Question

The maximum value of the directional derivative of \( \varphi = x^2 yz \) at the point \( (1, 4, 1) \) is:

• A. \( \sqrt{10} \)
• B. \( 3 \)
• C. \( 9 \)
• D. \( \dfrac{10}{3} \)

Hint:

The maximum directional derivative at a point is the magnitude of the gradient vector at that point.

Answer 1

The Correct Option is C

The maximum value of the directional derivative at a point is the magnitude of the gradient vector at that point. Given: \( \varphi = x^2 yz \) Gradient: \[ \nabla \varphi = \left( \frac{\partial \varphi}{\partial x}, \frac{\partial \varphi}{\partial y}, \frac{\partial \varphi}{\partial z} \right) = \left( 2x y z, x^2 z, x^2 y \right) \] At the point \( (1, 4, 1) \), we get: \[ \nabla \varphi = \left(2 \cdot 1 \cdot 4 \cdot 1, 1^2 \cdot 1, 1^2 \cdot 4\right) = (8, 1, 4) \] Magnitude of gradient: \[ |\nabla \varphi| = \sqrt{8^2 + 1^2 + 4^2} = \sqrt{64 + 1 + 16} = \sqrt{81} = 9 \]