Maximum rate of change
Problem 10.333 · medium
Find the directional derivative of \( \displaystyle f(x, y) = x^{2} + 3 y^{2} \) at \( \displaystyle P(1, 1) \) in the direction toward \( \displaystyle Q(4, 2) \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} + 3 y^{2}\right)\\\frac{\partial}{\partial y} \left(x^{2} + 3 y^{2}\right)\end{matrix}\right] = \left[\begin{matrix}2 x\\6 y\end{matrix}\right] \]∇f.✓ Proved
- \[ \left[\begin{matrix}2\\6\end{matrix}\right] \]∇f(1, 1).✓ Proved
- \[ \left[\begin{matrix}\frac{3 \sqrt{10}}{10}\\\frac{\sqrt{10}}{10}\end{matrix}\right] \]The unit vector from P toward Q.✓ Proved
- \[ \frac{6 \sqrt{10}}{5} \]D_u f = ∇f · u.✓ Proved
Answer \( D_{\mathbf u} f = \frac{6 \sqrt{10}}{5} \)
✓ 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 |
| 4 | ✓ 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 | difference quotient along the direction |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-04gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: pass 2026-10-04gpt-oss:20b: pass 2026-10-04
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/max_rate_of_change, checked 2026-10-04 with SymPy 1.14.0.