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: Help with Partial Derivatives and Implicit Differentiation

  1. Sep 24, 2006 #1
    The problem is as follows:

    Cartesian and polar coordinates are related by the formulas
    [tex]x = r\cos\theta[/tex]
    [tex]y = r\sin\theta[/tex]
    Determine [tex]\frac{\partial r}{\partial x}, \frac{\partial r}{\partial y}, \frac{\partial\theta}{\partial x}, and \frac{\partial\theta}{\partial x}[/tex]. Differentiate the equations above implcitly adn then solve the resulting system of 4 equations in the four unknowns [tex]\frac{\partial r}{\partial x}, \frac{\partial r}{\partial y}, \frac{\partial\theta}{\partial x}, and \frac{\partial\theta}{\partial x}[/tex].

    Now, this seems a whole lot harder than just differentiating [tex]r^2 = x^2 + y^2[/tex], but I did anyway and the 4 equations are:

    [tex]\frac{\partial x}{\partial x} = \frac{\partial r}{\partial x}\cos\theta - \sin\theta\frac{\partial\theta}{\partial x}r[/tex]

    [tex]\frac{\partial x}{\partial y} = \frac{\partial r}{\partial y}\cos\theta - \sin\theta\frac{\partial\theta}{\partial y}r[/tex]

    [tex]\frac{\partial y}{\partial x} = \frac{\partial r}{\partial x}\sin\theta + \cos\theta\frac{\partial\theta}{\partial x}r[/tex]

    [tex]\frac{\partial y}{\partial y} = \frac{\partial r}{\partial y}\sin\theta + \cos\theta\frac{\partial\theta}{\partial y}r[/tex]

    Now, if I try to solve for [tex]\frac{\partial r}{\partial x}[/tex], I know what the answer should be by differentiating [tex]r^2 = x^2 + y^2[/tex], and it's [tex]\frac xr[/tex] or [tex]\cos\theta[/tex], however I can't seem to get that from combining the 4 equations.

    What am I doing wrong?

    Thanks in advance.
    Last edited: Sep 24, 2006
  2. jcsd
  3. Sep 24, 2006 #2


    User Avatar
    Science Advisor

    Use the fact that dx/dx = 1 and dx/dy = 0, etc.
  4. Sep 24, 2006 #3
    I did... it still doesn't give me [tex]\cos\theta[/tex] as an answer.

    For example, to calculate dr/dx
    using eq 1
    I get [tex]1 = \frac{\partial r}{\partial x}\cos\theta[/tex]
    using eq 3
    [tex]0 = \frac{\partial r}{\partial x}\sin\theta[/tex]

    If I use both separatedly or try to combine them, I don't get the same answer as deriving r^2 = x^2 + y^2
  5. Sep 24, 2006 #4


    User Avatar
    Science Advisor

    You seem to have cut off the end parts of those two equations. Try multiplying eq. 1 by cos theta / sin theta and adding it to eq. 3.
  6. Sep 24, 2006 #5
    Why mutliply times tangent will help?
    I cut them off because I multiplied teh other stuff by zero.
    I can solve it using the Jacobian, but I just want to know the logic behind it.
  7. Sep 24, 2006 #6


    User Avatar
    Science Advisor

    You multiply eq. 1 by cos theta/sin theta so that when you add it to eq. 3, you cancel the terms involving dtheta/dx.

    When you say you cut them off because you "multiplied the other stuff by 0" you can't just multiply one term of an equation by 0 and assume it will still be correct. The stuff you cut off was not in general zero.
  8. Sep 24, 2006 #7
    Oh ****, you are right. Thanks a lot man, really appreciate it! =D
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook