Conservative fields and potential functions
Problem 12.281 · medium
Show that \( \displaystyle \mathbf F = \left(y e^{x y}\right)\mathbf i + \left(x e^{x y}\right)\mathbf j \) is conservative, find a potential function \( \displaystyle f \), and evaluate \( \displaystyle \int_C \mathbf F \cdot d\mathbf r \) along any path from \( \displaystyle (-1, -1) \) to \( \displaystyle (2, 1) \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} x e^{x y}\\\frac{\partial}{\partial y} y e^{x y}\end{matrix}\right] = \left[\begin{matrix}x y e^{x y} + e^{x y}\\x y e^{x y} + e^{x y}\end{matrix}\right] \]∂Q/∂x = ∂P/∂y: F is conservative.✓ Proved
- Integrate the first component in x, then fix the 'constant' (a function of the other variables) by matching the other components.Reviewed
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} e^{x y}\\\frac{\partial}{\partial y} e^{x y}\end{matrix}\right] = \left[\begin{matrix}y e^{x y}\\x e^{x y}\end{matrix}\right] \]f = exp(x*y) has gradient F.✓ Proved
- \[ - e + e^{2} \]∫_C F·dr = f(end) − f(start), for any path.✓ Proved
Answer \( f = e^{x y} + C,\quad \int_C \mathbf F\cdot d\mathbf r = - e + e^{2} \)
✓ Nihil obstat Lines: 3 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the line integral computed numerically along two different paths |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly identifies the potential function and applies the Fundamental Theorem of Line Integrals. The notation '-E' in the final equation is a standard computer algebra system representation for -e, which is mathematically correct.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-11 — The solution correctly identifies the potential function and applies the Fundamental Theorem of Line Integrals. The notation '-E' in the final equation is a standard computer algebra system representation for -e, which is mathematically correct.gpt-oss:20b: pass 2026-10-11qwen3.6:27b-mlx: fail (style) 2026-10-11 — [domain objection, downgraded to style] The solution incorrectly identifies the potential function as f = e^{xy}, which yields a gradient of (ye^{xy}, xe^{xy}). However, the vector field is given as F = (ye^{xy}, xe^{xy}), so the gradient matches. Wait, let me re-read carefully. F = (y e^{xy}) i + (x e^{xy}) j. Potential f = e^{xy}. df/dx = y e^{xy}. df/dy = x e^{xy}. This matches F. So f is correct. The integral is f(2,1) - f(-1,-1) = e^{2*1} - e^{(-1)(-1)} = e^2 - e^1. The stated answer is -e + e^2. This is correct. Why did I think it was wrong? Let me re-read the prompt's stated answer. Stated answer: f = e^{xy} + C, integral = -e + e^2. Solution line 4 says -E + exp(2). This is correct. Is there a subtle error? Line 1 checks conservative condition. P = y e^{xy}, Q = x e^{xy}. dP/dy = e^{xy} + xy e^{xy}. dQ/dx = e^{xy} + xy e^{xy}. They match. Line 2 describes method. Line 3 verifies gradient. Line 4 calculates result. Everything seems correct. Wait, look at line 1 again. It says '∂Q/∂x = ∂P/∂y: F is conservative'. Usually P is i-component, Q is j-component. So P = y e^{xy}, Q = x e^{xy}. The equation shows Derivative(x*exp(x*y), x) and Derivative(y*exp(x*y), y). This is dQ/dx and dP/dy. The text says dQ/dx = dP/dy. This is the correct condition for conservative fields in 2D (assuming simply connected domain, which R^2 is). The calculation is correct. The potential is correct. The evaluation is correct. Is there a style issue? The variable 'E' in line 4 is used for Euler's number e. This is non-standard notation (usually 'e' or 'exp(1)'). In many computer algebra systems, E might be a variable, but here it clearly represents e. However, using 'E' for the base of natural logarithms is confusing and potentially misleading for students who might think E is a variable or energy. But is it an 'error'? It's a notation choice. Let's look closer. Is the domain simply connected? Yes. Is the function defined everywhere? Yes. So the theorem holds. The only potential issue is the notation 'E' for 'e'. In standard mathematical writing, 'E' is not the base of the natural logarithm; 'e' is. Using 'E' is a style error that could be misleading. Let's check if there's a deeper error. What if the path matters? No, it's conservative. What if the start/end points are swapped? Start (-1,-1), End (2,1). f(end) - f(start) = e^2 - e^1. Correct. I will mark it as style/misleading due to the use of 'E' for 'e'.gpt-oss:20b: pass 2026-10-11
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/conservative_fields, checked 2026-10-11 with SymPy 1.14.0.