Chapter 13 - Functions of Several Variables - 13.7 Exercises - Page 933: 29

$$\eqalign{ & x - y + 2z = \frac{\pi }{2} \cr & {\text{Line: }}\frac{{\left( {x - 1} \right)}}{1} = \frac{{\left( {y - 1} \right)}}{{ - 1}} = \frac{{z - \left( {\pi /4} \right)}}{2} \cr} $$

$$\eqalign{ & z = \arctan \frac{y}{x},{\text{ }}\left( {1,1,\frac{\pi }{4}} \right) \cr & \arctan \frac{y}{x} - z = 0 \cr & {\text{Consider}} \cr & F\left( {x,y,z} \right) = \arctan \frac{y}{x} - z \cr & {\text{Calculating the partial derivatives}} \cr & {F_x}\left( {x,y,z} \right) = \frac{{ - y/{x^2}}}{{1 + {{\left( {y/x} \right)}^2}}} = \frac{{ - y/{x^2}}}{{1 + {y^2}/{x^2}}} = - \frac{y}{{{x^2} + {y^2}}} \cr & {F_y}\left( {x,y,z} \right) = \frac{{1/x}}{{1 + {{\left( {y/x} \right)}^2}}} = \frac{{1/x}}{{1 + {y^2}/{x^2}}} = \frac{x}{{{x^2} + {y^2}}} \cr & {F_z}\left( {x,y,z} \right) = - 1,{\text{ }} \cr & {\text{At the point }}\left( {1,1,\frac{\pi }{4}} \right) \cr & {F_x}\left( {1,1,\frac{\pi }{4}} \right) = - \frac{1}{{{1^2} + {1^2}}} = - \frac{1}{2}{\text{ }} \cr & {F_y}\left( {1,1,\frac{\pi }{4}} \right) = \frac{1}{{{1^2} + {1^2}}} = \frac{1}{2} \cr & {F_z}\left( {1,1,\frac{\pi }{4}} \right) = - 1,{\text{ }} \cr & {\text{Direction numbers: }} - \frac{1}{2},\frac{1}{2}, - 1 \cr & {\text{Multipliying by }} - {\text{2}} \cr & {\text{Direction numbers: }}1, - 1,2 \cr & {\text{An equation of the tangent plane at }}\left( {1,1,\frac{\pi }{4}} \right){\text{ is}} \cr & {F_x}\left( {1,1,\frac{\pi }{4}} \right)\left( {x - 3} \right) + {F_y}\left( {1,1,\frac{\pi }{4}} \right)\left( {y - 2} \right) + {F_z}\left( {1,1,\frac{\pi }{4}} \right)\left( {z - 5} \right) = 0 \cr & - \frac{1}{2}\left( {x - 1} \right) + \frac{1}{2}\left( {y - 1} \right) - \left( {z - \frac{\pi }{4}} \right) = 0 \cr & - \frac{1}{2}x + \frac{1}{2} + \frac{1}{2}y - \frac{1}{2} - z + \frac{\pi }{4} = 0 \cr & - \frac{1}{2}x + \frac{1}{2}y - z = - \frac{\pi }{4} \cr & - 2\left( { - \frac{1}{2}x + \frac{1}{2}y - z} \right) = - 2\left( { - \frac{\pi }{4}} \right) \cr & x - y + 2z = \frac{\pi }{2} \cr & \cr & {\text{The corresponding symmetric equations are:}} \cr & {\text{Line: }}\frac{{\left( {x - 1} \right)}}{1} = \frac{{\left( {y - 1} \right)}}{{ - 1}} = \frac{{z - \left( {\pi /4} \right)}}{2} \cr} $$