right)\left( {y + 3} \right) + {F_z}\left( {0, - 3,2} \right)\left( {z - 2} \right) = 0 \cr & 0\left( {x - 0} \right) - 6\left( {y + 3} \right) + 16\left( {z - 2} \right) = 0 \cr & {\text{Simplifying}} \cr & - 6y - 18 + 16z - 32 = 0 \cr & - 6y + 16z - 50 = 0 \cr & {\text{Divide by 2}} \cr & - 3y + 8z - 25 = 0 \cr & 3y - 8z + 25 = 0 \cr} $$