Home›Calculus 3›Arc length and curvature of space curves›Problem 9.807

Arc length and curvature of space curves

Problem 9.807 · medium

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

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

Answer \( \kappa = 1 \)

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) — Step 3 incorrectly computes the cross product r′×r″ as (0,0,1) instead of (0,0,−1). Although the magnitude is still 1, the sign error misleads a student about the direction of the cross product.
qwen3.6:27b-mlx: pass

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-19
gpt-oss:20b: fail (misleading) 2026-10-19 — Step 3 incorrectly computes the cross product r′×r″ as (0,0,1) instead of (0,0,−1). Although the magnitude is still 1, the sign error misleads a student about the direction of the cross product.
qwen3.6:27b-mlx: pass 2026-10-16
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: pass 2026-10-16
gpt-oss:20b: pass 2026-10-16

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 2026-10-19 with SymPy 1.14.0.