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

$$\nabla {\text{ }}p\left( {x,y} \right) = - \frac{1}{{\sqrt {12 - 4{x^2} - {y^2}} }}\left\langle {4x,y} \right\rangle {\text{ and }}\nabla p\left( { - 1, - 1} \right) = \frac{1}{{\sqrt 7 }}\left\langle {4,1} \right\rangle $$

$$\eqalign{ & p\left( {x,y} \right) = \sqrt {12 - 4{x^2} - {y^2}} ;\,\,\,\,\,\,P\left( { - 1, - 1} \right) \cr & {\text{compute the gradient of }}p\left( {x,y} \right),{\text{ }} \cr & {\text{first find the partial derivatives }}{p_x}\left( {x,y} \right){\text{ and }}{p_y}\left( {x,y} \right).{\text{ then}} \cr & {p_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\sqrt {12 - 4{x^2} - {y^2}} } \right] \cr & {\text{treat }}y{\text{ as a constant}}{\text{, use }}\frac{d}{{dx}}\left[ {\sqrt u } \right] = \frac{{u'}}{{2\sqrt u }} \cr & {p_x}\left( {x,y} \right) = \frac{{\frac{\partial }{{\partial x}}\left[ {12 - 4{x^2} - {y^2}} \right]}}{{2\sqrt {12 - 4{x^2} - {y^2}} }} \cr & {p_x}\left( {x,y} \right) = \frac{{ - 8x}}{{2\sqrt {12 - 4{x^2} - {y^2}} }} \cr & {p_x}\left( {x,y} \right) = - \frac{{4x}}{{\sqrt {12 - 4{x^2} - {y^2}} }} \cr & and \cr & {p_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\sqrt {12 - 4{x^2} - {y^2}} } \right] \cr & {\text{treat }}x{\text{ as a constant}} \cr & {p_y}\left( {x,y} \right) = \frac{{\frac{\partial }{{\partial y}}\left[ {12 - 4{x^2} - {y^2}} \right]}}{{2\sqrt {12 - 4{x^2} - {y^2}} }} \cr & {p_y}\left( {x,y} \right) = \frac{{ - 2y}}{{2\sqrt {12 - 4{x^2} - {y^2}} }} \cr & {p_y}\left( {x,y} \right) = - \frac{y}{{\sqrt {12 - 4{x^2} - {y^2}} }} \cr & {\text{using }}\nabla {\text{ }}p\left( {x,y} \right) = {p_x}\left( {x,y} \right){\bf{i}} + {p_y}\left( {x,y} \right){\bf{j}}.{\text{ then substituting the partial}} \cr & {\text{derivatives}} \cr & \nabla {\text{ }}p\left( {x,y} \right) = - \frac{{4x}}{{\sqrt {12 - 4{x^2} - {y^2}} }}{\bf{i}} - \frac{y}{{\sqrt {12 - 4{x^2} - {y^2}} }}{\bf{j}} \cr & or \cr & \nabla {\text{ }}p\left( {x,y} \right) = \left\langle { - \frac{{4x}}{{\sqrt {12 - 4{x^2} - {y^2}} }}, - \frac{y}{{\sqrt {12 - 4{x^2} - {y^2}} }}} \right\rangle \cr & \nabla {\text{ }}p\left( {x,y} \right) = - \frac{1}{{\sqrt {12 - 4{x^2} - {y^2}} }}\left\langle {4x,y} \right\rangle \cr & {\text{evaluate }}\nabla {\text{ }}p\left( {x,y} \right){\text{ at the point }}P \cr & \nabla p\left( { - 1, - 1} \right) = - \frac{1}{{\sqrt {12 - 4{{\left( { - 1} \right)}^2} - {{\left( 1 \right)}^2}} }}\left\langle {4\left( { - 1} \right),\left( { - 1} \right)} \right\rangle \cr & \nabla p\left( { - 1, - 1} \right) = \frac{1}{{\sqrt 7 }}\left\langle {4,1} \right\rangle \cr & \cr & \nabla {\text{ }}p\left( {x,y} \right) = - \frac{1}{{\sqrt {12 - 4{x^2} - {y^2}} }}\left\langle {4x,y} \right\rangle {\text{ and }}\nabla p\left( { - 1, - 1} \right) = \frac{1}{{\sqrt 7 }}\left\langle {4,1} \right\rangle \cr} $$