= Solution
The <determinant> of the <Jacobian matrix> is $s^2+t^2$. Wherever $(s,t)\ne(0,0)$, the <inverse function theorem> permits a local inverse of the <parabolic coordinates> transformation, and
$$
\frac{\partial(s,t)}{\partial(x,y)}
=J^{-1}
=\frac1{s^2+t^2}\begin{pmatrix}t&-s\\s&t\end{pmatrix}.
$$
Reading off its entries gives
$$
\boxed{s_x=\frac t{s^2+t^2},\quad s_y=-\frac s{s^2+t^2},\quad
t_x=\frac s{s^2+t^2},\quad t_y=\frac t{s^2+t^2}.}
$$
These are local formulas on a chosen branch: $(s,t)$ and $(-s,-t)$ have the same image, and at the origin the <Jacobian matrix> is singular, so this inverse-derivative formula does not apply there.
Back to article page
