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Continuity calc help

  1. Mar 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Let f(x,y) = { 2 if [tex]x^{2}[/tex]+[tex]y^{2}[/tex] < 1 , and 0 otherwise

    Using the definition of continuity to show that:

    (a) f is not continuous at each point ([tex]x_{0}[/tex],[tex]y_{0}[/tex]) such that [tex]x^{2}_{0}[/tex] = [tex]y^{2}_{0}[/tex] = 1

    (b) f is continuous at all other points ([tex]x_{0}[/tex],[tex]y_{0}[/tex]) in the plane

    2. Relevant equations

    None.

    3. The attempt at a solution

    I know the definition of continuity, but are you supposed to take some epsilon for the case in (a) and (b), and just show the normal definition or is the question asking something I'm not seeing.
     
  2. jcsd
  3. Mar 12, 2009 #2
    Re: Continuity

    I'm assuming you mean in a) that x_0^2 + y_0^2 = 1.

    Well, think about what the function is. You have f(x) = 2 inside the unit circle and 0 elsewhere. The definition of continuity (that is, the English interpretation of delta-epsilon) is that a function f is continuous at x if, given an arbitrary number epsilon > 0, you can find a ball around x such that all numbers in the ball have value which differs from the value of f at x by less than epsilon. So since strictly outside the ball it's 0 and inside the ball it's 2, it's obvious that we can always find a ball in which all values are arbitrarily close to 2 or 0 (since the function is constant, and the sets are open, so just take any ball contained in these respective sets). But why does this break down at x_0^2 + y_0^2 = 1? In fact, what IS the set x_0^2 + y_0^2 = 1? Think about it.
     
  4. Mar 12, 2009 #3
    Re: Continuity

    Well it is the boundary of the unit circle. On the boundary f(x) = 0, is that why it isn't continuous??
     
  5. Mar 12, 2009 #4
    Re: Continuity

    Interesting definition! :surprised

    Edit: Do we really need a ball with radius epsilon here phreak? if z=f(x,y) isn't z supposed to be linear here? so basically if

    [tex]\lim_{(x,y)\rightarrow\ (a,b)}f(x,y)=f(a,b)[/tex]

    Then this means that:

    [tex]\forall \epsilon>0,\exists D_{\delta}[/tex] such that (D-a circle with radius delta)


    [tex]|f(x,y)-f(a,b)|<\epsilon[/tex] whenever

    [tex] \sqrt{(x-a)^2+(y-b)^2}<\delta[/tex]

    Here it means that anytime the points (x,y) are within the circle centered at (a,b) with radius delta, the values of the function z=f(x,y) will lie between the parallel planes

    [tex]z=f(a,b)-\epsilon, z=f(a,b)+\epsilon[/tex]
     
    Last edited: Mar 12, 2009
  6. Mar 12, 2009 #5
    Re: Continuity

    O wait, for (a) is it not continuous because taking some ball around the boundary you get two different values for the function in that one ball, and given some epsilon the values in the ball might differ from x by a value greater than epsilon. Does that have any relevance??
     
  7. Mar 12, 2009 #6
    Re: Continuity

    You're definitely on the right track. The value of each boundary point is 0. So by definition, if f is continuous at any particular boundary point, we should be able to find a circle (in R^2) around it so that the value of each point in the circle differs from 0 by an arbitrarily small amount (which is what epsilon represents). For instance, take epsilon = 1/2. Can we find a circle around a boundary point such that all points in the circle have values between -1/2 and 1/2?
     
  8. Mar 12, 2009 #7
    Re: Continuity

    And the answer would be no, because all points in the circle are strictly 2. Thanks for the help. Didn't even think to look at open balls in a plane. I'll use the same notions to show (b) as well, thanks.
     
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