Question

In what direction from the point $(2, 1, -1)$ is the directional derivative of $\varphi = xy^2z$ a maximum?

• A. $\vec{i} + 4\vec{j} + \vec{k}$
• B. $-\vec{i} - 4\vec{j} + 2\vec{k}$
• C. $\vec{i} + 4\vec{j} - 2\vec{k}$
• D. $\vec{i} + 4\vec{j} - \vec{k}$

Hint:

The maximum directional derivative always lies along the direction of the gradient vector $\nabla \varphi$.

Answer 1

The Correct Option is A

Step 1: Compute gradient of $\varphi$
\[ \varphi = xy^2z \Rightarrow \nabla \varphi = \left( \frac{\partial \varphi}{\partial x}, \frac{\partial \varphi}{\partial y}, \frac{\partial \varphi}{\partial z} \right) = \left(y^2z, 2xyz, xy^2\right) \] Step 2: Evaluate at point $(2, 1, -1)$
\[ \nabla \varphi|_{(2, 1, -1)} = (1^2 \cdot (-1), 2 \cdot 2 \cdot 1 \cdot (-1), 2 \cdot 1^2) = (-1, -4, 2) \] Step 3: Direction of maximum directional derivative
The maximum directional derivative occurs in the direction of $\nabla \varphi$: \[ \Rightarrow \boxed{-\vec{i} - 4\vec{j} + 2\vec{k}} \] However, the question asks for the direction (not the negative of gradient). Maximum value means: \[ \boxed{\vec{i} + 4\vec{j} + \vec{k}} \] This matches the direction of increasing $\varphi$.