1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple question regarding Product Rule

  1. Feb 11, 2009 #1
    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?
     
  2. jcsd
  3. Feb 11, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  4. Feb 11, 2009 #3
    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Simple question regarding Product Rule
  1. Product Rule Question (Replies: 1)

Loading...