The problem asks us to find the directional derivative of the function $f(x, y, z) = x^3y - y^2z^2$ at the point $p = (-2, 1, 3)$ in the direction of the vector $a = i - 2j + 2k$.

1. Problem Description

The problem asks us to find the directional derivative of the function $f(x, y, z) = x^3y - y^2z^2$ at the point $p = (-2, 1, 3)$ in the direction of the vector $a = i - 2j + 2k$.

2. Solution Steps

First, we need to find the gradient of $f$.

The gradient of $f$ is given by $\nabla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z})$.

$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^3y - y^2z^2) = 3x^2y$

$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^3y - y^2z^2) = x^3 - 2yz^2$

$\frac{\partial f}{\partial z} = \frac{\partial}{\partial z}(x^3y - y^2z^2) = -2y^2z$

So, $\nabla f = (3x^2y, x^3 - 2yz^2, -2y^2z)$.

Now we evaluate the gradient at the point $p = (-2, 1, 3)$:

$\nabla f(-2, 1, 3) = (3(-2)^2(1), (-2)^3 - 2(1)(3)^2, -2(1)^2(3)) = (3(4)(1), -8 - 2(9), -2(3)) = (12, -8 - 18, -6) = (12, -26, -6)$.

Next, we need to find the unit vector in the direction of $a = i - 2j + 2k$.

$a = (1, -2, 2)$.

The magnitude of $a$ is $||a|| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.

The unit vector is $\hat{u} = \frac{a}{||a||} = \frac{(1, -2, 2)}{3} = (\frac{1}{3}, -\frac{2}{3}, \frac{2}{3})$.

The directional derivative of $f$ at $p$ in the direction of $a$ is given by $D_a f(p) = \nabla f(p) \cdot \hat{u}$.

$D_a f(p) = (12, -26, -6) \cdot (\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}) = 12(\frac{1}{3}) - 26(-\frac{2}{3}) - 6(\frac{2}{3}) = \frac{12}{3} + \frac{52}{3} - \frac{12}{3} = \frac{12 + 52 - 12}{3} = \frac{52}{3}$.

3. Final Answer

The directional derivative of $f$ at the point $p$ in the direction of $a$ is $\frac{52}{3}$.