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

• A. 12
• B. 24
• C. 36
• D. 48

Hint:

Use \( \nabla . \vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} \) and put in the given point.

Answer 1

The Correct Option is D

Another way to find the divergence at one specific point is to freeze one coordinate at its given value first, turn the field into a single-variable function, and then differentiate. This works because a partial derivative already asks how the function changes in one direction while the other coordinate stays fixed.

The field is $\vec{F} = x^2y\,\hat{i} + y^3x\,\hat{j}$, and we want the divergence at $(3,2)$, so $x=3$ and $y=2$.

x-component: fix $y=2$ inside $F_1=x^2y$, which turns it into $F_1(x) = 2x^2$. Differentiate with respect to $x$: $\frac{dF_1}{dx} = 4x$. Now put $x=3$: $4(3) = 12$.

y-component: fix $x=3$ inside $F_2=y^3x$, which turns it into $F_2(y) = 3y^3$. Differentiate with respect to $y$: $\frac{dF_2}{dy} = 9y^2$. Now put $y=2$: $9(4) = 36$.

The divergence is the sum of these two results, $\nabla . \vec{F} = 12 + 36 = 48$.

Let's summarize:

• Freezing the other coordinate before differentiating gives the same partial derivative, just found through an ordinary single-variable derivative instead of formal partial notation.
• The x-part contributes 12 and the y-part contributes 36, so neither option (A) 12 nor option (C) 36 alone is correct, since the question asks for their sum.
• Option (B) 24 does not match either term or any correct combination of them.

Adding both contributions gives $48$, so the divergence of $\vec{F}$ at $(3,2)$ is $48$.