∫Calc Practice

Calculus of vector-valued functions

Problem 9.411 · medium

For \( \displaystyle \mathbf r(t) = \left\langle e^{t}, e^{- t}, t \right\rangle \), find \( \displaystyle \mathbf r'(t) \) and the tangent vector at \( \displaystyle t = \pi \).
  1. \[ \left[\begin{matrix}\frac{d}{d t} e^{t}\\\frac{d}{d t} e^{- t}\\\frac{d}{d t} t\end{matrix}\right] = \left[\begin{matrix}e^{t}\\- e^{- t}\\1\end{matrix}\right] \]
    Differentiate each component.✓ Proved
  2. \[ \left[\begin{matrix}e^{\pi}\\- \frac{1}{e^{\pi}}\\1\end{matrix}\right] \]
    At t = pi.✓ Proved
Answer \( \mathbf r'(t) = \left\langle e^{t}, - e^{- t}, 1 \right\rangle,\ \mathbf r'(\pi) = \left\langle e^{\pi}, - \frac{1}{e^{\pi}}, 1 \right\rangle \)

✓ 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
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0central difference quotients of each component

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: pass — The solution correctly differentiates each component of the vector function and evaluates the result at t = pi. The steps are logically sound and the final answer matches the stated answer.
Every verdict on record (4)
  • gpt-oss:20b: pass 2026-10-08
  • qwen3.6:27b-mlx: pass 2026-10-08 — The solution correctly differentiates each component of the vector function and evaluates the result at t = pi. The steps are logically sound and the final answer matches the stated answer.
  • gpt-oss:20b: pass 2026-10-08
  • qwen3.6:27b-mlx: pass 2026-10-08 — The solution correctly differentiates each component of the vector function and evaluates the result at t = pi. The steps are logically sound and algebraically correct.

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 2026-10-08 with SymPy 1.14.0.