Question:
The velocity vector is given as $\overline{V} = 5xyi + 2y^{2}j + 3yz^{2}k$. The divergence of this vector at (1,1,1) is}
The velocity vector is given as $\overline{V} = 5xyi + 2y^{2}j + 3yz^{2}k$. The divergence of this vector at (1,1,1) is}
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Divergence results in a scalar field. It measures the "outwardness" of the vector field at a specific point.
Updated On: Jul 3, 2026
- 9
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- 15
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The Correct Option is D
Solution and Explanation
Step 1: Concept
The divergence of a vector $\overline{V} = V_1i + V_2j + V_3k$ is given by $\nabla \cdot \overline{V} = \frac{\partial V_1}{\partial x} + \frac{\partial V_2}{\partial y} + \frac{\partial V_3}{\partial z}$.
Step 2: Meaning
Here $V_1 = 5xy$, $V_2 = 2y^{2}$, and $V_3 = 3yz^{2}$. We calculate the partial derivatives: $\frac{\partial (5xy)}{\partial x} = 5y$, $\frac{\partial (2y^{2})}{\partial y} = 4y$, and $\frac{\partial (3yz^{2})}{\partial z} = 6yz$.
Step 3: Analysis
Summing the derivatives: $\nabla \cdot \overline{V} = 5y + 4y + 6yz = 9y + 6yz$.
Step 4: Conclusion
Substituting the point $(1,1,1)$: $9(1) + 6(1)(1) = 9 + 6 = 15$. Final Answer: (D)
The divergence of a vector $\overline{V} = V_1i + V_2j + V_3k$ is given by $\nabla \cdot \overline{V} = \frac{\partial V_1}{\partial x} + \frac{\partial V_2}{\partial y} + \frac{\partial V_3}{\partial z}$.
Step 2: Meaning
Here $V_1 = 5xy$, $V_2 = 2y^{2}$, and $V_3 = 3yz^{2}$. We calculate the partial derivatives: $\frac{\partial (5xy)}{\partial x} = 5y$, $\frac{\partial (2y^{2})}{\partial y} = 4y$, and $\frac{\partial (3yz^{2})}{\partial z} = 6yz$.
Step 3: Analysis
Summing the derivatives: $\nabla \cdot \overline{V} = 5y + 4y + 6yz = 9y + 6yz$.
Step 4: Conclusion
Substituting the point $(1,1,1)$: $9(1) + 6(1)(1) = 9 + 6 = 15$. Final Answer: (D)
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