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A= dv/dt

  1. Aug 4, 2009 #1
    I see in many text that a = dv/dt implies that

    dv = a dt

    How is that possible, can anybody please explain me. As far as i know dv/dt is a symbol for derivative of v w.r.t t and not ratio between dv and dt.
     
  2. jcsd
  3. Aug 4, 2009 #2
    What you wrote is in differential form.
     
  4. Aug 4, 2009 #3

    jgens

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    Rather than offer a sub-par explanation, this thread - with input from several knowledgeable members - should answer your questions: https://www.physicsforums.com/showthread.php?t=328193
     
  5. Aug 4, 2009 #4
    dv/dt is not a symbol. It is the mathematical formula for the derivative of v with respect to t. This is how one would show any derivative of a dependent variable with respect to an independent variable.
     
  6. Aug 4, 2009 #5
    And what did I say :-)
     
  7. Aug 4, 2009 #6
  8. Aug 4, 2009 #7

    arildno

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    Now, WHY can we utilize at times the dv=adt formula, in particular, WHERE is it usable?

    Answer:

    When doing integration with the technique called substitution of variables (i.e the "inverse" of the chain rule):

    Given a=dv/dt, we have, trivially:
    [tex]\int_{t_{1}}^{t_{2}}adt=\int_{t_{1}}^{t_{2}}\frac{dv}{dt}dt[/tex]
    But the right-hand side can, by the theorem of substitution of variables, be reformulated, giving the identity:
    [tex]\int_{t_{1}}^{t_{2}}adt=\int_{v(t_{1})}^{v(t_{2})}dv=\int_{v_{1}}^{v_{2}}dv[/tex]

    Now, by IGNORING that the limits of integration actually refers to the limits of DIFFERENT variables, we "may say" that the "integrands" are equal, i.e, adt=dv!


    Thus, adt=dv should, at this stage of your education, be regarded as notational garnish (or garbage, if you like!)
     
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