Chapter 12 - Functions of Several Veriables - Review Exercises - Page 961: 77

$$\eqalign{ & z = \frac{2}{3}x + \frac{1}{3}y - \frac{4}{3} + \ln 3 \cr & z = - \frac{1}{3}x - \frac{2}{3}y - \frac{4}{3} + \ln 3 \cr} $$

$$\eqalign{ & z = \ln \left( {1 + xy} \right);{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( {1,2,\ln 3} \right){\text{ and }}\left( { - 2, - 1,\ln 3} \right) \cr & f\left( {x,y} \right) = \ln \left( {1 + xy} \right) \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) = \frac{y}{{1 + xy}} \cr & {f_y}\left( {x,y} \right) = \frac{x}{{1 + xy}} \cr & {\text{Evaluate at the point }}\left( {1,2,\ln 3} \right){\text{ and }}\left( { - 2, - 1,\ln 3} \right){\text{at }}{f_x},{\mkern 1mu} {f_y} \cr & {f_x}\left( {1,2} \right) = \frac{2}{{1 + \left( 1 \right)\left( 2 \right)}} = \frac{2}{3} \cr & {f_y}\left( {1,2} \right) = \frac{1}{{1 + \left( 1 \right)\left( 2 \right)}} = \frac{1}{3} \cr & and \cr & {f_x}\left( { - 2, - 1} \right) = \frac{{ - 1}}{{1 + \left( { - 2} \right)\left( { - 1} \right)}} = - \frac{1}{3} \cr & {f_y}\left( { - 2, - 1} \right) = \frac{{ - 2}}{{1 + \left( { - 2} \right)\left( { - 1} \right)}} = - \frac{2}{3} \cr & {\text{An equation of the plane tangent to the surface is}} \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){\text{ at }}\left( {a,b,f\left( {a,b} \right)} \right) \cr & \cr & {\text{*For the point }}\left( {1,2,\ln 3} \right) \cr & z = \frac{2}{3}\left( {x - 1} \right) + \frac{1}{3}\left( {y - 2} \right) + \ln 3 \cr & z = \frac{2}{3}x - \frac{2}{3} + \frac{1}{3}y - \frac{2}{3} + \ln 3 \cr & z = \frac{2}{3}x + \frac{1}{3}y - \frac{4}{3} + \ln 3 \cr & \cr & {\text{*For the point }}\left( { - 2, - 1,\ln 3} \right) \cr & z = - \frac{1}{3}\left( {x + 2} \right) - \frac{2}{3}\left( {y + 1} \right) + \ln 3 \cr & z = - \frac{1}{3}x - \frac{2}{3} - \frac{2}{3}y - \frac{2}{3} + \ln 3 \cr & z = - \frac{1}{3}x - \frac{2}{3}y - \frac{4}{3} + \ln 3 \cr & \cr & {\text{Summary:}} \cr & z = \frac{2}{3}x + \frac{1}{3}y - \frac{4}{3} + \ln 3 \cr & z = - \frac{1}{3}x - \frac{2}{3}y - \frac{4}{3} + \ln 3 \cr} $$