# rubber band stretch problem

• Sep 26th 2007, 01:37 PM
mistykz
rubber band stretch problem
The question is, is it true that if you stretch a rubber band by moving one end to the right and the other to the left, some point of the band will end up in its original position?

What I wrote:

Yes. You may stretch it left and right, but the middle will remain in place assuming the two forces are equal.
Pretend the rubber band is a graph. It may be stretched horizontally, but the y-intercept will not change unless the graph is shifted (the y-intercept is equivalent to the center of the rubber band)
So the middle of the rubber band will remain in place as long as the rubber band itself is not moved or shifted other than the stretch, equal on both sides.

It was graded and I was told the reasoning was not exactly correct, I need to connect it to the intermediate value theorum. Can anyone explain how I would do so? Thanks!
• Sep 26th 2007, 02:12 PM
CaptainBlack
Quote:

Originally Posted by mistykz
The question is, is it true that if you stretch a rubber band by moving one end to the right and the other to the left, some point of the band will end up in its original position?

What I wrote:

Yes. You may stretch it left and right, but the middle will remain in place assuming the two forces are equal.
Pretend the rubber band is a graph. It may be stretched horizontally, but the y-intercept will not change unless the graph is shifted (the y-intercept is equivalent to the center of the rubber band)
So the middle of the rubber band will remain in place as long as the rubber band itself is not moved or shifted other than the stretch, equal on both sides.

It was graded and I was told the reasoning was not exactly correct, I need to connect it to the intermediate value theorum. Can anyone explain how I would do so? Thanks!

The band originaly may be thought of as the interval [a,b], and after
stretching [f(a), f(b)], f(a)<a, f(b)>b.

Now consider the function

g(x)=f(x)-x,

This is negative at a, and positive at b, and continuous, so by
the intermediate value theorem there is a point c in [a,b] such that g(c)=0
so f(c)=c, that is there is a fixed of invariant point on the band.

(you could do this without constructing g, from f directly, but I prefer
this way)

RonL
• Sep 26th 2007, 02:42 PM
F.A.P
CaptainBlack beat me to it... well well I post it anyway
Quote:

Originally Posted by mistykz
The question is, is it true that if you stretch a rubber band by moving one end to the right and the other to the left, some point of the band will end up in its original position?

"..."

It was graded and I was told the reasoning was not exactly correct, I need to connect it to the intermediate value theorum. Can anyone explain how I would do so? Thanks!

There was no specification of how far left or right the rubber band ends where moved so I figure the teacher didn't like your special case explanation with equal lengths(or forces)to the quite general problem statement.

The intermediate value theorem states that(from wolfram mathworld)...

"If f is continuous on a closed interval [a,b], and c is any number between f(a) and f(b) inclusive, then there is at least one number x in the closed interval such that f(x)=c."

Suggestion:
Let the rubber band itself constitute the x-axis. Then let f(x) be the distance point x have moved from it's original position when the rubber band is stretched. Now, the left endpoint a will move to the left and the right endpoint b to the right so that f(a)<0 and f(b)>0.
As f is continuous and f(a)<0<f(b) "there is at least one number x in the closed interval such that" f(x)=0."
• Sep 26th 2007, 03:54 PM
topsquark
I realize this is beyond what the original poster intended, but wouldn't this problem fall under the "fixed point theorem?"

-Dan
• Apr 19th 2011, 04:26 AM
Quote:

Originally Posted by F.A.P
There was no specification of how far left or right the rubber band ends where moved so I figure the teacher didn't like your special case explanation with equal lengths(or forces)to the quite general problem statement.

The intermediate value theorem states that(from wolfram mathworld)...

"If f is continuous on a closed interval [a,b], and c is any number between f(a) and f(b) inclusive, then there is at least one number x in the closed interval such that f(x)=c."

Suggestion:
Let the rubber band itself constitute the x-axis. Then let f(x) be the distance point x have moved from it's original position when the rubber band is stretched. Now, the left endpoint a will move to the left and the right endpoint b to the right so that f(a)<0 and f(b)>0.
As f is continuous and f(a)<0<f(b) "there is at least one number x in the closed interval such that" f(x)=0."

sir very nice proof
but sir still i have a doubt in my mind .how can we prove it practically
• Apr 19th 2011, 04:34 AM
TheChaz
Quote:

Originally Posted by ayushdadhwal
sir very nice proof
but sir still i have a doubt in my mind .how can we prove it practically

To see this principle in practice, you should acquire a rubber band.
#necropost