Ant and Rubber band

[Ghost Post]

Even if the rubber band can stretch infinitely, which I this is what the question was suggesting, stretching at 1 m/s allows the ant to reach the end just the same. The length the band increases by smaller intervals every time.
1 meter + 1 meter = 2 meters 100%
2 meters + 1 meter = 3 meters 33.3%
3 meters + 1 meter = 4 meters etc.
The intervals get small enough that the ant can complete his journey but not in 100 seconds.

[Ghost Post]

Think of it this way. If the ant started 1cm from his destination on the rubberband, or 99/100 of the way there, 99% of the stretching would be to the right of the ant, and 1% would be to the left. So the end of the rubberband would be moving away from the ant at 1meter * .01 = 1cm/s, and the ant would remain 99/100 of the way to his destination. If he starts there or anywhere farther from the end, he will never reach his destination. I think.

[tv snake]

Yeah...

I think you're both right...

After each second, the percentage of the rubber band that the ant has covered is

n = 1
sum (1/n)
t (number of seconds)

which tends to infinity... very slowly though!!! and very much greater then the 100% of rubber band there is.

[Andy]

A variation on this was covered quite thoroughly in an earlier thread. The answer is yes, the ant will reach the end of the rubber band (assuming also that the ant is practically immortal; it will take a very long time). I don't remember the details and haven't found the original thread.

[Wonko the Sane]

Look at it in the way it was explained in the original post.

Originanal: 1m
1s - 2m (100% growth)
2s - 3m (50% growth)
3s - 4m (33% growth)

and so on. At these points, the ant is walking 1cm, but the stretch occuring under him is over 1cm, however, once you get down to under 1%, the ant starts making headway relative to the beginning. No wait...this made perfect sense a second ago. Let me think...

Yeah...okay...let me try again. Basically, then you take those percentages and correlate them with the percentage of the way the ant has moved
0s: 0m, 0%
1s: .01m, .5%
2s: .02m, .67%
3s: .03m, .75%

Oh wait..forget it...the ant never makes it.

Look at it this way, the percentage of the way the ant has traveled along the length of the rubberband is the difference of the distance the ant has travelled over the lenght of the rubber band. Relate that to the time t and you get

%age = .01t/(t+1)

If the ant ever makes it, then at some point, this should equal 1 for 100%.

1 = .01t/(t+1)
t+1 = .01t
-.99t = 1
t = -(1/.99)
t = -1 - 1/99

Gentlemen...he can make it only if he travels back in time as he walks. Otherwise, this shows that he will NEVER make it. If that doesn't prove it, take the limit of that equation as t goes to infinite. It should be at least 1 if he makes it. However, it's 1%. Sorry guys...he's hopeless.

The problem you guys saw that he did make it in, the band was probably only being stretched at 1cm/s, in which case the ant would clock in when time ends. But the units are different. Can we bury this now?

------------------
It's not the size of the spork, it's whether or not it's made of #7 recyclable plastic.

[tv snake]

The way I saw the problem was (with teleporting ants and rubber bands)

1st second: ant moves one centimetre. The rubber band doubles in length, meaning that the length of rubber band behind the ant has doubled in length. So after 1 second, the ant has gone 2 cm.

2nd second: The ant teleports again (it's easier than trying to make the movement continuous - if i'm wrong it will be because of this assumption). Now it has gone 3cm, but this length, like the rubber band, is multiplied by 3/2. So the ant has gone 4.5cm. (1.5%)

Continue this pattern and you see the following pattern:

1st second: 1%
2nd......... 1.5%
3rd......... 1.83333...%
4th......... 2.08333...%

This sequence goes to infinity very slowly.

[Wonko the Sane]

Duh! Good call. I forgot about the stretchiness before where the ant has already been.

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It's not the size of the spork, it's whether or not it's made of #7 recyclable plastic.

[Quailman]

I don't know about the math involved, but it seems to me that stretching the rubber band is simply a way of changing the ant's velocity without telling the ant, but at no time does he stop or go backwards relative to where he is on the rubber band. Also, the length shouldn't matter. As long as with each step the ant takes, there is more of the rubber band behind him than there was before, he will get to the other end eventually.

[OcularGold]

The ant will reach the end in e^100 seconds.

[Ghost Post]

I got e^100 - 1.

[happymath]

shoover4 is right, and the intuitive idea that tv snake gave is pretty much on target. But of course, you need to remove that teleporting assumption, as tv snake said.
That's ok, because it's easier to integrate than to sum an infinite series. You just need to find the right differential equation.

Let me try to present the calculation that shoover4 must have done. Maybe that'll help the thread reach a final conclusion. Or maybe disagreement will continue, and
this will just be an annoyingly long post.

Oh well, here goes.

(All distance units in meters, all time units in seconds.) The situation starts at time t=0 with the ant at position 0 and the end of the band at position 1.

Let L(t) = position of end of band at time t.

So L(t) = t+1, since it increases by 1 each second.

At time t, and for a point at a distance x from the fixed end of the band, let p(x,t) = the proportion of the length of the band that is between position 0 and position x.
(So p is between 0 and 1 if x is actually somewhere on the band. And who cares otherwise.)

So p(x,t) = x/(t+1), since it's just the ratio of x to the length L(t) of the whole band.

Let v(x,t) = the velocity of the point on the band which is at position x at time t. (This is only meaningful if x is between 0 and t+1.) Since the band stretches evenly,
we have v(x,t) = p(x,t) * (velocity of endpoint of band). But the end of the band has velocity 1, so:

v(x,t) = p(x,t) = x/(t+1)

Finally, let a(t) = position of ant at time t. Its velocity a'(t) is the ant's velocity relative to the band (.01) plus the velocity of the band at that point. So:

a' = .01 + a/(t+1)

(I'm abbreviating a(t) as just a here.)

To solve this differential equation, bring the a/(t+1) to the other side, and multiply everything by the appropriate integrating factor, which in this case is 1/(t+1):

a'/(t+1) - a/(t+1)^2 = .01/(t+1)

Now anti-differentiate, bearing the product rule in mind:

a/(t+1) = .01*ln(t+1) + C

where C is some yet-to-be-determined constant, and ln denotes the natural logarithm. But a(0) = 0, so plugging in t=0
gives C = 0. SO:

a(t) = .01*(t+1)*ln(t+1)

Now when (if ever) does the ant reach the end? That would be when a(t)=L(t) (if ever). Solving:

.01*(t+1)*ln(t+1) = t+1

ln(t+1) = 100

t+1 = e^100

t = e^100 - 1

Voila. So provided the ant lives to be e^100 - 1 seconds (which is about a trillion trillion trillion years), it gets to the end
of the band.

[Quailman]

Wow! That looks pretty good. So, Ocular Gold, it appears that Shoover4's and happymath's ants got there ahead of yours.

[The Wise Fool]

quote:

---

Let me try to present the calculation that shoover4 must have done. Maybe that'll help the thread reach a final conclusion. Or maybe disagreement will continue, and
this will just be an annoyingly long post.

---

Or maybe we'll remain completly baffled....

[happymath]

In response to The Wise Fool:

For the intuitive idea, tv snake's first post (and Scott's too) on this thread is the way to see it.

Another way to see it is to imagine that you're watching the event through a zoom lens, and you're constantly zooming out so that the band always appears to be the same length. (So a marker attached to any point on the band would appear to be standing still, even though in real life it is moving.) The ant would appear to be moving forward, but slower and slower all the time.

So why does it reach the end? Well, if you think about it, the ant's apparent speed is
its actual speed divided by the actual length of the band. Or in other words, the apparent speed is .01/(t+1) meters per second.

And if you integrate that (you need calculus somewhere here) from t=0 to infinity, you get infinity, which means that the ant can go arbitrarily far, given enough time.

Note: just the fact that it's always moving forward relative to the band isn't quite enough; if that apparent speed above was something like .01/(t+1)^2, it'd never make it to the end.

[Ghost Post]

Acutally, I used a different approach that is, unfortunately, more complicated and less intuitive. I set up a differential equation as follows:

Let x be the position of the ant. Let y be the position of the end of the rubber band. So, x(0)=0 and y(0) = 1.

Now, dy/dt = 1 and y(t) = 1+t. dx/dt = 0.01 + x(t)/y(t) = 0.01 + x(t)/(1+t). [Note that the velocity of the ant is 0.01 m/s plus the contribution from stretching, which is the 1 m/s velocity of the end of the rubber band multiplied by the fraction of the rubber band’s length that the ant has crossed.]

Consider the transformation T = 1+t. dx/dT = dx/dt, so we can write dx/dT = 0.01 + x/T. Now, let z = x/T. Then x = zT and dx/dT = z + T dz/dT. Substituting into the differential equation gives z + T dz/dT = 0.01 + z, which is separable.

So, T dz/dT = 0.01 and dz = 0.01 dT / T. Integrating gives z = 0.01 ln(T) + C, but C=0 from the initial conditions. Then, x = 0.01 T ln(T).

Now, we want to find the time when x will equal y (the ant reaches the end of the rubber band). So, T = y = x = 0.01 T ln(T) gives ln(T) = 100, or T = e^100. Then t = T-1 = e^100 – 1.

[Ghost Post]

Quote: As long as with each step the ant takes, there is more of the rubber band behind him than there was before, he will get to the other end eventually.

(how do you do those tight quote lines?)

If the ant is converging on a certain point, then there is still more of the rubber band behind him than before but he will be unable to reach that point or beyond. Though I accept that the ant can get there in this case, its not always true that objects moving forward will arrive at any point in that direction.

[happymath]

Zakharov:

Right, it's not enough just to know that the ant is always moving forward relative to the band. The key thing is that the rate at which the percentage of the band already traveled is increasing is .01/(t+1). That means that the percentage of the band traveled is .01*ln(t+1) (from anti-differentiating and using the fact that the position is 0 at time 0). And ln(t+1) goes to infinity, albeit VERY slowly.

In answer to your earlier concern (from your first post on this thread): you're right that at the instant the ant is 99/100 of the way down the band, the end of the band is moving at exactly the same speed as the ant. However, percentage-wise, the ant is still making headway. So an instant later, the ant is something like 99.00000001/100 percent down the band. And although distance-wise she's no closer to the end than when she was 99/100 of the way, she's now going faster than the end of the band, and so she'll eventually catch up.

In fact, distance-wise, the ant is actually going slower than the end of the band up until that 99/100 point. And at that instant (which is time t=e^99 - 1 seconds, or a little more than a third of the total e^100 -1 time), the band is e^99 meters long, or about one octillion light years. So the ant spends the other two-thirds of her time making up that distance.

[Ghost Post]

I guess the equation could be generalised to, t=e^(1/v)-1, where v is ant's velocity. The reason I'm doing this is to find out how long does the ant take to complete the journey for various speeds. Apparently for v=1, t will be e-1, which is incorrect. I checked for other velocities as well, and I'm getting incorrect results. For instance the ant will reach the end in 2 seconds if it travelled at the rate of 66.6667 cm/sec. Am I doing something wrong? Can someone xplain.

[Ghost Post]

Somewhat surprisingly, the ant will eventually reach the end for a wide range of possibilities. In fact, as long as the ant has some (positive) velocity, he/she will reach the end. The expression is

t = [e^(v/V)-1] / v, where v is the velocity of the end of the rubber band and V is the walking speed of the ant. So, the ant can walk at 1 micron per year while the end is being pulled at the speed of light....and the any will eventually reach the end.

[This message has been edited by shoover4 (edited 12-12-2000).]

[This message has been edited by shoover4 (edited 12-13-2000).]

[happymath]

Yeah, you can even take the band to be whatever starting length you want, and the ant still makes it.

If L is the starting length of the band, V is the velocity of the end of the band, and w is the walking speed of the ant (relative to the band), then the time it takes to reach the end is:

(L/V) * [ e^(V/w) -1 ]

[MAT]

I'm not too sure about all this math stuff, but here is how I look at it.

1st Second
--------------------
Rubber Band: 2 Meters
--
Ant: 2 Centimeters

2nd Second
------------------------------
Rubber Band: 3 Meters
---
Ant: 3 Centimeters

3rd Second
----------------------------------------
Rubber Band: 4 Meters
----
Ant: 4 Centimeters

So you see, no matter how much time passes by the rubber band will always be ahead of the ant.

In Math Terms:

Ratio 1 = 200/2

Ratio 2 = 300/3

Ratio 3 = 400/4
~~~~~~~~~~~~~~~~~~~
Ratio 60 = 6100/61
~~~~~~~~~~~~~~~~~~~
Ratio 100 = 10100/101

Rubber Band = 10000% Ahead Of Ant

[Ghost Post]

@Mat: Don't forget that the ant, being on the rubber band, is moved along with it.

[Quesito]

With L as the length in meters of the band, and R as the relative position (0 - 1) of the ant,
L(t)=1+t
R'(t)=0.01t/L(t)
Solving the differential equation for R yields
R(t)=0.01t/(1+t) (which someone already mentioned)
then solving this equation for R(t)=1, I get 104.7 numerically.
The ant gets there.
If anyone sees an error in my work, I wouldn't be surprised, feel free to correct me.

[MAT]

Yes he is, but he is not moved with it enough to reach the end. 'Cause the bigger the rubber band get the less the ant is going to move with it so if he does get near the end (which he won't) the rubber band will stretch even more. Also remember that the ant only has 100 seconds.

--Mat--

--A watched pot never boils--

[This message has been edited by Mat (edited 12-22-2000).]

[ZenBeam]

[happymath]

Sorry to be corrective, but I gotta say my peace:

Quesito: I think you made a couple mistakes in your calculations. You're right that L(t) = 1+t. But then the RATE R'(t) at which the relative position of the ant is changing is R'(t) = 0.01/L(t). (Note: no t in the numerator. The ant's actual speed is constant, and R'(t) should be the actual speed divided by the current length of the band.)

Then it looks like you forgot to integrate; your R' and R were the same. With the R' above, we would get R(t)=0.01*ln(1+t).

To Mat: you said

[Ghost Post]

If you think the ant will never make it, please answer the following question:

Imagine not one ant, but ants lined up from one end of the band to the other. They all start walking as the band starts to stretch. Clearly, the ants near the destination will make it there soon, so by your theory, some ants will make it and some won't. The question is, "Where is the break point?" That is, "Which ants will make it and which won't?"

[Monocracy]

I wrote a little VB code to do the work. However, i used the assumption that the ant moves .01 m then the elastic stretches one meter. I'm sure the ant will reach the end eventually, because the ant's rate approaches 1.01 m/s, while the elastic is a constant 1 m/s. If there's any flaw in my code, let me know ... so far, after about 32 'days', the ant is moving about .15 m/s.

code:

---

  antplace = 0


  elaslen = 1


  Do


    antplace = antplace + 0.01


    elaslen = elaslen + 1


    antplace = antplace + (antplace / elaslen)


    DoEvents


  Loop Until antplace >= elaslen

---

[Ghost Post]

it sonds coll sigh me up ok

[tv snake]

Two things I have to say:

1. Wow! I was right! Even with all those bad assumptions!
2. I'd never thought of integrating it... and I couldn't have done it even if i wanted to Well done to the people who took the time to solve the puzzle properly.