Conservative fields and potential functions
Problem 12.289 · 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 \( \displaystyle (-1, -1) \) to \( \displaystyle (2, 0) \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(e^{x} + \cos{\left(y \right)}\right)\\\frac{\partial}{\partial y} y e^{x}\end{matrix}\right] = \left[\begin{matrix}e^{x}\\e^{x}\end{matrix}\right] \]∂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.
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(y e^{x} + \sin{\left(y \right)}\right)\\\frac{\partial}{\partial y} \left(y e^{x} + \sin{\left(y \right)}\right)\end{matrix}\right] = \left[\begin{matrix}y e^{x}\\e^{x} + \cos{\left(y \right)}\end{matrix}\right] \]f = y*exp(x) + sin(y) has gradient F.✓ Proved
- \[ 0 + e^{-1} + \sin{\left(1 \right)} = e^{-1} + \sin{\left(1 \right)} \]∫_C F·dr = f(end) − f(start), for any path.✓ Proved
Answer \( f = y e^{x} + \sin{\left(y \right)} + C,\quad \int_C \mathbf F\cdot d\mathbf r = e^{-1} + \sin{\left(1 \right)} \)
Lines: 3 proved, 1 not checked. 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 | Not checked | — | 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: fail (error) — The solution fails to explicitly calculate the potential at the starting point (-1, -1), which is f(-1, -1) = -e^{-1} - sin(1). Consequently, the final evaluation in step 4 is incorrect; it adds the starting value instead of subtracting it, yielding e^{-1} + sin(1) instead of the correct e^2 - e^{-1} + sin(1).
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-11 — The solution fails to explicitly calculate the potential at the starting point (-1, -1), which is f(-1, -1) = -e^{-1} - sin(1). Consequently, the final evaluation in step 4 is incorrect; it adds the starting value instead of subtracting it, yielding e^{-1} + sin(1) instead of the correct e^2 - e^{-1} + sin(1).gpt-oss:20b: pass 2026-10-11qwen3.6:27b-mlx: inconclusive 2026-10-11 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The evaluation of the line integral is incorrect. The potential function is f(x,y) = y e^x + sin(y). The value should be f(2,0) - f(-1,-1) = (0*e^2 +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.