Ant on the rubber band problem.
December 7th, 2010 at 9:32:17 AM
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I get asked this problem every once in a while, and I've never been able to solve it. Maybe somebody who is good with partial differential equations can help. Here is what you have:
1. There is a 1 km rubber band.
2. An ant starts at one end.
3. The ant moves towards the other end at a speed of 1 cm./sec..
4. The rubber band stretches at a rate of 1 km./sec..
How long does it take for the ant to reach the other end?
Here is about as far I get with this, which admittedly isn't much.
Let the ant's position at the beginning be 0 at time 0, with one end of the rubber band fixed to that point.
Let f(t) be the ant's distance from the starting point at time t.
f'(t) is the ant's speed at time t.
The speed is 1 + 100,000*ratio of progress (where the ratio of progress is how much of the rubber band the ant has crossed).
So, I think, we've got f'(t) = 1 + 100000*f(t)/(100000*(1+t))
My calculus skills are just too rusty to get past this point, not that they were ever outstanding to begin with. If I could solve for f(t) then I think I could get the rest of the way easily by finding the point at which the ant's position is equal to the end point (100000*(1+t)).
This is where I need to use my phone a friend. Can anybody get me further along the rubber band of this problem? ME? DG? Doc?
1. There is a 1 km rubber band.
2. An ant starts at one end.
3. The ant moves towards the other end at a speed of 1 cm./sec..
4. The rubber band stretches at a rate of 1 km./sec..
How long does it take for the ant to reach the other end?
Here is about as far I get with this, which admittedly isn't much.
Let the ant's position at the beginning be 0 at time 0, with one end of the rubber band fixed to that point.
Let f(t) be the ant's distance from the starting point at time t.
f'(t) is the ant's speed at time t.
The speed is 1 + 100,000*ratio of progress (where the ratio of progress is how much of the rubber band the ant has crossed).
So, I think, we've got f'(t) = 1 + 100000*f(t)/(100000*(1+t))
My calculus skills are just too rusty to get past this point, not that they were ever outstanding to begin with. If I could solve for f(t) then I think I could get the rest of the way easily by finding the point at which the ant's position is equal to the end point (100000*(1+t)).
This is where I need to use my phone a friend. Can anybody get me further along the rubber band of this problem? ME? DG? Doc?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 7th, 2010 at 9:53:01 AM
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Can it be proven he even gets to the other end? More specifically, does he ever reach a point where the end is receding from him more slowly than 1 cm/sec? Can you solve for this point?
December 7th, 2010 at 10:10:25 AM
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Quote: 7outlineawayCan it be proven he even gets to the other end? More specifically, does he ever reach a point where the end is receding from him more slowly than 1 cm/sec? Can you solve for this point?
Yes, the ant will reach the end, after a very long time. As a ratio of the rubber band traveled, the ant makes progress with every step.
That makes for a good side question about when the distance to the end starts to decrease with time. I would imagine it isn't until somewhere in the last minute or so.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 7th, 2010 at 10:25:53 AM
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Quote: WizardYes, the ant will reach the end, after a very long time. As a ratio of the rubber band traveled, the ant makes progress with every step. This ratio keeps increasing too, so eventually it must get to 1.
That makes for a good side question about when the distance to the end starts to decrease with time. I would imagine it isn't until somewhere in the last minute or so.
Isn't this just an infinite series that sums to 1?
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December 7th, 2010 at 10:54:28 AM
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I'm happy to say that I have ZERO IDEA of what you're talking about. VERY happy!
December 7th, 2010 at 11:40:48 AM
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Well, Wizard, it has been a heck of a long time since I have tried to solve a differential equation, and my brain has atrophied quite a bit. I had to go to a reference book for some help on the technique. This is what I came up with, and I still wish there were an easy way to post equations on this forum.
Starting from your equation, I simplify it and get:
I believe the solution to this differential equation is:
Then, considering the point when the ant reaches the end point, we can solve for the time required in seconds:
Even if this answer is correct, my calculator will not display the answer.
Starting from your equation, I simplify it and get:
I believe the solution to this differential equation is:
Then, considering the point when the ant reaches the end point, we can solve for the time required in seconds:
Even if this answer is correct, my calculator will not display the answer.
December 7th, 2010 at 12:26:27 PM
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Is the rubberband infinitely elastic? That is, will it (if it were a real object) eventually reach a maximum length, then stop, or for the purposes of this exercise, will it continue to stretch forever?
If infinite, I don't think the ant makes up the difference. Every 100,000 seconds, the ant will cover 1 km. But in that time, the other end of the rubberband will be another 99,999 km ahead of him. This growth is constant and infinite, so the ant will never catch up. What am I missing?
If infinite, I don't think the ant makes up the difference. Every 100,000 seconds, the ant will cover 1 km. But in that time, the other end of the rubberband will be another 99,999 km ahead of him. This growth is constant and infinite, so the ant will never catch up. What am I missing?
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December 7th, 2010 at 12:37:17 PM
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As the band stretches, the ant is moved along it as well... e.g it stays the same percentage on the band.
Why not solve for smaller numbers first?
Why not solve for smaller numbers first?
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December 7th, 2010 at 12:39:27 PM
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Ayecarumba, I think what you are missing is that as the ant crawls, the piece of rubber he is standing on is moving because of the stretch. His 1 cm/s walk is in relation to that piece of the rubber band. The further he moves along it, the faster he moves. As he approaches the end, he is traveling nearly 1 km/sec. I think his final step has him traveling 1.00001 km/s, or 1 cm/s faster than the moving end of the rubber band.
It takes him a long, long time to get there, though.
An yes, the assumption is that there is infinite stretch available but that neither the weight of the rubber band nor the weight of the ant cause it to sag.
It takes him a long, long time to get there, though.
An yes, the assumption is that there is infinite stretch available but that neither the weight of the rubber band nor the weight of the ant cause it to sag.
December 7th, 2010 at 12:58:35 PM
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Basically the end of the band is receding, but the rate of change of that distance decreases.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
December 7th, 2010 at 1:57:46 PM
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Part of the problem people have with this is the knowledge that even if the rubber band stayed the same length, it's gonna take the ant a long time to make the trip.
So let's put it into perspective that most people can wrap their heads around.
Assume that instead of an ant, that it's a car, and the rubberband isn't stretching so fast.
The actual units don't matter, although metrics, being multiples of 10, make things easier to grasp with the original problem.
For now assume that the starting length of the rubberband is also how much it stretches in the alloted time, and also assume that the car travels that same distance in the same time.
At the end of that first time unit, is the car in the middle? No. It's further along than that. If the rubber band stretches in spurts, i.e. it instantly stretches, then stops during the time unit, then the car would travel half-way. But since it's stretching at the same time that the car is moving, the car gets a boost as a result.
People can easily accept that.
But then tell them the car is only going half as fast, and suddenly people argue that the car can never catch up.
So let's put it into perspective that most people can wrap their heads around.
Assume that instead of an ant, that it's a car, and the rubberband isn't stretching so fast.
The actual units don't matter, although metrics, being multiples of 10, make things easier to grasp with the original problem.
For now assume that the starting length of the rubberband is also how much it stretches in the alloted time, and also assume that the car travels that same distance in the same time.
At the end of that first time unit, is the car in the middle? No. It's further along than that. If the rubber band stretches in spurts, i.e. it instantly stretches, then stops during the time unit, then the car would travel half-way. But since it's stretching at the same time that the car is moving, the car gets a boost as a result.
People can easily accept that.
But then tell them the car is only going half as fast, and suddenly people argue that the car can never catch up.
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December 7th, 2010 at 3:40:02 PM
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Quote: thecesspitAs the band stretches, the ant is moved along it as well... e.g it stays the same percentage on the band.
Why not solve for smaller numbers first?
I like this observation.
Let's solve based on percentage of the band instead of absolute distance.
At any point in time, the rate of travel of the ant, in terms of percentage, is:
1 cm/sec / (1+t) km = 0.00001 /(1+t)
Now I think we just need to integrate this from time 0 to T, and solve for T when the integral = 1.0 (i.e. 100%)
This is now a fairly straightforward integral calculation, and no longer a differential equation.
Edit: this leads to the same answer that Doc got.
December 7th, 2010 at 3:50:44 PM
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PapaChubby, I didn't follow your strategy. Could you describe it again, or more fully? Perhaps I would follow if you did the integration to show your solution. Thanks.
December 7th, 2010 at 4:55:55 PM
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Quote: DocPapaChubby, I didn't follow your strategy. Could you describe it again, or more fully? Perhaps I would follow if you did the integration to show your solution. Thanks.
Wow, good stuff here! Thank you Doc and Papa. I've been working through both solutions on paper and am happy as a claim to have seen the light.
Papa's solution took me a while to wrap my head around, but once you accept his premise, the math is very easy. Let me see if I can explain it.
Let r(t) = Ratio of the rubber band the ant is covering at time t.
At time t the rubber band is 100,000*(1+t) cm. long.
The ant's speed is 1, not considering the expansion of the rubber band.
Think of yourself as the ant moving just 1 cm./sec. and both ends of the rubber band moving away from the ant (except for the starting and ending times). In other words, the rubber band ends are moving relative to the ant, not the other way around. So if you base everything on the ant's progress by his own legs the ratio of the rubber band he covers at time t is 1/(100,000*(1+t)). This took me a while to accept, but think I have. I don't know if I explained it well.
Let T be the time at which the ant has covered 100% of the rubber band.
Integral from 0 to T of r(t) = 1
Integral from 0 to T of 1/(100,000*(1+t)) = 1
10^5*ln(t+t) from 0 to T = 1
ln(1+T) = 10^5
1+T = e^100000
T = e^100000 -1
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 7th, 2010 at 5:01:36 PM
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You don't need calculus to solve this (the first time I heard this problem was in 10th grade, I think, we did know derivatives and integrals by then, but definitely not differential equations).
In the first second, the ant leaves behind 1/200000 part of the band, in the second second it is 1/300000, etc ...
The total part of the band left behind by Nth second is 1/100000 * (1/2 + 1/3 + ... 1/N+1). The end is reached when this exceeds 1.
The harmonic series in parenthesis converge to ln(N+1) with very good accuracy (the fixed deficit, known as Euler–Mascheroni constant is about 0.6).
So, solving ln(N+1) = 100000 for N gives an exacts (fro all practical purposes) answer: N = e^100000 - 1.
I am pretty sure this has been discussed before here, but can't find the thread ...
In the first second, the ant leaves behind 1/200000 part of the band, in the second second it is 1/300000, etc ...
The total part of the band left behind by Nth second is 1/100000 * (1/2 + 1/3 + ... 1/N+1). The end is reached when this exceeds 1.
The harmonic series in parenthesis converge to ln(N+1) with very good accuracy (the fixed deficit, known as Euler–Mascheroni constant is about 0.6).
So, solving ln(N+1) = 100000 for N gives an exacts (fro all practical purposes) answer: N = e^100000 - 1.
I am pretty sure this has been discussed before here, but can't find the thread ...
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December 7th, 2010 at 5:28:43 PM
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weaselman, I understand that you had calculus by the 10th grade (which I did not), but I don't see how you solved the problem without calculus. How did you determine that the series converged to ln(N+1) without calculus? Without calculus, how does one even understand the significance of natural logarithms?
December 7th, 2010 at 6:07:46 PM
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Quote: Docweaselman, I understand that you had calculus by the 10th grade (which I did not), but I don't see how you solved the problem without calculus. How did you determine that the series converged to ln(N+1) without calculus? Without calculus, how does one even understand the significance of natural logarithms?
Yeah, I guess, you need *some* calculus after all :) What I meant to say was that you don't need differential equations. Like I said, we did know derivatives and integrals by then. (My 10th grade was in a different country. Age-wise, it is, I think, equivalent to 11th grade in the US, but math curriculum is more extensive over there).
"When two people always agree one of them is unnecessary"
December 7th, 2010 at 6:12:44 PM
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Quote: WizardWow, good stuff here! Thank you Doc and Papa. I've been working through both solutions on paper and am happy as a claim to have seen the light.
Papa's solution took me a while to wrap my head around, but once you accept his premise, the math is very easy. Let me see if I can explain it.
I got the idea while trying to convince myself that there actually was a finite solution. If the end of the rubber band is moving away at 1 km/sec, and the ant is only moving at 1 cm/sec, how will he ever reach the end?
Well consider the possibility that the ant somehow makes it 25% of the way down the band, then stops. Despite the speed of expansion of the band, the ant will remain at the 25% point. He's not losing anything in terms of the proportion of the band that he's covered. And when he takes his next step, the proportion will be 25% + delta. Every step moves him closer to 100%. So now the problem is just to figure out what delta is as a function of time, and integrate it up until you get to 100%.
Delta is just the size of his step (1 cm/sec) divided by the current length of the band (1+t km).
BTW, it's the same thing as Weaselman's answer.
December 7th, 2010 at 6:16:20 PM
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Yep, there are probably a variety of ways to analyze this problem. Since the Wizard had already structured it as a differential equation in the original post, that was the approach I tried to solve. My brain is exercised and well-conditioned even less than my flabby bod, so coming up with the DiffEq solution was all it could handle -- I certainly wasn't up to devising alternate approaches. After all this strenuous mental activity, I think I'll take a nap.Quote: weaselmanWhat I meant to say was that you don't need differential equations.
December 7th, 2010 at 6:38:10 PM
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Quote: weaselmanYou don't need calculus to solve this.
I would say you were using calculus, just not calling it that.
Thanks again Doc and PC. I just posted this on my mathproblems.info site. I'll also make an "Ask the Wizard" question out of it, although in this case it was more like "answer the wizard."
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 8th, 2010 at 2:11:23 PM
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Quote: Wizard
Integral from 0 to T of r(t) = 1
Integral from 0 to T of 1/(100,000*(1+t)) = 1
10^5*ln(t+t) from 0 to T = 1
ln(1+T) = 10^5
1+T = e^100000
T = e^100000 -1
Can this answer be generalized to t = e^y - x, for an ant that moves x cm/sec on a band that moves y cm/sec?
December 8th, 2010 at 2:37:49 PM
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Quote: 7outlineawayCan this answer be generalized to t = e^y - x, for an ant that moves x cm/sec on a band that moves y cm/sec?
No. Here is the general formula.
a = ant's speed
i = initial length of rubber band
r = rubber band's expansion speed
Then the ant will reach the other side in i × (e^(r/a)-1)/r units of time.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 8th, 2010 at 9:11:58 PM
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I just got around to looking at the math problems site. In your solution #1, as well as in the original post of this thread, you referred to this problem as involving partial differential equations. It does not. This problem is in the class of ordinary, linear, differential equations with variable coefficients. The "ordinary" and "linear" characteristics are things we lazy, slow folks hope for. The "variable coefficients" aspect makes it a little more difficult (perhaps very difficult for an atrophied brain like mine), but it is not nearly so difficult as the general category of partial differential equations.Quote: WizardI just posted this on my mathproblems.info site.
In case it is not completely clear, this is an ordinary differential equation because there is only one independent variable: time. The ant's position and the length of the rubber band are dependent variables.
In contrast, suppose we structured the problem so that the 1 km/s rate of stretch was also an independent variable (varying but not a function of time). Similarly, the ant's walking speed could be an independent variable. In those cases, we would likely be dealing with partial differential equations. Please don't ask me for help with that one. We'd both be in serious trouble.
December 8th, 2010 at 11:28:30 PM
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Quote: DocIn case it is not completely clear, this is an ordinary differential equation because there is only one independent variable: time. The ant's position and the length of the rubber band are dependent variables.
In contrast, suppose we structured the problem so that the 1 km/s rate of stretch was also an independent variable (varying but not a function of time). Similarly, the ant's walking speed could be an independent variable. In those cases, we would likely be dealing with partial differential equations. Please don't ask me for help with that one. We'd both be in serious trouble.
Thank you. Good catch. Not only is my calculus rusty, but my terminology too.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 9th, 2010 at 12:29:11 AM
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Can someone confirm that my maths might be correct, in that if the rubber band was 10cm that it would take the ant 12368 seconds, or about 3 hours 26 minutes
http://wizardofvegas.com/forum/off-topic/general/10042-woes-black-sheep-game-ii/#post151727
December 9th, 2010 at 6:47:29 AM
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Quote: WizardofEnglandCan someone confirm that my maths might be correct, in that if the rubber band was 10cm that it would take the ant 12368 seconds, or about 3 hours 26 minutes
No. If the only thing you change is the initial length of the rubber band (i.e. expansion rate remains 1 km/sec), the answer still include e^100,000, which is an enormous number. Changing the initial length only modifies the result by a factor of 10,000, which is pretty insignificant in this case.
