# 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. $f(x) \geq 0$ on the interval $0\leq x\leq 1$;
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.

$\mathbf{\color{red}{\text{Contest results:}}}$

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?

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 $\delta f(0)=\delta f(1)=0$, becomes

for all variations of $f$, $\delta f$, so that the variation of area is $0$

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

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

which has a length of

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

A Sequence of Approximations

where

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.