Answer
The directional derivative at $P = \left( {1,1,1} \right)$ in the direction of ${\bf{v}} = \left( {2, - 1,2} \right)$ is $\frac{4}{3}$.
Work Step by Step
We are given $f\left( {x,y,z} \right) = zx - x{y^2}$ and ${\bf{v}} = \left( {2, - 1,2} \right)$.
First, we normalize the direction vector:
${\bf{u}} = \frac{{\bf{v}}}{{||{\bf{v}}||}} = \frac{{\left( {2, - 1,2} \right)}}{{\sqrt {\left( {2,1,2} \right)\cdot\left( {2,1,2} \right)} }} = \left( {\frac{2}{3}, - \frac{1}{3},\frac{2}{3}} \right)$
Next, we compute the gradient at $P = \left( {1,1,1} \right)$:
$\nabla f = \left( {z - {y^2}, - 2xy,x} \right)$, ${\ \ \ \ }$ $\nabla {f_P} = \left( {0, - 2,1} \right)$
Using Theorem 3 of Section 15.5, we obtain the directional derivative in the direction of ${\bf{v}} = \left( {2, - 1,2} \right)$ at $P = \left( {1,1,1} \right)$:
${D_{\bf{u}}}f\left( P \right) = \nabla {f_P}\cdot{\bf{u}}$
${D_{\bf{u}}}f\left( P \right) = \left( {0, - 2,1} \right)\cdot\left( {\frac{2}{3}, - \frac{1}{3},\frac{2}{3}} \right) = \frac{4}{3}$
So, the directional derivative at $P = \left( {1,1,1} \right)$ in the direction of ${\bf{v}} = \left( {2, - 1,2} \right)$ is $\frac{4}{3}$.
