1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?
     
  2. jcsd
  3. Aug 26, 2012 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor

    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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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

    User Avatar
    Homework Helper
    Gold Member

    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.
     

    Attached Files:

  13. Aug 26, 2012 #12

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    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].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: DE book does not justify this claim
  1. Justify exp(-x^2) (Replies: 7)

Loading...