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

Question

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.

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$\mathbf{\color{red}{\text{Contest results:}}}$ $$ \begin{array}{c|ll} \hline \text{Rank} & \text{User} & {} & {} & \text{Arc length} \\ \hline \text{1} & \text{robjohn $\blacklozenge$} & {} & {} & 2.78540 \\ \text{2} & \text{Glen O} & {} & {} & 2.78567 \\ \text{3} & \text{mickep} & {} & {} & 2.81108 \\ \text{4} & \text{mstrkrft} & {} & {} & 2.91946 \\ \text{5} & \text{MathNoob} & {} & {} & 3.00000 \\\hline \text{-} & \text{xanthousphoenix} & {} & {} & 2.78540 \\ \text{-} & \text{Narasimham} & {} & {} & 2.78 \\ \end{array}$$

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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?

$$ \begin{array}{c|ll} \hline \text{Rank} & \text{Function} & {} & {} & \text{Arc length} \\ \hline \text{1} & 1.10278[\sin(\pi x)]^{0.153764} & {} & {} & 2.78946 \\ \text{2} & (8/\pi)\sqrt{x-x^2} & {} & {} & 2.91902 \\ \text{3} & 1.716209468\sqrt{x}\,\mathrm{arccos}(x) & {} & {} & 2.91913 \\ \text{4} & (8/\pi)x\,\mathrm{arccos}(x) & {} & {} & 3.15180 \\ \text{5} & (15/4)x\sqrt{1-x} & {} & {} & 3.17617 \\ \text{6} & -4x\ln x & {} & {} & 3.21360 \\ \text{7} & 10x(1-\sqrt{x}) & {} & {} & 3.22108 \\ \text{8} & -6x^2+6x & {} & {} & 3.24903 \\ \text{9} & 9.1440276(2^x-x^2-1) & {} & {} & 3.25382 \\ \text{10} & (-12/5)(x^3+x^2-2x) & {} & {} & 3.27402 \\ \end{array}$$

Answer 1

Find the Shape of the Graph

We wish to minimize $$ \int_0^1\sqrt{f'(x)^2+1}\,\mathrm{d}x\tag{1} $$ while keeping $$ \int_0^1f(x)\,\mathrm{d}x=1\tag{2} $$ This means that we wish to find an $f$ so that the variation of length is $0$ $$ \int_0^1\frac{f'(x)\,\delta f'(x)}{\sqrt{f'(x)^2+1}}\,\mathrm{d}x=0\tag{3} $$ which, after integration by parts, noting that $\delta f(0)=\delta f(1)=0$, becomes $$ \int_0^1\frac{f''(x)\,\delta f(x)}{\sqrt{f'(x)^2+1}^{\,3}}\,\mathrm{d}x=0\tag{4} $$ 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.

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

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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.

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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.

Answer 2

The absolute least value you can get is a rectangle topped by a half circle (the circle has the best area to arc length ratio of any shape) with a total arc length of $2 \big(1 - \frac{\pi}{8}\big) + \frac{\pi}{2} \approx 2.78539$. If you use Fourier approximation, you can come arbitrarily close to this limit. (I assume the fun of this challenge is to find an arbitrarily "low-term" function.)

Answer 3

Without a deeper thought or analysis, I thought it could be fun to look at parts of (translated) superellipses, and maybe make top 10 with it. And indeed it worked.

Thus, I defined $g(x,n)=(1-|x|^n)^{1/n}$, and then $$ f(x,n)=g(2x-1,n) = (1-|2x-1|^n)^{1/n}. $$ Normalizing $c_n=1/\int_0^1 f(x,n)\,dx$ and then calculating the length of $c_n f(x,n)$, it looked like the optimum choice was $n=4$.

The constant $c_4\approx 1.07871$. The arc length of $$ 1.07871(1-|2x-1|^4)^{1/4} $$ was numerically calculated to be $$ 2.81108, $$ which I leave as my contribution.

The graph of $c_4f(x,4)$ is shown below:

Answer 4

A nice solution can be obtained by modifying the "exact" solution. The "exact" solution is $$ f(x) = \frac{8-\pi}8 + \sqrt{x(1-x)} $$ which has an arc length of $\frac{8+\pi}4$. As such, I propose a solution of the form $$ f(x) = \sqrt{x(1-x)}(1+g(x)) $$ where the "exact" solution uses $g(x)=(8-\pi)/(8\sqrt{x(1-x)})$. We want a solution similar to this, but with a finite value at $x=0$ and $x=1$. As such, I propose a simple modification. $$ f(x) = \sqrt{x(1-x)}\left(1+\frac{A}{\sqrt{(x+B)(1+B-x)}}\right) $$ Note that we recover the "exact" solution if $B=0$ and $A=\frac{8-\pi}8$. We can thus get arbitrarily close to this solution by selecting appropriate values for $A$ and $B$. Although a closed-form expression relating the two parameters isn't obvious, values can be chosen numerically. For example, for $B=0.0001$, we have $A\approx\frac{8-\pi}8+0.00058333971346\approx0.60788425801473$. For these, we have $$ \int_0^1 \sqrt{1+f'(x)^2}dx\approx 2.78567 \approx \frac{8+\pi}4 + 2.67\times10^{−4} $$ In this case, the expression works out to be $$ f(x)=\sqrt{x(1-x)}\left(1+\frac{0.60788425801473}{\sqrt{(x+0.0001)(1.0001-x)}}\right) $$ Note that this can also be expressed as $$ f(x)=\sqrt{x(1-x)}\left(1+\frac{0.60788425801473}{\sqrt{x(1-x)+0.00010001}}\right)\tag{$\dagger$} $$ Here is the graph of the $f(x)$ given in $(\dagger)$:

Answer 5

I would not be content without a proper derivation of Dido's problem of variational calculus with constraints of moving boundary considered. When properly done I expect the curvature would be proportional to the square or cube or some other function of $y$-coordinate.

For time being proceeding purely on squared variation hypothesis for curvature as:

$$ k_g = - y^2 / a^3, $$

where $a$ is a constant, I obtained the above stationary closed loop.

Numerically adjusting constant $a$ and initial $y_i (a = 0.7925, y_i = 1.143)$, it is close to the results listed here. The constants are such that perturbation causes the loops to get either progressive or regressive. The area is not very accurately $1.0$ ($\sim 0.98$ only) satisfied, length is approximately $2.78$. Improvement of numerical accuracy possible, but proper theoretical basis is necessary. In this hypothetical case, hyper-Elliptic Integrals are involved.

Answer 6

An easy approach is to simply construct an ellipse with its upper half satisfying the above conditions.

An ellipse is defined via $2$ numbers $a$ and $b$ which are each the half of the major and minor axis of the ellipse.

Then all points $(x,y)$ which suffice the following equation are on the ellipse:

$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$

Or to get the upper half of the eclipse as a function:

$$ y = b \, \sqrt{\left(1 - \frac{x^2}{a^2}\right) } $$

The area $A$ of the complete ellipse is given via $A = \pi\,a\,b$ and therefore our first condition translates to:

$$ \frac{1}{2}\,\pi\,a\,b = 1$$

Also, as we want that $f(0) = 0 = f(1)$, we have:

$$ 2\,a = 1 $$

That already gives us

$$ a = \frac{1}{2} \\ b = \frac{4}{\pi} $$

and therefore an ellipse with the correct size. However, this results in an ellipse which intersects the $x$-Axis at $x_1=-0.5$ and $x_2=0.5$. To meet our conditions, we move the ellipse $0.5$ to the right and get:

$$ y = \frac{4}{\pi} \, \sqrt{1 - 4\,(x-0.5)^2 }\tag{$\dagger$} $$

Now we simply let Wolfram Alpha do the computation for the arc length. The result is $$ 2.919463, $$ and the graph in $(\dagger)$ appears below:

Answer 7

Consider the following: $$f(x)=\begin{cases} cx & \text{ if } 0 \le x \le \frac{1}{c} \\ -cx+c & \text{ if } 1-c < x \le 1 \\ 1 & \text{ otherwise } \\ \end{cases}$$

If we take $c\to\infty$ we get that it is an arc length of 3.

Answer 8

As has already been explained at least twice, the best functions follow this pattern: a continuous function $f$ with $f(0)=f(1)=0$ that approximates $y = h(x) = 1-\frac\pi8+\sqrt{x-x^2}$ for $0<x<1.$

I propose a family of functions for $n$ a positive integer, $$f_n(x) = \sqrt{x-x^2} + \left(1-\frac\pi8\right)g_n(x),$$ where $$g_n(x) = \left(1 + \frac{1}{2n}\right)\left(1-(1-2x)^{2n}\right).$$ Since $$\int_0^1 1-(1-2x)^{2n}\; dx = \frac{2n}{2n+1},$$

we have $\int_0^1 g_n(x)\;dx = 1,$ and therefore $\int_0^1 f_n(x)\; dx = 1.$

The path integral is more difficult to compute than the area integral, but $1-(1-2x)^{2n}$ takes on its maximum value, $1$, at $x=\frac12$. So if we set $h_n(x) = h(x) + \frac{1}{2n}\left(1-\frac\pi8\right)$ we ensure that $f(x) \leq h_n(x)$ for $0 \leq x \leq 1.$ I claim the path length is less than the length of the bounding curve consisting of the graph of $h_n(x)$ from $x=0$ to $x=1$ and the two segments joining $(0,0)$ to $(0,h_n(0))$ and $(1,h_n(1))$ to $(1,0)$. The length of that bounding path is $$2+\frac\pi4 + \frac1n\left(1-\frac\pi8\right) < 2+\frac\pi4 + \frac{0.607301}{n}.$$ Therefore if we pick, say $n = 1000000,$ the resulting path exceeds the theoretical minimum by less than $6.074 \times 10^{-7},$ which is less than one part in $4.5 \times 10^6.$

To within the accuracy possible in any visual graph I could present here, the graph of $f_n(x)$ for large $n$ is the same as the graph of every other near-theoretical-minimum solution:

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Alternatively, stealing an idea from robjohn, we have $$\int_0^1 (x-x^2)^{1/n} = B\left(1+\frac1n, 1+\frac1n\right) = \frac{\Gamma\left(1+\frac1n\right)^2}{\Gamma\left(2+\frac2n\right)},$$ so we can set $$g_n(x) = \frac{\Gamma\left(2+\frac2n\right)}{\Gamma\left(1+\frac1n\right)^2}(x-x^2)^{1/n}$$ and proceed as before. This $n$th-root approach seems to converge faster than my $2n$th-power approach.

Answer 9

$p_0 = (0, 0), p_i = (i/n, y_n), p_n = (1,0)$

$A = \frac{1}{2} \cdot \sum_{i=1}^n{(y_i + y_{i-1})} \cdot \frac{1}{n} = 1$

$L = \sum{|p_{i+1} - p_i|}$

$\frac{\partial A}{\partial y_i} = \frac{\partial }{\partial y_i} \frac{1}{2n} \cdot (y_i + y_{i-1} + y_{i+1} + y_i) = \frac{1}{n}$

$\frac{\partial L}{\partial y_i} = \frac{\partial }{\partial y_i} \bigg[\sqrt{\frac{1}{n^2} + (y_i - y_{i-1})^2} + \sqrt{\frac{1}{n^2} + (y_{i+1} - y_i)^2}\bigg] = $

$ = \frac{y_i - y_{i-1}}{\sqrt{\frac{1}{n^2} + (y_i - y_{i-1})^2}} + \frac{y_i - y_{i+1}}{\sqrt{\frac{1}{n^2} + (y_{i+1} - y_i)^2}} = \lambda / n$

Lagrange multiplier

Does anyone know how make this a differential equation? $n \rightarrow \infty, 1/n \rightarrow dx$

$\frac{\delta_i}{\sqrt{(dx)^2 + (\delta_i)^2}} - \frac{\Delta_i}{\sqrt{(dx)^2 + (\Delta_i)^2}} = \lambda \cdot dx$

$\frac{\delta_i^2}{{(dx)^2 + (\delta_i)^2}} = \lambda^2 \cdot (dx)^2 + \frac{\Delta_i^2}{{(dx)^2 + (\Delta_i)^2}} + 2 \frac{\Delta_i}{\sqrt{(dx)^2 + (\Delta_i)^2}} \lambda \cdot dx $

$\bigg[\frac{\delta_i^2}{{(dx)^2 + (\delta_i)^2}} - \lambda^2 \cdot (dx)^2 - \frac{\Delta_i^2}{{(dx)^2 + (\Delta_i)^2}}\bigg]^2 = 4 \frac{\Delta_i^2}{{(dx)^2 + (\Delta_i)^2}} \lambda^2 \cdot (dx)^2 $

Answer 10

The answer is that you will need to have a constant curvature, which is the partial circle solution by robjohn. If you do want the curve within (0,1) then the rectangle + 1/2 circle solution by both rob and xan.

Why is that? it is actually a physics problem. The solution is a shape of a membrane under pressure.

Parametric function:

$$x= \begin{cases} 1& 0\leq s\leq h\\ \frac{1}{2}+\frac{1}{2}\cos \left( 2\left( s-h \right) \right) & h<s<h+\frac{2}{\pi}\\ 0& h+\frac{2}{\pi}<s<2h+\frac{2}{\pi}\\ \end{cases} $$ $$y= \begin{cases} s& 0\leq s\leq h\\ h+\frac{1}{2}\sin \left( 2\left( s-h \right) \right)& h<s<h+\frac{2}{\pi}\\ \left( 2h+\frac{\pi}{2}-s \right)& h+\frac{\pi}{2}<s<2h+\frac{2}{\pi}\\ \end{cases} $$ Area: $$1=\int_{0}^{1}{y}dx$$ $$ 1=\int_{h+\frac{\pi}{2}}^h{\left[ h+\frac{1}{2}\sin \left( 2\left( s-h \right) \right) \right]}d\!\left[ \frac{1}{2}+\frac{1}{2}\cos \left( 2\left( s-h \right) \right) \right] $$

Take: $$\theta=2(s-h)$$ $$4=\int_{\pi}^0{\left[ 2h+\sin \left( \theta \right) \right]}d\!\left( \cos \left( \theta \right) \right) =4h+\frac{\pi}{2}$$ $$h=1-{\pi \over 8}$$ $${\rm Length}=s_{\max}=2h+\frac{\pi}{2}=2+\frac{\pi}{4}=2.785398163...$$

Answer 11

This is a wonderful challenge for my Calculus class. A student suggested the implicit equation $(x-0.5)^2+b f(x)^k=0.5^2$. Values of $k>1$ and $0.14\lt b\lt 0.15$ seem to give $L\lt 3$.

When $k=5$ and $b=0.1439862439244$, we got $L=2.7463527480137$. However, the area under the curve is $0.9999999999999$ which is not quite $1$ (but very close). The graph of the function is plotted here. graph

Answer 12

Surprised this is still an open problem.

This is the classical Dido's isoperimetric problem in Calculus of Variations. Same is the circle segment solution whether area is to be maximized for given perimeter length or for a given area the perimeter is to be minimized. We compute in the major circle segment.

If $R$ is the radius and $a =0.5$ the semi-chord length, then area $$ A =1= a \sqrt {R^2-a^2} +R^2 (\pi-\sin^{-1} (a/R)) $$

By numerical iteration ( Newton-Raphson etc. )

$$ R= 0.6131365125223184 $$

and perimeter of major arc

$$ 2 R (\pi - \sin^{-1}(a/R)) = 2.6831297778598477 $$

coinciding with robjohn's earlier solution. The part of his answer after "Problem" is ..

Answer 13

The most recent overkill contribution - which is at least somewhat interesting, though not the optimal - is the following behemoth: $f(x)=-0.00010156(b^{x-1/2}+b^{-(x-1/2)})+1.120357$ where $b=121716670.4$ You can think of this as a hyperbolic cosine with a different base. This has arc length $2.8788364$. The above answer does beat polynomials of the form $p(x)=a(x-1/2)^n+b$. Such polynomials can do no better than 2.8940115.