Recently I was playing around with some calculus and stumbled upon a curious idea. Perhaps someone can help me understand this.
Consider a smooth curve $f(x)$. Take a point $x=a$ and consider the arc length of the curve from $x = a - \Delta$ to $x = a$. Call this $L$. Next consider the arc length from $x = a$ to $x = a + \Delta$. Call this $R$. I am wondering if the ratio $\frac{R}{L}$ has some interesting interpretation in terms of elementary calculus concepts. Also, while working on this I came up with the following conjecture:
$$ e \stackrel{?}{=} \lim_{\Delta\to 0} \bigg( \dfrac{ \displaystyle\int_{\frac{1}{2}}^{\frac{1}{2}+\Delta} \sqrt{1+4x^2} \, dx } { \displaystyle\int_{\frac{1}{2}-\Delta}^{\frac{1}{2}} \sqrt{1+4x^2} \, dx } \bigg)^{1/\Delta} $$ I have no idea if this is true but would like to know if it is (and how one might prove it). This is coming from considering the idea above for the curve $f(x)=x^2$ with $a=\frac{1}{2}$ and doing some very questionable series expansions.
I would also still like to know in general if the ratio $\frac{R}{L}$ has any interesting interpretation in the limit that $\Delta\to 0$. It feels like it might, sort of measuring the relative rate at which the arc length is locally changing around $x=a$ or something like that? But it might also be trivially equal to 1. I'm unsure.
Thanks for reading. Any comments would be appreciated, and I'm totally open to the possibility that I'm being silly and there's nothing interesting going on here.
I think the limit of $\frac{R}{L}$ should just be 1. Here is both an intuitive and what I hope is a correct and rigorous reason:
Intuitive: As $\Delta \to 0$, the segment of the curve $y=f(x)$ above the interval $[x-\Delta, x+\Delta]$ looks more and more like the tangent line at $x$. And certainly for a line, the lengths $L$ and $R$ are the same.
Rigorous: $R$ and $L$ are really functions of $\Delta$:
$$ R(\Delta) = \int_x^{x+\Delta} \sqrt{1+f'(x)^2} ~ dx $$
and similarly for $L(\Delta)$. Since $R$ and $L$ are arc length, then definitely $R(\Delta) \to 0$ and $L(\Delta) \to 0$ as $\Delta \to 0$. This means the limit $\lim_{\Delta \to 0} \frac{R(\Delta)}{L(\Delta)}$ is an indeterminate form and you can apply L'Hopital's rule:
$$ \lim_{\Delta \to 0} \frac{R(\Delta)}{L(\Delta)} = \lim_{\Delta \to 0} \frac{R'(\Delta)}{L'(\Delta)} $$
By the fundamental theorem of calculus, $R'(\Delta) = \sqrt{1+f'(x+\Delta)^2}$ and $L'(\Delta) = \sqrt{1+f'(x-\Delta)^2}$. These are continuous, so you can just plug in $\Delta = 0$ to find the limit, and they are both $\sqrt{1+f'(x)^2}$, and the limit is 1.