Tangent planes
Problem 10.31 · medium
Find the tangent plane to \( \displaystyle z = x^{2} - y^{2} \) at the point \( \displaystyle (1, -1, 0) \).
- z = f(a,b) + f_x(a,b)(x − a) + f_y(a,b)(y − b).
- \[ \left. \frac{\partial}{\partial x} \left(x^{2} - y^{2}\right) \right|_{\substack{ x=1\\ y=-1 }} = 2 \]f_x at the point.✓ Proved
- \[ \left. \frac{\partial}{\partial y} \left(x^{2} - y^{2}\right) \right|_{\substack{ x=1\\ y=-1 }} = 2 \]f_y at the point.✓ Proved
- \[ 2 x + 2 y \]The plane.
Answer \( z = 2 x + 2 y \)
Lines: 2 proved, 2 not checked. 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. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 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 | Not checked | — | both sides are the same expression: the line states a value, and the working that produced it is not on the line |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the gap between surface and plane shrinks quadratically near the point |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The partial derivative with respect to y is -2y, which evaluates to 2 at (1, -1). However, the tangent plane equation requires the term f_y(a,b)(y - b), which is 2(y - (-1)) = 2(y + 1) = 2y + 2. The solution incorrectly simplifies this to just 2y, omitting the constant term and resulting in the wrong plane equation.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-09-28 — The partial derivative with respect to y is -2y, which evaluates to 2 at (1, -1). However, the tangent plane equation requires the term f_y(a,b)(y - b), which is 2(y - (-1)) = 2(y + 1) = 2y + 2. The solution incorrectly simplifies this to just 2y, omitting the constant term and resulting in the wrong plane equation.gpt-oss:20b: pass 2026-09-28qwen3.6:27b-mlx: fail (error) 2026-09-27 — The partial derivative with respect to y is -2y, which evaluates to 2 at (1, -1). However, the tangent plane formula requires f_y(y - b), so the term should be 2(y - (-1)) = 2y + 2. The solution incorrectly simplifies this to 2y, omitting the constant term and resulting in a plane that does not pass through the point (1, -1, 0).gpt-oss:20b: pass 2026-09-27qwen3.6:27b-mlx: fail (error) 2026-09-26 — The partial derivative with respect to y is -2y, which evaluates to 2 at (1, -1). However, the tangent plane equation should be z = 0 + 2(x-1) + 2(y+1), which simplifies to z = 2x + 2y. Wait, let me re-evaluate. f_x = 2x -> 2. f_y = -2y -> -2(-1) = 2. Point is (1, -1, 0). Equation: z - 0 = 2(x - 1) + 2(y - (-1)) => z = 2x - 2 + 2y + 2 => z = 2x + 2y. The stated answer is Eq(z, 2*x + 2*y). The solution steps are actually correct. Let me double check the 'proved' lines. Line 3 says f_y is 2. This is correct. Line 4 says the plane is 2x + 2y. This is correct. Why did I think it was wrong? Ah, usually students make a sign error on the y term. Here f_y = -2y. At y=-1, f_y = 2. The term in the plane equation is f_y(y-b) = 2(y - (-1)) = 2(y+1) = 2y + 2. The x term is 2(x-1) = 2x - 2. Sum = 2x + 2y. The constant terms cancel. So the answer is correct. The solution is sound.gpt-oss:20b: pass 2026-09-26
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/tangent_plane, checked 2026-10-13 with SymPy 1.14.0.