Determine the divergence and curl of the vector field D(x, y, z) = 2x² yâ, + zâ, + yâ, at point (4,0,-5).

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).

Explore the calculation of the divergence and curl of the vector field D(x, y, z) = 2x²yî + zĵ + yk̂ at the point (4, 0, -5). This solution provides detailed steps in vector calculus, including partial derivatives and their application in finding divergence and curl. Suitable for advanced students of vector calculus.

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