Calculus of vector-valued functions
Problem 9.257 · medium
Find the unit tangent vector \( \displaystyle \mathbf T \) of \( \displaystyle \mathbf r(t) = \left\langle t \cos{\left(t \right)}, t \sin{\left(t \right)}, t \right\rangle \) at \( \displaystyle t = \frac{\pi}{2} \).
- \[ \left[\begin{matrix}\frac{d}{d t} t \cos{\left(t \right)}\\\frac{d}{d t} t \sin{\left(t \right)}\\\frac{d}{d t} t\end{matrix}\right] = \left[\begin{matrix}- t \sin{\left(t \right)} + \cos{\left(t \right)}\\t \cos{\left(t \right)} + \sin{\left(t \right)}\\1\end{matrix}\right] \]r′(t).✓ Proved
- \[ \sqrt{2 + \frac{\pi^{2}}{4}} = \frac{\sqrt{8 + \pi^{2}}}{2} \]‖r′(pi/2)‖.✓ Proved
- \[ \left[\begin{matrix}- \frac{\pi}{2 \sqrt{2 + \frac{\pi^{2}}{4}}}\\\frac{1}{\sqrt{2 + \frac{\pi^{2}}{4}}}\\\frac{1}{\sqrt{2 + \frac{\pi^{2}}{4}}}\end{matrix}\right] = \left[\begin{matrix}- \frac{\pi}{\sqrt{8 + \pi^{2}}}\\\frac{2}{\sqrt{8 + \pi^{2}}}\\\frac{2}{\sqrt{8 + \pi^{2}}}\end{matrix}\right] \]T = r′/‖r′‖.✓ Proved
Answer \( \mathbf T(\frac{\pi}{2}) = \left\langle - \frac{\pi}{\sqrt{8 + \pi^{2}}}, \frac{2}{\sqrt{8 + \pi^{2}}}, \frac{2}{\sqrt{8 + \pi^{2}}} \right\rangle \)
✓ 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
| 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 | a numerical velocity, normalised |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly computes the derivative, evaluates it at the specified point, calculates the magnitude, and normalizes the vector. The algebraic simplifications are correct.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly computes the derivative, evaluates it at the specified point, calculates the magnitude, and normalizes the vector. The algebraic simplifications are correct.gpt-oss:20b: pass 2026-10-05qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly computes the derivative, evaluates it at t=pi/2, calculates the magnitude, and normalizes the vector. The algebraic simplifications are correct.gpt-oss:20b: pass 2026-10-05
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 2026-10-05 with SymPy 1.14.0.