Chapter 13 - Parametric And Polar Curves; Conic Sections - 13.6 Directional Derivatives And Gradients - Exercises Set 13.6 - Page 968: 9

$${D_{\bf{u}}}f\left( {1,2} \right) = \frac{{144}}{5}$$

$$\eqalign{ & f\left( {x,y} \right) = 4{x^3}{y^2};\,\,\,\,\,P\left( {1,2} \right);\,\,\,\,\,\,{\bf{a}} = 4{\bf{i}} - 3{\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( {4{x^3}{y^2}} \right) \cr & {\text{Treat }}y{\text{ as a constant}} \cr & {f_x}\left( {x,y} \right) = 4{y^2}\left( {3{x^2}} \right) \cr & {f_x}\left( {x,y} \right) = 12{x^2}{y^2} \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {4{x^3}{y^2}} \right) \cr & {\text{Treat }}x{\text{ as a constant}} \cr & {f_y}\left( {x,y} \right) = 4{x^3}\left( {2y} \right) \cr & {f_y}\left( {x,y} \right) = 8{x^3}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) = 12{x^2}{y^2}{\bf{i}} + 8{x^3}y{\bf{j}} \cr & {\text{evaluate the gradient at the given point }}P\left( {1,2} \right) \cr & \nabla f\left( {1,2} \right) = 12{\left( 1 \right)^2}{\left( 2 \right)^2}{\bf{i}} + 8{\left( 1 \right)^3}\left( 2 \right){\bf{j}} \cr & \nabla f\left( {1,2} \right) = 48{\bf{i}} + 64{\bf{j}} \cr & \cr & {\text{Note that }}{\bf{a}}{\text{ is not a unit vector}}{\text{, bu since }}\left| {\bf{a}} \right| = 5,{\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{{4{\bf{i}} - 3{\bf{j}}}}{5} = \frac{4}{5}{\bf{i}} - \frac{3}{5}{\bf{j}} \cr & {\text{Calculate the directional derivative }}{D_{\bf{u}}}f\left( {x,y} \right){\text{ at }}P\left( {1,2} \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( {1,2} \right) = \nabla f\left( {1,2} \right) \cdot {\bf{u}} \cr & {D_{\bf{u}}}f\left( {1,2} \right) = \left( {48{\bf{i}} + 64{\bf{j}}} \right) \cdot \left( {\frac{4}{5}{\bf{i}} - \frac{3}{5}{\bf{j}}} \right) \cr & {\text{solving the dot product}} \cr & {D_{\bf{u}}}f\left( {1,2} \right) = \left( {48} \right)\left( {\frac{4}{5}} \right) + \left( {64} \right)\left( { - \frac{3}{5}} \right) \cr & {D_{\bf{u}}}f\left( {1,2} \right) = \frac{{192}}{5} - \frac{{192}}{5} \cr & {D_{\bf{u}}}f\left( {1,2} \right) = 0 \cr} $$