Line integrals and work
Problem 12.10 · medium
Find the work done by \( \displaystyle \mathbf{F} = \langle x + y, x - y \rangle \) along the parabola y = x² from (0, 0) to (1, 1).
- Parametrize: r(t) = ⟨t, t**2⟩ for 0 ≤ t ≤ 1.
- \[ \left[\begin{matrix}\frac{d}{d t} t\\\frac{d}{d t} t^{2}\end{matrix}\right] = \left[\begin{matrix}1\\2 t\end{matrix}\right] \]r'(t).✓ Proved
- \[ t^{2} + 2 t \left(- t^{2} + t\right) + t = t \left(- 2 t \left(t - 1\right) + t + 1\right) \]F(r(t)) · r'(t).✓ Proved
- \[ \int\limits_{0}^{1} t \left(- 2 t \left(t - 1\right) + t + 1\right)\, dt = 1 \]W = ∫ F · dr.✓ Proved
Answer \( 1 \)
Lines: 3 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
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 |
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 2026-09-26 with SymPy 1.14.0.