Arc length and curvature of space curves
Problem 9.664 · medium
Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle 4 \cos{\left(t \right)}, 4 \sin{\left(t \right)}, t \right\rangle \) at \( \displaystyle t = 1 \).
- \[ \left[\begin{matrix}\left. \frac{d}{d t} 4 \cos{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d}{d t} 4 \sin{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d}{d t} t \right|_{\substack{ t=1 }}\end{matrix}\right] = \left[\begin{matrix}- 4 \sin{\left(1 \right)}\\4 \cos{\left(1 \right)}\\1\end{matrix}\right] \]r′(1).✓ Proved
- \[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} 4 \cos{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d^{2}}{d t^{2}} 4 \sin{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d^{2}}{d t^{2}} t \right|_{\substack{ t=1 }}\end{matrix}\right] = \left[\begin{matrix}- 4 \cos{\left(1 \right)}\\- 4 \sin{\left(1 \right)}\\0\end{matrix}\right] \]r″(1).✓ Proved
- \[ \left[\begin{matrix}4 \cos{\left(1 \right)} 0 - \left(-1\right) 4 \sin{\left(1 \right)}\\1 \left(-4\right) \cos{\left(1 \right)} - 1 \left(-4\right) \sin{\left(1 \right)} 0\\4 \cos{\left(1 \right)} 4 \cos{\left(1 \right)} - 4 \sin{\left(1 \right)} \left(-4\right) \sin{\left(1 \right)}\end{matrix}\right] = \left[\begin{matrix}4 \sin{\left(1 \right)}\\- 4 \cos{\left(1 \right)}\\16 \cos^{2}{\left(1 \right)} + 16 \sin^{2}{\left(1 \right)}\end{matrix}\right] \]r′ × r″.✓ Proved
- \[ \frac{\sqrt{16 \cos^{2}{\left(1 \right)} + 16 \sin^{2}{\left(1 \right)} + \left(16 \cos^{2}{\left(1 \right)} + 16 \sin^{2}{\left(1 \right)}\right)^{2}}}{\left(1 + 16 \cos^{2}{\left(1 \right)} + 16 \sin^{2}{\left(1 \right)}\right)^{\frac{3}{2}}} = \frac{4}{17} \]κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved
Answer \( \kappa = \frac{4}{17} \)
✓ 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 — The solution correctly applies the curvature formula κ = ‖r′ × r″‖ / ‖r′‖³. The algebraic simplifications in the intermediate steps are correct, and the final result matches the stated answer.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly applies the curvature formula κ = ‖r′ × r″‖ / ‖r′‖³. The algebraic simplifications in the intermediate steps are correct, and the final result matches the stated answer.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly applies the curvature formula κ = ‖r′ × r″‖ / ‖r′‖³. The algebraic simplifications in the final step are correct: ‖r′‖ = √17 and ‖r′ × r″‖ = 4, yielding 4/17.gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly applies the curvature formula κ = ‖r′ × r″‖ / ‖r′‖³. The derivatives, cross product, and magnitude calculations are algebraically correct and lead to the right result.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.