Will the Bug Reach the Other End of the Rubber Band?

[gabee]

Homework Statement

http://www.feynmanlectures.info/exercises/bug_on_band.html

An infinitely stretchable rubber band has one end nailed to a wall, while the other end is pulled away from the wall at the rate of 1 m/s; initially the band is 1 meter long. A bug on the rubber band, initially near the wall end, is crawling toward the other end at the rate of 0.001 cm/s. Will the bug ever reach the other end? If so, when?

Homework Equations

Differential ones!

The Attempt at a Solution

I solved this one in what I think is a sort of novel way. I imagine that we are viewing the situation in a "stretching" frame--as if we view the stretching band with a camera and we continuously zoom out to keep the image of the band on the film exactly 1 meter wide. Then, the velocity of the bug on the film is described by

$$v_{bug\,in\,frame} = \frac{l_0}{l} v_{bug\,0}$$

where $l_0$ is the initial length of the band, $l$ is the length of the band as a function of time, and $v_{bug\,0}$ is the initial velocity of the bug as seen on the film (which is the same as the real velocity, 1E-5 m/s).

In effect, the 'image on the film' becomes a representation for the fraction of the band traversed by the bug.

So, using the information given in the problem, this equation becomes

$$v_{bug\,in\,frame} = \frac{1}{1+t} 1 \times 10^{-5}$$

which we can integrate with respect to time to find the x position of the bug on the film:

$$\int v\,dt = \int \frac{1}{1+t} 1 \times 10^{-5}\,dt$$

$$x = 1 \times 10^{-5} \ln(1+t)$$

When x = 1, the bug has reached the end of the band:

$$1 = 1 \times 10^{-5} ln(1+t)$$, and with a little algebra,

$$t = e^{100000} - 1$$.
My original plan, however, was to use the following differential equation to describe the actual distance of the bug from the wall:

$$\frac{dx}{dt} = \frac{x}{l} v_{end} + v_{bug}$$

where $l$ (which is 1+t) is a function of t that describes the length of the band, $v_{end}$ is the velocity of the end of the band (dl/dt=1 m/s) and $v_{bug}$ is the velocity of the bug by itself (1E-5 m/s). But I didn't know how to solve this differential equation (I haven't taken a diff eq course yet and separation of variables won't work). I'm wondering--is there a general solution to this form of DE?

 

[dt19]

I think I can solve it, providing I have understood what you wrote in the right way.

This is what I think your differential equation is:
$\frac{dx}{dt} - \frac{x}{1+t} = 1 \times 10^{-5}$

This is a differential equation of the form
$\frac{dx}{dt} + P x = Q$

where P and Q are functions of t.

The way to solve this is as follows. Notice that, if we multiply through by $\frac{1}{1+t}$, we will end up with

$\frac{1}{1+t} \frac{dx}{dt} - \frac{x}{(1+t)^2} = \frac{1 \times 10^{-5}}{1 + t}$

The LHS of this equation is what you get when you differentiate $\frac{x}{1+t}$ w.r.t. t. So you can then write

$\frac{x}{1 + t} = \int \frac{1 \times 10^{-5}}{1+t} dt$

Which I think you can finish off from there. So if I have understood your original equation correctly, there is a general solution.

 

[gabee]

Ah, beautiful. Thanks a lot dt19!