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Domain, why are they both not defined?

  1. Dec 17, 2007 #1
    I'm currently self-studying ODE, text by Morris Tenenbaum, and I am confused by these 2 problems.

    [tex]z=\sqrt{-(x^2 + y^2)}[/tex]

    D: (0,0)

    [tex]z=\sqrt{-(x^2 + y^2 +1)}[/tex]

  2. jcsd
  3. Dec 17, 2007 #2
    the second one obviously, the first one no clue
  4. Dec 17, 2007 #3
    I assume that x and y are real valued as is the value z. So, for the first equation, you can only have x=0 and y=0 as domain because you would end up with [tex]\sqrt{0}=0[/tex] which is a real value. In all other cases you would have a square root of a negative number giving no solution for the real. For the second one it is always negative due to the squares and the adding of 1.

    I assume this is correct, however I have a feeling there is a a catch to it...
  5. Dec 17, 2007 #4


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    No, coomast, you are correct. These are obviously assumed to be real valued functions of real numbers.

    In [itex]z= \sqrt{-(x^2+ y^2)}[/itex], if x= y= 0, then z= 0. If either x or y is non-zero, then [itex]x^2+ y^2[/itex] is positive and so [itex]-(x^2+ y^2)[/itex] is negative. We cannot take the square root of a negative number so z is defined only for x= y= 0.

    In [itex]z= \sqrt{-x^2+ y^2+ 1)}[/itex], if x= y= 0 then [itex]z= \sqrt{-1}[/itex] which is undefined for real numbers. If either x or y is non-zero, then [itex]x^2+ y^2+ 1[/itex] is larger than 1 so [itex]-(x^2+ y^2+ 1)[/itex] is less than -1, still negative and z is not defined for any x or y.
  6. Dec 17, 2007 #5
    o wow i misread the problem
  7. Dec 17, 2007 #6
    AH!!! I get it.

    The first one, the only time it is defined is if it's restricted to (0,0).

    For the second, since x^2 and y^2 will always give positive values, thus -(x^2 + y^2 + 1) will never be 0.

    I appreciate the help, but be prepared for more! Lol, I'm on winter-break so I don't have a teacher to go to :-]
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