Motion along a stretching band

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This was given as a problem in a calculus textbook I'm working through (apologies if this should have gone in the physics forum)

1. Homework Statement

An ant crawls at 1foot/second along a rubber band whose
original length is 2 feet. The band is being stretched at 1
foot/second by pulling the other end. At what time T, if ever,
does the ant reach the other end?
One approach: The band's length at time t is t + 2. Let y(t)
be the fraction of that length which the ant has covered, and
explain
(a) y' = 1/(t + 2) (b)y =ln(t + 2) -ln 2 (c) T = 2e -2.

Homework Equations


∫1/x dx = ln(x)

The Attempt at a Solution


Given a, I can get to b by integrating and finding the constant, and then to c by solving for y=1, but I'm stumped on how to get to explain a. y' seems to be the ant's speed over the length of the band, by I don't understand why that is so.
 
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Calculus by Gilbert Strang? :woot: I read that one too.

y' is the rate of change of the fraction which has been traversed. Suppose the ant was not crawling: when the rubber band is being stretched, the fraction of the rubber band behind the ant would not change (because it also stretches).
 
Yup.

I see that y' is the rate of change of the fraction, but I'm still not sure why why it equals what it does.
 
Consider two special cases to try to get an intuition:

The case where the ant is not crawling (but starts out at some initial fraction y0).
The case where the rubber band is a fixed length.

What would the equation for y' be in each of these (separate) cases?
 
Got it, Thanks.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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