Find the directional derivative of the function at P in the direction of $\overrightarrow { \mathbf { v } } \text {. }$ $f ( x , y , z ) = x y + y z + x z , P ( 1,1,1 ) , \quad \overrightarrow { \mathbf { v } } = 4 \hat { \mathbf { i } } + 5 \hat { \mathbf { j } } - 3 \hat { \mathbf { k } }$
A) $4 / 5 \sqrt { 2 }$
B) $- 8 / 5 \sqrt { 2 }$
C) $12 / 5 \sqrt { 2 }$
D) $24 / 5 \sqrt { 2 }$
E) $- 24 / 5 \sqrt { 2 }$

Question

Quizplus · Accepted Answer

The directional derivative of $f$ at a point $P$ in the direction of a vector $\overrightarrow{\mathbf{v}}$ is given by $\nabla f(P) \cdot \frac{\overrightarrow{\mathbf{v}}}{|\overrightarrow{\mathbf{v}}|}$. First, calculate the gradient of $f$, $\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right) = (y+z, x+z, x+y)$. At $P(1,1,1)$, $\nabla f(P) = (2,2,2)$. The magnitude of $\overrightarrow{\mathbf{v}}$ is $|\overrightarrow{\mathbf{v}}| = \sqrt{4^2 + 5^2 + (-3)^2} = \sqrt{50}$. Thus, the unit vector in the direction of $\overrightarrow{\mathbf{v}}$ is $\frac{1}{\sqrt{50}}(4,5,-3)$. The directional derivative is $\nabla f(P) \cdot \frac{\overrightarrow{\mathbf{v}}}{|\overrightarrow{\mathbf{v}}|} = (2,2,2) \cdot \frac{1}{\sqrt{50}}(4,5,-3) = \frac{1}{\sqrt{50}}(8+10-6) = \frac{12}{\sqrt{50}} = \frac{12}{5\sqrt{2}}$.

Find the Directional Derivative of the Function at P in the Direction