Arc length and curvature of space curves
Problem 9.665 · medium
Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle t, t^{2}, 0 \right\rangle \) at \( \displaystyle t = 0 \).
- \[ \left[\begin{matrix}\left. \frac{d}{d t} t \right|_{\substack{ t=0 }}\\\left. \frac{d}{d t} t^{2} \right|_{\substack{ t=0 }}\\\left. \frac{d}{d t} 0 \right|_{\substack{ t=0 }}\end{matrix}\right] = \left[\begin{matrix}1\\0\\0\end{matrix}\right] \]r′(0).✓ Proved
- \[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} t \right|_{\substack{ t=0 }}\\\left. \frac{d^{2}}{d t^{2}} t^{2} \right|_{\substack{ t=0 }}\\\left. \frac{d^{2}}{d t^{2}} 0 \right|_{\substack{ t=0 }}\end{matrix}\right] = \left[\begin{matrix}0\\2\\0\end{matrix}\right] \]r″(0).✓ Proved
- \[ \left[\begin{matrix}0 \cdot 0 + 2 \cdot 0\\0 \cdot 0 + 0\\0 \cdot 0 + 1 \cdot 2\end{matrix}\right] = \left[\begin{matrix}0\\0\\2\end{matrix}\right] \]r′ × r″.✓ Proved
- \[ \frac{\sqrt{0^{2} + 0^{2} + 2^{2}}}{\left(\sqrt{0^{2} + 0^{2} + 1^{2}}\right)^{3}} = 2 \]κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved
Answer \( \kappa = 2 \)
✓ 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. How we verify
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 |
| 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 | the circle through three points 10⁻⁵ apart on the curve |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-17gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: pass 2026-10-15gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: pass 2026-10-14gpt-oss:20b: pass 2026-10-14
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/curvature, checked 2026-10-17 with SymPy 1.14.0.