Conservative fields and potential functions
Problem 12.225 · medium
Show that \( \displaystyle \mathbf F = \left(y e^{x}\right)\mathbf i + \left(e^{x} + \cos{\left(y \right)}\right)\mathbf j \) is conservative, find a potential function \( \displaystyle f \), and evaluate \( \displaystyle \int_C \mathbf F \cdot d\mathbf r \) along any path from \( \displaystyle (0, 2) \) to \( \displaystyle (0, 1) \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(e^{x} + \cos{\left(y \right)}\right)\\\frac{\partial}{\partial y} y e^{x}\end{matrix}\right] = \left[\begin{matrix}e^{x}\\e^{x}\end{matrix}\right] \]∂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
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(y e^{x} + \sin{\left(y \right)}\right)\\\frac{\partial}{\partial y} \left(y e^{x} + \sin{\left(y \right)}\right)\end{matrix}\right] = \left[\begin{matrix}y e^{x}\\e^{x} + \cos{\left(y \right)}\end{matrix}\right] \]f = y*exp(x) + sin(y) has gradient F.✓ Proved
- \[ -2 - \sin{\left(2 \right)} + \sin{\left(1 \right)} + 1 = -1 - \sin{\left(2 \right)} + \sin{\left(1 \right)} \]∫_C F·dr = f(end) − f(start), for any path.✓ Proved
Answer \( f = y e^{x} + \sin{\left(y \right)} + C,\quad \int_C \mathbf F\cdot d\mathbf r = -1 - \sin{\left(2 \right)} + \sin{\left(1 \right)} \)
Lines: 3 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.
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 verifies the conservative nature of the field, derives the potential function, and applies the Fundamental Theorem of Line Integrals with correct arithmetic.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly verifies the conservative nature of the field, derives the potential function, and applies the Fundamental Theorem of Line Integrals with correct arithmetic.qwen3.6:27b-mlx: inconclusive 2026-10-07 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The solution incorrectly identifies the components of F in the conservative test. It sets P = y e^x and Q = e^x + cos(y), but then checks ∂Q/∂x = e^xgpt-oss:20b: pass 2026-10-07
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 2026-10-07 with SymPy 1.14.0.