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Homework Help: DE book does not justify this claim

  1. Aug 26, 2012 #1
    1. The problem statement, all variables and given/known data

    My DE book claims that

    [itex]\frac{d\mu(t)/dt}{\mu(t)}[/itex]

    is equivalent to

    [itex]\frac{d}{dt}ln|\mu(t)|[/itex]

    But says nothing to justify this claim.

    Can you?
     
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  3. Aug 26, 2012 #2

    gabbagabbahey

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    Why would it? Most Authors would assume that if a student is studying differential equations, they have already mastered the basics of single variable calculus, such as computing derivatives and anti-derivatives.

    This is just an application of the chain rule, combined with the fact that [itex]\frac{d}{dx} \ln|x| = \frac{1}{x}[/itex].
     
  4. Aug 26, 2012 #3
    I see how from the bottom, the top follows by direct application of the chain rule.

    But given the top, I don't know how to come to the bottom expression.
     
  5. Aug 26, 2012 #4

    Bacle2

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    Re: DE Book does not justify its steps

    But that seems just like the chain rule , (though I don't know why they dropped the

    absolute value): d/dt f(g(t))= f'(g(t))g'(t) , where f(t)=ln|t| g(t)=u(t). Or

    maybe you can do a triple chain rule f(g(h)) , with g(t)=|t|, h(t)=u(t), and

    f(t)=ln(t).
     
  6. Aug 26, 2012 #5

    gabbagabbahey

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    Just compute the anti-derivative:

    [tex]\int \frac{ \frac{d\mu}{dt} }{ \mu(t) }dt = \int \frac{d\mu}{\mu} = \ln|\mu| + C[/tex]

    More importantly though, given how common a function the natural logarithm is, you need to be able to recognize its derivative when you see it. A big part of problem solving with differential equations requires you to have experience with calculating common derivatives (which is why most calculus textbooks have all those practice problems in them :wink:) and applying that experience, by recognizing derivatives of common functions when you see them.
     
    Last edited: Aug 26, 2012
  7. Aug 26, 2012 #6
    Re: DE Book does not justify its steps

    Remember that:
    [itex]\large \frac{δ}{δt}ln|x|=\frac{1}{x}[/itex]

    but since you have a another function inside the log, you need to apply the chain rule, so:
    [itex]\large \frac{δ}{δt}ln|\mu(t)|=\frac{1}{\mu(t)}\cdot\frac{δ\mu(t)}{δt}[/itex]
     
  8. Aug 26, 2012 #7
    That's not really my issue. I just like to see it written out even if it is recognizable - I can integrate and differentiate until the cows come home.

    Nonetheless:

    [itex]\frac{d\mu(t)/dt}{\mu(t)} = q[/itex]

    Integrating both sides

    [itex]ln|\mu(t)| = qt[/itex]

    Then taking the derivative of both sides

    [itex]\frac{dln|\mu(t)|}{dt} = q[/itex]

    Therefore:

    [itex]\frac{d\mu(t)/dt}{\mu(t)} = \frac{dln|\mu(t)|}{dt} = q[/itex]

    Is satisfactory.
     
  9. Aug 26, 2012 #8

    vela

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    This all seems kind of pointless. Integrating ##\frac{d\mu(t)/dt}{\mu(t)}## to get ##\ln|\mu(t)|##, you've used the fact you're trying to show, so why bother? Also, the righthand side is only equal to qt if q is a constant, which it most likely isn't.
     
  10. Aug 26, 2012 #9
    q is constant, I just put it there so I could visualize it thoroughly.
     
  11. Aug 26, 2012 #10

    gabbagabbahey

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    But you defined [itex]q[/itex] as being equal to [itex]\frac{ \frac{d\mu(t)}{dt} }{\mu(t)}[/itex], so it is only constant if [itex]\frac{ \frac{d\mu(t)}{dt} }{\mu(t)}[/itex] is constant, which it isn't, in general.
     
  12. Aug 26, 2012 #11
    It is constant, here.

    If it weren't, instead of qt it would be an arbitrary integral notation of q. It doesn't matter.
     

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  13. Aug 26, 2012 #12

    gabbagabbahey

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    Re: DE Book does not justify its steps

    Try computing [itex]\frac{d}{dx}\ln(x)[/itex] when [itex]x<0[/itex]. There's a reason that calculus books give the antiderivative of [itex]\frac{1}{x}[/itex] as [itex]\ln|x|+C[/itex] instead of [itex]\ln(x)+C[/itex].
     
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