Chapter 14 - Section 14.6 - Directional Derivatives and the Gradient Vector - 14.6 Exercise - Page 1005: 16

$ - \frac{{18}}{5}$

$$\eqalign{ & g\left( {u,v} \right) = {u^2}{e^{ - v}},{\text{ }}\left( {3,0} \right),{\text{ }}{\bf{v}} = 3{\bf{i}} + 4{\bf{j}} \cr & {\text{Calculate the partial derivatives }}{g_u}\left( {u,v} \right){\text{ and }}{g_v}\left( {u,v} \right) \cr & {g_u}\left( {u,v} \right) = \frac{\partial }{{\partial u}}\left[ {{u^2}{e^{ - v}}} \right] \cr & {g_u}\left( {u,v} \right) = 2u{e^{ - v}} \cr & and \cr & {g_v}\left( {u,v} \right) = \frac{\partial }{{\partial v}}\left[ {{u^2}{e^{ - v}}} \right] \cr & {g_v}\left( {u,v} \right) = - {u^2}{e^{ - v}} \cr & \cr & {\text{The gradient of the function is given by }} \cr & \nabla f\left( {x,y} \right) = {f_x}\left( {x,y} \right){\bf{i}} + {f_y}\left( {x,y} \right){\bf{j}} \cr & {\text{Then}} \cr & g\left( {u,v} \right) = {g_u}\left( {u,v} \right){\bf{i}} + {g_v}\left( {u,v} \right){\bf{j}} \cr & {\text{Substituting the partial derivatives}}{\text{, we obtain}} \cr & g\left( {u,v} \right) = 2u{e^{ - v}}{\bf{i}} - {u^2}{e^{ - v}}{\bf{j}} \cr & {\text{Calculate the gradient at the point }}\left( {3,0} \right) \cr & \nabla g\left( {3,0} \right) = 2\left( 3 \right){e^0}{\bf{i}} - {\left( 3 \right)^2}{e^0}{\bf{j}} \cr & \nabla g\left( {3,0} \right) = 6{\bf{i}} - 9{\bf{j}} \cr & \cr & {\text{The directional derivative of }}f\left( {x,y} \right){\text{ at }}\left( {{x_0},{y_0}} \right){\text{ in the}} \cr & {\text{direction of a unit vector }}{\bf{u}}{\text{ is}} \cr & {D_{\bf{u}}}f\left( {x,y} \right) = \nabla f\left( {x,y} \right) \cdot {\bf{u}} \cr & {\text{Then}} \cr & {D_{\bf{u}}}g\left( {s,t} \right) = \nabla g\left( {s,t} \right) \cdot {\bf{u}} \cr & {\text{Express }}{\bf{v}} = 3{\bf{i}} + 4{\bf{j}}{\text{ as a unit vector }}{\bf{u}} \cr & {\bf{u}} = \frac{{\bf{v}}}{{\left| {\bf{v}} \right|}} = \frac{{3{\bf{i}} + 4{\bf{j}}}}{{\left| {3{\bf{i}} + 4{\bf{j}}} \right|}} = \frac{{3{\bf{i}}}}{{\sqrt {9 + 16} }} + \frac{{4{\bf{j}}}}{{\sqrt {9 + 16} }} \cr & {\bf{u}} = \frac{3}{5}{\bf{i}} + \frac{4}{5}{\bf{j}} \cr & {\text{Therefore at the point }}\left( {3,0} \right){\text{ }} \cr & {D_{\bf{u}}}g\left( {3,0} \right) = \nabla g\left( {3,0} \right) \cdot {\bf{u}} \cr & {D_{\bf{u}}}g\left( {3,0} \right) = \left( {6{\bf{i}} - 9{\bf{j}}} \right) \cdot \left( {\frac{3}{5}{\bf{i}} + \frac{4}{5}{\bf{j}}} \right) \cr & {D_{\bf{u}}}g\left( {3,0} \right) = \left( 6 \right)\left( {\frac{3}{5}} \right) + \left( { - 9} \right)\left( {\frac{4}{5}} \right) \cr & {\text{Simplifying}} \cr & {D_{\bf{u}}}g\left( {3,0} \right) = \frac{{18}}{5} - \frac{{36}}{5} \cr & {D_{\bf{u}}}g\left( {3,0} \right) = - \frac{{18}}{5} \cr} $$