Line integrals and work
Problem 12.148 · medium
Find the work done by \( \displaystyle \mathbf{F} = \langle - y, x \rangle \) along the line segment from (0, 0) to (3, 1).
- Parametrize: r(t) = ⟨3*t, t⟩ for 0 ≤ t ≤ 1.
- \[ \left[\begin{matrix}\frac{d}{d t} 3 t\\\frac{d}{d t} t\end{matrix}\right] = \left[\begin{matrix}3\\1\end{matrix}\right] \]r'(t).✓ Proved
- \[ 0 \]F(r(t)) · r'(t).✓ Proved
- \[ \int\limits_{0}^{1} 0\, dt = 0 \]W = ∫ F · dr.✓ Proved
Answer \( 0 \)
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 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 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 | a 20,000-chord polygon along the path gives the same work |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The calculation of the dot product F(r(t)) · r'(t) is incorrect. F(r(t)) = <-t, 3t>, so F · r' = -3t + 3t = 0 is actually correct, wait. Let me re-evaluate. F = <-y, x>. r(t) = <3t, t>. So y=t, x=3t. F(r(t)) = <-t, 3t>. r'(t) = <3, 1>. Dot product: (-t)(3) + (3t)(1) = -3t + 3t = 0. The integral of 0 is 0. The solution is actually correct. Why did I think it was wrong? Let me check the problem statement again. F = <-y, x>. Path from (0,0) to (3,1). Parametrization r(t) = <3t, t>. F(r(t)) = <-t, 3t>. r'(t) = <3, 1>. Dot product is 0. Work is 0. The solution is correct. Wait, is the field conservative? curl F = d/dx(x) - d/dy(-y) = 1 - (-1) = 2. Not conservative. But the work along this specific line segment is 0. The solution is correct. I will mark it as pass.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-03 — The calculation of the dot product F(r(t)) · r'(t) is incorrect. F(r(t)) = <-t, 3t>, so F · r' = -3t + 3t = 0 is actually correct, wait. Let me re-evaluate. F = <-y, x>. r(t) = <3t, t>. So y=t, x=3t. F(r(t)) = <-t, 3t>. r'(t) = <3, 1>. Dot product: (-t)(3) + (3t)(1) = -3t + 3t = 0. The integral of 0 is 0. The solution is actually correct. Why did I think it was wrong? Let me check the problem statement again. F = <-y, x>. Path from (0,0) to (3,1). Parametrization r(t) = <3t, t>. F(r(t)) = <-t, 3t>. r'(t) = <3, 1>. Dot product is 0. Work is 0. The solution is correct. Wait, is the field conservative? curl F = d/dx(x) - d/dy(-y) = 1 - (-1) = 2. Not conservative. But the work along this specific line segment is 0. The solution is correct. I will mark it as pass.gpt-oss:20b: pass 2026-10-03qwen3.6:27b-mlx: fail (error) 2026-10-03 — The calculation of the dot product F(r(t)) · r'(t) is incorrect. F(r(t)) is < -t, 3t >, so the dot product with <3, 1> is 3t, not 0. The work is 3/2, not 0.gpt-oss:20b: pass 2026-10-03
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/line_integral_work, checked 2026-10-03 with SymPy 1.14.0.