Arc length and curvature of space curves
Problem 9.659 · medium
Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle t, \sin{\left(t \right)}, 0 \right\rangle \) at \( \displaystyle t = 1 \).
- \[ \left[\begin{matrix}\left. \frac{d}{d t} t \right|_{\substack{ t=1 }}\\\left. \frac{d}{d t} \sin{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d}{d t} 0 \right|_{\substack{ t=1 }}\end{matrix}\right] = \left[\begin{matrix}1\\\cos{\left(1 \right)}\\0\end{matrix}\right] \]r′(1).✓ Proved
- \[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} t \right|_{\substack{ t=1 }}\\\left. \frac{d^{2}}{d t^{2}} \sin{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d^{2}}{d t^{2}} 0 \right|_{\substack{ t=1 }}\end{matrix}\right] = \left[\begin{matrix}0\\- \sin{\left(1 \right)}\\0\end{matrix}\right] \]r″(1).✓ Proved
- \[ \left[\begin{matrix}\cos{\left(1 \right)} 0 + 0 \left(- \sin{\left(1 \right)}\right)\\0 \cdot 0 + 0\\1 \left(- \sin{\left(1 \right)}\right) - 1 \cos{\left(1 \right)} 0\end{matrix}\right] = \left[\begin{matrix}0\\0\\- \sin{\left(1 \right)}\end{matrix}\right] \]r′ × r″.✓ Proved
- \[ \frac{\sqrt{0^{2} + 0^{2} + \left(- \sin{\left(1 \right)}\right)^{2}}}{\left(\sqrt{0^{2} + \cos^{2}{\left(1 \right)} + 1^{2}}\right)^{3}} = \frac{\sin{\left(1 \right)}}{\left(\cos^{2}{\left(1 \right)} + 1\right)^{\frac{3}{2}}} \]κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved
Answer \( \kappa = \frac{\sin{\left(1 \right)}}{\left(\cos^{2}{\left(1 \right)} + 1\right)^{\frac{3}{2}}} \)
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: fail (misleading) — The curvature formula should involve the absolute value of the cross‑product norm, i.e. κ = |sin 1|/(1+cos² 1)^{3/2}. The solution omits the absolute value, which could mislead a student into thinking the curvature could be negative.qwen3.6:27b-mlx: inconclusive — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly simplifies ||r' x r''|| = sqrt(sin^2(1)) to sin(1). Since sin(1) > 0, this specific numerical instance is correct, but the
Every verdict on record (6)
qwen3.6:27b-mlx: inconclusive 2026-10-17 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly simplifies ||r' x r''|| = sqrt(sin^2(1)) to sin(1). Since sin(1) > 0, this specific numerical instance is correct, but thegpt-oss:20b: fail (misleading) 2026-10-17 — The curvature formula should involve the absolute value of the cross‑product norm, i.e. κ = |sin 1|/(1+cos² 1)^{3/2}. The solution omits the absolute value, which could mislead a student into thinking the curvature could be negative.qwen3.6:27b-mlx: inconclusive 2026-10-15 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution computes the magnitude of the cross product as sin(1), but the magnitude must be non-negative. Since sin(1) > 0, this specific instancegpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution computes the norm of the cross product as sin(1), but the norm is |sin(1)|. Since sin(1) > 0, this specific numerical result is correct, but the algebraic step sqrt(sin(1)^2) = sin(1) is technically an equality only because sin(1) is positive; generally, curvature formulas require absolute values or careful sign handling. More importantly, the final answer provided in the problem statement is sin(1)/(...), which is positive, so the calculation is numerically correct. However, looking closely at step 4, it simplifies sqrt((-sin(1))^2) to sin(1). This is valid for t=1. Is there a deeper error? Let's re-read the prompt constraints. 'Find what is wrong'. The solution is actually correct for t=1. Wait, let me double check the cross product. r' = <1, cos t, 0>, r'' = <0, -sin t, 0>. r' x r'' = <0, 0, -sin t>. Norm is |sin t|. At t=1, |sin 1| = sin 1. Denominator is (1 + cos^2 t)^(3/2). The result is correct. Why would it fail? Perhaps the 'unchecked' equations are just placeholders and I should focus on the sentences? There are no sentences. The prompt says 'Each line... is either an EQUATION... or a SENTENCE'. The solution provided ONLY has equations. The instructions say 'The sentences are what you are here for.' If there are no sentences, is it a pass? Or is the lack of setup a failure? Usually, these reviews check the logic. The logic is sound. However, often in these adversarial reviews, if the solution is just a list of equations without explanatory text linking them to the theorem, it might be considered 'style' or 'misleading' if it doesn't state the formula used. But step 4 explicitly states '-- κ = ‖r′ × r″‖ / ‖r′‖³'. This is a comment, not a formal sentence in the list? The prompt format lists them as [equation, unchecked]. The text after the equation is likely part of the equation line or a comment. If the system considers the comments as 'sentences', then the logic is fine. Let's look for a subtle error. r(t) = <t, sin t, 0>. r'(t) = <1, cos t, 0>. r''(t) = <0, -sin t, 0>. Cross product is <0, 0, -sin t>. Magnitude is |sin t|. At t=1, sin(1) is positive. So |sin(1)| = sin(1). The denominator is (sqrt(1+cos^2(1)))^3. The result matches. Is there any reason to fail? Maybe the prompt implies that the solution *must* have sentences? No, it says 'Each line... is either...'. If there are no sentences, there is nothing to check regarding 'setup models the problem' in natural language. However, usually, a 'pass' requires the solution to be complete and correct. It is. Let's look at the 'Stated answer'. It matches. I will mark it as pass unless I find a hidden trap. Trap: Did it evaluate at t=1 correctly? Yes. Did it use the right formula? Yes. Is the algebra correct? Yes. Verdict: pass.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 2026-10-17 with SymPy 1.14.0.