Chapter 13 - Parametric And Polar Curves; Conic Sections - 13.6 Directional Derivatives And Gradients - Exercises Set 13.6 - Page 968: 11

$${D_{\bf{u}}}f\left( {1,4} \right) = - \frac{{16}}{{\sqrt 2 }}$$

$$\eqalign{ & f\left( {x,y} \right) = {y^2}\ln x;\,\,\,\,\,P\left( {1,4} \right);\,\,\,\,\,\,{\bf{a}} = - 3{\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( {{y^2}\ln x} \right) \cr & {\text{Treat }}y{\text{ as a constant}} \cr & {f_x}\left( {x,y} \right) = {y^2}\left( {\frac{1}{x}} \right) \cr & {f_x}\left( {x,y} \right) = \frac{{{y^2}}}{x} \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {{y^2}\ln x} \right) \cr & {\text{Treat }}x{\text{ as a constant}} \cr & {f_y}\left( {x,y} \right) = \ln x\left( {2y} \right) \cr & {f_y}\left( {x,y} \right) = 2y\ln x \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) = \frac{{{y^2}}}{x}{\bf{i}} + 2y\ln x{\bf{j}} \cr & {\text{evaluate the gradient at the given point }}P\left( {1,4} \right) \cr & \nabla f\left( {1,4} \right) = \frac{{{{\left( 4 \right)}^2}}}{1}{\bf{i}} + 2\left( 4 \right)\ln \left( 1 \right){\bf{j}} \cr & \nabla f\left( {1,4} \right) = 16{\bf{i}} + 0{\bf{j}} \cr & \cr & {\text{Note that }}{\bf{a}}{\text{ is not a unit vector}}{\text{, but since }}\left| {\bf{a}} \right| = 3\sqrt 2 ,{\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{{ - 3{\bf{i}} + 3{\bf{j}}}}{{3\sqrt 2 }} = - \frac{1}{{\sqrt 2 }}{\bf{i}} + \frac{1}{{\sqrt 2 }}{\bf{j}} \cr & {\text{Calculate the directional derivative }}{D_{\bf{u}}}f\left( {x,y} \right){\text{ at }}P\left( {1,4} \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( {1,4} \right) = \nabla f\left( {1,4} \right) \cdot {\bf{u}} \cr & {D_{\bf{u}}}f\left( {1,4} \right) = \left( {16{\bf{i}} + 0{\bf{j}}} \right) \cdot \left( { - \frac{1}{{\sqrt 2 }}{\bf{i}} + \frac{1}{{\sqrt 2 }}{\bf{j}}} \right) \cr & {\text{solving the dot product}} \cr & {D_{\bf{u}}}f\left( {1,4} \right) = \left( {16} \right)\left( { - \frac{1}{{\sqrt 2 }}} \right) + \left( 0 \right)\left( {\frac{1}{{\sqrt 2 }}} \right) \cr & {D_{\bf{u}}}f\left( {1,4} \right) = - \frac{{16}}{{\sqrt 2 }} \cr} $$