2 points Divergence of V = e*" î – cos y ĵ + (sin z)² & is ye*y – sin y + 2 sin z cos z O a ye*y + sin y + 2 sin z cos z ye*y + cos y + 2 sin z cos z

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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7. What's answer of this MCQ? don't need explanation. 

 

## Divergence of a Vector Field

### Problem Statement

Calculate the divergence of the vector field \( \mathbf{V} = e^{xy} \mathbf{\hat{i}} - \cos y \mathbf{\hat{j}} + (\sin z)^2 \mathbf{\hat{k}} \).

### Answer Choices

- **a)** \( ye^{xy} - \sin y + 2 \sin z \cos z \)
  
- **b)** \( ye^{xy} + \sin y + 2 \sin z \cos z \)
  
- **c)** \( ye^{xy} + \cos y + 2 \sin z \cos z \)
  
- **d)** \( 0 \)

### Explanation

The divergence of a vector field \( \mathbf{V} \) with components \( V_x, V_y, V_z \) is given by:
\[
\nabla \cdot \mathbf{V} = \frac{\partial V_x}{\partial x} + \frac{\partial V_y}{\partial y} + \frac{\partial V_z}{\partial z}
\]

- \( V_x = e^{xy} \), so \( \frac{\partial V_x}{\partial x} = ye^{xy} \)
- \( V_y = -\cos y \), so \( \frac{\partial V_y}{\partial y} = \sin y \)
- \( V_z = (\sin z)^2 \), so \( \frac{\partial V_z}{\partial z} = 2 \sin z \cos z \)

Combine these results to find the divergence:
\[
\nabla \cdot \mathbf{V} = ye^{xy} + \sin y + 2 \sin z \cos z
\]

The correct answer that matches this expression is **b)**.
Transcribed Image Text:## Divergence of a Vector Field ### Problem Statement Calculate the divergence of the vector field \( \mathbf{V} = e^{xy} \mathbf{\hat{i}} - \cos y \mathbf{\hat{j}} + (\sin z)^2 \mathbf{\hat{k}} \). ### Answer Choices - **a)** \( ye^{xy} - \sin y + 2 \sin z \cos z \) - **b)** \( ye^{xy} + \sin y + 2 \sin z \cos z \) - **c)** \( ye^{xy} + \cos y + 2 \sin z \cos z \) - **d)** \( 0 \) ### Explanation The divergence of a vector field \( \mathbf{V} \) with components \( V_x, V_y, V_z \) is given by: \[ \nabla \cdot \mathbf{V} = \frac{\partial V_x}{\partial x} + \frac{\partial V_y}{\partial y} + \frac{\partial V_z}{\partial z} \] - \( V_x = e^{xy} \), so \( \frac{\partial V_x}{\partial x} = ye^{xy} \) - \( V_y = -\cos y \), so \( \frac{\partial V_y}{\partial y} = \sin y \) - \( V_z = (\sin z)^2 \), so \( \frac{\partial V_z}{\partial z} = 2 \sin z \cos z \) Combine these results to find the divergence: \[ \nabla \cdot \mathbf{V} = ye^{xy} + \sin y + 2 \sin z \cos z \] The correct answer that matches this expression is **b)**.
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