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Math Help - Integration method

  1. #1
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    malaca
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    Integration method

    Please help me to clarify is this method valid? if it is, is it something new?
    I consider the equation as in attachment below and try to convert it to another form.
    Attached Thumbnails Attached Thumbnails Integration method-int6.jpg  
    Last edited by chuackl; July 27th 2013 at 01:59 PM.
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  2. #2
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    Re: Integration method

    You have the equation V_2= V+ \frac{1}{3}v and so on the very next line write P_2V_2 as P2(V+ \frac{1}{3}dv). How did "v" in one formula becomes "dv" in the other? Also on the next line, you write that as (V+ \frac{1}{3}dv)(P+ \frac{1}{3}dp. (In fact, you have "dv" instead of dp but that's obviously a typo.) Where did that come from? Do you have another formula that says P2= P+\frac{1}{3}dp? After multiplying out, you drop the \frac{1}{9}dvdp saying "term of dvdp is ignored as to small". Where does that come from? In Calculus, if we have something like "dpdp" we will drop that, but "dpdv", with different variables is not generally dropped.
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  3. #3
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    Re: Integration method

    a)I do forgot to explain v is a variable so when the changes is small ie v= dv, V2 will become (V+1/3dv) so the equation can be integrate.
    b)For P2, i have tried subtitute P2 with coefficient of dp different than 1/3 and none of it work and it seem that the conversion of the original formula to the new form only work if coefficient of dp is chosen to be same as dv.( this is the part i need help due to lack of
    evidence). Any reference will be appreciate.
    c)For the dpdv is dropped, i did refer to the link Product rule - Wikipedia, the free encyclopedia
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