Thank you very much for your reply, Tide. I have graduated from college and took calc in both high school and college (pursuing a CS minor). I can't even follow your logic much less develop a solution to this problem myself. But I want to, so I appreciate your extensive explanation. I feel like I'm close, and I have a lot of the tools and a lot of the concepts floating around in my head, but they're not combining to make
understanding if you know what I mean. I WANT to understand this. I think I had a really lousey calculus teacher in high school, and I know I wasn't as intent on learning the material then as I am now...
w(x) = v \frac {x}{L}
This was a key concept that I totally overlooked. Each point on the rubber band moves with a velocity proportionate to its position on the rubber band. The point at L moves at v. The point at \frac {L}{2} moves at \frac {v}{2}. Excellent! Furthermore, this is not a "position" equation because it's not a function of time. Instead it reveals the velocity of any given point as a function of that point's position on the band.
\frac {dx}{dt} = u + w(x)
This is the first place I have trouble. Why is the ant's velocity \frac {dx}{dt}? Thinking out loud, I think, "Okay, we know the velocity of any point x on the rubber band is w(x). And we know that the ant's speed on the rubber band at any given point x is the ant's velocity, u, plus that point's velocity, w(x). So therefore the ant's velocity is a function of x, so let's say a(x) = u + w(x)." But where does taking the differential (\frac {dx}{dt}) come into this line of thinking? You can determine the ant's velocity at any point on the band with a(x). Why differentiate with respect to t? Essentially I'm saying that I'd be stuck at this point. I can understand up to this point and I can understand creating a(x) to tell you the ant's velocity at a given point, but I wouldn't know where to go from there. It looks like you knew and you introduced \frac {dx}{dt}, but what prompted you to do that?
My guess is that \frac {dx}{dt} = u + w(x) is another way to say, "Over an infinitismal distance, during an infinitismal period of time, the distance along the band that the ant walks equals u + w(x)." You could also write \frac {dx}{dt} = w(x) without the term u. But this is only another way to say, "Over an infinitismal distance, during an infinitismal period of time, the distance in space that a point on the band travels equals w(x)." But we knew that. And trying to find the instantaneous velocity of a point on the band is pointless because we know that the velocity of any point on the band
is constant, right? So differentiating w(x) with respect to t gives you \frac {v}{L}.
Is this correct? If so then I've just learned something new! If not back to the drawing board.
\frac {dx}{dt} = u + v \frac {x}{L}
Marching on and taking for granted that you're still reading this post (if so THANK YOU), and taking for granted that I haven't thought myself into a corner, this simply substitutes w(x). Thank god, I don't have trouble with this step.
Now, L depends only on time so I decided to turn it into the independent variable
This is a tough leap and it seems very intuitive. How did you know where to go from here, \frac {dx}{dt} = u + v \frac {x}{L}? What do we gain from the transformation that happens after
\frac {dx}{dt} = \frac {dL}{dt} \frac {dx}{dL}
? Frankly I'm almost totally lost, but I'll venture a guess at saying that this equation says, "The infinitismal distance that the ant walks up the band over an infinitismal time period equals the velocity of the band times the velocity of the point x on the band."
v = \frac {dL}{dt}
This is another intuitive step, and I'm not sure where we are after we substitute it into get
v \frac {dx}{dL} = u + v \frac {x}{L}
Are we trying to find the point when the ant's velocity (u + v \frac {x}{L}) equals the tip of the rubber band's velocity (v \frac {dx}{dL})?
If you're still reading this I'm surprised but very thankful. I appreciate it, Tide.
