1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: 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'?


    [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


    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook