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Continuity and Polar Coordinates

  1. Jul 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Show that the function f(x,y)= xy/sqrt(x^2+y^2) is continuous at the origin using polar coordinates. f(x,y)=0 if (x,y)=(0,0)


    2. Relevant equations
    r=sqrt(x^2+y^2)
    x=rcos(theta)
    y=rsin(theta)


    3. The attempt at a solution
    So, converting this equation to polar coordinates, I get rsin(theta)cos(theta). However, after this I'm stumped as to how I prove that this is continuous at the origin.
     
  2. jcsd
  3. Jul 20, 2010 #2

    rock.freak667

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    At the origin in polar coordinates (r,θ), shouldn't r=0 and θ=0 ?
     
  4. Jul 21, 2010 #3

    HallsofIvy

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    Staff Emeritus
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    Not exactly- at the origin r= 0 and [itex]\theta[/itex] is undefined.

    However, Schmidt7100, the point is that r alone measures the distance of a point from the origin- [itex]\theta[/itex] is irrelevant. That means that, if [itex]\lim_{r\to 0}f(r,\theta)[/itex] does not depend on [itex]\theta[/itex], then that is the limit of [itex]f(r, \theta)[/itex] as [itex](r, \theta)[/itex] goes to the origin.

    Note that is NOT the case for Cartesian coordinates. Since the distance from (x, y) to (0, 0) depends upon both x and y. I am sure you have seen examples of functions in (x, y) that give different limits as you go to (0, 0) along different paths.
     
  5. Jul 21, 2010 #4

    hunt_mat

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    You don't have to use polar co-ordinates, let
    [tex]
    x_{n}=\frac{1}{n},\quad y_{n}=\frac{1}{n}
    [/tex]
    Insert these into your equation for f and let [tex]n\rightarrow\infty[/tex], compute the limit and if it's the same as f(0,0) you've proved continuity.
     
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