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Bug moving on a rubber band

  1. Aug 21, 2014 #1

    Nathanael

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    1. The problem statement, all variables and given/known data

    "You have a 1 km long rubber band with one end attached to the wall, and the other in your hand. The bug begins to crawl towards you on the rubber band, starting from the wall, at a rate of 1 cm/sec. As he crawls the first centimeter you extend the rubber band 1 km; when he crawls the second centimeter you extend the rubber band another 1 km, and so on, every second. The question is: Does the bug ever reach you, and if so, in how much time?"


    2. Relevant equations

    [itex]L=1km[/itex]
    [itex]v=1km/s[/itex]
    [itex]u=1 cm/s[/itex]
    x = distance between the bug and the wall


    3. The attempt at a solution

    I came up with the following equation:

    [itex]dx=(u+\frac{x}{L+vt})dt[/itex]

    but I don't know how to solve it. I'm thinking of integrating it from [itex]0[/itex] to [itex]T[/itex] and then setting it equal to [itex]L+vT[/itex] and then solving for [itex]T[/itex] but I don't know how to do this (I don't have experience with differential equations)



    Is it possible to get a value for [itex]T[/itex] with this equation, and if so, how would I do it?

    Thanks.
     
    Last edited: Aug 21, 2014
  2. jcsd
  3. Aug 21, 2014 #2

    DrClaude

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    Have you made a dimensional analysis to see if this equation makes sense?



    I'm not sure how to interpret the question. The blue highlighted part I interpret to mean that the rubber band is stretched continuously, while the red I interpret to mean that the rubber band is stretched instantaneously every second.
     
  4. Aug 21, 2014 #3

    Orodruin

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    Watch out here, you are adding a velocity and a dimensionless number.

    I would suggest instead working with the fraction of the rubber band that the bug has behind him. When it reaches 1 the bug has arrived. The problem will then require only basic integration.

    @DrClaude: In the classical formulation of this problem, the bug crawls 1 cm/s and the band is being extended at a rate of 1 km/s. The only difference is whether to consider the problem a difference equation or a differential equation.
     
  5. Aug 21, 2014 #4

    Nathanael

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    Yes, it was just a typo. I meant to put [itex]dx=(u+v\frac{x}{L+vt})dt[/itex] (Sorry, I should type more carefully.)
    That's the ratio of the distance behind him to the entire length, multiplied by the speed of stretching, so the dimensions are now correct.

    It does seem a bit contradictory. It doesn't matter though, (it's not homework), so consider it to be continuously stretched. (That's the case that my equation applies to.)

    Let's say [itex]r=\frac{x}{L+vt}[/itex] then I get:

    [itex]dr=\frac{u}{L+vt}dt+\frac{vr}{L+vt}dt[/itex]

    Then I would set the integral equal to 1.
    I don't see how this helps. Maybe that's not what you meant, I'm tired. I think I'll go to bed.
     
  6. Aug 21, 2014 #5

    Orodruin

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    [STRIKE]##dx/dt## is not equal to ##u## and the second term should have a minus sign since ##L+vt## was in the denominator. It also has a contribution from the stretch of the rubber band as you described in your first post (in fact you should be able to insert your expression for it here). An alternative is just making the change of variables ##r(L+vt) = x## in your (corrected) original equation.[/STRIKE]

    Edit: I realize now that the above might not have been what you did. When you do the differentiation of the right-hand side of your substitution, remember that ##L+vt## is not a constant, but that ##d(L+vt) = v\, dt##.
     
    Last edited: Aug 21, 2014
  7. Aug 21, 2014 #6

    Nathanael

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    The problem is I don't know how to calculate the integral. It seems recursive. (I could write it as an infinite chain of fractions.)
    x depends on the integral, which depends on x, (which depends on the integral, and so on and so forth indefinitely).

    I've come across these infinite fraction chains before but I don't know what to do with them. I'm guessing it doesn't help to treat it as a continued fraction? And there's probably a simpler way to evaluate it? I just don't know how.
     
  8. Aug 21, 2014 #7

    D H

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    Your first derivative, [itex]\frac{dx}{dt} = u + v \frac x{L+vt}[/itex], is fairly easy to integrate. Clearing the denominator and rearranging yields [itex](L+vt)\frac {dx}{dt} - u(L+vt) = vx[/itex]. That's a first order ODE in normal linear form. Use an integrating factor and you're done.

    Your second derivative in post #4 is even easier. You computed the derivative incorrectly. With [itex]r = \frac x {L+vt}[/itex], the derivative is [itex]\frac {dr}{dt} = \frac{\dot x (L+vt) - vx}{(L+vt)^2}[/itex]. See if you can take it from there.
     
  9. Aug 21, 2014 #8

    BvU

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    We've had this one several times before; can't find it because I don't know how to operate the PF advanced search (help!).

    there's even a Wiki on it here
     
  10. Aug 21, 2014 #9

    Nathanael

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    Sorry about that

    Thanks for your help but I still haven't made it to the answer. I'm sure it is simple for you guys but I'm still learning. I think I will save this problem for later while I learn more mathematics.
     
  11. Aug 22, 2014 #10

    D H

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    If you are in the US, you'll typically learn how to solve [itex]\frac {dx} {dt} = u + v \frac {x(t)} {L+vt}[/itex] in your second or third college calculus class.

    Solving for [itex]r(t)[/itex] where [itex]r(t) = \frac {x(t)} {L+vt}[/itex] only requires high school AP calculus. You should be able to do that. Your problem is that you made an error in calculating the derivative of [itex]r(t)[/itex]. You didn't use the quotient rule.
     
  12. Aug 23, 2014 #11
    Could you please tell me how you got this equation?
     
  13. Aug 23, 2014 #12

    Nathanael

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    I assumed that the rubber band is stretched uniformly. So the "speed of stretching" of a piece of the rubber band is propotional to the length of that piece, and to the speed that the entire rubber band is being stretched.
    [[[EDIT:]]]
    [[[I should've said it's proportional to the ratio of the piece's length with the entire length (not just to the piece's length)]]]
    (I call the speed that the entire rubber band is being stretched "v" and the length of the entire band "L")

    I guess the mathematical way to say this idea (of "uniform stretching") would be [itex]dv=v\frac{dL}{L}[/itex]

    For example, a piece of the rubber band of length [itex]k[/itex] would be stretched with a speed of [itex]v\frac{k}{L}[/itex]


    Let's now call the position of the bug [itex]x[/itex] (as measured from the wall)

    The length of rubber band as a function of time is [itex]L+vt[/itex] (where [itex]L[/itex] is the initial length)

    The speed of the bug ([itex]\frac{dx}{dt}[/itex]) would be the walking speed of the bug ([itex]u[/itex]) plus the speed the rubber band behind the bug being stretched.

    So you get:

    [itex]\frac{dx}{dt}=u+v\frac{x}{L+vt}[/itex]


    I hope this was clear (I am sometimes bad at explaining)
     
    Last edited: Aug 24, 2014
  14. Aug 24, 2014 #13
    Thank you Nathanael.I got that.
    [
     
  15. Aug 24, 2014 #14

    Nathanael

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    I haven't officially taken calculus, yet, (my highschool didn't have it) and I've only read the first 5 chapters of a calculus text (strang's). I've been intending for a little while to finish reading Calculus, but I've made little progress because physics is far more fun. My understanding of calculus (however unthrorough) comes from physics problems.

    What didn't make sense to me in solving this, is that, I only know x(t) as "the solution to that other equation" so how am I supposed to integrate an ambiguous function?
    [[[EDIT:]]]
    [[[It makes more sense to me how it's possible, because although x(t) is ambiguous, the integral of x(t) (from 0 to the "T" which satisfies the problem) is known to be L+vT; but I still don't know how to use this knowledge]]]
    I've written it as the limit of a series, but there are two problems for figuring it out. The series is:

    [itex]lim_{\Delta t\rightarrow 0} [^{(k/\Delta t)}_{(n=0)}\Sigma(x_n)]=vk\Delta t+L[/itex]
    Where [itex]x_n=nu\Delta t+\frac{x_{n-1}}{L+nv\Delta t}[/itex]

    The first problem is the way I've defined k (the upper limit). It is not explicitly stated (I couldn't think of a way to write it explicitly, so I just wrote the relationship). This makes it quite confusing to deal with.

    The second problem is, (as I touched on in an earlier post,) the sequence is recursive. The definition of [itex]x_n[/itex] depends on [itex]x_{n-1}[/itex]. This isn't theoretically a problem (because we know [itex]x_0=0[/itex]) but it makes it too complicated for me. The only "solvable" (non recursive) way I can think of to write it is, as long chain of fractions (which becomes infinite in the limit).

    That is beyond my capabilities, so I will just save this problem for a future day.



    EDIT:
    I probably should've written " k " as " T " because that's what it represents in the equation (it is the solution of the problem)
     
    Last edited: Aug 24, 2014
  16. Aug 24, 2014 #15

    Nathanael

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    You're welcome



    P.S. (to D H)

    If you don't mind, can you show me how you would solve this?
    Even if I don't entirely understand, it could possibly give me something to think about.
    (I like learning math from physical ideas like this, where you can actually picture what's happening)
     
    Last edited: Aug 24, 2014
  17. Aug 24, 2014 #16
    Diffrential of form
    [itex]\frac { dy }{ dx }[/itex] +Py=Q
    (where P is constant and Q can be constant as well as function of y)
    can be solved as

    [itex]y{ e }^{ \int { Pdx } }[/itex]=[itex] \int { \{ Q.{ e }^{ Pdx } } \} dx\quad [/itex] +C

    [itex]\frac{dx}{dt} -v\frac{x}{L+vt}[/itex] =u

    Can it be solved like this?

    [itex]y{ e }^{ \int { \frac { v dt }{ L+vt } } }[/itex]=[itex]\int { u.{ e }^{ \int { \frac { v dt }{ L+vt } } } } dt\quad [/itex] +C
     
  18. Aug 24, 2014 #17

    ehild

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    Nathanael, I understand that you do not want to apply Maths methods you do not know and understand yet, but you might be curious about the solution, are you not? You can cheat a bit then and look at wolframalpha.com, but first bring the equation to a simpler form, by introducing dimensionless quantities where possible.

    So choose y=x/L and denote a=u/L and b=v/L. You get

    [itex]\frac{dy}{dt}=a+\frac{by}{1+bt}[/itex] with a= u/L and b=v/L and the initial condition y=0 at t=0.

    I usually solve first order linear differential equations y'+P(t)y+Q(t) =0 by a method I learned when I was a student.
    I assume that y is a product of two functions y=fg. y'=f'g+fg' . Substituting back,
    f'g+fg'+P(t)fg+Q(t)=0. I have the liberty to choose f and g so f'g+P(t)fg=0, that is f'+P(t)f=0. That is a separable equation, it can be written as

    [itex]\frac{df}{dt}=-P(t)f[/itex] and the remaining part is [itex]fg'+Q(t)=0[/itex]

    I rearrange it (it does not seem legal, but it will be)

    [tex]\frac{df}{f}=-P(t) dt[/tex]

    and integrate

    [tex]\int {\frac{df}{f}}=\int{-P(t) dt}[/tex]

    Do not bother with the integration constant. Substitute f into the remaining part of the equation:
    [itex]fg'+Q(t)=0[/itex]: [itex]g'=-Q(t)/f=0[/itex]
    Integrate, but add the integration constant now. Substitute back into y=fg.

    The method is equivalent with the integrating factor method in principle, but I never can remember that :redface:

    Try, just for fun.


    ehild
     
  19. Aug 24, 2014 #18
    Was my method correct,ehild?
     
  20. Aug 24, 2014 #19

    ehild

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    Yes, it can be solved by the integrating factor method, but what is∫(vdt/(L+vt))?


    ehild
     
  21. Aug 24, 2014 #20
    Substituting (L+vt)=a.So da=vdt.

    =[itex]\int { \frac { da }{ a } }[/itex] =[itex]\log { a }[/itex] =[itex]\log { (L+vt) } [/itex]
     
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