# Is this a hyperboloid?

1. Sep 30, 2008

### imsoconfused

Is 1 + e^(-x^2+y^2) a two-sheeted hyperboloid?

thanks

2. Sep 30, 2008

### Dick

It's just an expression so far. Do you mean z=1+e^(-x^2+y^2), that's a surface.

3. Sep 30, 2008

### imsoconfused

sorry, yes. and I made a typo--it's z= 1+ e^(-x^2-y^2).

4. Sep 30, 2008

### Dick

It's not a hyperboloid. A hyperboloid is a quadratic form. That's not, it has an e^ in it. And it doesn't have two sheets. In cylindrical coordinates it's z=1+e^(-r^2). Can you picture what that looks like?

5. Sep 30, 2008

### imsoconfused

I don't understand what you mean by cylindrical coordinates, I've never heard that term. I have trouble picturing things, too, until I start plotting points. Is that what I should do, just begin plotting points until I start to see it?

6. Sep 30, 2008

### Dick

Better to think about it a bit before you start plotting points. In z=1+e^(-r^2) I'm taking r=sqrt(x^2+y^2). So r is just the distance from (0,0) to (x,y) in the x-y plane. Can you picture it now?

7. Sep 30, 2008

### imsoconfused

I can't quite see it yet. why wouldn't you just let r=-x^2-y^2?

8. Sep 30, 2008

### Dick

Your choice. I usually like to pick r>=0. But look, at (x,y)=(0,0), z=2. As the distance of (x,y) from (0,0) gets larger and larger, -x^2-y^2 gets larger and larger in a negative way. So e^(-x^2-y^2) get closer to 0. So z->1 at infinity.

9. Sep 30, 2008

### imsoconfused

ok, I see how you get z=2, at least. I feel so stupid in that I only get about half of the second part and I absolutely cannot see what this thing looks like! I'm thinking of z as a level--is that wrong? looking at this from the xy plane, does it look like a function like (1/x)^2?

I have nothing in my notes about any surfaces that have e in them, nor are there any examples like this in the text--that's why I'm so lost.

10. Sep 30, 2008

### Dick

'z' is the height above (or below) the x-y plane for a given value of x and y. Maybe plotting some points isn't that bad an idea. Try some.

11. Sep 30, 2008

### imsoconfused

and now comes the inane question about how to plot some points. just pick random x's and y's?

12. Sep 30, 2008

### imsoconfused

oh wait, I just need to choose a z and use logarithms to find y in terms of x. correct?

13. Sep 30, 2008

### Dick

You are asking before you are thinking about it. How about (x,y)=(1,0),(2,0),(3,0) etc. Then (x,y)=(0,1),(0,2),(0,3)... Or (x,y)=(1,1),(2,2),(3,3).... You don't actually have to plot them all, just think about what would happen if you did.

14. Sep 30, 2008

### Dick

No, you actually have to think about what the surface would look like if you did plot a bunch of points. This isn't that hard.

15. Sep 30, 2008

### imsoconfused

when x and y grow larger, z approaches 1. that is why I thought it looked like (1/x)^2, because that decreases exponentially towards an asymptote. the difference is that this, instead of being a line is a surface. correct?

16. Sep 30, 2008

### imsoconfused

oh, I know it shouldn't be that hard, but I really appreciate your coaching me through this. just think where I'd be without you!

17. Sep 30, 2008

### Dick

What it really looks like is 1+e^(-x^2). But, yes, certainly, it decreases exponentially towards an asymptote. It doesn't have two sheets and it's not a hyperboloid.

18. Sep 30, 2008

### Dick

I tremble to think. :)

19. Sep 30, 2008

### imsoconfused

ok. so if I were to draw this thing, it would basically look like a flat plane with a hump in it near the origin? that's what brought on this question; I'm supposed to be drawing it.

20. Oct 1, 2008

### Dick

Yes, an almost flat sheet at infinity with a hump at the origin.