Chapter 13 - Parametric And Polar Curves; Conic Sections - 13.6 Directional Derivatives And Gradients - Exercises Set 13.6 - Page 968: 6

$${D_{\bf{u}}}f\left( {2, - 1,1} \right) = \frac{{45}}{7}$$

$$\eqalign{ & f\left( {x,y,z} \right) = y{e^{xz}} + {z^2};\,\,\,\,\,\,P\left( {0,2,3} \right);\,\,\,\,\,\,\,{\bf{u}} = \frac{2}{7}{\bf{i}} - \frac{3}{7}{\bf{j}} + \frac{6}{7}{\bf{k}} \cr & \cr & {\text{Calculate the partial derivatives }}{f_x}\left( {x,y,z} \right){\text{,}}\,{f_y}\left( {x,y,z} \right)\,{\text{ and}}{\text{,}}\,\,\,{f_z}\left( {x,y,z} \right) \cr & {f_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left( {y{e^{xz}} + {z^2}} \right) \cr & {f_x}\left( {x,y,z} \right) = yz{e^{xz}} + 0 \cr & {f_x}\left( {x,y,z} \right) = yz{e^{xz}} \cr & \cr & {f_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left( {y{e^{xz}} + {z^2}} \right) \cr & {f_y}\left( {x,y,z} \right) = {e^{xz}} + 0 \cr & {f_y}\left( {x,y,z} \right) = {e^{xz}} \cr & \cr & and \cr & {f_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left( {y{e^{xz}} + {z^2}} \right) \cr & {f_z}\left( {x,y,z} \right) = xy{e^{xz}} + 2z \cr & \cr & {\text{Calculate the gradient of }}f\left( {x,y,z} \right) \cr & \nabla f\left( {x,y,z} \right) = {f_x}\left( {x,y,z} \right){\bf{i}} + {f_y}\left( {x,y,z} \right){\bf{j}} + {f_y}\left( {x,y,z} \right){\bf{k}} \cr & \nabla f\left( {x,y,z} \right) = yz{e^{xz}}{\bf{i}} + {e^{xz}}{\bf{j}} + \left( {xy{e^{xz}} + 2z} \right){\bf{k}} \cr & {\text{evaluate the gradient at the given point }}P\left( {0,2,3} \right) \cr & \nabla f\left( {0,2,3} \right) = \left( 2 \right)\left( 3 \right){e^{\left( 0 \right)\left( 3 \right)}}{\bf{i}} + {e^{\left( 0 \right)\left( 3 \right)}}{\bf{j}} + \left( {\left( 0 \right)\left( 2 \right){e^{\left( 0 \right)\left( 3 \right)}} + 2\left( 3 \right)} \right){\bf{k}} \cr & \nabla f\left( {0,2,3} \right) = 6{\bf{i}} + {\bf{j}} + 6{\bf{k}} \cr & \cr & {\text{Calculate the directional derivative }}{D_{\bf{u}}}f\left( {x,y,z} \right){\text{ at }}P\left( {0,2,3} \right){\text{ in the direction of }}{\bf{u}} \cr & {D_{\bf{u}}}f\left( {x,y,z} \right) = \nabla f\left( {x,y,z} \right) \cdot {\bf{u}} \cr & {D_{\bf{u}}}f\left( {0,2,3} \right) = \nabla f\left( {2, - 1,1} \right) \cdot {\bf{u}} \cr & {D_{\bf{u}}}f\left( {0,2,3} \right) = \left( {6{\bf{i}} + {\bf{j}} + 6{\bf{k}}} \right) \cdot \left( {\frac{2}{7}{\bf{i}} - \frac{3}{7}{\bf{j}} + \frac{6}{7}{\bf{k}}} \right) \cr & {\text{solving the dot product}} \cr & {D_{\bf{u}}}f\left( {2, - 1,1} \right) = \left( 6 \right)\left( {\frac{2}{7}} \right) + \left( 1 \right)\left( { - \frac{3}{7}} \right) + \left( 6 \right)\left( {\frac{6}{7}} \right) \cr & {D_{\bf{u}}}f\left( {2, - 1,1} \right) = \frac{{12}}{7} - \frac{3}{7} + \frac{{36}}{7} \cr & {D_{\bf{u}}}f\left( {2, - 1,1} \right) = \frac{{45}}{7} \cr} $$