Chapter 12 - Functions of Several Veriables - 12.6 Directional Derivatives and the Gradient - 12.6 Exercises - Page 926: 23

$$ - 2$$

$$\eqalign{ & f\left( {x,y} \right) = 3{x^2} + 2y + 5;\,\,\,\,\,\,P\left( {1,2} \right);\,\,\,\,\,\left\langle { - 3,4} \right\rangle \cr & {\bf{u}} = \frac{{\left\langle { - 3,4} \right\rangle }}{{\sqrt {{{\left( { - 3} \right)}^2} + {{\left( 4 \right)}^2}} }} = \left\langle { - \frac{3}{5},\frac{4}{5}} \right\rangle \cr & {\bf{u}}{\text{ is a unit vector in the direction of }}\left\langle { - 3,4} \right\rangle \cr & {\text{The gradient of }}f\left( {x,y} \right){\text{ is}} \cr & {f_x}\left( {x,y} \right) = 6x \cr & {f_y}\left( {x,y} \right) = 2 \cr & \nabla f\left( {x,y} \right) = 6x{\bf{i}} + 2{\bf{j}} \cr & \cr & {\text{Calculate the gradient at the point }}P\left( {1,2} \right) \cr & \nabla f\left( {1,2} \right) = 6x{\bf{i}} + 2{\bf{j}} \cr & \nabla f\left( {1,2} \right) = 6\left( 1 \right){\bf{i}} + 2{\bf{j}} \cr & \nabla f\left( {1,2} \right) = \left\langle {6,2} \right\rangle \cr & \cr & {\text{Computing the directional derivatives of }}f{\text{ at }}\left( {1,2} \right) \cr & \operatorname{in} {\text{ the direction of the vector }}{\bf{u}} = \left\langle { - \frac{3}{5},\frac{4}{5}} \right\rangle \cr & {D_{\bf{u}}}f\left( {a,b} \right) = \nabla f\left( {a,b} \right) \cdot {\bf{u}} \cr & {D_{\bf{u}}}f\left( {1,2} \right) = \left\langle {6,2} \right\rangle \cdot \left\langle { - \frac{3}{5},\frac{4}{5}} \right\rangle \cr & {D_{\bf{u}}}f\left( {1,2} \right) = - \frac{{18}}{5} + \frac{8}{5} \cr & {D_{\bf{u}}}f\left( {1,2} \right) = - 2 \cr} $$