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Wrong answer on Differentiation

  1. Feb 7, 2009 #1
    1. The problem statement, all variables and given/known data

    In polar coords: y = rsin([tex]\theta[/tex])
    y' = dy/dt = r'sin([tex]\theta[/tex]) + rcos([tex]\theta[/tex]) [tex]\theta[/tex]'

    If I want [tex]\partial r' / \partial y' [/tex] can I just solve for r' and take the derivative of y'?

    so

    [tex]\partial r' / \partial y' = 1 / sin \theta[/tex]

    I am hoping this is incorrect because I am getting the wrong answer on another part of a problem...

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 7, 2009 #2

    Mark44

    Staff: Mentor

    Re: Differentiation

    Starting with y = r sin([itex]\theta[/itex]),
    [itex]\partial y/\partial r = \partial (r sin(\theta)) = sin(\theta)[/itex]

    So [itex]\partial r/\partial y = 1/sin(\theta)[/itex]

    Why do you have the primes on your partial derivative? I'm pretty sure that's an oversight on your part.

    If you're trying to calculate dy/dt, and y is a function of both r and [itex]\theta[/itex], and r and [itex]\theta[/itex] are functions of t alone, then
    dy/dt = [itex]\partial y/\partial r * dr/dt + \partial y/\partial \theta * d \theta/dt [/itex]
     
  4. Feb 7, 2009 #3
    Re: Differentiation

    Thanks for your reply,

    The primes actually are indicating time derivatives, let me expand upon my question

    I'm actually trying to do the following

    [tex]\frac{\partial L}{\partial y'} = \frac{\partial L}{\partial r} \frac{\partial r}{\partial y'} + \frac{\partial L}{\partial \theta} \frac{\partial \theta}{ \partial y'}+ \frac{\partial L }{\partial r'} \frac{\partial r'}{\partial y'} + \frac{\partial L}{\partial \theta '} \frac{\partial \theta '}{\partial \y '}[/tex]

    I am having a hard time find the various partial derivs with respect to y' (or maybe I'm not).

    L is actually L = T-U where T = KE = 1/2 m (x'^2 + y'^2) and U doesn't matter in this case
    so in other words, y' and x' are the velocities in y and x.

    where of course
    x = rcos(theta)
    y = rsin(theta)
     
    Last edited: Feb 7, 2009
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