Answer
The rate of change of $f\left( {x,y} \right)$ at $\left( {0,0} \right)$ is $3\sqrt 2 $.
Work Step by Step
We are given $P = \left( {0,0} \right)$ and $\nabla {f_P} = \left( {2,4} \right)$. According to Theorem 3, the rate of change of $f\left( {x,y} \right)$ at $P$ is
${D_{\bf{u}}}f\left( P \right) = \nabla {f_P}\cdot{\bf{u}}$,
where ${\bf{u}}$ in this case is the unit vector in the direction making an angle of $45^\circ $ with the $x$-axis. So,
${\bf{u}} = \left( {\cos 45^\circ ,\sin 45^\circ } \right) = \left( {\frac{1}{2}\sqrt 2 ,\frac{1}{2}\sqrt 2 } \right)$
${D_{\bf{u}}}f\left( P \right) = \left( {2,4} \right)\cdot\left( {\frac{1}{2}\sqrt 2 ,\frac{1}{2}\sqrt 2 } \right) = 3\sqrt 2 $
So, the rate of change of $f\left( {x,y} \right)$ at $\left( {0,0} \right)$ is $3\sqrt 2 $.
