∫Calc Practice

Arc length and curvature of space curves

Problem 9.660 · medium

Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle 4 \cos{\left(t \right)}, 4 \sin{\left(t \right)}, 0 \right\rangle \) at \( \displaystyle t = 1 \).
  1. \[ \left[\begin{matrix}\left. \frac{d}{d t} 4 \cos{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d}{d t} 4 \sin{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d}{d t} 0 \right|_{\substack{ t=1 }}\end{matrix}\right] = \left[\begin{matrix}- 4 \sin{\left(1 \right)}\\4 \cos{\left(1 \right)}\\0\end{matrix}\right] \]
    r′(1).✓ Proved
  2. \[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} 4 \cos{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d^{2}}{d t^{2}} 4 \sin{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d^{2}}{d t^{2}} 0 \right|_{\substack{ t=1 }}\end{matrix}\right] = \left[\begin{matrix}- 4 \cos{\left(1 \right)}\\- 4 \sin{\left(1 \right)}\\0\end{matrix}\right] \]
    r″(1).✓ Proved
  3. \[ \left[\begin{matrix}4 \cos{\left(1 \right)} 0 + 0 \left(-4\right) \sin{\left(1 \right)}\\0 \left(-4\right) \cos{\left(1 \right)} - 1 \left(-4\right) \sin{\left(1 \right)} 0\\4 \cos{\left(1 \right)} 4 \cos{\left(1 \right)} - 4 \sin{\left(1 \right)} \left(-4\right) \sin{\left(1 \right)}\end{matrix}\right] = \left[\begin{matrix}0\\0\\16 \cos^{2}{\left(1 \right)} + 16 \sin^{2}{\left(1 \right)}\end{matrix}\right] \]
    r′ × r″.✓ Proved
  4. \[ \frac{1}{\sqrt{16 \cos^{2}{\left(1 \right)} + 16 \sin^{2}{\left(1 \right)}}} = \frac{1}{4} \]
    κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved
Answer \( \kappa = \frac{1}{4} \)

✓ 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
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0the 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 2026-10-17
  • gpt-oss:20b: pass 2026-10-17
  • 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, and the algebraic simplifications 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 2026-10-17 with SymPy 1.14.0.