Can You Derive F = gamma^3*ma from Newton's Second Law?

[beecher]

Homework Statement

Newton's second law is given by F = dp/dt. If the force is always parallel to the velocity, show that F = gamma³ma

Homework Equations

p = gamma*mv
gamma = 1/(1-v²/c²)^1/2

The Attempt at a Solution

I really have no clue where to begin. This is what I've done so far, but I don't think I'm even on the right track.

F = dp/dt = gamma³ma = [1/(1-v²/c²)^1/2]³ma
=1/(1-v²/c²)^3/2ma
=1/(1 - 3v²/c² + 3v⁴/c⁴ - v⁶/c⁶)^1/2ma

Thats as far as i get on the right side.

For the left, all i can manage is dp/dt = d/dx p = d/dx gamma*mv
This is where I get confused. gamma and m are constant, so does dp/dt = gamma*m*dv/dt?

Some clarification would be great.

Thanks!

 

[LUphysics]

Are you in Modern Physics and answering this for your Russian Prof?

 

[RoyalCat]

You're taking the hard route, you're trying to integrate the given expression, into the derivative. Just take the derivative, and rearrange.

Some pointers:

Express $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=({1-\frac{v^2}{c^2}})^{-\tfrac{1}{2}$$

That should make taking the derivative much more simple.

Another bit that should help you is: $$\frac{d({1-\frac{v^2}{c^2}})}{dt}=\frac{-2v\cdot\dot v}{c^2}$$