I am working on something using general 2D curvilinear coordinates $u(x,y)$ and $v(x,y)$ (which are locally orthogonal). I have trouble understanding why the following equality holds. I found it in the notes I was looking up but there is no explanation accompanying it: \begin{equation} \sqrt{\left( \displaystyle\frac{\partial v}{\partial x} \right)^2 + \left( \displaystyle\frac{\partial v}{\partial y} \right)^2} = \frac{1}{\sqrt{\left( \displaystyle\frac{\partial x}{\partial v} \right)^2 + \left( \displaystyle\frac{\partial y}{\partial v} \right)^2}} \end{equation}
Note 1: the right-hand side is actually $h_v^{-1}$, with $h_v$ the Lamé coefficient for $v$. You can find more information on this here: https://en.wikipedia.org/wiki/Curvilinear_coordinates#Relation_to_Lamé_coefficients)
Note 2: I suppose the same equality holds if you replace $v$ by $u$ since there is no specified preference for the $v$ coordinate
I have tried putting both square roots in the left-hand side and proving their product is equal to $1$. I used the chain rule like $\frac{\partial x}{\partial x} = \frac{\partial x}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial x}{\partial v} \frac{\partial v}{\partial x} = 1$ and $\frac{\partial x}{\partial y} = \frac{\partial x}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial x}{\partial v} \frac{\partial v}{\partial y} = 0$, but got nowhere...
I have verified the equation using polar coordinates (using $u:=r=\sqrt{x^2 + y^2}$ and $v:=\theta=\arctan(\frac{y}{x})$) and indeed found it to hold, with both sides equal to $\frac{1}{r}$.
Thanks in advance for any help with this.
I found the explanation. The left-hand side is equal to $||\nabla v||$ whereas the denominator of the right-hand side is $||\frac{\partial \pmb{r}}{\partial v}||$ with $\pmb{r}$ the position vector.
$\pmb{b}^v := \nabla v$ is actually a covariant basis vector with respect to the coordinate $v$, of which the corresponding contravariant basis vector is $\pmb{b}_v := \frac{\partial \pmb{r}}{\partial v}$.
There is a result that says that $\pmb{b}^v \cdot \pmb{b}_v = 1$ must hold (the proof is given on the Wikipedia page mentioned in the question). The equality I asked about in the question directly follows from this result.