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Problem simplifying v*dv/dt to get d/dt(V^2/2)

  1. Sep 11, 2010 #1
    1. The problem statement, all variables and given/known data
    we know dV/dt = a from the definition of acceleration
    multiplying by sides by V, velocity yields
    V dV/dt = aV
    Left side can be reduced to d/dt(V^2/2)

    so to be more clear, just looking at the left side
    V dV/dt = d/dt( V^2/2)

    I want to know what calculus topic that's addressed in, and how it works because I don't see it.

    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 12, 2010 #2

    Pengwuino

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    Gold Member

    It's just a chain rule. Let f(t) = v(t) and f'(t) = a(t). Then [tex]{d \over dt} {f(t)^2 \over 2} = f(t)f'(t)[/tex]

    All you're doing is assigning physical concepts (velocity, acceleration) to mathematical ideas (functions of a parameter, t)
     
  4. Sep 12, 2010 #3
    Thank you very much pengwuino. I would never have thought of that.
     
  5. Jul 17, 2012 #4
    I'm probably missing something obvious here. Can't find anything in the chain rule that explains this simplification. Can somebody elaborate this more?
     
  6. Jul 17, 2012 #5
    You can explain it as either the chain rule or the product rule.

    Chain rule: [itex]\frac{ds}{dt} = \frac{ds}{dv} \frac{dv}{dt}[/itex]. Let [itex]s = v^2[/itex], which gives [itex]\frac{ds}{dt} = (2v)\frac{dv}{dt}[/itex].

    Product rule: [itex]\frac{d}{dt} ab = \frac{da}{dt} b + a \frac{db}{dt}[/itex]. Let [itex]a=b=v[/itex].

    In general, this technique of manipulating an expression to turn it into a derivative of something else is useful because you can often use that to write some kind of conservation law. In this case, dealing with [itex]\frac{d}{dt} \frac{v^2}{2}[/itex] was probably leading up to invoking conservation of energy (only a factor of mass is missing).
     
  7. Mar 4, 2014 #6
    could you also think of it as: since you added a v to the right you must make the left match by first taking the anti derivative of v>>>(1/2)v^2?
     
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