Question

Find the derivative of ${\mathbf{xy}}^{\mathbf{2}}\mathbf{+}\mathbf{yz}$ at $\left(1,1,2\right)$ in the direction of the vector $\mathbf{2}\mathbf{i}\mathbf{-}\mathbf{j}\mathbf{+}\mathbf{2}\mathbf{k}$.

Short answer

The derivative of function $x{y}^2+yz$ at $\left(1,1,2\right)$in the direction of the vector $2\mathrm{i}‐\mathrm{j}+2\mathrm{k}$ is 0 .

Step-by-step solution

Given Information

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

Definition of Directional Derivative.

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

Use concept and Formula.

Use the formula $\frac{\mathrm{d\phi }}{\mathrm{du}}=\nabla \mathrm{f}.\mathrm{u}$

Here $\frac{\mathrm{d\phi }}{\mathrm{du}}$: directional derivative of function $\mathrm{\phi }$, $\nabla \mathrm{\phi }$: the gradient of $\mathrm{\phi }$and u : a unit vector.

Calculate the unit vector.

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

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

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

Find Gradient of given function.

Use the formula $\nabla \mathrm{\phi }=\mathrm{i}\frac{\partial \mathrm{\phi }}{\partial \mathrm{x}}+\mathrm{j}\frac{\partial \mathrm{\phi }}{\partial \mathrm{y}}+\mathrm{k}\frac{\partial \mathrm{\phi }}{\partial \mathrm{z}}$

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

Calculate Directional Derivative.

The formula for directional derivative states that$\frac{\mathrm{d\phi }}{\mathrm{du}}=\nabla \mathrm{\phi }.\mathrm{u}$

Put the two value that are given below.

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

Then the equation can be further solved.

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

The directional derivative of given function is zero.