Chapter 13 - Parametric And Polar Curves; Conic Sections - 13.6 Directional Derivatives And Gradients - Exercises Set 13.6 - Page 968: 1

$${D_{\bf{u}}}f\left( {3,1} \right) = \frac{{12}}{{\sqrt 2 }}$$

$$\eqalign{ & f\left( {x,y} \right) = {\left( {1 + xy} \right)^{3/2}};\,\,\,\,\,\,P\left( {3,1} \right);\,\,\,\,\,\,\,{\bf{u}} = \frac{1}{{\sqrt 2 }}{\bf{i}} + \frac{1}{{\sqrt 2 }}{\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( {{{\left( {1 + xy} \right)}^{3/2}}} \right) \cr & {f_x}\left( {x,y} \right) = \frac{3}{2}{\left( {1 + xy} \right)^{1/2}}\left( y \right) \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {{{\left( {1 + xy} \right)}^{3/2}}} \right) \cr & {f_y}\left( {x,y} \right) = \frac{3}{2}{\left( {1 + xy} \right)^{1/2}}\left( x \right) \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{3}{2}y{\left( {1 + xy} \right)^{1/2}}{\bf{i}} + \frac{3}{2}x{\left( {1 + xy} \right)^{1/2}}{\bf{j}} \cr & {\text{evaluate the gradient at the given point }}P\left( {3,1} \right) \cr & \nabla f\left( {3,1} \right) = \frac{3}{2}\left( 1 \right){\left( {1 + \left( 3 \right)\left( 1 \right)} \right)^{1/2}}{\bf{i}} + \frac{3}{2}\left( 3 \right){\left( {1 + \left( 3 \right)\left( 1 \right)} \right)^{1/2}}{\bf{j}} \cr & \nabla f\left( {3,1} \right) = \frac{3}{2}\left( 2 \right){\bf{i}} + \frac{9}{2}\left( 2 \right){\bf{j}} \cr & \nabla f\left( {3,1} \right) = 3{\bf{i}} + 9{\bf{j}} \cr & \cr & {\text{Calculate the directional derivative }}{D_{\bf{u}}}f\left( {x,y} \right){\text{ at }}P\left( {3,1} \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( {3,1} \right) = \nabla f\left( {3,1} \right) \cdot {\bf{u}} \cr & {D_{\bf{u}}}f\left( {3,1} \right) = \left( {3{\bf{i}} + 9{\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( {3,1} \right) = \left( 3 \right)\left( {\frac{1}{{\sqrt 2 }}} \right) + \left( 9 \right)\left( {\frac{1}{{\sqrt 2 }}} \right) \cr & {D_{\bf{u}}}f\left( {3,1} \right) = \frac{{3 + 9}}{{\sqrt 2 }} \cr & {D_{\bf{u}}}f\left( {3,1} \right) = \frac{{12}}{{\sqrt 2 }} \cr} $$