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Homework Help: Is this a hyperboloid?

  1. Sep 30, 2008 #1
    Is 1 + e^(-x^2+y^2) a two-sheeted hyperboloid?

    thanks
     
  2. jcsd
  3. Sep 30, 2008 #2

    Dick

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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.
     
  4. Sep 30, 2008 #3
    sorry, yes. and I made a typo--it's z= 1+ e^(-x^2-y^2).
     
  5. Sep 30, 2008 #4

    Dick

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    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?
     
  6. Sep 30, 2008 #5
    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?
     
  7. Sep 30, 2008 #6

    Dick

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    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?
     
  8. Sep 30, 2008 #7
    I can't quite see it yet. why wouldn't you just let r=-x^2-y^2?
     
  9. Sep 30, 2008 #8

    Dick

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    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.
     
  10. Sep 30, 2008 #9
    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.
     
  11. Sep 30, 2008 #10

    Dick

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    '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.
     
  12. Sep 30, 2008 #11
    and now comes the inane question about how to plot some points. just pick random x's and y's?
     
  13. Sep 30, 2008 #12
    oh wait, I just need to choose a z and use logarithms to find y in terms of x. correct?
     
  14. Sep 30, 2008 #13

    Dick

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    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.
     
  15. Sep 30, 2008 #14

    Dick

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    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.
     
  16. Sep 30, 2008 #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?
     
  17. Sep 30, 2008 #16
    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!
     
  18. Sep 30, 2008 #17

    Dick

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    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.
     
  19. Sep 30, 2008 #18

    Dick

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    I tremble to think. :)
     
  20. Sep 30, 2008 #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.
     
  21. Oct 1, 2008 #20

    Dick

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