Arc length of parametric curves
Problem 8.93 · medium
Find the length of the curve \( \displaystyle x = 6 t^{2} \), \( \displaystyle y = 4 t^{3} \), \( \displaystyle 0 \le t \le 1 \).
- \[ \left[\begin{matrix}\frac{d}{d t} 6 t^{2}\\\frac{d}{d t} 4 t^{3}\end{matrix}\right] = \left[\begin{matrix}12 t\\12 t^{2}\end{matrix}\right] \]Velocity components.✓ Proved
- \[ 144 t^{4} + 144 t^{2} = 144 t^{2} \left(t^{2} + 1\right) \](dx/dt)² + (dy/dt)², simplified.✓ Proved
- \[ \int\limits_{0}^{1} 12 \sqrt{t^{2} + 1} \left|{t}\right|\, dt = -4 + 8 \sqrt{2} \]Integrate the speed.✓ Proved
Answer \( -4 + 8 \sqrt{2} \approx 7.31371 \)
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: inconclusive — reviewer returned a non-object
Every verdict on record (4)
qwen3.6:27b-mlx: inconclusive 2026-10-05 — reviewer returned a non-objectgpt-oss:20b: pass 2026-10-05qwen3.6:27b-mlx: inconclusive 2026-10-05 — reviewer returned a non-objectgpt-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.