Chapter 13 - Parametric And Polar Curves; Conic Sections - 13.6 Directional Derivatives And Gradients - Exercises Set 13.6 - Page 968: 13

$${D_{\bf{u}}}f\left( { - 2,2} \right) = \frac{1}{{2\sqrt 2 }}$$

$$\eqalign{ & f\left( {x,y} \right) = {\tan ^{ - 1}}\left( {y/x} \right);\,\,\,\,\,P\left( { - 2,2} \right);\,\,\,\,\,\,{\bf{a}} = - {\bf{i}} - {\bf{j}} \cr & \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{\partial }{{\partial x}}\left( {{{\tan }^{ - 1}}\left( {y/x} \right)} \right) \cr & {\text{Treat }}y{\text{ as a constant and use the rule }}\frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{1 + {u^2}}}\left( {\frac{{du}}{{dx}}} \right) \cr & {f_x}\left( {x,y} \right) = \frac{1}{{1 + {{\left( {y/x} \right)}^2}}}\left( { - \frac{y}{{{x^2}}}} \right) \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {{{\tan }^{ - 1}}\left( {y/x} \right)} \right) \cr & {\text{Treat }}x{\text{ as a constant and use the rule }}\frac{d}{{dy}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{1 + {u^2}}}\left( {\frac{{du}}{{dy}}} \right) \cr & {f_y}\left( {x,y} \right) = \frac{1}{{1 + {{\left( {y/x} \right)}^2}}}\left( {\frac{1}{x}} \right) \cr & \cr & {\text{Calculate the gradient of }}f\left( {x,y} \right) \cr & \nabla f\left( {x,y} \right) = {f_x}\left( {x,y} \right){\bf{i}} + {f_y}\left( {x,y} \right){\bf{j}} \cr & \nabla f\left( {x,y} \right) = \frac{1}{{1 + {{\left( {y/x} \right)}^2}}}\left( { - \frac{y}{{{x^2}}}} \right){\bf{i}} + \frac{1}{{1 + {{\left( {y/x} \right)}^2}}}\left( {\frac{1}{x}} \right){\bf{j}} \cr & {\text{Evaluate the gradient at the given point }}P\left( { - 2,2} \right) \cr & \nabla f\left( { - 2,2} \right) = \frac{1}{{1 + {{\left( {2/\left( { - 2} \right)} \right)}^2}}}\left( { - \frac{2}{{{{\left( { - 2} \right)}^2}}}} \right){\bf{i}} + \frac{1}{{1 + {{\left( {2/\left( { - 2} \right)} \right)}^2}}}\left( {\frac{1}{{ - 2}}} \right){\bf{j}} \cr & \nabla f\left( { - 2,2} \right) = \frac{1}{2}\left( { - \frac{2}{4}} \right){\bf{i}} + \frac{1}{2}\left( {\frac{1}{{ - 2}}} \right){\bf{j}} \cr & \nabla f\left( { - 2,2} \right) = - \frac{1}{4}{\bf{i}} - \frac{1}{4}{\bf{j}} \cr & \cr & {\text{Note that }}{\bf{a}}{\text{ is not a unit vector}}{\text{, but since }}\left| {\bf{a}} \right| = \sqrt 2 ,{\text{ the unit vector in the direction }} \cr & {\text{of }}{\bf{a}}{\text{ is}} \cr & \,\,\,\,\,\,\,{\bf{u}} = \frac{{\bf{a}}}{{\left| {\bf{a}} \right|}} = \frac{{ - {\bf{i}} - {\bf{j}}}}{{\sqrt 2 }} = - \frac{1}{{\sqrt 2 }}{\bf{i}} - \frac{1}{{\sqrt 2 }}{\bf{j}} \cr & {\text{Calculate the directional derivative }}{D_{\bf{u}}}f\left( {x,y} \right){\text{ at }}P\left( { - 2,2} \right){\text{ in the direction of }}{\bf{u}} \cr & {D_{\bf{u}}}f\left( {x,y} \right) = \nabla f\left( {x,y} \right) \cdot {\bf{u}} \cr & {D_{\bf{u}}}f\left( { - 2,2} \right) = \nabla f\left( { - 2,2} \right) \cdot {\bf{u}} \cr & {D_{\bf{u}}}f\left( { - 2,2} \right) = \left( { - \frac{1}{4}{\bf{i}} - \frac{1}{4}{\bf{j}}} \right) \cdot \left( { - \frac{1}{{\sqrt 2 }}{\bf{i}} - \frac{1}{{\sqrt 2 }}{\bf{j}}} \right) \cr & {\text{solving the dot product}} \cr & {D_{\bf{u}}}f\left( { - 2,2} \right) = \left( { - \frac{1}{4}} \right)\left( { - \frac{1}{{\sqrt 2 }}} \right) + \left( { - \frac{1}{4}} \right)\left( { - \frac{1}{{\sqrt 2 }}} \right) \cr & {D_{\bf{u}}}f\left( { - 2,2} \right) = \frac{1}{{4\sqrt 2 }} + \frac{1}{{4\sqrt 2 }} \cr & {D_{\bf{u}}}f\left( { - 2,2} \right) = \frac{1}{{2\sqrt 2 }} \cr} $$