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Ant on a rubber string, mathematical series

  1. Apr 15, 2014 #1
    1. The problem statement, all variables and given/known data
    Ideal rubber stirng with lenght L=1km.Ant is takng a walk on the string with speed v=1cm/s
    After every minute(Δt=60s) ,string is getting longer by ΔL=1km.
    1)Will ant get to the end of string?
    2)If yes,then how long it will take ?


    2. Relevant equations
    1) So i used series expansion
    [itex]\frac{s}{l}[/itex]=[itex]\frac{vΔt}{L}[/itex]+[itex]\frac{vΔt}{L+ΔL}[/itex]+..=
    vΔt∑[itex]\frac{1}{L+iΔL}[/itex]

    The ant will get to the end when s/l=1
    So i got that.

    2)But I have problem figuring out how to get the time needed to get there.
    I have tip to use integral test,but i do not understand how to use it,beacuse as far as i know,tests are used to test if series converges.
    Please,help here ?



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 15, 2014 #2

    BvU

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    Your series expansion can be simplified to
    $$\frac{s}{l}=\frac{vΔt}{L}\Bigl ( {1\over 1} + {1\over 2} + {1\over 3}+ {1\over 4} ... \Bigr )$$
    If you (mentally) replace the summation by an integration you get a hunch that this sum is very close to a logarithm.
    The expression in brackets is a harmonic number link
    Mathematicians will claim the ant gets there.
    Physicists are much more realistic: they take the lifetime of the universe into consideration and claim it doesn't get there.
     
  4. Apr 15, 2014 #3
    Are you sure about that? That series does not converge, I don't think.
     
  5. Apr 16, 2014 #4

    BvU

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    There is no question of convergence: the summation doesn't have to be continued ad infinitum. In fact, how far it has to be continued is OP part 2) of the exercise.
     
  6. Apr 16, 2014 #5
    I would like to offer a different analysis of this problem.

    Let L represent the total length of the string at time t, and let r represent the rate at which the length is increasing (1 km/min). Then,
    [tex]L=L_0+rt[/tex]
    where L0 is the initial length (1 km). The velocity of the far end of the string is dL/dt = r. The velocity at location y along the string is
    [tex]v_s(y)=r\frac{y}{L}=r\frac{y}{L_0+rt}[/tex]
    When the ant is at location y, its velocity is
    [tex]\frac{dy}{dt}=r_A+v_s(y)=r_A+\frac{ry}{L_0+rt}[/tex]
    where rA is the velocity of the ant relative to the string.

    The solution to this differential equation subject to the initial conditions is:
    [tex]\frac{y}{L_0+rt}=\frac{r_A}{r}\ln \left(\frac{L_0+rt}{L_0}\right)[/tex]
    The ant reaches the end of the string when [itex]y=L_0+rt[/itex]. Therefore, the time required is obtained from:

    [tex]t=\frac{L_0}{r}\left(\exp(\frac{r}{r_A})-1\right)[/tex]

    Chet
     
  7. Apr 17, 2014 #6

    BvU

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    Interesting factor ##\gamma## between the "("faster!") discrete case and the continuous case !

    With ##\gamma = 0.57721566490153286060651209008240243104215933593992## the discrete ant gets there 3*10723 minutes before the continuous ant !

    But I'm afraid we've lost our prehisto friend here. Still listening in ?
     
  8. Apr 21, 2014 #7
    I red the first post and that was enough for me at the time. And forgot about this thread :(
    But thanks guys.
    Now Im looking in to latest posts and they are interesting!
     
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