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I was wondering if the factor $\sqrt{(x')^2 + (y')^2}$ in the line integral formula $$\int_a^b\ f(x,\ y)\ \sqrt{(x')^2 + (y')^2}\ dt$$ can also be thought of as a Jacobian determinant, due to the resemblance of parameterization to changing variables in multiple integrals.

For this to be the case, there must be at least one solution to the partial differential equation $$(x')^2 + (y')^2 = a(x,\ y)^2d(x,\ y)^2 - 2a(x,\ y)b(x,\ y)c(x,\ y)d(x,\ y) + b(x,\ y)^2c(x,\ y)^2,$$ which would correspond to the Jacobian $$\begin{bmatrix}a(x,\ y) \ c(x,\ y) \\ b(x,\ y) \ d(x,\ y)\end{bmatrix}.$$

It should be noted that I don't know any methods for solving PDEs other than guessing, but I doubt analytic methods exist for anything this multivariate. My initial attempt was to let $a^2d^2 = x'$ and $b^2c^2 = y'$, but it follows that $-2abcd = 0$, so at least one of $a$, $b$, $c$, and $d$, must equal $0$, which messes up at least one of the other terms. I am aware that the Jacobian in a change of variables is interchangeable with a certain differential manipulation involving the exterior product, but since $dt\ \Lambda\ dt = 0$, its not clear that this process is more easily reversible than the Jacobian method. Is there a way to solve this PDE and/or the equivalent exterior algebra problem?

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