Answer
The directional derivative at $P = \left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right)$ in the direction of ${\bf{v}} = \left( {3, - 4} \right)$ is $ - \frac{{\sqrt 2 }}{5}{\rm{e}}$.
Work Step by Step
We are given $f\left( {x,y} \right) = {{\rm{e}}^{{x^2} + {y^2}}}$ and ${\bf{v}} = \left( {3, - 4} \right)$.
First, we normalize the direction vector:
${\bf{u}} = \frac{{\bf{v}}}{{||{\bf{v}}||}} = \frac{{\left( {3, - 4} \right)}}{{\sqrt {\left( {3, - 4} \right)\cdot\left( {3, - 4} \right)} }} = \left( {\frac{3}{5}, - \frac{4}{5}} \right)$
Next, we compute the gradient at $P = \left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right)$:
$\nabla f = \left( {2x{{\rm{e}}^{{x^2} + {y^2}}},2y{{\rm{e}}^{{x^2} + {y^2}}}} \right)$, ${\ \ \ \ }$ $\nabla {f_P} = \left( {\sqrt 2 {\rm{e}},\sqrt 2 {\rm{e}}} \right)$
Using Theorem 3 of Section 15.5, we obtain the directional derivative in the direction of ${\bf{v}} = \left( {3, - 4} \right)$ at $P = \left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right)$:
${D_{\bf{u}}}f\left( P \right) = \nabla {f_P}\cdot{\bf{u}}$
${D_{\bf{u}}}f\left( P \right) = \left( {\sqrt 2 {\rm{e}},\sqrt 2 {\rm{e}}} \right)\cdot\left( {\frac{3}{5}, - \frac{4}{5}} \right) = - \frac{{\sqrt 2 }}{5}{\rm{e}}$
So, the directional derivative at $P = \left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2}} \right)$ in the direction of ${\bf{v}} = \left( {3, - 4} \right)$ is $ - \frac{{\sqrt 2 }}{5}{\rm{e}}$.
