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Derivatives in polar coordinates

  1. Aug 21, 2010 #1
    I appologise in the lack of distinction between curly d's and infinitesimals! All derivatives are partial and anything outside of brackets is an infinitesimal.

    also, I sincerely apologise for any dodgy terminology, but I am for the most part self taught (regarding calculus) :/

    (also, 0 is my poor attempt to represent theta :P)

    1. The problem statement, all variables and given/known data

    More my obsessive compulsiveness when it's really something I'm just supposed to remember, but here goes...

    Express (dz/dx) in polar coordinates for some arbitrary z = z(r, 0)

    2. Relevant equations

    the answer given by my book (Steiner - The Chemistry Maths Book):

    (dz/dx) = (dz/dr)cos(0) - sin(theta)(1/r)(dz/d0)

    3. The attempt at a solution

    dz = (dz/dr)dr + (dz/d0)d0

    (dz/dx) = (dz/dr)(dr/dx) + (dz/d0)(d0/dx)

    essentially, my problem is in evaluating (dr/dx) and (d0/dx) (0 is theta)


    x = r cos(0) => r = x/cos(0)

    => (dr/dx) = 1/cos(0)

    which is... wrong :/. I've also tried so much more silly sh*t like implicit differentiation, and inverting dx/dr, always arriving at the same answer ):


    x = r cos(0) => 0 = arccos(x/r)

    and I have no idea how to start evaluating that...

    Any help anyone could give would be greatly appreciated! (:


    I am an idiot... after 3 days of ploughing through with this and 10 minutes typing the problem to physicsforums, I just worked out that r can be represented in terms of x and y as sqrt(x^2 + y^2) and theta as arctan(y/x)... the rest writes itself lol :P

    I hate my life. :P
    Last edited: Aug 21, 2010
  2. jcsd
  3. Aug 21, 2010 #2


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    good that you've got it now, and sounds like a much easier method, but just to point out the reason this fell over, is becauser theta is a function of x & y also,
    so you must consider 0(x,y) when you take the derivative w.r.t. x
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