In page-479 of Visual Complex Analysis, Tirstan Needham derives the flux of a vector field in Geometric form:
Here $z$ is the point to which the shaded region $R$ will ultimately be collapsed in order to find the divergence there, $S$ and $P$ are the streamline and orthogonal trajectory through $z$ , and $s$ and $p$ are arclength along $S$ and $P$, the direction of increasing $p$ being choosen to make a positive right angle with $X$
$$ \nabla \cdot X = \partial_s |X| + \kappa_p |X|$$
The $\partial_S$ is a derivative along streamlines of the vector field $X$ and $\kappa_S$ is the curvature of the streamline at the point we are taking divergence at.
In the derivation of the formula using inifinitesimals, the following identity is used:
$$ \delta ( dp)= \kappa_p ds dp= \kappa_p dA$$
I am trying to derive it.
My attempt:
$$ dp = r_p d \theta$$
Where $r_p$ is radius of osculating circle at $p$ of the orthogonal trajectory of $S$ and $d \theta$ is the angle in common with circle circle and arclength of the orthogonal trajectory.
Assuming that the $d \theta$ doesn't change from the labelled dp to the red marked edge , we have:
$$ \delta dp = dr_p d \theta \tag{1}$$
But we can consider the shaded region as a differential area of a circle:
$$ dA= d(r^2) \frac{ d \theta}{2} = r_p dr_p d \theta$$
Rerranging $$ \frac{dA}{r_p} = \kappa_p dA = dr_p d \theta$$. Plugging that into (1) we have:
$$ \delta (dp) = \kappa_p dA$$
Does my derivation look like what was intended?