Velocity, acceleration and speed in space
Problem 9.396 · 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 \( \displaystyle t = 0 \).
- \[ \left[\begin{matrix}\frac{d}{d t} t \cos{\left(t \right)}\\\frac{d}{d t} t \sin{\left(t \right)}\\\frac{d}{d t} t\end{matrix}\right] = \left[\begin{matrix}- t \sin{\left(t \right)} + \cos{\left(t \right)}\\t \cos{\left(t \right)} + \sin{\left(t \right)}\\1\end{matrix}\right] \]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
- \[ \sqrt{2} \]Speed = ‖v(0)‖.✓ Proved
Answer \( \mathbf v = \left\langle 1, 0, 1 \right\rangle,\ \mathbf a = \left\langle 0, 2, 0 \right\rangle,\ \text{speed} = \sqrt{2} \)
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 explicitly evaluate the velocity and acceleration vectors at t=0. While the final numerical answers are correct, the steps shown only provide the general symbolic expressions for v(t) and a(t), omitting the crucial substitution step required by the problem statement.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution fails to explicitly evaluate the velocity and acceleration vectors at t=0. While the final numerical answers are correct, the steps shown only provide the general symbolic expressions for v(t) and a(t), omitting the crucial substitution step required by the problem statement.gpt-oss:20b: pass 2026-10-08gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution fails to explicitly evaluate the velocity and acceleration vectors at t=0 before stating the final answer. While the algebraic derivatives are correct, the step from the general derivative expressions to the specific numerical vectors <1, 0, 1> and <0, 2, 0> is missing, making the derivation incomplete and potentially confusing for a student.
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 2026-10-08 with SymPy 1.14.0.