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Velocity with respect to acceleration

  1. Jul 3, 2009 #1
    Is it possible to differentiate a function with respect to acceleration where the function is expressed in terms of velocity?

    [tex]\frac{dy}{da} = \frac{d}{da}{\frac{1}{\sqrt{1 - v^2/c^2}}}[/tex]
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  3. Jul 3, 2009 #2


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    Well, if say the acceleration/time-relationship is invertible, so that time may be regarded as a function of the acceleration, we would have:

    Thus, the derivative of velocity wrt. to acceleration is the fraction between the acceleration itself and its rate of change.
  4. Jul 3, 2009 #3
    Thanks for your reply.
    I tried to keep my question simple but I think that that was a mistake. My maths is extremely rusty and I definitely feel uncomfortable with it.

    The situation that I am dealing with is the relationship between the energy of a body and its acceleration.

    I want to determine the relationship dE/da (E is energy, a is acceleration)
    I have arrived at the expression;
    [tex]\frac{dE}{da} = \frac{d}{da}{mc^2\frac{1}{\sqrt{1 - v^2/c^2}}}[/tex]

    and I am not sure how to proceed from this point.
  5. Jul 3, 2009 #4


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    [tex]mc^2(1- v^2/c^2)^{-1/2}[/tex]
    Now use the chain rule.
  6. Jul 3, 2009 #5
    The velocity v is the only variable in the equation. Surely I need to express v in terms of acceleration ‘a’ before I can differentiate the expression using the chain rule?

    If I was resolving dE/dv, I believe that I could go ahead and differentiate the expression using the chain rule but I am trying to resolve dE/da.
  7. Jul 3, 2009 #6


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    dE/da = dE/dv*dv/da

    aka the chain rule. Go for it
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