Date: Tue, 28 Mar 95 05:38:06 CST
From: rusin (Dave Rusin)
To: shackle@alpha2.csd.uwm.edu
Subject: Re: Minimum arc length problem
Newsgroups: sci.math

In article <3l6vdh$o73@uwm.edu> you write:
>Larry Riddle chiseled onto a granite tablet:
>>I gave my calculus II students the following contest problem:
>>Find three examples of a function f : [0,1] -> R satisfying
>>(1) f(x) >= 0
>>(2) f(0)=f(1)=0
>>(3) area under the graph of f is equal to 1
>>Calculate the arc length of the graph of each function. The goal 
>>is to find a function with an arc length as small as possible.
>
>I'm guessing that the function would be
>y=\sqrt{1/4 - (x-1/2)^2}

Careful - I almost made the same mistake. You forgot condition (3).
Of course what we want to do is to use the fact that the region with
a given area and minimal circumference is the circle; but the added
condition that there be a line segment of length  1  on the perimeter
is not consistent with this, nor is it possible to remove the line
segment with a symmetry argument since the required area is not pi/8.

I don't do calculus of variations so I don't remember how to solve this one.
The best symmetric linear function meeting the requirements has perimeter
5.123; the best parabola has perimeter 4.25; the best half-ellipse has perimetr
3.919; the best circular-arc-on-a-box has area 3+pi/4 = 3.78.

dave

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Newsgroups: sci.math
From: lew@usgp3.ih.att.com (-Mammel,L.H.)
Subject: Re: Minimum arc length problem
Date: Tue, 28 Mar 1995 18:44:03 GMT

Looking in Chapter 2 of Goldstein's CLASSICAL MECHANICS, we
refresh our memory of the variational equation for the extrema
of Int { f(y,y',x) dx } :

df/dy - d/dx(df/dy') = 0

( df/dy and df/dy' are partial derivatives )

Goldstein even applies this to arc length. In our case though
we have the constraint, Int {y dx} = const., and the variation
of this is just delta_y. When we add a Lagrange undetermined
multiplier times this variation ( which is zero due to the constraint, )
and remembering f = sqrt( 1 + y'^2 ), we just get:

d/dx( y'/sqrt(1+y'^2) ) + lambda = 0

or y'/sqrt(1+y'2) = -lambda*x + C == a*(x-r)

This can be inverted and integrated easily, and it comes out to
be a circular arc whose center and radius must be adjusted to
satisfy the constraint and boundary conditions. My opinion is that
we get a semicircular cap atop a rectangle for the problem as
posed. That still doesn't actually satisfy f(0)=f(1)=0, but we
can lean the vertical walls in by epsilon to meet this condition.

The "physical" constraints which lead to this solution are vertical
walls at x=0 and x=1, a floor at y=0, and a "rubber band"
stretched between (0,0) and (1,0).  We then inject a unit volume
of incompressible fluid under the rubber band. It will fill to a
semicircle and then extend upwards keeping a semicircular cap.
Without the vertical walls we would just get a circular bubble
with a chord for the floor.

Lew Mammel, Jr.

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From: fkauf@fstgds06.tu-graz.ac.at (Friedrich Kaufmann)
Newsgroups: sci.math
Subject: Re:Minimum arc length
Date: 30 Mar 1995 21:29:00 GMT

>Find three examples of a function f : [0,1] -> R satisfying
>(1) f(x) >= 0
>(2) f(0)=f(1)=0
>(3) area under the graph of f is equal to 1
>Calculate the arc length of the graph of each function. The goal
>is to find a function with an arc length as small as possible.

You will get f(x) = y:

      1
     /
    |       /      2 \
l = |  sqrt| 1 + y'   | dx      ---> min
    |       \        /
   /
   0


considering:

        1
       /
      |
0 =   |   (y - A) dx
      |
     /
    0

here A is the area:


More generally you can say:

      b
     /
I = |  F(y,y') dx
   /
   a

considering:

       b
      /
0 =  |  G(y) dx
    /
    a

Using Euler's equation you will get:
                     d
F  + lamda * G   -  --- F    = 0
 y            y     dx   y'


using:
         /      2 \
F = sqrt| 1 + y'   |          G = y - A
         \        /


you will get:

 d        y'
---   --------------  =  lamda
dx    sqrt(1 + y'^2)

integrating this, you will get:

      1
y = ----- sqrt(1 - (lamda * x + C1)^2) + C2
    lamda

Now there are 3 unknowns to determine and there are 3 conditions:

y(0) = 0
y(1) = 0
A = 1

Using the first and second, you will get:

     1     /                                                       \
y = ----- |  sqrt(1 - lamda^2 * (x - 0.5)^2) - sqrt(1 - lamda^2/4)  |
    lamda  \                                                       /

Looking at the formula, you will see, that this is a circle segment. The
area is a continuous function of lamda, but lamda is limited by 1, because
otherwise there is no real solution in the interval [0,1]. The maximum
area is therefore:

Amax = pi/8 for lamda = 1

Since the area should be 1, I suggest following solution:
Take a rectangular and put a halve circle on it and you will get following
function graph:

   f(x)
  ^
  |
  |          *     *     *
  |      *                   *
  |   *                        *
  | *                            *
  |*                              *
  *                                * ---
  *                                *    ^
  *                                *    |  1 - pi/8
  *                                *    |
  *                                *    v
  ********************************** ------------------------------> x
  0                                1

  The length of the arc is: 2 + pi/4 = 2.78539

  f(x) = sqrt(0.25 - (x-0.5)^2) + 1 - pi/8  for 0 < x < 1
  f(x) = 0  for x=0 and x=1

  The function f(x) isn't continuous at x=0 and at x=1 and the question is
  now, what is the arc length of a discontinuous function.


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