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Polar Coordinates and Conservative Vector Fields

  1. Nov 24, 2011 #1
    1. The problem statement, all variables and given/known data

    Let F = <-y/(x2+y2, x/(x2+y2>. Recall that F was not conservative on R2 - (0,0). In this problem, we show that F is conservative on R2 minus the non-positive x-axis. Let D be all of R2 except points of the form (-x,0), where x≥0.

    a) If (x,y) is included on D, show that it is possible to express (x,y) in polar coordinates (r,θ) such that x=r*cosθ and y=r*sinθ, where θ can be chosen such that θ equals,

    {arctan(y)/x if x>0
    {pi/2 if x=0
    {pi + arctan(y)/x if x<0.

    3. The attempt at a solutionIt seems to me that if arctan(y) is bounded between -pi/2 and pi/2, then by taking the limits as x goes to "positive" zero and infinity, we see that θ is bounded, by the first definition, between 0 and infinity.

    θ = pi/2 if x=0 simply includes all points along the y-axis.

    Similarly, by taking limits, we find that, by the third definition, is bounded between pi and negative infinity. Since (x,y) necessarily lies on D, we need not worry that (x,y) is on the non-positive x-axis.

    So what we have here is a θ capable of ranging from negative infinity to infinity. If we choose an arbitrary r, then our range of θ will allow us to reach any (x,y).

    Personally, I think that this is either wrong or that there is more convincing way of doing the problem.
     
  2. jcsd
  3. Nov 24, 2011 #2

    vela

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    You seem to have made numerous typos. It shouldn't be arctan(y)/x. It should be arctan(y/x). Additionally, is θ suppose to equal pi/2 regardless of the sign of y?
     
  4. Nov 24, 2011 #3
    I think part of the issue here is that the problem, for me, was ambiguously worded. All it said was: theta equals

    {arctan y/x if x>0
    {pi/2 if x=0
    {pi + arctan y/x if x<0.

    I interpreted this as arctan (y)/x, but I can see how arctan (x/y) could also be an interpretation. In any case, I still don't really know how to begin the problem, especially since my original interpretation is, apparently, incorrect. There is no mention of y-signs in the problem.
     
  5. Nov 24, 2011 #4

    vela

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    I think the correct definition of [itex]\theta[/itex] should be
    [tex]\theta = \begin{cases}
    \arctan (y/x) & \text{if }x>0 \\
    \arctan (y/x)+\pi & \text{if }x<0 \\
    \pi/2 & \text{if } x=0 \text{ and } y>0 \\
    -\pi/2 & \text{if } x=0 \text{ and } y<0
    \end{cases}[/tex]The reason I think you need to consider the sign of y when x=0 is because r is typically assumed to be positive. If r>0 and θ=+π/2 when x=0, there's no way to express a point like (0,-1) in polar coordinates.
     
  6. Nov 24, 2011 #5
    Okay, I think that makes sense. I guess if you let r be negative, then it could no longer have the interpretation as the radius. In any case, could you please tell how to begin to show that theta can be selected in this way? I don't think my method of taking limits was correct or legitimate.
     
  7. Nov 24, 2011 #6
    I think I have a counter-example to this problem. I will show that F is not conservative on the closed path r(t) = <cos(t) + 2, sin(t) + 2>. This path lies entirely in the first quadrant, and therefore meets the condition that the path does not pass over the non-positive x-axis. I will show that that the integral over r(t) of F(x(t),y(t)) dot r'(t) dt does not equal zero.

    r'(t) = <-sin(t),cos(t)> and F(x(t),y(t)) = <-sin(t),cos(t)>.

    Taking the dot product of the simply gives 1.

    The integral from 0 to 2*pi of 1 dt = 2*pi.

    Since 2*pi does not equal zero, F is not conservative, even when the non-positive x-axis is excluded.

    Gah! I don't get this problem at all!
     
  8. Nov 24, 2011 #7

    vela

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    Your expression for [itex]\vec{F}[/itex] should be [itex]\vec{F} = \langle -\frac{\sin \theta}{r},\frac{\cos \theta}{r}\rangle[/itex].
     
  9. Nov 24, 2011 #8
    But I'm not parameterizing F by x=r*cos(theta). I'm using x=cos(t). Why shouldn't this parameterization work properly?
     
  10. Nov 24, 2011 #9

    vela

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    I'm not sure exactly what you're supposed to show. You probably need to show that for any (x,y) in D, you can calculate θ and that r cos θ equals x and r sin θ equals y with an appropriately chosen r.
     
  11. Nov 24, 2011 #10

    vela

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    You have
    \begin{align*}
    x(t) &= 2+\cos t = r(t) \cos\theta(t) \\
    y(t) &= 2+\sin t = r(t) \sin\theta(t)
    \end{align*}The parameter t isn't the same as θ.
     
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