91影视

The given function is $x{y}^2+yz$ with respect to reference point $\left(1,1,2\right)$ and vector $2\mathrm{i}-\mathrm{j}+2\mathrm{k}$.

Directional derivative represents the rate at which value of the function changes at fixed point and direction.

Use the formula $\frac{\mathrm{诲蠁}}{\mathrm{du}}=鈭�\mathrm{f}.\mathrm{u}$

Here $\frac{\mathrm{诲蠁}}{\mathrm{du}}$: directional derivative of function $\mathrm{蠁}$, $鈭�\mathrm{蠁}$: the gradient of $\mathrm{蠁}$and u : a unit vector.

Given that $\mathrm{蠁}={\mathrm{xy}}^2+\mathrm{yz}$and point $\left(1,1,2\right)$, let$\mathrm{A}=2\mathrm{i}鈥�\mathrm{j}+2\mathrm{k}$

Then, unit vector $\mathrm{u}=\frac{\mathrm{A}}{\left|\mathrm{A}\right|}$

$$\begin{gathered} \mathrm{u}=\frac{2\mathrm{i}鈥�\mathrm{j}+2\mathrm{k}}{\sqrt{{\left(2\right)}^2+{\left(鈥�1\right)}^2+{\left(2\right)}^2}} \\ \mathrm{u}=\frac{2\mathrm{i}鈥�\mathrm{j}+2\mathrm{k}}{\sqrt{9}} \\ \mathrm{u}=\frac{2\mathrm{i}鈥�\mathrm{j}+2\mathrm{k}}{3} \end{gathered}$$

Use the formula $鈭�\mathrm{蠁}=\mathrm{i}\frac{鈭�\mathrm{蠁}}{鈭�\mathrm{x}}+\mathrm{j}\frac{鈭�\mathrm{蠁}}{鈭�\mathrm{y}}+\mathrm{k}\frac{鈭�\mathrm{蠁}}{鈭�\mathrm{z}}$

$$\begin{gathered} 鈭�\mathrm{蠁}=\mathrm{i}\frac{鈭�\mathrm{蠁}}{鈭�\mathrm{x}}+\mathrm{j}\frac{鈭�\mathrm{蠁}}{鈭�\mathrm{y}}+\mathrm{k}\frac{鈭�\mathrm{蠁}}{鈭�\mathrm{z}} \\ 鈭�\mathrm{蠁}=\mathrm{i}\frac{鈭�}{鈭�\mathrm{x}}\left({\mathrm{xy}}^2+\mathrm{yz}\right)+\mathrm{j}\frac{鈭�}{鈭�\mathrm{y}}+\mathrm{k}\frac{鈭�}{鈭�\mathrm{z}}\left({\mathrm{xy}}^2+\mathrm{yz}\right) \\ 鈭�{\mathrm{蠁}}_{\left(1,1,2\right)}=\mathrm{i}{\left(1\right)}^2+\mathrm{j}\left(2\left(1\right)\left(1\right)+\left(2\right)\right)+\mathrm{k}\left(1\right) \\ 鈭�{\mathrm{蠁}}_{\left(1,1,2\right)}=\mathrm{i}+4\mathrm{j}+\mathrm{k} \end{gathered}$$

The formula for directional derivative states that$\frac{\mathrm{诲蠁}}{\mathrm{du}}=鈭�\mathrm{蠁}.\mathrm{u}$

Put the two value that are given below.

$$\begin{gathered} 鈭�\mathrm{蠁}=\mathrm{i}+4\mathrm{j}+\mathrm{k} \\ \mathrm{u}=\frac{1}{3}\left(2\mathrm{i}-\mathrm{j}+2\mathrm{k}\right) \end{gathered}$$

Then the equation can be further solved.

$$\begin{gathered} \frac{\mathrm{诲蠁}}{\mathrm{du}}=\left(\mathrm{i}+4\mathrm{j}+\mathrm{k}\right)\frac{1}{3}\left(2\mathrm{i}-\mathrm{j}+2\mathrm{k}\right) \\ \frac{\mathrm{诲蠁}}{\mathrm{du}}=\frac{1}{3}\left(2-4+2\right) \\ \frac{\mathrm{诲蠁}}{\mathrm{du}}=0 \end{gathered}$$

The directional derivative of given function is zero.