Chapter 13 - Functions of Several Variables - 13.7 Exercises - Page 933: 10

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

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