Calculus of vector-valued functions
Problem 9.254 · 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 \( \displaystyle t = 1 \).
- \[ \left[\begin{matrix}\frac{d}{d t} t \cos{\left(t \right)}\\\frac{d}{d t} t \sin{\left(t \right)}\\\frac{d}{d t} t\end{matrix}\right] = \left[\begin{matrix}- t \sin{\left(t \right)} + \cos{\left(t \right)}\\t \cos{\left(t \right)} + \sin{\left(t \right)}\\1\end{matrix}\right] \]r′(t).✓ Proved
- \[ \sqrt{\left(- \sin{\left(1 \right)} + \cos{\left(1 \right)}\right)^{2} + 1 + \left(\cos{\left(1 \right)} + \sin{\left(1 \right)}\right)^{2}} = \sqrt{3} \]‖r′(1)‖.✓ Proved
- \[ \left[\begin{matrix}\frac{- \sin{\left(1 \right)} + \cos{\left(1 \right)}}{\sqrt{\left(- \sin{\left(1 \right)} + \cos{\left(1 \right)}\right)^{2} + 1 + \left(\cos{\left(1 \right)} + \sin{\left(1 \right)}\right)^{2}}}\\\frac{\cos{\left(1 \right)} + \sin{\left(1 \right)}}{\sqrt{\left(- \sin{\left(1 \right)} + \cos{\left(1 \right)}\right)^{2} + 1 + \left(\cos{\left(1 \right)} + \sin{\left(1 \right)}\right)^{2}}}\\\frac{1}{\sqrt{\left(- \sin{\left(1 \right)} + \cos{\left(1 \right)}\right)^{2} + 1 + \left(\cos{\left(1 \right)} + \sin{\left(1 \right)}\right)^{2}}}\end{matrix}\right] = \left[\begin{matrix}\frac{\sqrt{6} \cos{\left(\frac{\pi}{4} + 1 \right)}}{3}\\\frac{\sqrt{6} \sin{\left(\frac{\pi}{4} + 1 \right)}}{3}\\\frac{\sqrt{3}}{3}\end{matrix}\right] \]T = r′/‖r′‖.✓ Proved
Answer \( \mathbf T(1) = \left\langle \frac{\sqrt{6} \cos{\left(\frac{\pi}{4} + 1 \right)}}{3}, \frac{\sqrt{6} \sin{\left(\frac{\pi}{4} + 1 \right)}}{3}, \frac{\sqrt{3}}{3} \right\rangle \)
✓ 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
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-05gpt-oss:20b: pass 2026-10-05qwen3.6:27b-mlx: pass 2026-10-05gpt-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.