∫Calc Practice

Arc length and curvature of space curves

Problem 9.751 · medium

Find the curvature of \( \displaystyle \mathbf r(t) = \left\langle 3 \cos{\left(t \right)}, 3 \sin{\left(t \right)}, t \right\rangle \) at \( \displaystyle t = 1 \).
  1. \[ \left[\begin{matrix}\left. \frac{d}{d t} 3 \cos{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d}{d t} 3 \sin{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d}{d t} t \right|_{\substack{ t=1 }}\end{matrix}\right] = \left[\begin{matrix}- 3 \sin{\left(1 \right)}\\3 \cos{\left(1 \right)}\\1\end{matrix}\right] \]
    r′(1).✓ Proved
  2. \[ \left[\begin{matrix}\left. \frac{d^{2}}{d t^{2}} 3 \cos{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d^{2}}{d t^{2}} 3 \sin{\left(t \right)} \right|_{\substack{ t=1 }}\\\left. \frac{d^{2}}{d t^{2}} t \right|_{\substack{ t=1 }}\end{matrix}\right] = \left[\begin{matrix}- 3 \cos{\left(1 \right)}\\- 3 \sin{\left(1 \right)}\\0\end{matrix}\right] \]
    r″(1).✓ Proved
  3. \[ \left[\begin{matrix}3 \cos{\left(1 \right)} 0 - \left(-1\right) 3 \sin{\left(1 \right)}\\1 \left(-3\right) \cos{\left(1 \right)} - 1 \left(-3\right) \sin{\left(1 \right)} 0\\3 \cos{\left(1 \right)} 3 \cos{\left(1 \right)} - 3 \sin{\left(1 \right)} \left(-3\right) \sin{\left(1 \right)}\end{matrix}\right] = \left[\begin{matrix}3 \sin{\left(1 \right)}\\- 3 \cos{\left(1 \right)}\\9 \cos^{2}{\left(1 \right)} + 9 \sin^{2}{\left(1 \right)}\end{matrix}\right] \]
    r′ × r″.✓ Proved
  4. \[ \frac{\sqrt{9 \cos^{2}{\left(1 \right)} + 9 \sin^{2}{\left(1 \right)} + \left(9 \cos^{2}{\left(1 \right)} + 9 \sin^{2}{\left(1 \right)}\right)^{2}}}{\left(1 + 9 \cos^{2}{\left(1 \right)} + 9 \sin^{2}{\left(1 \right)}\right)^{\frac{3}{2}}} = \frac{3}{10} \]
    κ = ‖r′ × r″‖ / ‖r′‖³.✓ Proved
Answer \( \kappa = \frac{3}{10} \)

✓ 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 — The solution correctly applies the curvature formula using the cross product of the first and second derivatives. The algebraic simplification is correct, as the magnitude of the cross product is 3 and the magnitude of the velocity vector cubed is 10.
Every verdict on record (6)
  • qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly applies the curvature formula using the cross product of the first and second derivatives. The algebraic simplification is correct, as the magnitude of the cross product is 3 and the magnitude of the velocity vector cubed is 10.
  • gpt-oss:20b: pass 2026-10-18
  • qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly applies the curvature formula for a space curve, κ = ‖r′ × r″‖ / ‖r′‖³. The derivatives, cross product, and norm calculations are algebraically correct and simplify to the stated answer.
  • gpt-oss:20b: pass 2026-10-16
  • qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly applies the curvature formula κ = ‖r′ × r″‖ / ‖r′‖³. The derivatives, cross product, and magnitude calculations are algebraically correct, leading to the right final answer.
  • 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.