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

Answer

Find the Shape of the Graph

We wish to minimize
10f(x)2+1dx
while keeping
10f(x)dx=1
This means that we wish to find an f so that the variation of length is 0
10f(x)δf(x)f(x)2+1dx=0
which, after integration by parts, noting that δf(0)=δf(1)=0, becomes
10f(x)δf(x)f(x)2+13dx=0
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

\frac{f”(x)}{\sqrt{f'(x)^2+1}^{\,3}}=\lambda\tag{6}

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

2r\sin\left(\frac\theta2\right)=1\tag{7}

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

r^2\left[\frac\theta2-\sin\left(\frac\theta2\right)\cos\left(\frac\theta2\right)\right]=1\tag{8}

Square (7) to get

2r^2(1-\cos(\theta))=1\tag{9}

and rewrite (8) to get

\frac12r^2(\theta-\sin(\theta))=1\tag{10}

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

\theta=4.3760724130128873845\tag{11}

and then (7) gives

r=0.61313651252231835636\tag{12}

This would lead to a minimum length of

L=r\theta=2.6831297778598481320\tag{13}


Problem

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

y=1-\frac\pi8+\sqrt{x-x^2}\tag{14}

enter image description here

which has a length of

2+\frac\pi4=2.7853981633974483096\tag{15}

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


A Sequence of Approximations


f_n(x)=\frac1{c_n}\left(1-\frac\pi8+\sqrt{x-x^2}\right)\left(x-x^2\right)^{1/n}\tag{16}

where

c_n=\left(1-\frac\pi8\right)\frac{\Gamma\left(1+\frac1n\right)^2}{\Gamma\left(2+\frac2n\right)}+\frac{\Gamma\left(\frac32+\frac1n\right)^2}{\Gamma\left(3+\frac2n\right)}\tag{17}

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.

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

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