Cockroach Traveling Along an Elastic TightropeDate: 05/15/98 at 01:17:05 From: FieldsCL Subject: Fwd: Question regarding geometric series Dear Dr. Math, I have a project that is as follows: Suppose a cockroach starts at one end of a 1000 meter elastic tightrope and runs towards the other end at a speed of one meter per second. At the end of every second, the tightrope stretches uniformly and instantaneously, increasing its length by 1000 meters each time throughout the whole string. Does the roach ever reach the other end and how long does it take? A hint was given that b + c = d (the total distance initially) and b = b * (b + c + d)/(b + c) is the new length from one end and c = c * (b + c + d)/(b + c) is the new length from the other end. How would you express this as an infinite series in which you would have to take the limit in order to find the answer? Thank you so much! Sincerely, C. Fields Date: 05/15/98 at 10:47:17 From: Doctor Schwa Subject: Re: Fwd: Question regarding geometric series Great question, and very clearly stated. The hint you supplied just had the effect of confusing me, so I'm not going to respond to it right now. If you do want to make use of that hint, then send us another email! In the meantime, here's how I would approach this problem. It sure seems like the roach would never get anywhere, with the rope stretching so fast and the roach running so slowly. But let's see what we can do with a little math. After 1 second, the roach travels 1 meter, or 1/1000 of the length. Then the rope stretches. But since it stretches uniformly, the roach has still made it 1/1000 of the way along the new 2000 foot rope. In the next second, he travels 1 meter, which is 1/2000 of the length. Then the rope stretches. But since it stretches uniformly, the roach is still 1/1000 + 1/2000 of the total distance along the rope. Continue on with this pattern. I don't see a geometric series here, but rather a series with a different name, the harmonic series. Nonetheless, it should help you to solve your problem. Feel free to send another email if you'd like to keep talking about this problem! -Doctor Schwa, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 05/16/98 at 14:25:12 From: FieldsCL Subject: Fwd: Question regarding geometric series Dear Dr. Math, Thank you so much for helping me with the cockroach question. Would the proper way to express the harmonic series of: 1/1000 + 1/2000 + 1/3000 + ... be the following: 1/(n*10^3) Since the cockroach seems to be getting nowhere, the limit of this series should diverge. Right? Thank you so much! Sincerely, C. Fields Date: 05/18/98 at 20:01:26 From: Doctor Schwa Subject: Re: Fwd: Question regarding geometric series Yes, that would be the formula giving the fraction of the rope that the cockroach covers in the nth second. Now wait a minute! Yes, the cockroach *seems* to be getting nowhere. That's what makes this such a fascinating problem. You are right that this series diverges. It's (1/1000) * (1/n), or 1/1000, times the harmonic series, and the harmonic series diverges (very slowly). But this series is representing the fraction of the tightrope that the cockroach has covered. So since it diverges, it will eventually (after approximately e^1000 seconds, which is a BIG number!) reach 1. And when it reaches 1, the cockroach reaches the end of the very long string. If the series converged to something less than 1, that would show that the cockroach never reached the end. But since the series gets bigger than 1, the cockroach *does* eventually make it. This result really surprised me! -Doctor Schwa, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 05/20/98 at 00:27:55 From: FieldsCL Subject: Re: Question regarding geometric series Dear Dr. Math, Thank you so much for your help, but I have one last question, I think. How did you come up with e^1000 seconds as the number to make it reach 1 and the cockroach reaches the end? Did you take the limit of 1/1000 * 1/n to come up with 1/1000 * 1/infinity? Doesn't 1/ infinity = 0? Maybe I am misunderstanding the series. I appreciate your help! Sincerely, C. Fields Date: 05/20/98 at 16:56:54 From: Doctor Schwa Subject: Re: Question regarding geometric series Thanks, I'm glad my explanation was helpful. You're right that I wasn't very clear about how I got e^1000. Let me try that again. In the first second, the roach moves 1/1000 of the length of the string. In the second second, the roach moves 1/2000 of the length of the string, for a total progress of 1/1000 + 1/2000. In the third second, the roach moves 1/3000 of the length of the string, and so has now covered a total of 1/1000 + 1/2000 + 1/3000 of the length. That is, 1/1000 (1/1 + 1/2 + 1/3). So the roach will reach the end when (1/ 1 + 1/2 + 1/3 + 1/4 + ... + 1/n) reaches 1000. It's hard to calculate that exactly, since n has to get really big for that to happen. But you can approximate it. I used calculus to find that the sum in parentheses is approximately equal to the natural log of n. (Actually, it differs by a constant called Euler's constant, but that's another story). So I figured natural log of n equal to 1000 is a good approximation, and thus n is about e^1000. Quite a long time, but definitely finite. -Doctor Schwa, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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