Chapter 13 - Parametric And Polar Curves; Conic Sections - 13.6 Directional Derivatives And Gradients - Exercises Set 13.6 - Page 968: 3

$${D_{\bf{u}}}f\left( {0,0} \right) = - \frac{3}{{\sqrt {10} }}$$

$$\eqalign{ & f\left( {x,y} \right) = \ln \left( {1 + {x^2} + y} \right);\,\,\,\,\,\,P\left( {0,0} \right);\,\,\,\,\,\,\,{\bf{u}} = - \frac{1}{{\sqrt {10} }}{\bf{i}} - \frac{3}{{\sqrt {10} }}{\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( {\ln \left( {1 + {x^2} + y} \right)} \right) \cr & {f_x}\left( {x,y} \right) = \frac{1}{{1 + {x^2} + y}}\left( {0 + 2x + 0} \right) \cr & {f_x}\left( {x,y} \right) = \frac{{2x}}{{1 + {x^2} + y}} \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {\ln \left( {1 + {x^2} + y} \right)} \right) \cr & {f_y}\left( {x,y} \right) = \frac{1}{{1 + {x^2} + y}}\left( {0 + 0 + 1} \right) \cr & {f_y}\left( {x,y} \right) = \frac{1}{{1 + {x^2} + y}} \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{{2x}}{{1 + {x^2} + y}}{\bf{i}} + \frac{1}{{1 + {x^2} + y}}{\bf{j}} \cr & {\text{evaluate the gradient at the given point }}P\left( {0,0} \right) \cr & \nabla f\left( {0,0} \right) = \frac{{2\left( 0 \right)}}{{1 + {{\left( 0 \right)}^2} + \left( 0 \right)}}{\bf{i}} + \frac{1}{{1 + {{\left( 0 \right)}^2} + \left( 0 \right)}}{\bf{j}} \cr & \nabla f\left( {0,0} \right) = 0{\bf{i}} + {\bf{j}} \cr & \cr & {\text{Calculate the directional derivative }}{D_{\bf{u}}}f\left( {x,y} \right){\text{ at }}P\left( {0,0} \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( {0,0} \right) = \nabla f\left( {3,1} \right) \cdot {\bf{u}} \cr & {D_{\bf{u}}}f\left( {0,0} \right) = \left( {0{\bf{i}} + {\bf{j}}} \right) \cdot \left( { - \frac{1}{{\sqrt {10} }}{\bf{i}} - \frac{3}{{\sqrt {10} }}{\bf{j}}} \right) \cr & {\text{solving the dot product}} \cr & {D_{\bf{u}}}f\left( {0,0} \right) = \left( 0 \right)\left( { - \frac{1}{{\sqrt {10} }}} \right) + \left( 1 \right)\left( { - \frac{3}{{\sqrt {10} }}} \right) \cr & {D_{\bf{u}}}f\left( {0,0} \right) = - \frac{3}{{\sqrt {10} }} \cr} $$