Chapter 13 - Functions of Several Variables - Review Exercises - Page 961: 56

$$3y + 4z - 25 = 0$$

$$\eqalign{ & f\left( {x,y} \right) = \sqrt {25 - {y^2}} ,{\text{ }}\left( {2,3,4} \right) \cr & z = \sqrt {25 - {y^2}} \cr & z - \sqrt {25 - {y^2}} = 0 \cr & {\text{Considering }} \cr & F\left( {x,y,z} \right) = z - \sqrt {25 - {y^2}} \cr & {\text{Calculate the partial derivatives }} \cr & {F_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {z - \sqrt {25 - {y^2}} } \right] = 0 \cr & {F_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {z - \sqrt {25 - {y^2}} } \right] = - \frac{{ - 2y}}{{2\sqrt {25 - {y^2}} }} = \frac{y}{{\sqrt {25 - {y^2}} }} \cr & {F_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {z - \sqrt {25 - {y^2}} } \right] = 1 \cr & {\text{At the point }}\left( {2,3,4} \right){\text{ the partial derivatives are}} \cr & {F_x}\left( {2,3,4} \right) = 0 \cr & {F_y}\left( {2,3,4} \right) = \frac{3}{{\sqrt {25 - {{\left( 3 \right)}^2}} }} = \frac{3}{4} \cr & {F_z}\left( {2,3,4} \right) = 1 \cr & {\text{An equation of the tangent plane at }}\left( {{x_0},{y_0},{z_0}} \right){\text{ is}} \cr & {F_x}\left( {{x_0},{y_0},{z_0}} \right)\left( {x - {x_0}} \right) + {F_y}\left( {{x_0},{y_0},{z_0}} \right)\left( {y - {y_0}} \right) \cr & + {F_z}\left( {{x_0},{y_0},{z_0}} \right)\left( {z - {z_0}} \right) = 0 \cr & {\text{Substituting}} \cr & {F_x}\left( {2,3,4} \right)\left( {x - 2} \right) + {F_y}\left( {2,3,4} \right)\left( {y - 3} \right) \cr & + {F_z}\left( {2,3,4} \right)\left( {z - 4} \right) = 0 \cr & 0\left( {x - 2} \right) + \frac{3}{4}\left( {y - 3} \right) + \left( {z - 4} \right) = 0 \cr & {\text{Simplifying}} \cr & \frac{3}{4}y - \frac{9}{4} + z - 4 = 0 \cr & 3y - 9 + 4z - 16 = 0 \cr & 3y + 4z - 25 = 0 \cr} $$