# Velocity with respect to acceleration

1. Jul 3, 2009

### Zman

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

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

2. Jul 3, 2009

### arildno

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

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

3. Jul 3, 2009

### Zman

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

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

4. Jul 3, 2009

### HallsofIvy

Staff Emeritus
$$mc^2(1- v^2/c^2)^{-1/2}$$
Now use the chain rule.

5. Jul 3, 2009

### Zman

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.

6. Jul 3, 2009

### Office_Shredder

Staff Emeritus
dE/da = dE/dv*dv/da

aka the chain rule. Go for it