To evaluate the curl and divergence of the vector field $\mathbf{F} = \langle x^2 + 2yz, y^2 - z^3, 4xyz \rangle$ at the point $(-1, 4, 2)$, let's proceed with the following steps:

Step 1: Calculate the Curl of $\mathbf{F}$

The curl of a vector field $\mathbf{F} = \langle F_1, F_2, F_3 \rangle$ is given by: $\nabla \times \mathbf{F} = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} \right) \mathbf{i} - \left( \frac{\partial F_3}{\partial x} - \frac{\partial F_1}{\partial z} \right) \mathbf{j} + \left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) \mathbf{k}$

1. Partial derivatives:

   • $\frac{\partial F_3}{\partial y} = \frac{\partial}{\partial y} (4xyz) = 4xz$
   • $\frac{\partial F_2}{\partial z} = \frac{\partial}{\partial z} (y^2 - z^3) = -3z^2$
   • $\frac{\partial F_3}{\partial x} = \frac{\partial}{\partial x} (4xyz) = 4yz$
   • $\frac{\partial F_1}{\partial z} = \frac{\partial}{\partial z} (x^2 + 2yz) = 2y$
   • $\frac{\partial F_2}{\partial x} = \frac{\partial}{\partial x} (y^2 - z^3) = 0$
   • $\frac{\partial F_1}{\partial y} = \frac{\partial}{\partial y} (x^2 + 2yz) = 2z$

2. Components of the Curl:

   • $(\nabla \times \mathbf{F})_x = \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} = 4xz + 3z^2$
   • $(\nabla \times \mathbf{F})_y = \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x} = 2y - 4yz = 2y(1 - 2z)$
   • $(\nabla \times \mathbf{F})_z = \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} = 0 - 2z = -2z$

3. Evaluate at $(-1, 4, 2)$:

   • $(\nabla \times \mathbf{F})_x = 4(-1)(2) + 3(2)^2 = -8 + 12 = 4$
   • $(\nabla \times \mathbf{F})_y = 2(4)(1 - 2 \cdot 2) = 8 \cdot (-3) = -24$
   • $(\nabla \times \mathbf{F})_z = -2 \cdot 2 = -4$

So, $\nabla \times \mathbf{F} = \langle 4, -24, -4 \rangle$

Step 2: Calculate the Divergence of $\mathbf{F}$

The divergence of $\mathbf{F} = \langle F_1, F_2, F_3 \rangle$ is given by: $\nabla \cdot \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}$

1. Partial derivatives:

   • $\frac{\partial F_1}{\partial x} = \frac{\partial}{\partial x} (x^2 + 2yz) = 2x$
   • $\frac{\partial F_2}{\partial y} = \frac{\partial}{\partial y} (y^2 - z^3) = 2y$
   • $\frac{\partial F_3}{\partial z} = \frac{\partial}{\partial z} (4xyz) = 4xy$

2. Evaluate at $(-1, 4, 2)$:

   • $\frac{\partial F_1}{\partial x} = 2(-1) = -2$
   • $\frac{\partial F_2}{\partial y} = 2(4) = 8$
   • $\frac{\partial F_3}{\partial z} = 4(-1)(4) = -16$

So, $\nabla \cdot \mathbf{F} = -2 + 8 - 16 = -10$

Final Answer

• The curl of $\mathbf{F}$ at $(-1, 4, 2)$ is $\langle 4, -24, -4 \rangle$.
• The divergence of $\mathbf{F}$ at $(-1, 4, 2)$ is $-10$.

Would you like further clarification on any of these steps?

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1. How does the curl of a vector field indicate rotation in the field?
2. What are the applications of divergence in physics?
3. How would the curl and divergence change if the vector field had different functions for each component?
4. Can the curl and divergence of a vector field both be zero? If so, what does that imply?
5. How would you visualize the curl and divergence of a 3D vector field?

Tip: The curl of a vector field describes its rotational tendency, while divergence indicates the rate of expansion or contraction at a point.