0

What curve is the longest distance between points a and b such that (i) one gets closer to point b at all times; (ii) it can be graphed as a function; and (iii) it is symmetrical?

I would think each side approaches the interior perimeter of a circle as a curvilinear asymptote.

Is it a segment of an ellipse? Something else?

Aldante
  • 1
  • What do you mean by "symmetrical", exactly? – Arthur May 19 '19 at 20:00
  • I mean that the portion of the curve up to its midpoint is the mirror image of the portion of the curve thereafter (as would be the case, for example, with a parabola). – Aldante May 19 '19 at 20:13
  • I guess it is two $60^\circ$ circular arcs glued together. –  May 19 '19 at 20:22
  • Another way to describe it is what is the name of the curve that would form an arch connecting two points on the ground such that as you proceed along the arch from one point you are always getting closer to the other, and the arch is the maximum total length possible. Could be a particular type of catenary for example. Or an ellipse segment. Or even a parabolic segment perhaps. Don’t know. – Aldante May 19 '19 at 20:24
  • Not quite Rahul. In your shape for the entire portion of the journey that traces the perimeter of a circle you would not be getting closer, but rather only maintaining your distance. The arc needs to approach the one you described as a limit. Think of the endpoints being the perimeter of a circle and its radius. Starting from the perimeter, you would want to deviate from the arc of the perimeter slightly until you hit the maximum that would be above the midpoint between the two points, then do the mirror image of that curve on the other side. Does the complete curve have a name? – Aldante May 19 '19 at 20:33
  • Maybe we can take $a=(-1,0)\in\mathbb{R}^2$ and $b=(1,0)\in\mathbb{R}^2$, and let $f: [-1,1] \to \mathbb{R}$ be a function with the symmetry $f(-x)=f(x)$ (an even function), and with $f(-1)=0=f(1)$ (i.e. the graph of $f$ contains $a$ and $b$). We have the restriction that if $x_1 < x_2$ then $(x_1,f(x_1))$ is closer to $a$ than $(x_2,f(x_2))$ is to $a$. If $f$ is "nice" enough that we can talk of the length of its graph, say $f$ is piecewise $C^1$ or something, is there a maximum of the possible length the graph of $f$ can have? What $f$ attains that maximum? – Jeppe Stig Nielsen May 19 '19 at 20:47
  • Then there is no optimal curve that attains the longest distance, because for any curve you can make a longer one that goes closer to the circumference of the circle. It is like asking for the largest real number strictly less than 1. –  May 19 '19 at 21:30
  • Thank you all. I agree that we are talking about something that infinitely approximates what looks like a Gothic arch, with two 60-degree arcs meeting at a point. So at best it would be a non-smooth, albeit continuous function. And I accept that this “arch” is no particular kind of curve, at least not one of a common variety (such as a catenary). The question at this point is, given that there certainly are equations that describe curves that approach limits, is there an equation that can describe this? Much appreciated. – Aldante May 20 '19 at 02:15
  • If the points are at $a$ and $b$ on the $x$-axis, then $y=\min\left(\sqrt{(b-a)^2-(x-a)^2},\sqrt{(a-b)^2-(x-b)^2}\right)$. –  May 20 '19 at 10:34
  • Thank you very much Rahul, that appears to be the equation for the two 60 degree arcs that meet at a point. What further tweaks or notations would be required to reflect that the shape we have in mind is not that per se, but one that approaches it as a limit from inside? So for example it would encompass two 59.5 degree arcs but not two 60.5 degree arcs. Really appreciate your skill. – Aldante May 20 '19 at 14:20

0 Answers0