Chapter 13 - Parametric And Polar Curves; Conic Sections - 13.6 Directional Derivatives And Gradients - Exercises Set 13.6 - Page 968: 4

$${D_{\bf{u}}}f\left( {3,4} \right) = \frac{{ - 7c - 7d}}{5}$$

$$\eqalign{ & f\left( {x,y} \right) = \frac{{cx + dy}}{{x - y}};\,\,\,\,\,\,P\left( {3,4} \right);\,\,\,\,\,\,\,{\bf{u}} = \frac{4}{5}{\bf{i}} + \frac{3}{5}{\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( {\frac{{cx + dy}}{{x - y}}} \right) \cr & {\text{use the quotient rule and treat }}y{\text{ as a constant}} \cr & {f_x}\left( {x,y} \right) = \frac{{\left( {x - y} \right)\left( c \right) - \left( {cx + dy} \right)\left( 1 \right)}}{{{{\left( {x - y} \right)}^2}}} \cr & {f_x}\left( {x,y} \right) = \frac{{xc - cy - cx - dy}}{{{{\left( {x - y} \right)}^2}}} \cr & {f_x}\left( {x,y} \right) = \frac{{ - cy - dy}}{{{{\left( {x - y} \right)}^2}}} \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {\frac{{cx + dy}}{{x - y}}} \right) \cr & {\text{use the quotient rule and treat }}x{\text{ as a constant}} \cr & {f_y}\left( {x,y} \right) = \frac{{\left( {x - y} \right)\left( d \right) - \left( {cx + dy} \right)\left( { - 1} \right)}}{{{{\left( {x - y} \right)}^2}}} \cr & {f_y}\left( {x,y} \right) = \frac{{dx - dy + cx + dy}}{{{{\left( {x - y} \right)}^2}}} \cr & {f_y}\left( {x,y} \right) = \frac{{cx + dx}}{{{{\left( {x - y} \right)}^2}}} \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{{ - cy - dy}}{{{{\left( {x - y} \right)}^2}}}{\bf{i}} + \frac{{cx + dx}}{{{{\left( {x - y} \right)}^2}}}{\bf{j}} \cr & {\text{evaluate the gradient at the given point }}P\left( {3,4} \right) \cr & \nabla f\left( {3,4} \right) = \frac{{ - 4c - 4d}}{{{{\left( {3 - 4} \right)}^2}}}{\bf{i}} + \frac{{3c + 3d}}{{{{\left( {3 - 4} \right)}^2}}}{\bf{j}} \cr & \nabla f\left( {3,4} \right) = \left( { - 4c - 4d} \right){\bf{i}} + \left( {3c + 3d} \right){\bf{j}} \cr & \cr & {\text{Calculate the directional derivative }}{D_{\bf{u}}}f\left( {x,y} \right){\text{ at }}P\left( {3,4} \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( {3,4} \right) = \nabla f\left( {3,4} \right) \cdot {\bf{u}} \cr & {D_{\bf{u}}}f\left( {3,4} \right) = \left( { - 4c - 4d} \right){\bf{i}} + \left( {3c + 3d} \right){\bf{j}} \cdot \left( {\frac{4}{5}{\bf{i}} + \frac{3}{5}{\bf{j}}} \right) \cr & {\text{solving the dot product}} \cr & {D_{\bf{u}}}f\left( {3,4} \right) = \left( { - 4c - 4d} \right)\left( {\frac{4}{5}} \right) + \left( {3c + 3d} \right)\left( {\frac{3}{5}} \right) \cr & {D_{\bf{u}}}f\left( {3,4} \right) = \frac{{ - 16c - 16d}}{5} + \frac{{9c + 9d}}{5} \cr & {D_{\bf{u}}}f\left( {3,4} \right) = \frac{{ - 16c - 16d + 9c + 9d}}{5} \cr & {D_{\bf{u}}}f\left( {3,4} \right) = \frac{{ - 7c - 7d}}{5} \cr} $$