The potential energy function associated with the force `vecF=4xyhati+2x^2hatj` is

Options

A. (a) `U=-2x^2y`
B. (b) `U=-2x^2y+const ant`
C. (c) `U=2x^2y+const ant`
D. (d) Not defined

Explanation

To find the potential energy function associated with the force vector \(\vec{F} = 4xy \hat{i} + 2x^2 \hat{j}\), we need to follow these steps:

### Step 1: Understand the relationship between force and potential energy
The potential energy \(U\) is related to the force \(\vec{F}\) by the equation:
\[ \vec{F} = -\nabla U \]
This means that the force is the negative gradient of the potential energy function. In Cartesian coordinates, this can be expressed as:
\[ \vec{F} = -\left( \frac{\partial U}{\partial x} \hat{i} + \frac{\partial U}{\partial y} \hat{j} \right) \]

### Step 2: Write down the components of the force
From the given force vector:
\[ \vec{F} = 4xy \hat{i} + 2x^2 \hat{j} \]
we can identify the components:
\[ F_x = 4xy \quad \text{and} \quad F_y = 2x^2 \]

### Step 3: Set up the equations for potential energy
Using the relationship between force and potential energy, we have:
\[ -\frac{\partial U}{\partial x} = 4xy \quad \text{(1)} \]
\[ -\frac{\partial U}{\partial y} = 2x^2 \quad \text{(2)} \]

### Step 4: Integrate the first equation
From equation (1):
\[ \frac{\partial U}{\partial x} = -4xy \]
Integrating with respect to \(x\):
\[ U = -4xy^2 + g(y) \]
where \(g(y)\) is an arbitrary function of \(y\).

### Step 5: Differentiate with respect to \(y\) and use the second equation
Now we differentiate \(U\) with respect to \(y\):
\[ \frac{\partial U}{\partial y} = -8xy + g'(y) \]
Setting this equal to equation (2):
\[ -8xy + g'(y) = -2x^2 \]
Rearranging gives:
\[ g'(y) = -2x^2 + 8xy \]

### Step 6: Solve for \(g(y)\)
Since \(g'(y)\) must not depend on \(x\), we can set the coefficient of \(x\) to zero:
\[ 8y = 0 \implies y = 0 \quad \text{(not useful)} \]
Thus, we can assume \(g(y)\) is a function of \(y\) alone. Integrating \(g'(y)\) gives:
\[ g(y) = -2y^2 + C \]
where \(C\) is a constant of integration.

### Step 7: Combine results to find the potential energy function
Now substituting \(g(y)\) back into our expression for \(U\):
\[ U = -4xy^2 - 2y^2 + C \]

### Final Result
Thus, the potential energy function associated with the force \(\vec{F}\) is:
\[ U(x, y) = -4xy^2 - 2y^2 + C \]