∫Calc Practice

Calculus of vector-valued functions

Problem 9.499 · medium

Find the unit tangent vector \( \displaystyle \mathbf T \) of \( \displaystyle \mathbf r(t) = \left\langle t^{2}, t^{3}, t \right\rangle \) at \( \displaystyle t = 2 \).
  1. \[ \left[\begin{matrix}\frac{d}{d t} t^{2}\\\frac{d}{d t} t^{3}\\\frac{d}{d t} t\end{matrix}\right] = \left[\begin{matrix}2 t\\3 t^{2}\\1\end{matrix}\right] \]
    r′(t).✓ Proved
  2. \[ \sqrt{161} \]
    ‖r′(2)‖.✓ Proved
  3. \[ \left[\begin{matrix}\frac{4 \sqrt{161}}{161}\\\frac{12 \sqrt{161}}{161}\\\frac{\sqrt{161}}{161}\end{matrix}\right] \]
    T = r′/‖r′‖.✓ Proved
Answer \( \mathbf T(2) = \left\langle \frac{4 \sqrt{161}}{161}, \frac{12 \sqrt{161}}{161}, \frac{\sqrt{161}}{161} \right\rangle \)

✓ 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 1.3.0a numerical velocity, normalised

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: pass — The solution correctly computes the derivative, evaluates its magnitude at t=2, and normalizes the vector. The steps are logically sound and the final answer is correct.
Every verdict on record (4)
  • gpt-oss:20b: pass 2026-10-11
  • qwen3.6:27b-mlx: pass 2026-10-11 — The solution correctly computes the derivative, evaluates its magnitude at t=2, and normalizes the vector. The steps are logically sound and the final answer is correct.
  • qwen3.6:27b-mlx: pass 2026-10-11 — The solution correctly computes the derivative, evaluates its magnitude at t=2, and normalizes the vector to find the unit tangent vector. The algebraic steps are correct and the final answer matches the stated answer.
  • gpt-oss:20b: pass 2026-10-11

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/vector_function_calculus, checked 2026-10-11 with SymPy 1.14.0.