∫Calc Practice

Conservative fields and potential functions

Problem 12.196 · 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 \( \displaystyle (1, 0) \) to \( \displaystyle (0, 2) \).
  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
  2. Integrate the first component in x, then fix the 'constant' (a function of the other variables) by matching the other components.
  3. \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(y e^{x} + \sin{\left(y \right)}\right)\\\frac{\partial}{\partial y} \left(y e^{x} + \sin{\left(y \right)}\right)\end{matrix}\right] = \left[\begin{matrix}y e^{x}\\e^{x} + \cos{\left(y \right)}\end{matrix}\right] \]
    f = y*exp(x) + sin(y) has gradient F.✓ Proved
  4. \[ 0 + \sin{\left(2 \right)} + 2 = \sin{\left(2 \right)} + 2 \]
    ∫_C F·dr = f(end) − f(start), for any path.✓ Proved
Answer \( f = y e^{x} + \sin{\left(y \right)} + C,\quad \int_C \mathbf F\cdot d\mathbf r = \sin{\left(2 \right)} + 2 \)

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
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
2Not checked—a sentence; read, not computed
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0the line integral computed numerically along two different paths

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: fail (error) — The solution fails to explicitly calculate the potential function f(x,y) in step 3, jumping straight to a verification that a specific function works. While the verification is correct, the prompt asks to 'find' the potential function, and the solution does not show the integration steps (e.g., integrating P with respect to x to get y*e^x + g(y), then differentiating with respect to y to find g'(y) = cos(y)). Furthermore, step 4 calculates f(0,2) - f(1,0) but does not explicitly state the values f(0,2) = sin(2) + 2 and f(1,0) = 0, making the arithmetic jump opaque.
Every verdict on record (4)
  • qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution fails to explicitly calculate the potential function f(x,y) in step 3, jumping straight to a verification that a specific function works. While the verification is correct, the prompt asks to 'find' the potential function, and the solution does not show the integration steps (e.g., integrating P with respect to x to get y*e^x + g(y), then differentiating with respect to y to find g'(y) = cos(y)). Furthermore, step 4 calculates f(0,2) - f(1,0) but does not explicitly state the values f(0,2) = sin(2) + 2 and f(1,0) = 0, making the arithmetic jump opaque.
  • gpt-oss:20b: pass 2026-10-05
  • qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution fails to explicitly calculate the potential function f(x,y) via integration, jumping directly to the result in step 3 without showing the work. Furthermore, step 4 incorrectly implies that f(start) is 0 without stating f(1,0) = 1*e^1 + sin(0) = e, leading to the wrong final answer (the correct answer is sin(2) + 2 - e).
  • gpt-oss:20b: pass 2026-10-05

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 2026-10-05 with SymPy 1.14.0.