Chapter 13 - Parametric And Polar Curves; Conic Sections - 13.6 Directional Derivatives And Gradients - Exercises Set 13.6 - Page 968: 5

$${D_{\bf{u}}}f\left( {2, - 1,1} \right) = - 320$$

$$\eqalign{ & f\left( {x,y,z} \right) = 4{x^5}{y^2}{z^3};\,\,\,\,\,\,P\left( {2, - 1,1} \right);\,\,\,\,\,\,\,{\bf{u}} = \frac{1}{3}{\bf{i}} + \frac{2}{3}{\bf{j}} - \frac{2}{3}{\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( {4{x^5}{y^2}{z^3}} \right) \cr & {f_x}\left( {x,y,z} \right) = 5{y^2}{z^3}\left( {5{x^4}} \right) \cr & {f_x}\left( {x,y,z} \right) = 20{x^4}{y^2}{z^3} \cr & \cr & {f_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left( {4{x^5}{y^2}{z^3}} \right) \cr & {f_y}\left( {x,y,z} \right) = 4{x^5}{z^3}\left( {2y} \right) \cr & {f_y}\left( {x,y,z} \right) = 8{x^5}y{z^3} \cr & \cr & and \cr & {f_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left( {4{x^5}{y^2}{z^3}} \right) \cr & {f_z}\left( {x,y,z} \right) = 4{x^5}{y^2}\left( {3{z^2}} \right) \cr & {f_z}\left( {x,y,z} \right) = 12{x^5}{y^2}{z^2} \cr & \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) = 20{x^4}{y^2}{z^3}{\bf{i}} + 8{x^5}y{z^3}{\bf{j}} + 12{x^5}{y^2}{z^2}{\bf{k}} \cr & {\text{evaluate the gradient at the given point }}P\left( {2, - 1,1} \right) \cr & \nabla f\left( {2, - 1,1} \right) = 20{\left( 2 \right)^4}{\left( { - 1} \right)^2}{\left( 1 \right)^3}{\bf{i}} + 8{\left( 2 \right)^5}\left( { - 1} \right){\left( 1 \right)^3}{\bf{j}} + 12{\left( 2 \right)^5}{\left( { - 1} \right)^2}{\left( 1 \right)^2}{\bf{k}} \cr & \nabla f\left( {2, - 1,1} \right) = 320{\bf{i}} - 256{\bf{j}} + 348{\bf{k}} \cr & \cr & {\text{Calculate the directional derivative }}{D_{\bf{u}}}f\left( {x,y,z} \right){\text{ at }}P\left( {2, - 1,1} \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( {2, - 1,1} \right) = \nabla f\left( {2, - 1,1} \right) \cdot {\bf{u}} \cr & {D_{\bf{u}}}f\left( {2, - 1,1} \right) = \left( {320{\bf{i}} - 256{\bf{j}} + 348{\bf{k}}} \right) \cdot \left( {\frac{1}{3}{\bf{i}} + \frac{2}{3}{\bf{j}} - \frac{2}{3}{\bf{k}}} \right) \cr & {\text{solving the dot product}} \cr & {D_{\bf{u}}}f\left( {2, - 1,1} \right) = \left( {320} \right)\left( {\frac{1}{3}} \right) + \left( { - 256} \right)\left( {\frac{2}{3}} \right) + \left( {384} \right)\left( { - \frac{2}{3}} \right) \cr & {D_{\bf{u}}}f\left( {2, - 1,1} \right) = \frac{{320}}{3} - \frac{{512}}{3} - 256 \cr & {D_{\bf{u}}}f\left( {2, - 1,1} \right) = - 320 \cr} $$