Chapter 13 - Parametric And Polar Curves; Conic Sections - 13.6 Directional Derivatives And Gradients - Exercises Set 13.6 - Page 968: 7

$${D_{\bf{u}}}f\left( { - 1,2,4} \right) = - \frac{{314}}{{741}}$$

$$\eqalign{ & f\left( {x,y,z} \right) = \ln \left( {{x^2} + 2{y^2} + 3{z^2}} \right);\,\,\,\,\,\,P\left( { - 1,2,4} \right);\,\,\,\,\,\,\,{\bf{u}} = - \frac{3}{{13}}{\bf{i}} - \frac{4}{{13}}{\bf{j}} - \frac{{12}}{{13}}{\bf{k}} \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( {\ln \left( {{x^2} + 2{y^2} + 3{z^2}} \right)} \right) \cr & {f_x}\left( {x,y,z} \right) = \frac{1}{{{x^2} + 2{y^2} + 3{z^2}}}\frac{\partial }{{\partial x}}\left( {{x^2} + 2{y^2} + 3{z^2}} \right) \cr & {f_x}\left( {x,y,z} \right) = \frac{1}{{{x^2} + 2{y^2} + 3{z^2}}}\left( {2x} \right) \cr & {f_x}\left( {x,y,z} \right) = \frac{{2x}}{{{x^2} + 2{y^2} + 3{z^2}}} \cr & \cr & {f_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left( {\ln \left( {{x^2} + 2{y^2} + 3{z^2}} \right)} \right) \cr & {f_y}\left( {x,y,z} \right) = \frac{1}{{{x^2} + 2{y^2} + 3{z^2}}}\frac{\partial }{{\partial y}}\left( {{x^2} + 2{y^2} + 3{z^2}} \right) \cr & {f_y}\left( {x,y,z} \right) = \frac{1}{{{x^2} + 2{y^2} + 3{z^2}}}\left( {4y} \right) \cr & {f_y}\left( {x,y,z} \right) = \frac{{4y}}{{{x^2} + 2{y^2} + 3{z^2}}} \cr & \cr & and \cr & {f_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left( {\ln \left( {{x^2} + 2{y^2} + 3{z^2}} \right)} \right) \cr & {f_z}\left( {x,y,z} \right) = \frac{1}{{{x^2} + 2{y^2} + 3{z^2}}}\frac{\partial }{{\partial y}}\left( {6z} \right) \cr & {f_z}\left( {x,y,z} \right) = \frac{{6z}}{{{x^2} + 2{y^2} + 3{z^2}}} \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) = \frac{{2x}}{{{x^2} + 2{y^2} + 3{z^2}}}{\bf{i}} + \frac{{4y}}{{{x^2} + 2{y^2} + 3{z^2}}}{\bf{j}} + \frac{{6z}}{{{x^2} + 2{y^2} + 3{z^2}}}{\bf{k}} \cr & \nabla f\left( {x,y,z} \right) = \frac{2}{{{x^2} + 2{y^2} + 3{z^2}}}\left( {x{\bf{i}} + 2y{\bf{j}} + 3z{\bf{k}}} \right) \cr & {\text{evaluate the gradient at the given point }}P\left( { - 1,2,4} \right) \cr & \nabla f\left( { - 1,2,4} \right) = \frac{2}{{{{\left( { - 1} \right)}^2} + 2{{\left( 2 \right)}^2} + 3{{\left( 4 \right)}^2}}}\left( { - {\bf{i}} + 2\left( 2 \right){\bf{j}} + 3\left( 4 \right){\bf{k}}} \right) \cr & \nabla f\left( { - 1,2,4} \right) = \frac{2}{{57}}\left( { - {\bf{i}} + 4{\bf{j}} + 12{\bf{k}}} \right) \cr & \nabla f\left( { - 1,2,4} \right) = - \frac{2}{{57}}{\bf{i}} + \frac{8}{{57}}{\bf{j}} + \frac{{24}}{{57}}{\bf{k}} \cr & \cr & {\text{Calculate the directional derivative }}{D_{\bf{u}}}f\left( {x,y,z} \right){\text{ at }} \cr & P\left( { - 1,2,4} \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( { - 1,2,4} \right) = \nabla f\left( { - 1,2,4} \right) \cdot {\bf{u}} \cr & {D_{\bf{u}}}f\left( { - 1,2,4} \right) = \left( { - \frac{2}{{57}}{\bf{i}} + \frac{8}{{57}}{\bf{j}} + \frac{{24}}{{57}}{\bf{k}}} \right) \cdot \left( { - \frac{3}{{13}}{\bf{i}} - \frac{4}{{13}}{\bf{j}} - \frac{{12}}{{13}}{\bf{k}}} \right) \cr & {\text{solving the dot product}} \cr & {D_{\bf{u}}}f\left( { - 1,2,4} \right) = \left( { - \frac{2}{{57}}} \right)\left( { - \frac{3}{{13}}} \right) + \left( {\frac{8}{{57}}} \right)\left( { - \frac{4}{{13}}} \right) + \left( {\frac{{24}}{{57}}} \right)\left( { - \frac{{12}}{{13}}} \right) \cr & {D_{\bf{u}}}f\left( { - 1,2,4} \right) = \frac{2}{{247}} - \frac{{32}}{{741}} - \frac{{96}}{{247}} \cr & {D_{\bf{u}}}f\left( { - 1,2,4} \right) = - \frac{{314}}{{741}} \cr} $$