Simple question regarding Product Rule

[bitrex]

Homework Statement

As part of a first order differential equation I need to find \frac{d}{dt}(mv)
where v=\frac{dy}{dt}

Homework Equations

Product Rule.

The Attempt at a Solution

\frac{d}{dt}(m\frac{dy}{dt}) = m*\frac{d}{dt}\frac{dy}{dt} + \frac{dm}{dt}\frac{dy}{dt} = ??

I know I shouldn't have to deal with a second derivative, at least in this equation. Is there a way to simplify the equation first that I'm not seeing?

 

[Dick]

Why do you think there's no second derivative? There is and you did it correctly. The second derivative would be the acceleration. If m*v represents a momentum you can often take the mass m to be constant. That would simplify it.

 

[bitrex]

Well, the equation I'm dealing with in full is \frac{d}{dt}(mv) = {mg }, where m is a function 4/3{\pi}kt^3. Ahh, I see how to set the equation up. After I get m\frac{d^2y}{d^2t} + \frac{dm}{dt}\frac{dy}{dt} I substitute \frac{dv}{dt} for \frac{d^2y}{d^2t} and v for \frac{dy}{dt}. I then solve the differential equation for v(t) with initial condition v(0) = 0. I have to prove that in this case \frac{d^2y}{d^2t} is proportional to \frac{g}{4}, so once I know velocity I can differentiate the function to get acceleration and see if that works out. I think I can do this now - I just needed my brain jogged a bit. Thanks!