# Domain, why are they both not defined?

1. Dec 17, 2007

### 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

2. Dec 17, 2007

### ice109

the second one obviously, the first one no clue

3. Dec 17, 2007

### 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...

4. Dec 17, 2007

### 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.

5. Dec 17, 2007

### ice109

o wow i misread the problem

6. Dec 17, 2007

### 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 :-]