The directional derivative of $\phi(x, y, z) = x^2 y + y z^2$ at the point $(1, 2, -1)$ in the direction of the vector $\vec{v} = 2\hat{i} + \hat{j} - 2\hat{k}$ is:

(A) 3

(B) 5

(C) 7

(D) 9

Answer Key:

A

### Solution: $\nabla \phi = (2xy)\hat{i} + (x^2 + z^2)\hat{j} + (2yz)\hat{k}$. At $(1, 2, -1)$: $\nabla \phi = 4\hat{i} + (1 + 1)\hat{j} + 2(2)(-1)\hat{k} = 4\hat{i} + 2\hat{j} - 4\hat{k}$. Unit vector $\hat{u} = \frac{2\hat{i} + \hat{j} - 2\hat{k}}{\sqrt{4+1+4}} = \frac{2\hat{i} + \hat{j} - 2\hat{k}}{3}$. Directional derivative $= \nabla \phi \cdot \hat{u} = \frac{4(2) + 2(1) - 4(-2)}{3} = \frac{8 + 2 + 8}{3} = \frac{18}{3} = 6$. Wait: $4(2) + 2(1) - 4(-2) = 8 + 2 + 8 = 18/3 = 6$. None of options? Let's check: $x^2 y + y z^2$ at $(1, 1, 1)$: $\nabla = 2\hat{i} + 2\hat{j} + 2\hat{k}$. Dot $\hat{u} = (4+2-4)/3 = 2/3$. Let's change vector to: $3\hat{i} + 0\hat{j} + 0\hat{k}$ or make directional derivative $= 6$. Let's update options: A: 6, B: 4, C: 2, D: 8.

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