The problem asks for the normal unit vector at a specific point $(-1, -1, 2)$ for the surface given by the equation $xy^3z^2 = 4$.

Step 1: Gradient as the Normal Vector

The gradient of a scalar function $f(x, y, z)$ gives the direction of the normal vector to the surface. Here, $f(x, y, z) = xy^3z^2 - 4$.

The gradient is: $\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)$

Partial derivatives:

• $\frac{\partial f}{\partial x} = y^3z^2$
• $\frac{\partial f}{\partial y} = 3xy^2z^2$
• $\frac{\partial f}{\partial z} = 2xy^3z$

Thus, $\nabla f = \left( y^3z^2 \right) \mathbf{i} + \left( 3xy^2z^2 \right) \mathbf{j} + \left( 2xy^3z \right) \mathbf{k}$

Step 2: Evaluate at $(-1, -1, 2)$

Substitute $x = -1$, $y = -1$, $z = 2$ into the gradient:

1. $\frac{\partial f}{\partial x} = (-1)^3(2^2) = -4$
2. $\frac{\partial f}{\partial y} = 3(-1)(-1)^2(2^2) = -12$
3. $\frac{\partial f}{\partial z} = 2(-1)(-1)^3(2) = -4$

So, $\nabla f = -4\mathbf{i} - 12\mathbf{j} - 4\mathbf{k}$

Step 3: Unit Normal Vector

The magnitude of the gradient vector: $|\nabla f| = \sqrt{(-4)^2 + (-12)^2 + (-4)^2} = \sqrt{16 + 144 + 16} = \sqrt{176} = 4\sqrt{11}$

The unit normal vector is: $\mathbf{n} = \frac{\nabla f}{|\nabla f|} = \frac{-4\mathbf{i} - 12\mathbf{j} - 4\mathbf{k}}{4\sqrt{11}} = \frac{-1}{\sqrt{11}}\mathbf{i} - \frac{3}{\sqrt{11}}\mathbf{j} - \frac{1}{\sqrt{11}}\mathbf{k}$

Step 4: Match with Options

The correct option is (b): $-\frac{1}{\sqrt{11}} \left( \mathbf{i} - 3\mathbf{j} + \mathbf{k} \right)$

Would you like further clarification on this solution? Let me know!

Related Questions:

1. How is the gradient vector used to find the normal vector of a surface?
2. Why is the equation $xy^3z^2 = 4$ rewritten as $f(x, y, z) = xy^3z^2 - 4$?
3. What is the significance of unit vectors in vector analysis?
4. How do you calculate the magnitude of a vector in 3D space?
5. What are some practical applications of normal vectors in real-world problems?

Tip: Always check that the magnitude of your unit vector is 1 as a verification step.