 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2001/paper-2/9/b/solution

Differentiate both algebraic constraints at fixed

y

or fixed

x

z_{x} = \frac{2 x}{z}, z_{y} = \frac{y}{z}, u_{x} = \frac{x}{u}, u_{y} = - \frac{y}{u} .

ϕ (x, y) = ψ (z, u)

therefore gives

(\frac{\partial ϕ}{\partial x})_{y} = \frac{2 x}{z} (\frac{\partial ψ}{\partial z})_{u} + \frac{x}{u} (\frac{\partial ψ}{\partial u})_{z}, (\frac{\partial ϕ}{\partial y})_{x} = \frac{y}{z} (\frac{\partial ψ}{\partial z})_{u} - \frac{y}{u} (\frac{\partial ψ}{\partial u})_{z} .

Divide the first identity by

x

and the second by

y

, and add. The

ψ_{u}

terms cancel, leaving

\frac{1}{x} (\frac{\partial ϕ}{\partial x})_{y} + \frac{1}{y} (\frac{\partial ϕ}{\partial y})_{x} = \frac{3}{z} (\frac{\partial ψ}{\partial z})_{u} .

(x, y) \mapsto (z, u)

- 3 x y / (z u)

, so the calculation is valid on real branches with

x yz u \neq = 0

. This explicitly supplies the local independence assumption underlying the change of coordinates.

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