Answer
The directional derivative at $P = \left( {3, - 1} \right)$ in the direction of ${\bf{v}} = 2{\bf{i}} + {\bf{j}}$ is $ - \frac{{54}}{{\sqrt 5 }}$.
Work Step by Step
We are given $f\left( {x,y} \right) = {x^3}{y^4}$ and ${\bf{v}} = 2{\bf{i}} + {\bf{j}}$.
First, we normalize the direction vector:
${\bf{u}} = \frac{{\bf{v}}}{{||{\bf{v}}||}} = \frac{{\left( {2,1} \right)}}{{\sqrt {\left( {2,1} \right)\cdot\left( {2,1} \right)} }} = \left( {\frac{2}{{\sqrt 5 }},\frac{1}{{\sqrt 5 }}} \right)$
Next, we compute the gradient at $P = \left( {3, - 1} \right)$:
$\nabla f = \left( {3{x^2}{y^4},4{x^3}{y^3}} \right)$, ${\ \ \ }$ $\nabla {f_P} = \left( {27, - 108} \right)$
Using Theorem 3 of Section 15.5, we obtain the directional derivative in the direction of ${\bf{v}}$ at $P = \left( {3, - 1} \right)$:
${D_{\bf{u}}}f\left( P \right) = \nabla {f_P}\cdot{\bf{u}}$
${D_{\bf{u}}}f\left( P \right) = \left( {27, - 108} \right)\cdot\left( {\frac{2}{{\sqrt 5 }},\frac{1}{{\sqrt 5 }}} \right) = - \frac{{54}}{{\sqrt 5 }}$
So, the directional derivative at $P = \left( {3, - 1} \right)$ in the direction of ${\bf{v}} = 2{\bf{i}} + {\bf{j}}$ is $ - \frac{{54}}{{\sqrt 5 }}$.
