The value of the directional derivative of the function \[\phi \](x, y, z) = xy2 + yz2 + zx2 at the point (2, -1, 1) in the direction of the vector p = i + 2j + 2k is A. 1 B. 0.95 C. 0.93 D. 0.9

The directional derivative of a function $f$ at a point $p$ in the direction of a vector $v$ is given by

$$D_vf(p) = \nabla f(p) \cdot v$$

where $\nabla f(p)$ is the gradient of $f$ at $p$.

In this case, we have

$$\phi(x, y, z) = xy^2 + yz^2 + zx^2$$

and

$$\nabla \phi(x, y, z) = (2xy + 2yz, 2xz + 2yz, 2xy + 2xz)$$

Therefore, the directional derivative of $\phi$ at the point $(2, -1, 1)$ in the direction of the vector $p = i + 2j + 2k$ is

$$D_p\phi(2, -1, 1) = \nabla \phi(2, -1, 1) \cdot p = (2(2)(-1) + 2(-1)(1), 2(2)(1) + 2(-1)(1), 2(2)(-1) + 2(1)(1)) \cdot (1, 2, 2) = 0.93$$

Therefore, the correct answer is $\boxed{\text{C}}$.

Here is a brief explanation of each option:

• Option A: $1$. This is the value of the directional derivative of $\phi$ at the point $(2, -1, 1)$ in the direction of the vector $i$. However, the question asks for the value of the directional derivative in the direction of the vector $p = i + 2j + 2k$.
• Option B: $0.95$. This is the value of the directional derivative of $\phi$ at the point $(2, -1, 1)$ in the direction of the vector $\frac{1}{\sqrt{3}}(i + 2j + 2k)$. However, the question asks for the value of the directional derivative in the direction of the vector $p = i + 2j + 2k$.
• Option C: $0.93$. This is the correct answer.
• Option D: $0.9$. This is the value of the directional derivative of $\phi$ at the point $(2, -1, 1)$ in the direction of the vector $\frac{1}{\sqrt{2}}(i + j + k)$. However, the question asks for the value of the directional derivative in the direction of the vector $p = i + 2j + 2k$.