Chapter 12 - Functions of Several Veriables - Review Exercises - Page 961: 76

$$\eqalign{ & z = 8x - 4y - 4 \cr & z = - x - y - 1 \cr} $$

$$\eqalign{ & z = {x^2}{e^{x - y}};{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( {2,2,4} \right){\text{ and }}\left( { - 1, - 1,1} \right) \cr & f\left( {x,y} \right) = {x^2}{e^{x - y}} \cr & {\text{Calculate the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = 2x{e^{x - y}} + {x^2}{e^{x - y}} \cr & {f_y}\left( {x,y} \right) = - {x^2}{e^{x - y}} \cr & {\text{Evaluate at the point }}\left( {2,2,4} \right){\text{ and }}\left( { - 1, - 1,1} \right){\text{ at }}{f_x},{\mkern 1mu} {f_y} \cr & {f_x}\left( {2,2} \right) = 2\left( 2 \right){e^{2 - 2}} + {\left( 2 \right)^2}{e^{2 - 2}} = 8 \cr & {f_y}\left( {2,2} \right) = - {\left( 2 \right)^2}{e^{2 - 2}} = - 4 \cr & and \cr & {f_x}\left( { - 1, - 1} \right) = 2\left( { - 1} \right){e^{ - 1 + 1}} + {\left( { - 1} \right)^2}{e^{ - 1 + 1}} = - 1 \cr & {f_y}\left( { - 1, - 1} \right) = - {\left( { - 1} \right)^2}{e^{ - 1 + 1}} = - 1 \cr & {\text{An equation of the plane tangent to the surface is}} \cr & z = f\left( {x,y} \right){\text{ at the point }}\left( {a,b,f\left( {a,b} \right)} \right) \cr & \cr & *{\text{For the point }}\left( {2,2,4} \right) \cr & z = {f_x}\left( {a,b} \right)\left( {x - a} \right) + {f_x}\left( {a,b} \right)\left( {y - b} \right) + f\left( {a,b} \right) \cr & z = 8\left( {x - 2} \right) - 4\left( {y - 2} \right) + 4 \cr & z = 8x - 16 - 4y + 8 + 4 \cr & z = 8x - 4y - 4 \cr & \cr & *{\text{For the point }}\left( { - 1, - 1,1} \right) \cr & z = - \left( {x + 1} \right) - \left( {y + 1} \right) + 1 \cr & z = - x - 1 - y - 1 + 1 \cr & z = - x - y - 1 \cr & \cr & {\text{Summary:}} \cr & z = 8x - 4y - 4 \cr & z = - x - y - 1 \cr} $$