Chapter 13 - Parametric And Polar Curves; Conic Sections - 13.6 Directional Derivatives And Gradients - Exercises Set 13.6 - Page 968: 12

$${D_{\bf{u}}}f\left( {0,\pi /4} \right) = \frac{7}{2}\sqrt {\frac{2}{{29}}} $$

$$\eqalign{ & f\left( {x,y} \right) = {e^x}\cos y;\,\,\,\,\,P\left( {0,\pi /4} \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( {{e^x}\cos y} \right) \cr & {\text{Treat }}y{\text{ as a constant}} \cr & {f_x}\left( {x,y} \right) = {e^x}\cos y \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {{e^x}\cos y} \right) \cr & {\text{Treat }}x{\text{ as a constant}} \cr & {f_y}\left( {x,y} \right) = {e^x}\left( { - \sin y} \right) \cr & {f_y}\left( {x,y} \right) = - {e^x}\sin y \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) = {e^x}\cos y{\bf{i}} - {e^x}\sin y{\bf{j}} \cr & {\text{evaluate the gradient at the given point }}P\left( {0,\pi /4} \right) \cr & \nabla f\left( {0,\pi /4} \right) = {e^0}\cos \left( {\pi /4} \right){\bf{i}} - {e^0}\sin \left( {\pi /4} \right){\bf{j}} \cr & \nabla f\left( {0,\pi /4} \right) = \frac{{\sqrt 2 }}{2}{\bf{i}} - \frac{{\sqrt 2 }}{2}{\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,\pi /4} \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,\pi /4} \right) = \nabla f\left( {0,\pi /4} \right) \cdot {\bf{u}} \cr & {D_{\bf{u}}}f\left( {0,\pi /4} \right) = \left( {\frac{{\sqrt 2 }}{2}{\bf{i}} - \frac{{\sqrt 2 }}{2}{\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,\pi /4} \right) = \left( {\frac{{\sqrt 2 }}{2}} \right)\left( {\frac{5}{{\sqrt {29} }}} \right) + \left( { - \frac{{\sqrt 2 }}{2}} \right)\left( { - \frac{2}{{\sqrt {29} }}} \right) \cr & {D_{\bf{u}}}f\left( {0,\pi /4} \right) = \frac{5}{2}\sqrt {\frac{2}{{29}}} + \sqrt {\frac{2}{{29}}} \cr & {D_{\bf{u}}}f\left( {0,\pi /4} \right) = \frac{7}{2}\sqrt {\frac{2}{{29}}} \cr} $$