Arc length and curvature of space curves
Problem 9.687 · medium
Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle t, t^{2}, 0 \right\rangle \) at \( \displaystyle t = 2 \).
- \[ \left[\begin{matrix}\left. \frac{d}{d t} t \right|_{\substack{ t=2 }}\\\left. \frac{d}{d t} t^{2} \right|_{\substack{ t=2 }}\\\left. \frac{d}{d t} 0 \right|_{\substack{ t=2 }}\end{matrix}\right] = \left[\begin{matrix}1\\4\\0\end{matrix}\right] \]r′(2).✓ Proved
- \[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} t \right|_{\substack{ t=2 }}\\\left. \frac{d^{2}}{d t^{2}} t^{2} \right|_{\substack{ t=2 }}\\\left. \frac{d^{2}}{d t^{2}} 0 \right|_{\substack{ t=2 }}\end{matrix}\right] = \left[\begin{matrix}0\\2\\0\end{matrix}\right] \]r″(2).✓ Proved
- \[ \left[\begin{matrix}4 \cdot 0 + 2 \cdot 0\\0 \cdot 0 + 0\\0 \left(-4\right) + 1 \cdot 2\end{matrix}\right] = \left[\begin{matrix}0\\0\\2\end{matrix}\right] \]r′ × r″.✓ Proved
- \[ \frac{\sqrt{0^{2} + 0^{2} + 2^{2}}}{\left(\sqrt{0^{2} + 1^{2} + 4^{2}}\right)^{3}} = \frac{2 \sqrt{17}}{289} \]κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved
Answer \( \kappa = \frac{2 \sqrt{17}}{289} \)
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. 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 — The solution correctly applies the curvature formula using the cross product of the first and second derivatives evaluated at t=2. The calculations for r'(2), r''(2), their cross product, and the final magnitude ratio are all correct.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly applies the curvature formula using the cross product of the first and second derivatives evaluated at t=2. The calculations for r'(2), r''(2), their cross product, and the final magnitude ratio are all correct.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: fail (error) 2026-10-15 — The labels for lines 1 and 2 are incorrect: line 1 computes the first derivative r'(2), not the second derivative as labeled, and line 2 computes the second derivative r''(2), not the first. While the numerical values used in subsequent steps are correct, the text descriptions are factually wrong and would confuse a student regarding which vector is which.gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: fail (error) 2026-10-14 — The labels for lines 1 and 2 are incorrect. Line 1 computes the second derivative vector r''(2), not the first derivative r'(2). Line 2 computes the third derivative vector r'''(2), not the second derivative r''(2). Although the numerical values used in the final curvature formula happen to correspond to the correct r' and r'' vectors, the intermediate steps are mislabeled and mathematically inconsistent with their descriptions.gpt-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 2026-10-18 with SymPy 1.14.0.