Arc length of parametric curves
Problem 8.124 · medium
Find the length of the curve \( \displaystyle x = 1 - 3 t \), \( \displaystyle y = - 4 t - 2 \), \( \displaystyle 0 \le t \le 4 \).
- \[ \left[\begin{matrix}\frac{d}{d t} \left(1 - 3 t\right)\\\frac{d}{d t} \left(- 4 t - 2\right)\end{matrix}\right] = \left[\begin{matrix}-3\\-4\end{matrix}\right] \]Velocity components.✓ Proved
- \[ 25 \](dx/dt)² + (dy/dt)², simplified.✓ Proved
- \[ \int\limits_{0}^{4} 5\, dt = 20 \]Integrate the speed.✓ Proved
Answer \( 20 \approx 20.00000 \)
✓ Nihil obstat 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. Reviewers found nothing wrong with the explanation.
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: pass
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: pass 2026-10-06gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly computes the derivatives, the magnitude of the velocity vector, and the definite integral for arc length. The steps are logically sound and the final answer is correct.
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 2026-10-06 with SymPy 1.14.0.