Question:
Let \( \hat{i} \) and \( \hat{j} \) be the unit vectors in the x and y direction, respectively. The divergence \( (\nabla . \vec{F}) \) of \( \vec{F} = x^2y\hat{i} + y^3x\hat{j} \) at the point \( (3,2) \) is
Let \( \hat{i} \) and \( \hat{j} \) be the unit vectors in the x and y direction, respectively. The divergence \( (\nabla . \vec{F}) \) of \( \vec{F} = x^2y\hat{i} + y^3x\hat{j} \) at the point \( (3,2) \) is
Show Hint
Use \( \nabla . \vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} \) and put in the given point.
Updated On: Aug 17, 2026
- 12
- 24
- 36
- 48
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The Correct Option is D
Solution and Explanation
Step 1: Recall the formula for divergence.
For a vector field \( \vec{F} = F_1\hat{i} + F_2\hat{j} \) in two dimensions, the divergence is \( \nabla . \vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} \).
It tells us how much the field spreads out from a point.
Step 2: Identify the components.
Here \( F_1 = x^2y \) and \( F_2 = y^3x \).
Step 3: Differentiate each component.
Differentiate \( F_1 \) with respect to x, treating y as constant:
\[ \frac{\partial F_1}{\partial x} = \frac{\partial}{\partial x}(x^2y) = 2xy \]
Differentiate \( F_2 \) with respect to y, treating x as constant:
\[ \frac{\partial F_2}{\partial y} = \frac{\partial}{\partial y}(y^3x) = 3xy^2 \]
Step 4: Add the two partial derivatives.
\[ \nabla . \vec{F} = 2xy + 3xy^2 \]
Step 5: Substitute the point (3,2).
Put \( x=3 \) and \( y=2 \):
\[ 2(3)(2) + 3(3)(2)^2 = 12 + 3(3)(4) = 12 + 36 = 48 \]
Step 6: Check the wrong options.
Option A (12) is only the first term \( 2xy \), missing the second term.
Option B (24) does not match either term or their correct sum.
Option C (36) is only the second term \( 3xy^2 \), missing the first term.
We need both terms added together to get the full divergence.
Final Answer:
The divergence of \( \vec{F} \) at (3,2) is 48.
\[ \boxed{48} \]
For a vector field \( \vec{F} = F_1\hat{i} + F_2\hat{j} \) in two dimensions, the divergence is \( \nabla . \vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} \).
It tells us how much the field spreads out from a point.
Step 2: Identify the components.
Here \( F_1 = x^2y \) and \( F_2 = y^3x \).
Step 3: Differentiate each component.
Differentiate \( F_1 \) with respect to x, treating y as constant:
\[ \frac{\partial F_1}{\partial x} = \frac{\partial}{\partial x}(x^2y) = 2xy \]
Differentiate \( F_2 \) with respect to y, treating x as constant:
\[ \frac{\partial F_2}{\partial y} = \frac{\partial}{\partial y}(y^3x) = 3xy^2 \]
Step 4: Add the two partial derivatives.
\[ \nabla . \vec{F} = 2xy + 3xy^2 \]
Step 5: Substitute the point (3,2).
Put \( x=3 \) and \( y=2 \):
\[ 2(3)(2) + 3(3)(2)^2 = 12 + 3(3)(4) = 12 + 36 = 48 \]
Step 6: Check the wrong options.
Option A (12) is only the first term \( 2xy \), missing the second term.
Option B (24) does not match either term or their correct sum.
Option C (36) is only the second term \( 3xy^2 \), missing the first term.
We need both terms added together to get the full divergence.
Final Answer:
The divergence of \( \vec{F} \) at (3,2) is 48.
\[ \boxed{48} \]
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