Arc length contest! Minimize the arc length of f(x)f(x) when given three conditions.

Contest: Give an example of a continuous function f that satisfies three conditions:

  1. f(x)0 on the interval 0x1;
  2. f(0)=0 and f(1)=0;
  3. the area bounded by the graph of f and the x-axis between x=0 and x=1 is equal to 1.

Compute the arc length, L, for the function f. The goal is to minimize L given the three conditions above.

Contest results:
RankUserArc length1robjohn 2.785402Glen O2.785673mickep2.811084mstrkrft2.919465MathNoob3.00000xanthousphoenix2.78540Narasimham2.78

Original question after contest statement: The contest question was inspired by this paper. Can anyone come up with a different entry than those listed in the table below?

RankFunctionArc length11.10278[sin(πx)]0.1537642.789462(8/π)xx22.9190231.716209468xarccos(x)2.919134(8/π)xarccos(x)3.151805(15/4)x1x3.1761764xlnx3.21360710x(1x)3.2210886x2+6x3.2490399.1440276(2xx21)3.2538210(12/5)(x3+x22x)3.27402


Find the Shape of the Graph

We wish to minimize
while keeping
This means that we wish to find an f so that the variation of length is 0
which, after integration by parts, noting that δf(0)=δf(1)=0, becomes
for all variations of f, \delta f, so that the variation of area is 0

\int_0^11\,\delta f(x)\,\mathrm{d}x=0\tag{5}

This means that \frac{f”(x)}{\sqrt{f'(x)^2+1}^{\,3}} is perpendicular to all \delta f that 1 is. This is so only when there is a \lambda so that


However, (6) just says that the curvature of the graph of f is \lambda. That is, the graph of f is an arc of a circle.

Find the Length of the Arc

Since the length of the chord of the circle we want is 1, we have


Since the area cut off by this chord is 1, we have


Square (7) to get


and rewrite (8) to get


Solve 4(1-\cos(\theta))=\theta-\sin(\theta) to get


and then (7) gives


This would lead to a minimum length of



Unfortunately, since \theta\gt\pi, the minimizing curve is an arc that cannot be represented by a function.

enter image description here

The minimizing curve that is closest to the graph of a function is the curve that joins (0,0) and (1,0) to the endpoints of


enter image description here

which has a length of


However, this curve is not the graph of a function.

A Sequence of Approximations




As n\to\infty, the length of f_n approaches 2+\frac\pi4.

At n=100, we get a length of L=2.7857313936, less than \frac1{3000} above the minimum:

enter image description here

At n=1000, we get a length of L=2.7854017568, less than \frac1{250000} above the minimum.

Source : Link , Question Author : Daniel W. Farlow , Answer Author : robjohn

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