When discussing finding the arc length of some curve $C$ defined by parametric equations $x = f(\theta)$, $y = g(\theta)$, my professor said the following:
Given $f', g'$ are continuous on $[\alpha, \beta]$ and $C$ is traversed exactly once as $\theta$ increases from $\alpha$ to $\beta$, then the length of $C$ is $$L = \int^\beta_\alpha \sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}d\theta = \int^\beta_\alpha dS$$
I understand how this formula was derived, but I am confused as to why
The curve must only be traversed once for this to be true - I'd think this only matters if you're making a distinction between distance a particle travels and length of path.
Why it is useful to substitute in $dS$, and why that substitution is dependent on the single-traverse condition - we'd already derived this formula before this particular substitution was introduced and the single-traverse condition wasn't mentioned until this point.
Thank you!
Consider if $C$ passes a bit multiple times. Then the actual length of the curve would not increase, but the integral does. Thus we need to assume that the curve is traversed exactly once. The notation $\int_C dS = \int_C 1\,dS$ is not really a Substitution, but rather the definition of the curve integral. Basically this says that the length of the curve is the integral of $1$ over the curve (similar to how the volume of a $n$-dimensional set is the integral of $1$ over that set).
The single-traverse condition is not related to the definition of the integral, it’s just (as I already said) there as the length of the curve should only depend on its trace, and thus not increase by traversing it multiple times.