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

- f(x)≥0 on the interval 0≤x≤1;
- f(0)=0 and f(1)=0;
- 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 O2.785673mickep2.811084mstrkrft2.919465MathNoob3.00000–xanthousphoenix2.78540–Narasimham2.78

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)√x−x22.9190231.716209468√xarccos(x)2.919134(8/π)xarccos(x)3.151805(15/4)x√1−x3.176176−4xlnx3.21360710x(1−√x)3.221088−6x2+6x3.2490399.1440276(2x−x2−1)3.2538210(−12/5)(x3+x2−2x)3.27402

**Answer**

**Find the Shape of the Graph**

We wish to minimize

∫10√f′(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.

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}

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:

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*