Cockroach Traveling Along an Elastic Tightrope

Date: 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/