Find the directional derivative of $f ( x , y ) = 2 \sqrt { x } - y ^ { 3 }$ at the point $( 1,3 )$ in the direction toward the point $( 3,1 )$.
Select the correct answer.
A) $7 \sqrt { 2 }$
B) $14 \sqrt { 2 }$
C) $\sqrt { 2 }$
D) 28
E) none of these

Question

Quizplus · Accepted Answer

We need to find the unit vector in the direction of $\overrightarrow{PQ}$, where $P=(1,3)$ and $Q=(3,1)$:

$$\overrightarrow{PQ}=\begin{pmatrix}3-1\1-3\end{pmatrix}=\begin{pmatrix}2\-2\end{pmatrix}$$

So the unit vector in that direction is:

$$\vec{u}=\frac{\begin{pmatrix}2\-2\end{pmatrix}}{\left\| \begin{pmatrix}2\-2\end{pmatrix} ight\|}=\frac{\begin{pmatrix}1\-1\end{pmatrix}}{\sqrt{2}}$$

Now we need to find the gradient of $f$ at $(1,3)$:

$$
abla f(x,y)=\begin{pmatrix}\frac{1}{\sqrt{x}}\-3y^2\end{pmatrix}$$

So:

$$
abla f(1,3)=\begin{pmatrix}1\-27\end{pmatrix}$$

Finally, the directional derivative of $f$ at $(1,3)$ in the direction of $\vec{u}$ is:

$$D_{\vec{u}}f(1,3)=
abla f(1,3)\cdot \vec{u}=\begin{pmatrix}1\-27\end{pmatrix}\cdot \frac{\begin{pmatrix}1\-1\end{pmatrix}}{\sqrt{2}}=14\sqrt{2}$$

So the answer is (B).

Find the Directional Derivative Of f(x,y)=2x−y3f ( x , y ) = 2 \sqrt { x } - y ^ { 3 }f(x,y)=2x​−y3

Find the Directional Derivative Of $f ( x , y ) = 2 \sqrt { x } - y ^ { 3 }$