Let $f ( x , y , z ) = x ^ { 2 } y - x y z ^ { 2 }$ . Then the directional derivative of f at $P = ( 0,1,2 )$ in the direction of the unit vector $\mathbf { U }$ which is parallel to $\overrightarrow { P Q }$ where $Q = ( 1,4,3 )$ is

A) $\frac { \sqrt { 3 } + 1 } { 2 }$
B) $- \frac { 4 } { \sqrt { 26 } }$
C) $\frac { - 4 + 5 \sqrt { 3 } } { 2 }$
D) $- \frac { 5 } { 13 }$
E) $\frac { 78 } { 325 }$

Question

Let

f ( x , y , z ) = x ^ { 2 } y - x y z ^ { 2 }

. Then the directional derivative of f at

P = ( 0,1,2 )

in the direction of the unit vector

\mathbf { U }

which is parallel to

\overrightarrow { P Q }

where

Q = ( 1,4,3 )

is

A) $\frac { \sqrt { 3 } + 1 } { 2 }$
B) $- \frac { 4 } { \sqrt { 26 } }$
C) $\frac { - 4 + 5 \sqrt { 3 } } { 2 }$
D) $- \frac { 5 } { 13 }$
E) $\frac { 78 } { 325 }$

Quizplus · Accepted Answer

The directional derivative of $f$ at a point $P$ in the direction of a unit vector $\mathbf{U}$ is given by $\nabla f(P) \cdot \mathbf{U}$, where $\nabla f(P)$ is the gradient of $f$ at $P$, and $\cdot$ denotes the dot product. First, calculate the gradient of $f$:$$\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = (2xy - yz^2, x^2 - xz^2, -2xyz)$$At $P = (0,1,2)$:$$\nabla f(P) = (0 - 1\cdot2^2, 0 - 0, -2\cdot0\cdot1) = (-4, 0, 0)$$The direction vector from $P$ to $Q$ is $\overrightarrow{PQ} = Q - P = (1-0, 4-1, 3-2) = (1, 3, 1)$. To find the unit vector in the direction of $\overrightarrow{PQ}$, we normalize it:$$\mathbf{U} = \frac{\overrightarrow{PQ}}{||\overrightarrow{PQ}||} = \frac{(1, 3, 1)}{\sqrt{1^2 + 3^2 + 1^2}} = \frac{(1, 3, 1)}{\sqrt{11}}$$However, the magnitude calculation seems to have been mistaken in the normalization step. Correcting this, the magnitude of $\overrightarrow{PQ}$ is $\sqrt{1^2 + 3^2 + 1^2} = \sqrt{11}$, but the direction vector should actually be normalized correctly for the calculation, which is a step in finding the unit vector correctly. The mistake in the explanation does not affect the correct choice directly but indicates a need for careful calculation in these steps.The directional derivative is then:$$\nabla f(P) \cdot \mathbf{U} = (-4, 0, 0) \cdot \left(\frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{1}{\sqrt{11}}\right) = -\frac{4}{\sqrt{11}} \cdot 1 + 0 + 0 = -\frac{4}{\sqrt{11}}$$The correct calculation of the directional derivative involves using the correct unit vector, which is derived from the direction of $\overrightarrow{PQ}$. The mistake in normalization does not directly lead to any of the provided options, indicating a need to correct the explanation for finding the unit vector and then correctly applying it to find the directional derivative. The correct answer should be based on the correct application of these principles, but the provided options and calculations do not directly align with the mistake made in the explanation.

Let f(x,y,z)=x2y−xyz2f ( x , y , z ) = x ^ { 2 } y - x y z ^ { 2 }f(x,y,z)=x2y−xyz2

Let $f ( x , y , z ) = x ^ { 2 } y - x y z ^ { 2 }$