Chapter 13 - Parametric And Polar Curves; Conic Sections - 13.6 Directional Derivatives And Gradients - Exercises Set 13.6 - Page 968: 14

$${D_{\bf{u}}}f\left( {0,0} \right) = \frac{7}{{\sqrt {29} }}$$

$$\eqalign{ & f\left( {x,y} \right) = x{e^y} - y{e^x};\,\,\,\,\,P\left( {0,0} \right);\,\,\,\,\,\,{\bf{a}} = 5{\bf{i}} - 2{\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( {x{e^y} - y{e^x}} \right) \cr & {\text{Treat }}y{\text{ as a constant }} \cr & {f_x}\left( {x,y} \right) = {e^y} - y{e^x} \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {x{e^y} - y{e^x}} \right) \cr & {\text{Treat }}x{\text{ as a constant}} \cr & {f_y}\left( {x,y} \right) = x{e^y} - {e^x} \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) = \left( {{e^y} - y{e^x}} \right){\bf{i}} + \left( {x{e^y} - {e^x}} \right){\bf{j}} \cr & {\text{Evaluate the gradient at the given point }}P\left( {0,0} \right) \cr & \nabla f\left( {0,0} \right) = \left( {{e^0} - 0{e^0}} \right){\bf{i}} + \left( {0{e^0} - {e^0}} \right){\bf{j}} \cr & \nabla f\left( {0,0} \right) = {\bf{i}} - {\bf{j}} \cr & \cr & {\text{Note that }}{\bf{a}}{\text{ is not a unit vector}}{\text{, but since }}\left| {\bf{a}} \right| = \sqrt {29} ,{\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{{5{\bf{i}} - 2{\bf{j}}}}{{\sqrt {29} }} = \frac{5}{{\sqrt {29} }}{\bf{i}} - \frac{2}{{\sqrt {29} }}{\bf{j}} \cr & {\text{Calculate the directional derivative }}{D_{\bf{u}}}f\left( {x,y} \right){\text{ at }}P\left( {0,0} \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( {0,0} \right) = \nabla f\left( {0,0} \right) \cdot {\bf{u}} \cr & {D_{\bf{u}}}f\left( {0,0} \right) = \left( {{\bf{i}} - {\bf{j}}} \right) \cdot \left( {\frac{5}{{\sqrt {29} }}{\bf{i}} - \frac{2}{{\sqrt {29} }}{\bf{j}}} \right) \cr & {\text{Solving the dot product}} \cr & {D_{\bf{u}}}f\left( {0,0} \right) = \left( 1 \right)\left( {\frac{5}{{\sqrt {29} }}} \right) + \left( { - 1} \right)\left( { - \frac{2}{{\sqrt {29} }}} \right) \cr & {D_{\bf{u}}}f\left( {0,0} \right) = \frac{5}{{\sqrt {29} }} + \frac{2}{{\sqrt {29} }} \cr & {D_{\bf{u}}}f\left( {0,0} \right) = \frac{7}{{\sqrt {29} }} \cr} $$