Chapter 13 - Parametric And Polar Curves; Conic Sections - 13.6 Directional Derivatives And Gradients - Exercises Set 13.6 - Page 968: 10

$${D_{\bf{u}}}f\left( {1,0} \right) = \frac{{27}}{{\sqrt 2 }}$$

$$\eqalign{ & f\left( {x,y} \right) = 9{x^3} - 2{y^3};\,\,\,\,\,P\left( {1,0} \right);\,\,\,\,\,\,{\bf{a}} = {\bf{i}} - {\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( {9{x^3} - 2{y^3}} \right) \cr & {\text{Treat }}y{\text{ as a constant}} \cr & {f_x}\left( {x,y} \right) = 27{x^2} \cr & and \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {9{x^3} - 2{y^3}} \right) \cr & {\text{Treat }}x{\text{ as a constant}} \cr & {f_y}\left( {x,y} \right) = - 6{y^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) = 27{x^2}{\bf{i}} - 6{y^2}{\bf{j}} \cr & {\text{evaluate the gradient at the given point }}P\left( {1,0} \right) \cr & \nabla f\left( {1,0} \right) = 27{\left( 1 \right)^2}{\bf{i}} - 6{\left( 0 \right)^2}{\bf{j}} \cr & \nabla f\left( {1,0} \right) = 27{\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| = \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{{{\bf{i}} - {\bf{j}}}}{{\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( {1,0} \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( {1,0} \right) = \nabla f\left( {1,0} \right) \cdot {\bf{u}} \cr & {D_{\bf{u}}}f\left( {1,0} \right) = \left( {27{\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( {1,0} \right) = \left( {27} \right)\left( {\frac{1}{{\sqrt 2 }}} \right) + \left( 0 \right)\left( { - \frac{1}{{\sqrt 2 }}} \right) \cr & {D_{\bf{u}}}f\left( {1,0} \right) = \frac{{27}}{{\sqrt 2 }} \cr} $$