Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]

    undefined
     
  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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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 :-]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Domain, why are they both not defined?
Loading...