Chapter 12 - Functions of Several Veriables - Review Exercises - Page 961: 74

$$\eqalign{ & x + \frac{1}{2}y + \sqrt 3 z = 2 + \frac{{\sqrt 3 }}{6}\pi \cr & - \frac{1}{2}x - y - \sqrt 3 z = 2 - \frac{{5\sqrt 3 \pi }}{6} \cr} $$

$$\eqalign{ & xy\sin z - 1 = 0;{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( {1,2,\frac{\pi }{6}} \right){\text{ and }}\left( { - 2, - 1,\frac{{5\pi }}{6}} \right) \cr & {\text{Let }}F\left( {x,y,z} \right) = xy\sin z - 1 \cr & {\text{Calculate the partial derivatives }}{F_x}\left( {x,y,z} \right){\text{, }}{F_y}\left( {x,y,z} \right) \cr & {\text{and }}{F_z}\left( {x,y,z} \right) \cr & {F_x}\left( {x,y,z} \right) = y\sin z \cr & {F_y}\left( {x,y,z} \right) = x\sin z \cr & {F_z}\left( {x,y,z} \right) = xy\cos z \cr & {\text{Evaluate at the point }}\left( {1,2,\frac{\pi }{6}} \right){\text{ and }}\left( { - 2, - 1,\frac{{5\pi }}{6}} \right) \cr & {\text{at }}{F_x}\left( {x,y,z} \right),{\mkern 1mu} {F_y}\left( {x,y,z} \right){\text{ and }}{F_z}\left( {x,y,z} \right) \cr & {F_x}\left( {1,2,\frac{\pi }{6}} \right) = 1 \cr & {F_y}\left( {1,2,\frac{\pi }{6}} \right) = \frac{1}{2} \cr & {F_z}\left( {1,2,\frac{\pi }{6}} \right) = \sqrt 3 \cr & and \cr & {F_x}\left( { - 2, - 1,\frac{{5\pi }}{6}} \right) = - \frac{1}{2} \cr & {F_y}\left( { - 2, - 1,\frac{{5\pi }}{6}} \right) = - 1 \cr & {F_z}\left( { - 2, - 1,\frac{{5\pi }}{6}} \right) = - \sqrt 3 \cr & {\text{An equation of the plane tangent to the surface is}} \cr & {F_x}\left( {a,b,c} \right)\left( {x - a} \right) + {F_y}\left( {a,b,c} \right)\left( {y - b} \right) + {F_z}\left( {a,b,c} \right)\left( {z - c} \right) = 0 \cr & \cr & {\text{*For the point }}\left( {1,2,\frac{\pi }{6}} \right) \cr & \left( {x - 1} \right) + \frac{1}{2}\left( {y - 2} \right) + \sqrt 3 \left( {z - \frac{\pi }{6}} \right) = 0 \cr & x - 1 + \frac{1}{2}y - 1 + \sqrt 3 z - \frac{{\sqrt 3 }}{6}\pi = 0 \cr & x + \frac{1}{2}y + \sqrt 3 z = 2 + \frac{{\sqrt 3 }}{6}\pi \cr & \cr & {\text{*For the point }}\left( { - 2, - 1,\frac{{5\pi }}{6}} \right) \cr & - \frac{1}{2}\left( {x + 2} \right) - \left( {y + 1} \right) - \sqrt 3 \left( {z - \frac{{5\pi }}{6}} \right) = 0 \cr & - \frac{1}{2}x - 1 - y - 1 - \sqrt 3 z + \frac{{5\sqrt 3 \pi }}{6} = 0 \cr & - \frac{1}{2}x - y - \sqrt 3 z = 2 - \frac{{5\sqrt 3 \pi }}{6} \cr & \cr & {\text{Summary}} \cr & x + \frac{1}{2}y + \sqrt 3 z = 2 + \frac{{\sqrt 3 }}{6}\pi \cr & - \frac{1}{2}x - y - \sqrt 3 z = 2 - \frac{{5\sqrt 3 \pi }}{6} \cr} $$