Jacobians and change of variables
Problem 11.205 · easy
Find the Jacobian \( \displaystyle \dfrac{\partial(x, y)}{\partial(u, v)} \) of the transformation \( \displaystyle x = u \cos{\left(v \right)} \), \( \displaystyle y = u \sin{\left(v \right)} \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial u} u \cos{\left(v \right)} & \frac{\partial}{\partial v} u \cos{\left(v \right)}\\\frac{\partial}{\partial u} u \sin{\left(v \right)} & \frac{\partial}{\partial v} u \sin{\left(v \right)}\end{matrix}\right] = \left[\begin{matrix}\cos{\left(v \right)} & - u \sin{\left(v \right)}\\\sin{\left(v \right)} & u \cos{\left(v \right)}\end{matrix}\right] \]The matrix of partial derivatives.✓ Proved
- \[ u \sin^{2}{\left(v \right)} + u \cos^{2}{\left(v \right)} = u \]Its determinant: x_u·y_v − x_v·y_u.✓ Proved
- |J| is the factor by which the transformation stretches small areas: dA = |J| du dv.
Answer \( u \)
Lines: 2 proved, 1 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.
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 | Not checked | — | a sentence; read, not computed |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the image of a tiny square measured by the shoelace formula |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (misleading) — The problem asks for the Jacobian determinant, but step 3 introduces the absolute value |J| and discusses area scaling. While |J| is used for area, the Jacobian itself is signed (and can be negative), so equating the requested quantity with |J| is mathematically imprecise and potentially misleading.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (misleading) 2026-10-04 — The problem asks for the Jacobian determinant, but step 3 introduces the absolute value |J| and discusses area scaling. While |J| is used for area, the Jacobian itself is signed (and can be negative), so equating the requested quantity with |J| is mathematically imprecise and potentially misleading.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: fail (misleading) 2026-10-04 — The problem asks for the Jacobian determinant, but the solution concludes with the absolute value |J|. While |J| is used for area elements, the Jacobian itself is u, which can be negative. Stating |J| as the final answer or implying it is the Jacobian is mathematically incorrect.gpt-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/jacobian, checked 2026-10-04 with SymPy 1.14.0.