Velocity with respect to acceleration

**Zman** · Jul3-09, 06:46 AM

Is it possible to differentiate a function with respect to acceleration where the function is expressed in terms of velocity?

$LaTeX Code: \\frac{dy}{da} = \\frac{d}{da}{\\frac{1}{\\sqrt{1 - v^2/c^2}}}$

**arildno** · Jul3-09, 07:23 AM

Well, if say the acceleration/time-relationship is invertible, so that time may be regarded as a function of the acceleration, we would have:
$LaTeX Code: \\frac{dv}{da}=\\frac{dv}{dt}\\frac{dt}{da}=a\\frac{dt }{da}=\\frac{a}{\\frac{da}{dt}}$

Thus, the derivative of velocity wrt. to acceleration is the fraction between the acceleration itself and its rate of change.

**Zman** · Jul3-09, 08:17 AM

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;
$LaTeX Code: \\frac{dE}{da} = \\frac{d}{da}{mc^2\\frac{1}{\\sqrt{1 - v^2/c^2}}}$

and I am not sure how to proceed from this point.

**HallsofIvy** · Jul3-09, 08:25 AM

Originally Posted by Zman

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;
$LaTeX Code: \\frac{dE}{da} = \\frac{d}{da}{mc^2\\frac{1}{\\sqrt{1 - v^2/c^2}}}$

and I am not sure how to proceed from this point.

$LaTeX Code: mc^2(1- v^2/c^2)^{-1/2}$
Now use the chain rule.

**Zman** · Jul3-09, 08:36 AM

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.

**Office_Shredder** · Jul3-09, 08:09 PM

dE/da = dE/dv*dv/da

aka the chain rule. Go for it