Simple question regarding Product Rule

  • Thread starter bitrex
  • Start date
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1. The problem statement, all variables and given/known data
As part of a first order differential equation I need to find [tex]\frac{d}{dt}(mv)[/tex]
where [tex]v=\frac{dy}{dt}[/tex]

2. Relevant equations

Product Rule.


3. The attempt at a solution

[tex]\frac{d}{dt}(m\frac{dy}{dt}) = m*\frac{d}{dt}\frac{dy}{dt} + \frac{dm}{dt}\frac{dy}{dt} = ?? [/tex]

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

Science Advisor
Homework Helper
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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.
 
195
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Well, the equation I'm dealing with in full is [tex]\frac{d}{dt}(mv) = {mg }[/tex], where m is a function [tex]4/3{\pi}kt^3[/tex]. Ahh, I see how to set the equation up. After I get [tex]m\frac{d^2y}{d^2t} + \frac{dm}{dt}\frac{dy}{dt}[/tex] I substitute [tex]\frac{dv}{dt}[/tex] for [tex]\frac{d^2y}{d^2t}[/tex] and v for [tex]\frac{dy}{dt}[/tex]. I then solve the differential equation for v(t) with initial condition v(0) = 0. I have to prove that in this case [tex]\frac{d^2y}{d^2t}[/tex] is proportional to [tex]\frac{g}{4}[/tex], 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!
 

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