So I was a bit bored earlier today and decided I'd work some problems with projectiles.
I was kind of curious to see what solving for the arc length of a projectile launched horizontally would be like. Hence, I came up with a pretty general parametrisation:
$$\vec r(t)=\langle bt,-at^2+c\rangle$$
Decided my interval would be from its maximum until its height was zero, which gave the following bounds:
$$t\in\left[0,\sqrt\frac{c}{a}\right]$$
Last thing to do was compute some derivatives and set up a rectification integral.
$$x'=b$$
$$y'=-2a$$
Which gave me the following equation for arc length:
$$\int_0^{\sqrt\frac{c}{a}}\sqrt{b^2+4a^2t}\;dt$$
I'd recently been working on problems involving the area of hyperbolic cross sections and thought it was a bit strange that the length of one conic section (parabola) was equivalent to the area under another (hyperbola). Due to the relations between conics, I figured it might be more than just coincidence.
Is there any particular reason that the arclength of a parabola evaluates to the area of a hyperbola?
If I understand correctly, you are suggesting that given parameters $a,b$ and a range of $x$, that the area under the hyperbola is equal to the arc length of the parabola. Clearly, as we shall see below, they are structurally similar, but by no means equal under the assumptions stated above. I'm going to consider a very simple case to illustrate the point. With reference to the figure below,
$$ y_h^2=1+x^2\\ y_p=1-x^2\\ x\in[-1,1] $$
Thus,
$$ A_h=\int_{-1}^{1}y_h~dx=\int_{-1}^{1}\sqrt{1+x^2}~dx=\sqrt{2}+\sinh^{-1}(1)\\ s_p=\int_{-1}^{1}\sqrt{1+y'^2}~dx=\int_{-1}^{1}\sqrt{1+4x^2}~dx=\sqrt{5}+\frac{\sinh^{-1}(2)}{2} $$
Now, given free reign over the parameters $a,b$ I'm sure you could always find a case where the area under the hyperbola is equal to the arc length of the parabola. For example, in the case given above, just change $y_p$ to
$$y_p=1-\frac{x^2}{2}$$
to achieve the desired result.