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Polar and cartesian issue

  1. Oct 15, 2015 #1
    This is a problem that has been bugging me all day. While working with the well-known dydx = rdrdθ, where r is a function of θ I divided both sides of the equation by dxdθ to get dy/dθ = r(dr/dx)

    For the left side, I use y = rsinθ and derive with respect to θ to get dy/dθ = sinθdr/dθ + rcosθ. For the left side, I use r^2 = y^2 + x^2, and derive both sides dx, to get

    2r(dr/dx)=2y(dy/dx) + 2x, which simplifies to r (dr/dx)=y(dy/dx) + x.

    I then put both of these equalities in the equation to get

    sinθ(dr/dθ) + rcosθ = y(dy/dx) + x. Knowing that x = rcosθ, and that y/r = sinθ

    I subtract x then divide y from both sides to get

    (1/r)(dr/dθ) = (dy/dx).

    However, this contradicts the proof that

    dy/dx =((dr/dθ)sinθ+rcosθ)/((dr/dθ)cosθ-rsinθ).

    I just want to know what went wrong. (I have checked mathematically, these two functions will NOT give the same results at most points.)
     
    Last edited by a moderator: Oct 15, 2015
  2. jcsd
  3. Oct 15, 2015 #2

    andrewkirk

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    Where did you get this expression? In general it is not a valid expression. In certain tightly constrained contexts it might be able to be given a meaning specific to the context, in such a way that the equality holds. But we don't know from what context you took it, and whether the conditions of that context are preserved throughout the manipulations you do above.

    A good general principle is to not treat differentials like dx like numeric quantities by which one can multiply and divide. They are not, and one can't. In some cases one can use such multiplication or division as a shorthand, but one needs to have a sense of what the shorthand is short for, and be confident that the abbreviation is valid.
     
  4. Oct 15, 2015 #3
    This expression is the Jacobian which the differentials dydx in cartesians are translated into polar differentials rdrdθ, and I am keeping it in the context of coordinates in a plane where it is valid (as that is where I make the comparison to the proof y'=(r'sinθ+rcosθ)/(r'cosθ-rsinθ), which simply equates dy/dx to (dy/dθ)/(dx/dθ), an algebraic transformation). I think I may see your point in the division of two differential numbers at once, which may violate the conditions of shorthand and there is where it becomes illegal, but I wish I knew for sure if that was the case or not.
     
  5. Oct 15, 2015 #4
    After checking further, this is indeed the case, as when there are two differential numbers this doesn't become a legal operation. The initial statement dydx = rdrdθ is valid, but changing it to dy/dθ=rdr/dx is not. (by this logic this would mean (dx/dr)(1/r) would have to equal dθ/dy, which is most certainly not true or how derivatives work. Thank you.)
     
  6. Oct 15, 2015 #5

    andrewkirk

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    OK, then the context where we are using Jacobians for area is that we have the following equation:

    $$\iint_A f(x, y) \,dx \,dy = \iint_A f(r \cos \vartheta, r \sin \vartheta) \, r \, dr \, d\vartheta$$

    That is not the same as saying that:

    $$dx \,dy = r \, dr \, d\vartheta$$

    The manipulations need to be done within the context of the first equation. As soon as they are taken out of that context, they lose validity.
     
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