Question

Given two vectors \(\vec{U}\) and \(\vec{V}\), which of the following options is CORRECT for the expansion of divergence, \(\nabla \cdot (\vec{U} + \vec{V})\)?

• A. \(\nabla \times \vec{U} + \nabla \times \vec{V}\)
• B. \(\nabla \times \vec{U} - \nabla \times \vec{V}\)
• C. \(\nabla \cdot \vec{U} + \nabla \cdot \vec{V}\)
• D. \(\nabla \cdot \vec{U} - \nabla \cdot \vec{V}\)

Hint:

Divergence is a linear operator built from partial derivatives, so it distributes over vector addition without any sign change.

Answer 1

The Correct Option is C

Step 1: Recall the definition of divergence:
Divergence is a linear differential operator that acts on a vector field and produces a scalar field. For a vector field \(\vec{F} = (F_x, F_y, F_z)\), the divergence is defined as \(\nabla \cdot \vec{F} = \dfrac{\partial F_x}{\partial x} + \dfrac{\partial F_y}{\partial y} + \dfrac{\partial F_z}{\partial z}\).
Step 2: Apply linearity of the derivative:
Since differentiation is a linear operation, the derivative of a sum of two functions equals the sum of the derivatives of each function taken separately. This linearity extends directly to the divergence operator because divergence is built entirely out of partial derivatives.
Step 3: Expand the divergence of the sum:
Writing \(\vec{U} = (U_x, U_y, U_z)\) and \(\vec{V} = (V_x, V_y, V_z)\), we have \(\nabla \cdot (\vec{U} + \vec{V}) = \dfrac{\partial (U_x + V_x)}{\partial x} + \dfrac{\partial (U_y + V_y)}{\partial y} + \dfrac{\partial (U_z + V_z)}{\partial z}\). Splitting each partial derivative using linearity gives \(\left(\dfrac{\partial U_x}{\partial x} + \dfrac{\partial U_y}{\partial y} + \dfrac{\partial U_z}{\partial z}\right) + \left(\dfrac{\partial V_x}{\partial x} + \dfrac{\partial V_y}{\partial y} + \dfrac{\partial V_z}{\partial z}\right)\), which is exactly \(\nabla \cdot \vec{U} + \nabla \cdot \vec{V}\).
Step 4: Check the options:
Option A involves the curl operator, \(\nabla \times\), which produces a vector, not a scalar, so it cannot represent divergence, which is a scalar quantity. Option B has the same issue and also has an incorrect minus sign. Option C gives \(\nabla \cdot \vec{U} + \nabla \cdot \vec{V}\), which matches our derivation exactly. Option D has the correct divergence operator but the wrong sign, since divergence distributes over addition without any sign change.
Final Answer:
\[ \boxed{\nabla \cdot (\vec{U} + \vec{V}) = \nabla \cdot \vec{U} + \nabla \cdot \vec{V}} \]