∫Calc Practice

Velocity, acceleration and speed in space

Problem 9.473 · medium

A particle has position \( \displaystyle \mathbf r(t) = \left\langle t^{2}, t^{3}, t \right\rangle \). Find its velocity, acceleration and speed at \( \displaystyle t = 1 \).
  1. \[ \left[\begin{matrix}\frac{d}{d t} t^{2}\\\frac{d}{d t} t^{3}\\\frac{d}{d t} t\end{matrix}\right] = \left[\begin{matrix}2 t\\3 t^{2}\\1\end{matrix}\right] \]
    v = r′.✓ Proved
  2. \[ \left[\begin{matrix}\frac{d}{d t} 2 t\\\frac{d}{d t} 3 t^{2}\\\frac{d}{d t} 1\end{matrix}\right] = \left[\begin{matrix}2\\6 t\\0\end{matrix}\right] \]
    a = v′.✓ Proved
  3. \[ \sqrt{14} \]
    Speed = ‖v(1)‖.✓ Proved
Answer \( \mathbf v = \left\langle 2, 3, 1 \right\rangle,\ \mathbf a = \left\langle 2, 6, 0 \right\rangle,\ \text{speed} = \sqrt{14} \)

✓ 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.

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
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0difference quotients of the position

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: pass — The solution correctly computes the velocity and acceleration vectors by differentiating the position vector component-wise, and correctly evaluates the speed as the magnitude of the velocity vector at t=1. All steps are logically sound and algebraically correct.
Every verdict on record (4)
  • qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly computes the velocity and acceleration vectors by differentiating the position vector component-wise, and correctly evaluates the speed as the magnitude of the velocity vector at t=1. All steps are logically sound and algebraically correct.
  • gpt-oss:20b: pass 2026-10-10
  • qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly computes the velocity and acceleration vectors by differentiating the position vector component-wise, and correctly calculates the speed as the magnitude of the velocity vector at t=1. The algebraic steps are verified and the final values match the stated answer.
  • gpt-oss:20b: pass 2026-10-10

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.