Is 1 + e^(-x^2+y^2) a Two-Sheeted Hyperboloid?

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Homework Help Overview

The discussion revolves around the expression z = 1 + e^(-x^2 - y^2) and whether it represents a two-sheeted hyperboloid. Participants clarify the nature of the expression and explore its geometric implications.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the nature of the expression and its representation as a surface. Questions arise about cylindrical coordinates and how to visualize the surface. There are attempts to understand the behavior of z as x and y change, and whether plotting points would aid in comprehension.

Discussion Status

Participants are actively engaging with the problem, offering insights into the expression's behavior and discussing visualization strategies. Some guidance has been provided regarding plotting points and interpreting the surface, but there is no explicit consensus on the nature of the surface.

Contextual Notes

Some participants express confusion about terms like cylindrical coordinates and the lack of examples in their notes or texts, which contributes to their difficulty in visualizing the surface.

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

thanks
 
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It's just an expression so far. Do you mean z=1+e^(-x^2+y^2), that's a surface.
 
sorry, yes. and I made a typo--it's z= 1+ e^(-x^2-y^2).
 
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?
 
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?
 
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?
 
I can't quite see it yet. why wouldn't you just let r=-x^2-y^2?
 
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.
 
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
'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
and now comes the inane question about how to plot some points. just pick random x's and y's?
 
  • #12
oh wait, I just need to choose a z and use logarithms to find y in terms of x. correct?
 
  • #13
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
imsoconfused said:
oh wait, I just need to choose a z and use logarithms to find y in terms of x. correct?

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
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
Dick said:
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.

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
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
imsoconfused said:
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!

I tremble to think. :)
 
  • #19
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
Yes, an almost flat sheet at infinity with a hump at the origin.
 

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