Home›Calculus 3›Arc length and curvature of space curves›Problem 9.745

Arc length and curvature of space curves

Problem 9.745 · medium

Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle 3 \cos{\left(t \right)}, 3 \sin{\left(t \right)}, 0 \right\rangle \) at \( \displaystyle t = 1 \).

\[ \left[\begin{matrix}\left. \frac{d}{d t} 3 \cos{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d}{d t} 3 \sin{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d}{d t} 0 \right|_{\substack{ t=1 }}\end{matrix}\right] = \left[\begin{matrix}- 3 \sin{\left(1 \right)}\\3 \cos{\left(1 \right)}\\0\end{matrix}\right] \]
r′(1).✓ Proved
\[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} 3 \cos{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d^{2}}{d t^{2}} 3 \sin{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d^{2}}{d t^{2}} 0 \right|_{\substack{ t=1 }}\end{matrix}\right] = \left[\begin{matrix}- 3 \cos{\left(1 \right)}\\- 3 \sin{\left(1 \right)}\\0\end{matrix}\right] \]
r″(1).✓ Proved
\[ \left[\begin{matrix}3 \cos{\left(1 \right)} 0 + 0 \left(-3\right) \sin{\left(1 \right)}\\0 \left(-3\right) \cos{\left(1 \right)} - 1 \left(-3\right) \sin{\left(1 \right)} 0\\3 \cos{\left(1 \right)} 3 \cos{\left(1 \right)} - 3 \sin{\left(1 \right)} \left(-3\right) \sin{\left(1 \right)}\end{matrix}\right] = \left[\begin{matrix}0\\0\\9 \cos^{2}{\left(1 \right)} + 9 \sin^{2}{\left(1 \right)}\end{matrix}\right] \]
r′ × r″.✓ Proved
\[ \frac{1}{\sqrt{9 \cos^{2}{\left(1 \right)} + 9 \sin^{2}{\left(1 \right)}}} = \frac{1}{3} \]
κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved

Answer \( \kappa = \frac{1}{3} \)

✓ 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 2026-10-16
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly applies the curvature formula for a space curve, and the algebraic simplifications are correct.
gpt-oss:20b: pass 2026-10-15

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.