Velocity, acceleration and speed in space
Problem 9.478 · medium
A particle has position \( \displaystyle \mathbf r(t) = \left\langle t \cos{\left(t \right)}, t \sin{\left(t \right)}, t \right\rangle \). Find its velocity, acceleration and speed at \( \displaystyle t = \pi \).
- \[ \left[\begin{matrix}\frac{d}{d t} t \cos{\left(t \right)}\\\frac{d}{d t} t \sin{\left(t \right)}\\\frac{d}{d t} t\end{matrix}\right] = \left[\begin{matrix}- t \sin{\left(t \right)} + \cos{\left(t \right)}\\t \cos{\left(t \right)} + \sin{\left(t \right)}\\1\end{matrix}\right] \]v = r′.✓ Proved
- \[ \left[\begin{matrix}\frac{d}{d t} \left(- t \sin{\left(t \right)} + \cos{\left(t \right)}\right)\\\frac{d}{d t} \left(t \cos{\left(t \right)} + \sin{\left(t \right)}\right)\\\frac{d}{d t} 1\end{matrix}\right] = \left[\begin{matrix}- t \cos{\left(t \right)} - 2 \sin{\left(t \right)}\\- t \sin{\left(t \right)} + 2 \cos{\left(t \right)}\\0\end{matrix}\right] \]a = v′.✓ Proved
- \[ \sqrt{2 + \pi^{2}} \]Speed = ‖v(pi)‖.✓ Proved
Answer \( \mathbf v = \left\langle -1, - \pi, 1 \right\rangle,\ \mathbf a = \left\langle \pi, -2, 0 \right\rangle,\ \text{speed} = \sqrt{2 + \pi^{2}} \)
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.
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 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | difference quotients of the position |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution fails to evaluate the velocity and acceleration vectors at t = \pi, leaving them as functions of t. The final answer claims specific numerical vectors, but the derivation does not show the substitution step required to obtain them.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution fails to evaluate the velocity and acceleration vectors at t = \pi, leaving them as functions of t. The final answer claims specific numerical vectors, but the derivation does not show the substitution step required to obtain them.gpt-oss:20b: pass 2026-10-10gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution fails to explicitly evaluate the velocity and acceleration vectors at t = pi, which are required by the problem statement. It only provides the speed at t = pi, omitting the specific vector values for v and a.
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/motion_in_space, checked 2026-10-10 with SymPy 1.14.0.