1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted