Tangent planes
Problem 10.88 · hard
Find the tangent plane to \( \displaystyle z = e^{x - y} \) at the point \( \displaystyle (1, 2, e^{-1}) \).
- z = f(a,b) + f_x(a,b)(x − a) + f_y(a,b)(y − b).
- \[ \left. \frac{\partial}{\partial x} e^{x - y} \right|_{\substack{ x=1\\ y=2 }} = e^{-1} \]f_x at the point.✓ Proved
- \[ \left. \frac{\partial}{\partial y} e^{x - y} \right|_{\substack{ x=1\\ y=2 }} = - \frac{1}{e} \]f_y at the point.✓ Proved
- \[ \frac{x - 1}{e} - \frac{y - 2}{e} + e^{-1} = \frac{x}{e} - \frac{y}{e} + \frac{2}{e} \]The plane.✓ Proved
Answer \( z = \frac{x}{e} - \frac{y}{e} + \frac{2}{e} \)
Lines: 3 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
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 | ✓ 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 | the gap between surface and plane shrinks quadratically near the point |
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 2026-09-26 with SymPy 1.14.0.