Answer
The directional derivative in the direction of ${\bf{v}}$ at $P = \left( { - 2,1} \right)$ is $4\sqrt 2 $.
Work Step by Step
We are given $f\left( {x,y} \right) = {x^2}{y^3}$ and ${\bf{v}} = {\bf{i}} + {\bf{j}}$.
First, we normalize the direction vector:
${\bf{u}} = \frac{{\bf{v}}}{{||{\bf{v}}||}} = \frac{{\left( {1,1} \right)}}{{\sqrt {\left( {1,1} \right)\cdot\left( {1,1} \right)} }} = \left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)$
Next, we compute the gradient at $P = \left( { - 2,1} \right)$:
$\nabla f = \left( {2x{y^3},3{x^2}{y^2}} \right)$, ${\ \ \ }$ $\nabla {f_P} = \left( { - 4,12} \right)$
Using Theorem 3, we obtain the directional derivative in the direction of ${\bf{v}}$ at $P = \left( { - 2,1} \right)$:
${D_{\bf{u}}}f\left( P \right) = \nabla {f_P}\cdot{\bf{u}}$
${D_{\bf{u}}}f\left( P \right) = \left( { - 4,12} \right)\cdot\left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right) = \frac{8}{{\sqrt 2 }} = 4\sqrt 2 $
So, the directional derivative in the direction of ${\bf{v}}$ at $P = \left( { - 2,1} \right)$ is $4\sqrt 2 $.
