Velocity with respect to acceleration

  • Context: Graduate 
  • Thread starter Thread starter Zman
  • Start date Start date
  • Tags Tags
    Acceleration Velocity
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 2K views
Zman
Messages
96
Reaction score
0
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]
 
Physics news on Phys.org
Well, if say the acceleration/time-relationship is invertible, so that time may be regarded as a function of the acceleration, we would have:
[tex]\frac{dv}{da}=\frac{dv}{dt}\frac{dt}{da}=a\frac{dt}{da}=\frac{a}{\frac{da}{dt}}[/tex]

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

[tex]mc^2(1- v^2/c^2)^{-1/2}[/tex]
Now use the chain rule.
 
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.