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Math Help - Implicit Function Theorem for a System of Equations

  1. #1
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    Implicit Function Theorem for a System of Equations

    Consider the equations that relate cylindrical and Cartesian coordinates in \mathbb{R}^3;
    x=rcos\theta
    y=rsin\theta
    z=z

    a. Near which points of \mathbb{R}^3 can we solve for r, \theta, and z in terms of the Cartesian coordinates?

    b. Explain the geometry behind your answer in (a).

    I started off by saying

    F: x-rcos\theta=0
    G: y-rsin\theta=0

    but then what do you do for the third equation since z-z=0? If I use H:0 then that makes \Delta J=0 everywhere, meaning that it is not in general possible to solve for r, \theta, and z in terms of x, y, and z. That seems so odd that I assume it's wrong, and I have to think that my mistake is in H.

    Any thoughts?
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  2. #2
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    The equations F, G express the Cartesian coordinates x,y in terms of the cylindrical coordinates r, θ. But the question asks for how to do it the other way round: express the cylindrical coordinates in terms of the Cartesian coordinates.

    The third coordinate z (cylindrical) is the same as z (Cartesian), so there is no problem finding z. There is also no problem finding r, because it is given by the formula r = \sqrt{x^2+y^2}. So the question is really about discussing whether θ can be reconstructed in a neighbourhood of the point (x,y,z).

    Quote Originally Posted by davesface View Post
    what do you do for the third equation since z-z=0? If I use H:0 then that makes \Delta J=0 everywhere, meaning that it is not in general possible to solve for r, \theta, and z in terms of x, y, and z. That seems so odd that I assume it's wrong, and I have to think that my mistake is in H.
    Ah, I see, you're trying to use a more sophisticated approach, inverting a Jacobian matrix. (I should have realised that, from the title of the thread.) In that case, your third equation should be z_{\text{Cart}} - z_{\text{cyl}} = 0. The two z's are conceptually different, and the relation between them should result in a 1 somewhere in the Jacobian matrix.

    In this problem, the sophisticated method can obscure the simple geometry of the transformation, and a naive approach may give more insight. Maybe that is why part b. of the question asks you to think about the geometric meaning of your answer.
    Last edited by Opalg; March 18th 2010 at 01:33 AM.
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    That makes a lot of sense, because \frac{\delta H}{\delta z_{Cart}}=1. So then the determinate evaluated at P_0=(\theta_0, r_0, z_0) ends up being \theta_0 r_0 (cos^2(\theta_0)+sin^2(\theta_0))= \theta_0 r_0 \neq 0.

    So we can solve for (x,y,z) in terms of (r,\theta,z) near all points except \theta =0 or r=0.

    For part b, this would seem to correspond to being unable to solve for (r,\theta,z) only at the origin. This makes sense for \theta, since \theta=arctan(\frac{y}{x}), but r= \pm \sqrt{x^2+y^2} has no such restrictions. Is there something else wrong in my calculations somewhere?
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