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Simple Integral

  1. Jan 22, 2014 #1

    Radarithm

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    1. The problem statement, all variables and given/known data
    [tex]F=-mg-m\alpha v[/tex][tex]F=m\frac{dv}{dt} =[/tex]
    [tex]\int_0^v \frac{dv'}{g+\alpha v'} = -\int_0^t dt'[/tex]
    This equals:
    [tex]\ln \frac{1+\alpha v}{g} = -\alpha t[/tex]

    I don't understand why it does; I keep getting an incorrect answer. Can anyone tell me what I'm doing wrong?

    2. Relevant equations
    3. The attempt at a solution

    let [tex]u=g+\alpha v'[/tex][tex]du=g+vdv[/tex] so [itex]du-g=vdv[/itex]
    [tex]-g+\int_0^v udu = \ln (g+\alpha v)-g[/tex]
    This has to do with getting the velocity as a function of time; I don't know if it should be in the physics section or not but this is a problem I'm having with calculus.
     
    Last edited: Jan 22, 2014
  2. jcsd
  3. Jan 22, 2014 #2
    Is ##F=mg-m\alpha v## or ##F=mg+m\alpha v## ?
     
  4. Jan 22, 2014 #3

    Radarithm

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    [tex]F=-mg-m\alpha[/tex]
     
  5. Jan 22, 2014 #4

    HallsofIvy

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    This makes no sense. You chose to use v' as the variable of integration. Why does it become v when you find du? Also g is a constant so its derivative is 0. If [itex]u= g+ \alpha v'[/itex] then [itex]du= \alpha dv'[/itex] so that [itex]dv'= du/\alpha[/itex]
    [tex]\int_0^v \frac{dv'}{g+ \alpha v'}= \frac{1}{\alpha}\int_g^{g+\alpha v} \frac{du}{u}[/tex]

     
  6. Jan 22, 2014 #5

    Radarithm

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    Why do the limits of integration change from [itex]0 \to v(t)[/itex] to [itex]g \to g+\alpha v[/itex] ?
     
  7. Jan 22, 2014 #6
    Because the variable of integration has changed from v' to u .

    Use the relation between v' and u (u=g+αv') to find the corresponding new limits .

    When v' = 0 → u = g (lower limit)
    When v' = v → u = g+αv (upper limit)
     
  8. Jan 22, 2014 #7

    HallsofIvy

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    Because you changed variables from v' to [itex]u= g+ \alpha v'[/itex]. When v'= 0, [itex]u= g+ \alpha(0)= g[/itex] and when v'= v, [itex]u= g+ \alpha v[/itex].
     
  9. Jan 22, 2014 #8

    Radarithm

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    edit: double posted. sorry
     
  10. Jan 22, 2014 #9

    Radarithm

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    Thanks for the help guys. I appreciate it.
     
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