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Homework Help: Bug on band

  1. May 30, 2007 #1
    1. The problem statement, all variables and given/known data


    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?

    2. Relevant equations
    Differential ones!

    3. 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

    [tex]v_{bug\,in\,frame} = \frac{l_0}{l} v_{bug\,0}[/tex]

    where [itex]l_0[/itex] is the initial length of the band, [itex]l[/itex] is the length of the band as a function of time, and [itex]v_{bug\,0}[/itex] 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

    [tex]v_{bug\,in\,frame} = \frac{1}{1+t} 1 \times 10^{-5}[/tex]

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

    [tex]\int v\,dt = \int \frac{1}{1+t} 1 \times 10^{-5}\,dt[/tex]

    [tex]x = 1 \times 10^{-5} \ln(1+t)[/tex]

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

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

    [tex]t = e^{100000} - 1[/tex].

    My original plan, however, was to use the following differential equation to describe the actual distance of the bug from the wall:

    [tex]\frac{dx}{dt} = \frac{x}{l} v_{end} + v_{bug}[/tex]

    where [itex]l[/itex] (which is 1+t) is a function of t that describes the length of the band, [itex]v_{end}[/itex] is the velocity of the end of the band (dl/dt=1 m/s) and [itex]v_{bug}[/itex] 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?
    Last edited: May 30, 2007
  2. jcsd
  3. Jun 1, 2007 #2
    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:
    [itex]\frac{dx}{dt} - \frac{x}{1+t} = 1 \times 10^{-5}[/itex]

    This is a differential equation of the form
    [itex]\frac{dx}{dt} + P x = Q [/itex]

    where P and Q are functions of t.

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

    [itex]\frac{1}{1+t} \frac{dx}{dt} - \frac{x}{(1+t)^2} = \frac{1 \times 10^{-5}}{1 + t}[/itex]

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

    [itex]\frac{x}{1 + t} = \int \frac{1 \times 10^{-5}}{1+t} dt[/itex]

    Which I think you can finish off from there. So if I have understood your original equation correctly, there is a general solution.
  4. Jun 2, 2007 #3
    Ah, beautiful. Thanks a lot dt19!
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