Arc length of parametric curves
Problem 8.95 · medium
Find the length of the curve \( \displaystyle x = t - \sin{\left(t \right)} \), \( \displaystyle y = 1 - \cos{\left(t \right)} \), \( \displaystyle 0 \le t \le 2 \pi \).
- \[ \left[\begin{matrix}\frac{d}{d t} \left(t - \sin{\left(t \right)}\right)\\\frac{d}{d t} \left(1 - \cos{\left(t \right)}\right)\end{matrix}\right] = \left[\begin{matrix}1 - \cos{\left(t \right)}\\\sin{\left(t \right)}\end{matrix}\right] \]Velocity components.✓ Proved
- \[ \left(1 - \cos{\left(t \right)}\right)^{2} + \sin^{2}{\left(t \right)} = 4 \sin^{2}{\left(\frac{t}{2} \right)} \](dx/dt)² + (dy/dt)², simplified with 1 − cos t = 2 sin²(t/2).✓ Proved
- \[ \int\limits_{0}^{2 \pi} 2 \sin{\left(\frac{t}{2} \right)}\, dt = 8 \]Integrate the speed (sin(t/2) ≥ 0 on [0, 2π]).✓ Proved
Answer \( 8 \approx 8.00000 \)
Every line of this solution was proved by the computer algebra system SymPy. 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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ 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 | numerical quadrature of the speed |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution incorrectly assumes that the square root of the sum of squares equals 2*sin(t/2) without explicitly taking the square root of the expression in step 2. Step 2 establishes (dx/dt)^2 + (dy/dt)^2 = 4*sin^2(t/2), but the integrand for arc length is the square root of this quantity, i.e., 2*|sin(t/2)|. While the final integral value is correct because sin(t/2) is non-negative on [0, 2pi], the logical jump from the squared sum directly to the linear term in the integral setup is mathematically incomplete and skips the crucial square root operation.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution incorrectly assumes that the square root of the sum of squares equals 2*sin(t/2) without explicitly taking the square root of the expression in step 2. Step 2 establishes (dx/dt)^2 + (dy/dt)^2 = 4*sin^2(t/2), but the integrand for arc length is the square root of this quantity, i.e., 2*|sin(t/2)|. While the final integral value is correct because sin(t/2) is non-negative on [0, 2pi], the logical jump from the squared sum directly to the linear term in the integral setup is mathematically incomplete and skips the crucial square root operation.gpt-oss:20b: pass 2026-10-05qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution fails to take the square root of the sum of squared derivatives. The integrand should be sqrt((dx/dt)^2 + (dy/dt)^2) = 2|sin(t/2)|, but the solution integrates 2*sin(t/2) directly, implying (dx/dt)^2 + (dy/dt)^2 was already the speed, which is incorrect.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/parametric_arc_length, checked 2026-10-05 with SymPy 1.14.0.