Chapter 12 - Functions of Several Veriables - 12.6 Directional Derivatives and the Gradient - 12.6 Exercises - Page 925: 15

$$\nabla F\left( {x,y} \right) = - {e^{ - {x^2} - 2{y^2}}}\left\langle {2x,4y} \right\rangle {\text{ and }}\nabla F\left( { - 1,2} \right) = - {e^{ - 9}}\left\langle { - 2,8} \right\rangle $$

$$\eqalign{ & F\left( {x,y} \right) = {e^{ - {x^2} - 2{y^2}}};\,\,\,\,\,\,P\left( { - 1,2} \right) \cr & {\text{compute the gradient of }}F\left( {x,y} \right),{\text{ }} \cr & {\text{first find the partial derivatives }}{F_x}\left( {x,y} \right){\text{ and }}{F_y}\left( {x,y} \right).{\text{ then}} \cr & {F_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{e^{ - {x^2} - 2{y^2}}}} \right] \cr & {\text{treat }}y{\text{ as a constant}} \cr & {F_x}\left( {x,y} \right) = {e^{ - {x^2} - 2{y^2}}}\frac{\partial }{{\partial x}}\left[ { - {x^2} - 2{y^2}} \right] \cr & {F_x}\left( {x,y} \right) = {e^{ - {x^2} - 2{y^2}}}\left( { - 2x} \right) \cr & {F_x}\left( {x,y} \right) = - 2x{e^{ - {x^2} - 2{y^2}}} \cr & and \cr & {F_y}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{e^{ - {x^2} - 2{y^2}}}} \right] \cr & {\text{treat }}x{\text{ as a constant}} \cr & {F_y}\left( {x,y} \right) = {e^{ - {x^2} - 2{y^2}}}\frac{\partial }{{\partial y}}\left[ { - {x^2} - 2{y^2}} \right] \cr & {F_y}\left( {x,y} \right) = {e^{ - {x^2} - 2{y^2}}}\left( { - 4y} \right) \cr & {F_y}\left( {x,y} \right) = - 4y{e^{ - {x^2} - 2{y^2}}} \cr & {\text{using }}\nabla {\text{ }}F\left( {x,y} \right) = {F_x}\left( {x,y} \right){\bf{i}} + {F_y}\left( {x,y} \right){\bf{j}}.{\text{ then substituting the partial}} \cr & {\text{derivatives}} \cr & \nabla {\text{ }}F\left( {x,y} \right) = - 2x{e^{ - {x^2} - 2{y^2}}}{\bf{i}} - 4y{e^{ - {x^2} - 2{y^2}}}{\bf{j}} \cr & or \cr & \nabla F\left( {x,y} \right) = \left\langle { - 2x{e^{ - {x^2} - 2{y^2}}}, - 4y{e^{ - {x^2} - 2{y^2}}}{\bf{j}}} \right\rangle \cr & \nabla F\left( {x,y} \right) = - {e^{ - {x^2} - 2{y^2}}}\left\langle {2x,4y} \right\rangle \cr & {\text{evaluate }}\nabla {\text{ }}F\left( {x,y} \right){\text{ at the point }}P \cr & \nabla F\left( { - 1,2} \right) = - {e^{ - {{\left( { - 1} \right)}^2} - 2{{\left( 2 \right)}^2}}}\left\langle {2\left( { - 1} \right),4\left( 2 \right)} \right\rangle \cr & \nabla F\left( { - 1,2} \right) = - {e^{ - 9}}\left\langle { - 2,8} \right\rangle \cr & \cr & \nabla F\left( {x,y} \right) = - {e^{ - {x^2} - 2{y^2}}}\left\langle {2x,4y} \right\rangle {\text{ and }}\nabla F\left( { - 1,2} \right) = - {e^{ - 9}}\left\langle { - 2,8} \right\rangle \cr} $$