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A= dv/dt =>adt=dv

  1. Jun 24, 2010 #1
    I am new to calculus, and in some books I read
    a= dv/dt

    If dt means with respect to t, how is it possible multiply both sides of the equation by dt?
    Is there a theorem stating this?

  2. jcsd
  3. Jun 25, 2010 #2
    Re: adt=dv

    adt=dv means nothing but a= dv/dt; just a prevalent abuse of notation. Also, dt is not a number which can be multiplied.
  4. Jun 25, 2010 #3


    Staff: Mentor

    Re: adt=dv

    dv and dt are differentials. For more information, see http://en.wikipedia.org/wiki/Differential_(infinitesimal [Broken]).

    Differentials come up in the study of differential equations, a simple example of which is a = dv/dt. If a is a constant, we can separate this equation to dv = a dt, and integrate both sides with respect to t, to get v = at + C, where C is an arbitrary constant.

    If we realize that v = ds/dt, the time rate of change of position, then we have ds/dt = at + C, which implies that ds = (at + C)dt. Integrating again with respect to t, we get s = (1/2)at^2 + Ct + D, which gives us the displacement of an object moving with a constant acceleration as a function of t.
    Last edited by a moderator: May 4, 2017
  5. Jun 25, 2010 #4
    Re: adt=dv

    Thank you for the response.
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