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A calculus how do they get there from here question

  1. Aug 13, 2007 #1
    A calculus "how do they get there from here" question

    I'm trying to understand this step in my physics book:

    From this (K is a constant.)

    (1) [tex]{\frac {{\it dp}}{{\it dy}}}=Kp[/tex]

    To this

    (2) [tex]{\frac {{\it dp}}{p}}=K{\it dy}[/tex]

    and then to this

    (3) [tex]\int \!{p}^{-1}{dp}=\int \!K{dy}[/tex]

    How do they get from (1) to (2) How is it that they can break up the notation for the derivative of p with respect to y and treat it like a fraction where dp is divided by dy?
     
  2. jcsd
  3. Aug 13, 2007 #2

    Dick

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    It's a sloppy but handy shorthand. More correctly you get (1/p)*(dp/dy)=K. Now integrate both sides dy and do a change of variable y->p(y) on the left side.
     
  4. Aug 13, 2007 #3

    HallsofIvy

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    Well, my guess would be that they actually completed Calculus I and so had learned about "differentials" as opposed to "derivatives"!
     
  5. Aug 13, 2007 #4
    Dick,

    Thanks very much.

    Kenny
     
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