Domain, why are they both not defined?

[rocomath]

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

z=\sqrt{-(x^2 + y^2)}

D: (0,0)

z=\sqrt{-(x^2 + y^2 +1)}

undefined

 

[ice109]

the second one obviously, the first one no clue

 

[coomast]

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 \sqrt{0}=0 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...

 

[HallsofIvy]

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

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

In z= \sqrt{-x^2+ y^2+ 1)}, if x= y= 0 then z= \sqrt{-1} which is undefined for real numbers. If either x or y is non-zero, then x^2+ y^2+ 1 is larger than 1 so -(x^2+ y^2+ 1) is less than -1, still negative and z is not defined for any x or y.

 

[ice109]

o wow i misread the problem

 

[rocomath]

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 :-]