Given the vector field $\mathbf{D}(x, y, z) = 2x^2 y \hat{i} + z \hat{j} + y \hat{k}$, we will compute both the divergence and the curl at the point $(4, 0, -5)$.

1. Divergence of $\mathbf{D}$:

The divergence of a vector field $\mathbf{D} = D_x \hat{i} + D_y \hat{j} + D_z \hat{k}$ is given by: $\text{div}(\mathbf{D}) = \nabla \cdot \mathbf{D} = \frac{\partial D_x}{\partial x} + \frac{\partial D_y}{\partial y} + \frac{\partial D_z}{\partial z}$

In this case:

• $D_x = 2x^2 y$
• $D_y = z$
• $D_z = y$

We now compute the partial derivatives:

$\frac{\partial D_x}{\partial x} = \frac{\partial}{\partial x} (2x^2 y) = 4x y$ $\frac{\partial D_y}{\partial y} = \frac{\partial}{\partial y} (z) = 0$ $\frac{\partial D_z}{\partial z} = \frac{\partial}{\partial z} (y) = 0$

Thus, the divergence at point $(4, 0, -5)$ is: $\text{div}(\mathbf{D}) = 4x y + 0 + 0 = 4(4)(0) = 0$

2. Curl of $\mathbf{D}$:

The curl of a vector field $\mathbf{D} = D_x \hat{i} + D_y \hat{j} + D_z \hat{k}$ is given by: $\text{curl}(\mathbf{D}) = \nabla \times \mathbf{D} = \left( \frac{\partial D_z}{\partial y} - \frac{\partial D_y}{\partial z} \right) \hat{i} + \left( \frac{\partial D_x}{\partial z} - \frac{\partial D_z}{\partial x} \right) \hat{j} + \left( \frac{\partial D_y}{\partial x} - \frac{\partial D_x}{\partial y} \right) \hat{k}$

We compute the partial derivatives:

$\frac{\partial D_z}{\partial y} = \frac{\partial}{\partial y} (y) = 1$ $\frac{\partial D_y}{\partial z} = \frac{\partial}{\partial z} (z) = 1$ $\frac{\partial D_x}{\partial z} = \frac{\partial}{\partial z} (2x^2 y) = 0$ $\frac{\partial D_z}{\partial x} = \frac{\partial}{\partial x} (y) = 0$ $\frac{\partial D_y}{\partial x} = \frac{\partial}{\partial x} (z) = 0$ $\frac{\partial D_x}{\partial y} = \frac{\partial}{\partial y} (2x^2 y) = 2x^2$

Thus, the curl at point $(4, 0, -5)$ is: $\text{curl}(\mathbf{D}) = \left( 1 - 1 \right) \hat{i} + \left( 0 - 0 \right) \hat{j} + \left( 0 - 2x^2 \right) \hat{k} = -2x^2 \hat{k}$ At $(4, 0, -5)$: $\text{curl}(\mathbf{D}) = -2(4)^2 \hat{k} = -32 \hat{k}$

Final Answers:

• Divergence: $\text{div}(\mathbf{D}) = 0$
• Curl: $\text{curl}(\mathbf{D}) = -32 \hat{k}$

Would you like any further details or explanations?

Here are 5 related questions you might explore:

1. How would the divergence and curl change for a different vector field?
2. Can divergence and curl be used to determine if a field is conservative?
3. What physical phenomena do divergence and curl represent in vector fields?
4. How do we interpret the curl of a vector field in 3D space?
5. Can a vector field have both zero divergence and zero curl?

Tip: In physics, divergence often relates to sources (like charges in electromagnetism), while curl relates to rotational tendencies (like magnetic fields).