Re: Derivative of KE

[Hugh Haskell <hhaskell@MINDSPRING.COM>]

At 22:04 -0400 10/16/01, Tim O'Donnell wrote:
> As I was watching a video with my class (for the unteenth
> time) I made a startling observation/conclusion. As a look
> at the KE = 1/2mv^2 equation and take the derivative I
> believe I get  mv which is momentum (p).  Now clearly
> there has to be a relationship between KE and p simply
> because we are talking about the same mass and velocity.
> As I understand derivatives, if I was graphing the  delta
> KE of an object with respect to time, the slope of said
> graph would be the momentum of the object at that time.
Not quite. The derivative you took was with respect to *velocity* so
the derivative of KE wrt velocity is indeed the momentum. If you want
the derivative of KE wrt time, you get mv(dv/dt), which is mva.
Rearrange that to (ma)v, and you get Fv, or power, which is what the
rate of change of energy wrt time is.

Hugh
--

Hugh Haskell
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<mailto://hhaskell@mindspring.com>

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