Minimum arc length problem
Larry Riddle
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.
The winner of the contest would be the student whose function had the
smallest arc length among all those submitted.
Now here's my challenge to the sci.math readers. Is there a minimum arc
length? If so, what function would have the minimum arc length? I would
guess this is a calculus of variations problem, but it's been far too
long since I studied that topic. Thanks.
Larry Riddle
--
Larry Riddle, Chair | rid...@mathcs.emory.edu or
Dept of Mathematics | larry....@asc.scottlan.edu
Agnes Scott College | 404-638-6222 PHONE
Decatur, GA 30030 | 404-638-6177 FAX
David Henry Fetter
>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.
>The winner of the contest would be the student whose function had the
>smallest arc length among all those submitted.
>Now here's my challenge to the sci.math readers. Is there a minimum arc
>length?
Yes, there is a minimum arc length. You can show this by constructing the
appropriate compact set.
>If so, what function would have the minimum arc length? I would
I'm guessing that the function would be
\[
y=\sqrt{1/4 - (x-1/2)^2}
\]
and so the length would be $\pi$.
>guess this is a calculus of variations problem, but it's been far too
>long since I studied that topic. Thanks.
You're welcome :)
--
David Fetter # 1456 N. Farwell Ave. #23
sha...@alpha2.csd.uwm.edu # Milwaukee, WI 53202 USA
(414) 224-6192 # Ezekiel 25:17 (Q. Tarantino, trans.)
-Mammel,L.H.
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.
Carsten Saager
In article <3l6vdh$o...@uwm.edu>, sha...@alpha2.csd.uwm.edu (David Henry Fetter) writes:
|> 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.
|>
|> >If so, what function would have the minimum arc length? I would
|>
|> I'm guessing that the function would be
|> \[
|> y=\sqrt{1/4 - (x-1/2)^2}
|> \]
|>
|> and so the length would be $\pi$.
|>
Otherwise you are definitly right, because a circle
has the best volumne/surface ratio.
My guess:
\[
y=\frac{\pi}{4} \sqrt{1/4 - (x-1/2)^2}
\]
But I cannot crack up the damned integral for
calculating the arc length. If there is someone
who knows the arc length of an elliptic curve.
Further guesses:
y= 6x(1-x)
hyperbolic curves
The three functions describe the movement of
masses in gravitational fields. They
minimize the "Wirkung"( energy * time).
I think that's it.
Carsten
David Henry Fetter
>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.
>>The winner of the contest would be the student whose function had the
>>smallest arc length among all those submitted.
>>Now here's my challenge to the sci.math readers. Is there a minimum arc
>>length?
>Yes, there is a minimum arc length. You can show this by constructing the
>appropriate compact set.
I think this is still true. Perhaps it's not a unique curve, though...
>>If so, what function would have the minimum arc length? I would
>I'm guessing that the function would be
>\[
>y=\sqrt{1/4 - (x-1/2)^2}
>\]
>and so the length would be $\pi$.
I know I'm not supposed to follow up to my own article, but
I didn't read condition 3, so please ignore the function above.
Thanks to all who pointed this out.
Kurt Foster
: 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.
The shortest curve enclosing an area of 1 square unit is a circle of
radius 1/sqrt(pi). Its arc length is 2*sqrt(pi), so the length of the
curve in question cannot be less than 2*sqrt(pi) - 1, or
2.54490770181032, approximately.
Let C be the circle with radius r = 0.6131365125223184 (approx),
with center at (0.5 (exactly), 0.35487516535823 (approx)). The arc of C
for which y >= 0 has a length of 2.683129777859848 (approx), and that
arc, together with the interval [0,1] on the x-axis, encloses an area of
1 square unit.
The length of the arc is within 0.14 of being best possible. The
arc, unfortunately, violates the conditions of the problem, since it is
NOT a curve of the form y = f(x) for x on [0,1]. It is, also, almost
certainly NOT the shortest arc which, together with [0,1], encloses an
area of 1 square unit.
I'm sure someone can produce a shorter arc joining the points (0,0)
and (1,0) so that the arc together with the interval [0.1] on the x-axis
encloses an area of 1 square unit. I'd like to see someone produce a
shorter curve satisfying the conditions of the problem, though ;-)
Rick F. Hoselton
>Larry Riddle (rid...@mathcs.emory.edu) wrote:
>: 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 sure someone can produce a shorter arc joining the points (0,0)
>and (1,0) so that the arc together with the interval [0.1] on the x-axis
>encloses an area of 1 square unit. I'd like to see someone produce a
>shorter curve satisfying the conditions of the problem, though ;-)
How about the following "curve":
for 0 <= x <= .5, f(x)= 2*x
for .5 <= x <= 1, f(x)=2-2*x
if forms a triangle, with a base measure of one and a height of two, so
the area under the graph is equal to one. Since it is continuous,
it has an arclength, 2*sqrt(4.25) or about 4.123
_______________________________________
Rick Hoselton h...@univel.telescan.com
_______________________________________
Rick F. Hoselton
>From: h...@univel.telescan.com (Rick F. Hoselton)
>Subject: Re: Minimum arc length problem
>Date: Tue, 28 Mar 1995 14:13:28
>>Larry Riddle (rid...@mathcs.emory.edu) wrote:
>>: 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 sure someone can produce a shorter arc joining the points (0,0)
>>and (1,0) so that the arc together with the interval [0.1] on the x-axis
>>encloses an area of 1 square unit. I'd like to see someone produce a
>>shorter curve satisfying the conditions of the problem, though ;-)
>How about the following "curve":
>for 0 <= x <= .5, f(x)= 2*x
>for .5 <= x <= 1, f(x)=2-2*x
EXCUSE ME,
Jason Stratos Papadopoulos
f(x)=6x-6x^2. This was my first idea, too: an interpolating
parabola. A quick numerical check shows it has an arclength
of approx. 3.12409 ; I suspect that if you force the curve
to be differentiable then improvements could come in the form
of higher order polynomials, each of which would be more and
more "square"; thus it's my guess that the arclength would
tend to 3.
jasonp
PS: I'll update this after I look at cubics and better.
Jason Stratos Papadopoulos
: Someone mentioned a candidate for shortest arc length as being
: jasonp
Uh, I guess that post needed correcting. The approximate arclength
is actually 3.24903 or so, and a cubic can't satisfy the conditions
given, it being an odd function and all.
"Engage brain before speaking"
-a very wise man.
jasonp
Jason Stratos Papadopoulos
: This does not seem compact to me. There are numerous continuous functions
: which have no maximum on this set, such as the maximum of the function,
: and this arclength problem, if the following is correct:
: Suppose that one did not require the curve to be a function. Then by
: symmetry/pressure arguments, it is clear that the curve would have
: constant curvature (and would be convex). This is in the plane, so
: the curve would be an arc. However, this would not be a function.
: Given the constraint that the curve be a function, one can relax the
: constant curvature condition on the "sides" of the function. So an
: ideal solution would be a semi-circle connected to the x-axis by lines,
: but this is not a function. However, one can approximate this
: arbitrarily closely by functions, so one can find functions arbitrarily
: close to having the arclength of the ideal.
: Arithmetic: a semicircle with radius 1/2 has area pi/8 and arclength
: pi/2. So we need line segments supporting this semicircle of length
: 1-pi/8, and there are two of these. Thus, the arclength is 2+pi/4, so
: arbitrarily close functions are can be constructed.
I don't know that I have anything rigorous to add; I looked at in-
terpolating polynomials, assuming that differentiability was implied
in the requirements (this is a followup to a previous post). Assume
a quartic:
a0 + a1 x + a2 x^2 + a3 x^3 + a4 x^4
given f(0)=f(1)=0, Integral from 0 to 1 is 1, and the derivative
at zero is "c" (and is -c at 1). This yields the family
c x + (30 - 6 c) x^2 + (-60 + 10 c) x^3 + (30 - 5 c) x^4
where c>0 (otherwise fcn is negative) and <30 (otherwise x=1/2
yields negative value of f). I then made a really cheesy optimization
(a homemade integrator calculating the arclength for discrete
values of c). The best results were c=10.5 and arclength=2.958
(approx). Thus the best quartic (I would guess among all quartics)
is
f(x)= 10.5 x - 33 x^2 + 45 x^3 - 22.5 x^4
The value of c is probably within .01 of being optimal, and
was exactly 10.5 to almost the limit of double precision.
Whew!
jasonp
Douglas Zare
sha...@alpha2.csd.uwm.edu (David Henry Fetter) wrote:
> David Henry Fetter [wrote]:
> >Larry Riddle [wrote]:
> >>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.
> >>Now here's my challenge to the sci.math readers. Is there a minimum arc
> >>length?
>
> >Yes, there is a minimum arc length. You can show this by constructing the
> >appropriate compact set.
>
> I think this is still true. Perhaps it's not a unique curve, though...
This does not seem compact to me. There are numerous continuous functions
which have no maximum on this set, such as the maximum of the function,
and this arclength problem, if the following is correct:
Suppose that one did not require the curve to be a function. Then by
symmetry/pressure arguments, it is clear that the curve would have
constant curvature (and would be convex). This is in the plane, so
the curve would be an arc. However, this would not be a function.
Given the constraint that the curve be a function, one can relax the
constant curvature condition on the "sides" of the function. So an
ideal solution would be a semi-circle connected to the x-axis by lines,
but this is not a function. However, one can approximate this
arbitrarily closely by functions, so one can find functions arbitrarily
close to having the arclength of the ideal.
Arithmetic: a semicircle with radius 1/2 has area pi/8 and arclength
pi/2. So we need line segments supporting this semicircle of length
1-pi/8, and there are two of these. Thus, the arclength is 2+pi/4, so
arbitrarily close functions are can be constructed.
> [...]
> David Fetter # 1456 N. Farwell Ave. #23
> sha...@alpha2.csd.uwm.edu # Milwaukee, WI 53202 USA
> (414) 224-6192 # Ezekiel 25:17 (Q. Tarantino, trans.)
Douglas Zare
Robert Johnson
In article <3l5dvp$p...@cssun.mathcs.emory.edu>,
Larry Riddle <rid...@cssun.mathcs.emory.edu> wrote:
>I gave my calculus II students the following contest problem:
>Find three examples of a function f : [0,1] -> R satisfying
>(1) f(x) >= 0
>(2) f(0)=f(1)=0
>(3) area under the graph of f is equal to 1
>Calculate the arc length of the graph of each function. The goal
>is to find a function with an arc length as small as possible.
>
>The winner of the contest would be the student whose function had the
>smallest arc length among all those submitted.
>
>Now here's my challenge to the sci.math readers. Is there a minimum arc
>length?
Yes, there is a minimum arc length, but it can not be attained by the
graph of a continuous function.
If we do not require the curve to be the graph of a function, then the
minimum length is given by the arc above the x-axis of the circle with
radius .613136512521 and center (.5,.354875165355). It connects the
points (0,0) and (1,0), its area above the x-axis is 1, and its length
is 2.68312977784.
If we do not require the function to be continuous, but we add the size
of any discontinuities into the arc length, the following function gives
minimal arc length:
f(0) = f(1) = 0
and for x in (0,1)
f(x) = (1 - pi/8) + sqrt(x(1-x))
The area under this curve is 1. The length of the curve is pi/2 and the
sum of the sizes of the discontinuities is 2 - pi/4, giving an arc length
of 2 + pi/4 = 2.7853981634. This length can be approached but not attained
by continuous functions satisfying (1)-(3).
Rob Johnson
Apple Computer, Inc.
rjoh...@apple.com
Robert Johnson
In article <3l93et$p...@uwm.edu>,
David Henry Fetter <sha...@alpha2.csd.uwm.edu> wrote:
>David Henry Fetter chiseled onto a granite tablet:
>>I'm guessing that the function would be
>>\[
>>y=\sqrt{1/4 - (x-1/2)^2}
>>\]
>
>>and so the length would be $\pi$.
In article <3l8n4r$6...@linda.lion.de>,
Carsten Saager <saa...@linda.lion.de> wrote:
>Unhappily the integral of the above function is $\pi / 4$.
The function cited above, which is simply y = sqrt(x(1-x)), is the top half
of a circle whose center is (1/2,0) and whose radius is 1/2. Thus, it has
half the area of the full circle whose radius is 1/2; that is,
1/2 (pi (1/2)^2) = pi/8
This was implicitly used in my previous post to this thread.