Calculus of vector-valued functions
Problem 9.573 · medium
Find the unit tangent vector \( \displaystyle \mathbf T \) of \( \displaystyle \mathbf r(t) = \left\langle 3 t, 4 t^{2}, 2 t^{3} \right\rangle \) at \( \displaystyle t = 0 \).
- \[ \left[\begin{matrix}\frac{d}{d t} 3 t\\\frac{d}{d t} 4 t^{2}\\\frac{d}{d t} 2 t^{3}\end{matrix}\right] = \left[\begin{matrix}3\\8 t\\6 t^{2}\end{matrix}\right] \]r′(t).✓ Proved
- \[ 3 \]‖r′(0)‖.✓ Proved
- \[ \left[\begin{matrix}1\\0\\0\end{matrix}\right] \]T = r′/‖r′‖.✓ Proved
Answer \( \mathbf T(0) = \left\langle 1, 0, 0 \right\rangle \)
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. The reviewers disagree about how one step is explained; every verdict is in the receipt.
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 its magnitude at t=0, and normalizes to find the unit tangent vector. The steps are logically sound and algebraically correct.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-13qwen3.6:27b-mlx: pass 2026-10-13 — The solution correctly computes the derivative, evaluates its magnitude at t=0, and normalizes to find the unit tangent vector. The steps are logically sound and algebraically correct.qwen3.6:27b-mlx: fail (error) 2026-10-13 — The solution skips the crucial step of evaluating the derivative vector r'(t) at t=0 before normalizing. While the final numerical answer is correct, the logic presented in step 3 implies that the vector <1, 0, 0> is the result of normalizing <3, 8t, 6t^2> generally, or fails to explicitly show the substitution t=0 into r'(t) to get <3, 0, 0>. A student might incorrectly believe that one can normalize the symbolic derivative vector directly without substituting the parameter value first, or that the normalization of <3, 8t, 6t^2> yields <1, 0, 0> for all t. The step 'T = r'/||r'||' is applied to the symbolic vector rather than the evaluated vector at t=0.gpt-oss:20b: pass 2026-10-13
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 2026-10-13 with SymPy 1.14.0.