Calculus of vector-valued functions
Problem 9.493 · medium
Find the unit tangent vector \( \displaystyle \mathbf T \) of \( \displaystyle \mathbf r(t) = \left\langle e^{t}, e^{- t}, t \right\rangle \) at \( \displaystyle t = 2 \).
- \[ \left[\begin{matrix}\frac{d}{d t} e^{t}\\\frac{d}{d t} e^{- t}\\\frac{d}{d t} t\end{matrix}\right] = \left[\begin{matrix}e^{t}\\- e^{- t}\\1\end{matrix}\right] \]r′(t).✓ Proved
- \[ \sqrt{e^{-4} + 1 + e^{4}} = \frac{\sqrt{1 + e^{4} + e^{8}}}{e^{2}} \]‖r′(2)‖.✓ Proved
- \[ \left[\begin{matrix}\frac{e^{2}}{\sqrt{e^{-4} + 1 + e^{4}}}\\- \frac{1}{\sqrt{e^{-4} + 1 + e^{4}} e^{2}}\\\frac{1}{\sqrt{e^{-4} + 1 + e^{4}}}\end{matrix}\right] = \left[\begin{matrix}\frac{e^{4}}{\sqrt{1 + e^{4} + e^{8}}}\\- \frac{1}{\sqrt{1 + e^{4} + e^{8}}}\\\frac{e^{2}}{\sqrt{1 + e^{4} + e^{8}}}\end{matrix}\right] \]T = r′/‖r′‖.✓ Proved
Answer \( \mathbf T(2) = \left\langle \frac{e^{4}}{\sqrt{1 + e^{4} + e^{8}}}, - \frac{1}{\sqrt{1 + e^{4} + e^{8}}}, \frac{e^{2}}{\sqrt{1 + e^{4} + e^{8}}} \right\rangle \)
✓ 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 t=2, calculates the magnitude, and normalizes the vector. The algebraic simplification in step 3 is correct and leads to the stated answer.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-11qwen3.6:27b-mlx: pass 2026-10-11 — The solution correctly computes the derivative, evaluates it at t=2, calculates the magnitude, and normalizes the vector. The algebraic simplification in step 3 is correct and leads to the stated answer.gpt-oss:20b: pass 2026-10-11qwen3.6:27b-mlx: pass 2026-10-11 — The solution correctly computes the derivative, evaluates it at t=2, calculates the magnitude, and normalizes the vector. The algebraic simplification in step 3 is 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 2026-10-11 with SymPy 1.14.0.