Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 2 / 9

Solve the two algebraic constraints for the squared coordinates:On any nonsingular real branch, implicit differentiation at fixed $u$ gives $(\partial x/\partial z{)}_u=z/(3x)$ and $(\partial y/\partial z{)}_u=z/(3y)$. Differentiating $z^2=2x^2+y^2$ with the other original coordinate fixed gives $(\partial z/\partial x{)}_y=2x/z$ and $(\partial z/\partial y{)}_x=y/z$. ThereforeThis is the multivariable chain rule applied to $z(x(z,u),y(z,u))=z$. These formulas assume the relevant denominators are nonzero and the chosen variables are valid local coordinates.

At fixed $z$, differentiating $2x^2+y^2=z^2$ gives $(\partial x/\partial y{)}_z=-y/(2x)$. Also, $x^2=(z^2+u^2)/3$ gives $(\partial x/\partial u{)}_z=u/(3x)$. At fixed $y$, eliminate $x^2$ to write $u^2=(z^2-3y^2)/2$, giving $(\partial u/\partial z{)}_y=z/(2u)$. Finally, the previous part gave $(\partial y/\partial z{)}_u=z/(3y)$. ThusThe held-fixed subscripts matter: substituting a reciprocal partial derivative while silently changing what is held fixed would not preserve these cancellations.

Differentiate both algebraic constraints at fixed $y$ or fixed $x$:The multivariable chain rule for $\varphi (x,y)=\psi (z,u)$ therefore givesDivide the first identity by $x$ and the second by $y$, and add. The ${\psi }_u$ terms cancel, leavingThe Jacobian determinant of $(x,y)\mapsto (z,u)$ is $-3xy/(zu)$, so the calculation is valid on real branches with $xyzu\ne 0$. This explicitly supplies the local independence assumption underlying the change of coordinates.