The derivative of Kinetic Energy

  1. From my very limited knowledge of calculus I would have thought that if mass is assumed constant d/dt (1/2mv^2) = 1/2m2v. This seems to be corroborated by one website and seems to follow the general rule d/dt (x^n) = nx^(n-1)
    However in the Feynman lectures (Volume 1 13-1 (13.2)) 1 Ch 13 - Work and Potential Energy 1.pdf he states that if mass is assumed constant d/dt (1/2mv^2) = 1/2m2v(dv/dt).
    Could someone help explain what I have most likely misunderstood?
    Last edited: Sep 21, 2011
  2. jcsd
  3. Hootenanny

    Hootenanny 9,678
    Staff Emeritus
    Science Advisor
    Gold Member

    Welcome to Physics Forums.
    Your expression is incorrect and does not agree with either reference.

    You should note that in the first reference you quote, they are taking the derivative with respect to velocity, whilst in the second Feynman is taking the derivative with respect time time. In the former case you can indeed simply apply the "power rule". However, in the latter case because v depends on time, in the latter case, you need to use the chain rule in conjunction with the power rule.

    Simply put, the first reference finds the rate of change of kinetic energy with respect to velocity. Whilst, the second the reference find the rate of change of kinetic energy with respect to time.

    Does that make sense?
  4. Yes, I understand what I have overlooked now, thanks
  5. Ken G

    Ken G 3,571
    Gold Member

    Incidentally, the interesting derivative of kinetic energy is with respect to distance, not time. If you take d/dx (mv2/2) = mv dv/dx, and you note that v dv/dx = a if you imagine the curve v(x) and want the "a" at some "x", then we have:
    d/dx (mv2/2) = ma = F = -d/dx V
    where V is potential energy for a force F that can be written that way (a "conservative" force). This is entirely the basis of the concept of potential energy to track the changes in kinetic energy.
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