Home›Calculus 3›Arc length and curvature of space curves›Problem 9.690

Arc length and curvature of space curves

Problem 9.690 · medium

Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle \cos{\left(t \right)}, \sin{\left(t \right)}, 0 \right\rangle \) at \( \displaystyle t = 1 \).

\[ \left[\begin{matrix}\left. \frac{d}{d t} \cos{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d}{d t} \sin{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d}{d t} 0 \right|_{\substack{ t=1 }}\end{matrix}\right] = \left[\begin{matrix}- \sin{\left(1 \right)}\\\cos{\left(1 \right)}\\0\end{matrix}\right] \]
r′(1).✓ Proved
\[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} \cos{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d^{2}}{d t^{2}} \sin{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d^{2}}{d t^{2}} 0 \right|_{\substack{ t=1 }}\end{matrix}\right] = \left[\begin{matrix}- \cos{\left(1 \right)}\\- \sin{\left(1 \right)}\\0\end{matrix}\right] \]
r″(1).✓ Proved
\[ \left[\begin{matrix}\cos{\left(1 \right)} 0 + 0 \left(- \sin{\left(1 \right)}\right)\\0 \left(- \cos{\left(1 \right)}\right) - 1 \left(- \sin{\left(1 \right)}\right) 0\\\cos{\left(1 \right)} \cos{\left(1 \right)} + - \sin{\left(1 \right)} \left(- \sin{\left(1 \right)}\right)\end{matrix}\right] = \left[\begin{matrix}0\\0\\\cos^{2}{\left(1 \right)} + \sin^{2}{\left(1 \right)}\end{matrix}\right] \]
r′ × r″.✓ Proved
\[ \frac{1}{\sqrt{\cos^{2}{\left(1 \right)} + \sin^{2}{\left(1 \right)}}} = 1 \]
κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved

Answer \( \kappa = 1 \)

✓ 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: pass
qwen3.6:27b-mlx: pass

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-18
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: pass 2026-10-15
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly applies the curvature formula for a space curve using the cross product of the first and second derivatives. The algebraic simplifications and evaluations are correct.
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.