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Very hard calculus

  1. Sep 13, 2007 #1
    calculus(dv/(g-kv^2))
    I think it's incalculable.
    Could anyone approximate it ?
     
  2. jcsd
  3. Sep 13, 2007 #2

    dextercioby

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    U mean something like this

    [tex]\int \frac{dv}{g-kv^2} [/tex] ?

    Do you know the derivative of arctanh x ?
     
  4. Sep 13, 2007 #3

    arildno

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    You could start with a rewriting:
    [tex]\int\frac{dv}{g-kv^{2}}=\frac{1}{g}\int\frac{dv}{1-(\frac{v}{\sqrt{\frac{g}{k}}})^{2}}=\frac{1}{\sqrt{gk}}\int\frac{du}{1-u^{2}},v=u\sqrt{\frac{g}{k}}[/tex]

    see if you can do something about that, for example along dex's line.
    Alternatively, partial fractions decomposition can come to your aid.
     
  5. Sep 13, 2007 #4

    arildno

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    Since I assume you are modelling the behaviour of a mass particle in a constant gravity field including a quadratic air resistance law, I would just like to say that the proper form of resistive force R is [itex]R= -K|v|v[/itex], rather than [itex]R=-Kv^{2}[/itex]

    Think about it..
     
  6. Sep 13, 2007 #5
    because v is a vector?
     
  7. Sep 14, 2007 #6

    cks

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    I have found the final answer, it's v/g. Is your calculation same with mine?
     
  8. Sep 14, 2007 #7

    dextercioby

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    I can't be that one, it has to have either anctanh or some natural logarithms in it.
     
  9. Sep 14, 2007 #8
    No. The quadratic drag force must point in the opposite direction of the velocity. If the velocity is positive then the drag force direction should be negative and if the velocity is negative then the drag force direction should be positive. If you square the velocity you will always have a drag force in the negative direction, even when the velocity is negative. That is why the absolute value expression is necessary.
     
  10. Sep 14, 2007 #9

    arildno

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    Now, to help you along a bit:
    We have:
    [tex]\int\frac{du}{1-u^{2}}=\int\frac{du}{(1+u)(1-u)}=\frac{1}{2}\int(\frac{1}{(1+u)}+\frac{1}{(1-u)})du[/tex]
    Can you take it from here?
     
  11. Sep 15, 2007 #10

    cks

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    I used the Euler equation to solve it but, i don't know how to select the real part for my solution, i ended up in this way.
     

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  12. Sep 15, 2007 #11

    cks

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    It's pretty obvious that arildno's answer is straightfoward, but how about my unfinished solution. stucked in converting it from a complex part to real part.
     
  13. Sep 15, 2007 #12

    arildno

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    If you want to use complex integration here, you must be way more careful in choosing a proper contour than you have done. Your answer is therefore nonsense.
     
  14. Sep 15, 2007 #13

    HallsofIvy

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    You don't say what contour you are using, so I'm with arildno- it just doesn't mean anything.

    However, in case you have a problem like that in the future, the standard way of finding real and imaginary parts of a fraction like
    [tex]\frac{1}{a+ bi}[/itex]
    is to multiply both numerator and denominator by the complex conjugate of the denominator:
    [tex]\frac{a-bi}{(a+bi)(a-bi)}= \frac{a- bi}{a^2+ b^2}= \frac{a}{a^2+ b^2}+ i\frac{b}{a^2+ b^2}[/tex]
    The real part is a/(a2+ b2) and the imaginary part is -b/(a2+ b2).
     
  15. Sep 15, 2007 #14
    i'm having a hard time following your work

    [tex]\frac{1}{g}\int\frac{dv}{1-(\frac{v}{\sqrt{\frac{g}{k}}})^{2}}=\frac{1}{\sqrt{gk}}\int\frac{du}{1-u^{2}},v[/tex]

    so you factored out a 1\g

    then you rearranged it into a complex fraction and square root and put that whole variable and squared it to show it's equivalent. can you show me what else you did after that in more detail? also, you already have a 1\g as a constant?
     
  16. Sep 15, 2007 #15

    arildno

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    Okay, we set:

    [tex]u=\frac{v}{\sqrt{\frac{g}{k}}}[/tex]
    yielding:
    [tex]v=u\sqrt{\frac{g}{k}}[/tex]
    hence,
    [tex]\frac{dv}{du}=\sqrt{\frac{g}{k}}[/tex]

    Yielding the differential relation:
    [tex]{dv}=\sqrt{\frac{g}{k}}du[/tex]

    Now, our first integral can be rewritten as:
    [tex]\frac{1}{g}\int\frac{\sqrt{\frac{g}{k}}du}{1-u^{2}}[/tex]
    Factor out the constant, and get the one I have in front of my second integral.

    Remember that:
    [tex]\frac{1}{g}=\frac{1}{\sqrt{g^{2}}}[/tex]
    along with the rules for multiplying fractions&square roots together.
     
    Last edited: Sep 15, 2007
  17. Sep 16, 2007 #16

    cks

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    Thank you very much. Really appreciate that!

    However, so, what you mean is that when we want to substitute a cosine or sine function with the euler equation, e^i(theta)=cos(theta)+isin(theta), we have to choose a contour? this is the condition for substituting the euler equation?
     
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