Chapter 13 - Functions of Several Variables - 13.7 Exercises - Page 933: 15

$$3x + 4y - 25z = 25\left( {1 - \ln 5} \right)$$

$$\eqalign{ & h\left( {x,y} \right) = \ln \sqrt {{x^2} + {y^2}} ,{\text{ }}\left( {3,4,\ln 5} \right) \cr & h\left( {x,y} \right) = \ln {\left( {{x^2} + {y^2}} \right)^{1/2}} \cr & z = \frac{1}{2}\ln \left( {{x^2} + {y^2}} \right) \cr & z - \frac{1}{2}\ln \left( {{x^2} + {y^2}} \right) = 0 \cr & {\text{Considering }} \cr & F\left( {x,y,z} \right) = z - \frac{1}{2}\ln \left( {{x^2} + {y^2}} \right) \cr & {\text{Calculate the partial derivatives }} \cr & {F_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {z - \frac{1}{2}\ln \left( {{x^2} + {y^2}} \right)} \right] = - \frac{x}{{{x^2} + {y^2}}} \cr & {F_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {z - \frac{1}{2}\ln \left( {{x^2} + {y^2}} \right)} \right] = - \frac{y}{{{x^2} + {y^2}}} \cr & {F_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {z - {{\left( {x - y} \right)}^2}} \right] = 1 \cr & {\text{At the point }}\left( {3,4,\ln 5} \right){\text{ the partial derivatives are}} \cr & {F_x}\left( {3,4,\ln 5} \right) = - \frac{3}{{{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2}}} = - \frac{3}{{25}} \cr & {F_y}\left( {3,4,\ln 5} \right) = - \frac{4}{{{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2}}} = - \frac{4}{{25}} \cr & {F_z}\left( {3,4,\ln 5} \right) = 1 \cr & {\text{An equation of the tangent plane at }}\left( {{x_0},{y_0},{z_0}} \right){\text{ is}} \cr & {F_x}\left( {{x_0},{y_0},{z_0}} \right)\left( {x - {x_0}} \right) + {F_y}\left( {{x_0},{y_0},{z_0}} \right)\left( {y - {y_0}} \right) \cr & + {F_z}\left( {{x_0},{y_0},{z_0}} \right)\left( {z - {z_0}} \right) = 0 \cr & {\text{Substituting}} \cr & - \frac{3}{{25}}\left( {x - 3} \right) - \frac{4}{{25}}\left( {y - 4} \right) + \left( {z - \ln 5} \right) = 0 \cr & {\text{Simplifying}} \cr & 3\left( {x - 3} \right) + 4\left( {y - 4} \right) - 25\left( {z - \ln 5} \right) = 0 \cr & 3x - 9 + 4y - 16 - 25z + 25\ln 5 = 0 \cr & 3x + 4y - 25z = 25 - 25\ln 5 \cr & 3x + 4y - 25z = 25\left( {1 - \ln 5} \right) \cr} $$