Linear approximation in two variables
Problem 10.377 · hard
Find the linear approximation of \( \displaystyle f(x, y) = e^{x} \cos{\left(y \right)} \) at \( \displaystyle (0, 0) \), and use it to estimate \( \displaystyle f\left(- \frac{1}{50}, \frac{1}{20}\right) \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} e^{x} \cos{\left(y \right)}\\\frac{\partial}{\partial y} e^{x} \cos{\left(y \right)}\end{matrix}\right] = \left[\begin{matrix}e^{x} \cos{\left(y \right)}\\- e^{x} \sin{\left(y \right)}\end{matrix}\right] \]f_x and f_y.✓ Proved
- \[ \left[\begin{matrix}1\\1\\0\end{matrix}\right] \]f, f_x and f_y at the point.✓ Proved
- \[ x + 1 \]The linearization.✓ Proved
- \[ \frac{49}{50} \]The estimate; the true value is about 0.97897.✓ Proved
Answer \( L(x, y) = x + 1,\ f \approx \frac{49}{50} = 0.98000 \)
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 |
| 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: fail (error) — The solution incorrectly calculates the partial derivative f_y at (0,0) as 1 instead of 0. Since f_y = -e^x sin(y), f_y(0,0) = 0. Consequently, the linear approximation L(x,y) = 1 + x is missing the y-term (which is correct to be 0, but the intermediate step claiming f_y=1 is wrong), and the estimate 49/50 is derived from an incorrect intermediate claim, although coincidentally the final numerical estimate happens to be close because the y-term is zero. However, stating f_y(0,0)=1 is a factual error.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution incorrectly calculates the partial derivative f_y at (0,0) as 1 instead of 0. Since f_y = -e^x sin(y), f_y(0,0) = 0. Consequently, the linear approximation L(x,y) = 1 + x is missing the y-term (which is correct to be 0, but the intermediate step claiming f_y=1 is wrong), and the estimate 49/50 is derived from an incorrect intermediate claim, although coincidentally the final numerical estimate happens to be close because the y-term is zero. However, stating f_y(0,0)=1 is a factual error.gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution incorrectly calculates the partial derivative with respect to y at (0,0) as 1 instead of 0. Since f_y(0,0) = -e^0 sin(0) = 0, the linear approximation should be L(x,y) = 1 + x, not 1 + x + y. Although the final numerical estimate happens to be correct because the y-term vanishes at the specific point, the stated formula L(x,y) = x + 1 is missing the y-term explicitly, but more importantly, step 2 claims f_y(0,0)=1 which is factually wrong.
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.