Linear approximation in two variables
Problem 10.373 · hard
Find the linear approximation of \( \displaystyle f(x, y) = x^{2} y^{3} \) at \( \displaystyle (2, 1) \), and use it to estimate \( \displaystyle f\left(\frac{99}{50}, \frac{9}{10}\right) \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} x^{2} y^{3}\\\frac{\partial}{\partial y} x^{2} y^{3}\end{matrix}\right] = \left[\begin{matrix}2 x y^{3}\\3 x^{2} y^{2}\end{matrix}\right] \]f_x and f_y.✓ Proved
- \[ \left[\begin{matrix}4\\4\\12\end{matrix}\right] \]f, f_x and f_y at the point.✓ Proved
- \[ 4 x + 12 y - 16 \]The linearization.✓ Proved
- \[ \frac{68}{25} \]The estimate; the true value is about 2.85797.✓ Proved
Answer \( L(x, y) = 4 x + 12 y - 16,\ f \approx \frac{68}{25} = 2.72000 \)
✓ 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 | L agrees with f to within O(h²) in four directions at h = 1e−5 |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: pass 2026-10-06gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: pass 2026-10-06
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/linear_approx_2var, checked 2026-10-06 with SymPy 1.14.0.