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Material derivative

  1. Jan 13, 2017 #1

    joshmccraney

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    1. The problem statement, all variables and given/known data
    Show ##DF/Dt=0##. ##F = x-a-e^b\sin(a+t)## and ##a## is given implicitly as ##y=b-e^b\cos(a+t)## where ##a=f(y,t)## and ##b## is a constant. Also, velocity is $$u=e^b\cos(a+t)\\v=e^b\sin(a+t)$$

    2. Relevant equations
    ##DF/Dt=F_t+v\cdot\nabla F##

    3. The attempt at a solution
    ##F_t = -e^b\cos(a+t)##
    ##v\cdot \nabla F = e^b\cos(a+t) \cdot 1 + e^b\sin(a+t) \cdot 0 = e^b\cos(a+t)##.
    Then ##DF/Dt = -e^b\cos(a+t)+e^b\cos(a+t)=0##. Is this correct? It feels too easy.

    Thanks!
     
  2. jcsd
  3. Jan 14, 2017 #2

    Orodruin

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    What happened to Da/Dt?
     
  4. Jan 14, 2017 #3

    joshmccraney

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    So you're saying ##F_t=a'(t)-e^b\cos (a+t)\cdot (a'(t)+1)##. But then the convective term would also have the ##y## component, namely ##v\cdot (a'(y)-e^b\cos (a+t)\cdot a'(y)##?

    Also, ##x## velocity is given by ##u=e^b\cos(a+t)##. Then ##a## was not differentiated for ##u=x'(t)##.

    Sorry if this looks weird, I'm using the app for the first time and I can't see Tex output.
     
    Last edited: Jan 14, 2017
  5. Jan 18, 2017 #4

    joshmccraney

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    Orodruin, do you still think I should differentiate ##a## since they did not for the velocity term?
     
  6. Jan 18, 2017 #5

    Orodruin

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    You make it sound as if the velocity was not given on the form you have quoted. My understanding of what you posted is that you were given the velocity.
     
  7. Jan 18, 2017 #6

    joshmccraney

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    I apologize for the ambiguity. So the particle's position is given by: $$x=a+e^b\sin(a+t)\\y=b-e^b\cos(a+t)$$ Then the velocity as in the first post, which is given in the question stem. Does this clarify my question?
    Thanks for your patience!
     
  8. Jan 18, 2017 #7

    Orodruin

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    Could you quote the problem statement verbatim?
     
  9. Jan 18, 2017 #8

    joshmccraney

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    Definitely. It follows:
    A particle's flow path is described as
    $$x=a+e^b\sin(a+t)\\y=b-e^b\cos(a+t)$$
    Thus the spacial velocity is
    $$u=e^b\cos(a+t)\\v=e^b\sin(a+t)$$ Show that the kinematic boundary condition is satisfied along the curve derived from above by setting ##b=const## and ##a## is a parameter.
    Hint: One could consider this curve to be $$x-a(y,t)-e^b\sin(a(y,t)+t)=0$$.

    The kinematic boundary condition is ##DF/Dt=0## where ##F## is a curve of the boundary, presumably the hint's curve.
     
  10. Jan 19, 2017 #9

    joshmccraney

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    But ##u=\partial_tx##, and they did not implicitly differentiate ##a##, but treated it as a constant. If this is true for the time derivative, since ##a## is a functino of ##t## and ##y##, then ##D a/D t = 0##. Do you agree?
     
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