[Guest]

A ant starts at the beginning of a straight rubber band (not a loop) 10 inches long and moves at a speed of 1 inch/sec. After the ant starts its journey, the rubber band is stretched one inch every second. Assume that the placement of the ant is wherever its front legs are.

How long does it take the ant to reach the end and how long will the rubber band be when it does?

[Guest]

If you plot out your excel points they should form a hyperbolic curve (or a portion of). In order to solve for the area under that curve you need to use natural logs, which are done in base e, so that's where the e comes in. Otherwise your just added up "steps" under (or centered on) the curve.

It seems to me that since the band is being stretched instantaneously at discrete intervals, it won't be a smooth curve. Anyway, the inelegant, brute force method yeilds an answer that I think is

correct.

at 1 seconds, ant has traveled 1.00 of 10.00 inches

<streeeetch> still at 1 seconds, ant has traveled 1.10 of 11.00 inches

at 2 seconds, ant has traveled 2.10 of 11.00 inches

<streeeetch> still at 2 seconds, ant has traveled 2.29 of 12.00 inches

at 3 seconds, ant has traveled 3.29 of 12.00 inches

<streeeetch> still at 3 seconds, ant has traveled 3.57 of 13.00 inches

at 4 seconds, ant has traveled 4.57 of 13.00 inches

<streeeetch> still at 4 seconds, ant has traveled 4.92 of 14.00 inches

at 5 seconds, ant has traveled 5.92 of 14.00 inches

<streeeetch> still at 5 seconds, ant has traveled 6.34 of 15.00 inches

at 6 seconds, ant has traveled 7.34 of 15.00 inches

<streeeetch> still at 6 seconds, ant has traveled 7.83 of 16.00 inches

at 7 seconds, ant has traveled 8.83 of 16.00 inches

<streeeetch> still at 7 seconds, ant has traveled 9.38 of 17.00 inches

at 8 seconds, ant has traveled 10.38 of 17.00 inches

<streeeetch> still at 8 seconds, ant has traveled 10.99 of 18.00 inches

at 9 seconds, ant has traveled 11.99 of 18.00 inches

<streeeetch> still at 9 seconds, ant has traveled 12.66 of 19.00 inches

at 10 seconds, ant has traveled 13.66 of 19.00 inches

<streeeetch> still at 10 seconds, ant has traveled 14.38 of 20.00 inches

at 11 seconds, ant has traveled 15.38 of 20.00 inches

<streeeetch> still at 11 seconds, ant has traveled 16.14 of 21.00 inches

at 12 seconds, ant has traveled 17.14 of 21.00 inches

<streeeetch> still at 12 seconds, ant has traveled 17.96 of 22.00 inches

at 13 seconds, ant has traveled 18.96 of 22.00 inches

<streeeetch> still at 13 seconds, ant has traveled 19.82 of 23.00 inches

at 14 seconds, ant has traveled 20.82 of 23.00 inches

<streeeetch> still at 14 seconds, ant has traveled 21.73 of 24.00 inches

at 15 seconds, ant has traveled 22.73 of 24.00 inches

<streeeetch> still at 15 seconds, ant has traveled 23.67 of 25.00 inches

at 16 seconds, ant has traveled 24.67 of 25.00 inches

<streeeetch> still at 16 seconds, ant has traveled 25.66 of 26.00 inches

at 16.34 seconds, ant has traveled 26.00 of 26.00 inches

Edited January 9, 2009 by xucam

[bonanova]

Let the increments of distance and time be so small

that whether the ant's initial move happens before

or after the band's initial stretch doesn't matter.

L=10 [L=band's length]

x=time=0 [x=ant's position]

Loop:

time=time+delta

x=x+delta [dx/dt = 1 so time step is also distance step]

x=x [(L=L+delta)/L] [left-to right execution - multiply x by the ratio of new to old L values]

Until L<x

For delta=.00001

time=17.18281 [1 718 281 steps]

L=27.18281

x=27.18281031

L/x = 1.581977006

For delta=.000001

time=17.182818 [17 182 818 steps]

L=27.18281801

x=27.18281859

L/x=1.581976751

The red digits are the ones still changing as delta decreases.

Pretty safe to conclude that

L[final] / L[initial] = e

x[final] = L[final] = L[initial] x e = 10e

time = L[initial] x [e-1] = 10[e-1]

avg speed => e/(e-1) = 1.5819767068..... relative to the fixed end of the band.

[Guest]

Let the increments of distance and time be so small

that whether the ant's initial move happens before

or after the band's initial stretch doesn't matter.

L=10 [L=band's length]

x=time=0 [x=ant's position]

Loop:

time=time+delta

x=x+delta [dx/dt = 1 so time step is also distance step]

x=x [(L=L+delta)/L] [left-to right execution - multiply x by the ratio of new to old L values]

Until L<x

For delta=.00001

time=17.18281 [1 718 281 steps]

L=27.18281

x=27.18281031

L/x = 1.581977006

For delta=.000001

time=17.182818 [17 182 818 steps]

L=27.18281801

x=27.18281859

L/x=1.581976751

The red digits are the ones still changing as delta decreases.

Pretty safe to conclude that

L[final] / L[initial] = e

x[final] = L[final] = L[initial] x e = 10e

time = L[initial] x [e-1] = 10[e-1]

avg speed => e/(e-1) = 1.5819767068..... relative to the fixed end of the band.

What language is this, S-Plus or something? You have 2 assignment statements in a single line, pretty awesome but confusing. Translated into a lowbrow language

my $l = 10;

my $x = 0;

my $time = 0;

#my $delta = .00001;

my $delta = 1;

$time += $delta;
$x += $delta;
$x *= ($l + $delta)/$l;
$l += $delta;
}
print "len is $l, pos is $x, time is $time\n";

until ( $l < $x ) {

I wondered about the effect of swapping these 2 lines:

x=x+delta [dx/dt = 1 so time step is also distance step]

x=x [(L=L+delta)/L] [left-to right execution - multiply x by the ratio of new to old L values]

At small increments, the diff is small as you state:

len is 27.1828099993495, pos is 27.1828103058539, time is 17.1828099995187

len is 27.1828299993495, pos is 27.1828331230349, time is 17.1828299995187

However, with step size = 1 there is a large difference

len is 29, pos is 29.9478807649453, time is 19 [ expand first]

len is 27, pos is 27.6871894806634, time is 17 [ step first ]

The last solution strikes me as fitting the OP, step first and expand every second on the second. In this loop the ant overshoots, so the band actually stretched to 26 inches max, and the ant reached it in 16.3x seconds.

[bonanova]

What language is this, S-Plus or something? You have 2 assignment statements in a single line, pretty awesome but confusing.

APL2.

Amazingly quick for prototyping/simulations.

My "code" is a mix of actual code and description of code.

I reproduce both of your [delta=1] results.