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Stuck on this problem

  1. Aug 24, 2009 #1
    1. The problem statement, all variables and given/known data
    problem.jpg

    Not sure how to approach this one, does anyone have an idea?

    BTW, in the top line it should read xz-plane, not xy.
     
  2. jcsd
  3. Aug 24, 2009 #2

    tiny-tim

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    Hi diffusion! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    Forget all the stuff about the cylinder …

    the question is simply to prove that if you rotate z = e-x2 about the z-axis, you get z = e-x2-y2

    that's easy, isn't it? :wink:
     
  4. Aug 24, 2009 #3
    I'm sure it's easy once I've been set in the right direction. I can see that z = e-(x2-y2) by graphing it, I just don't know how to show mathematically that it does.
     
  5. Aug 24, 2009 #4

    tiny-tim

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    ok … what does "rotate z = f(x) about the z-axis" mean? … how would you express the instruction as an equation? :smile:
     
  6. Aug 24, 2009 #5
    Well you will rotate the function around the z-axis, so that z = f(x) becomes z = f(x,y). Not sure how to express it as an equation, besides telling you that the equation is a function of both x and y.
     
  7. Aug 24, 2009 #6

    tiny-tim

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    Try putting it in words first …

    what happens to each individual point on the original curve?​
     
  8. Aug 24, 2009 #7
    Produces a circle when rotated.
     
  9. Aug 24, 2009 #8

    tiny-tim

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    Yup! And the equation of a circle is … ? :smile:
     
  10. Aug 24, 2009 #9
    Yep, I recognized this before I posted the question. Again I just don't know how to show it, or justify it. I suppose I would say something like

    "Since each point on the graph z = e-r2 will generate a circle when rotated about the z-axis, and the equation for a circle is r2 = x2 + y2, we can make this substitution into our equation for r2, giving us z = e-(x2+y2)."

    I don't know, somehow this justification seems vague and inadequate.
     
  11. Aug 24, 2009 #10

    tiny-tim

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    ok, let's formalise that …

    the point (x0,0,z0) generates the circle (x,y,z), where … ? :smile:
     
  12. Aug 24, 2009 #11
    At z0? Sorry, not sure if I really understand what you're asking. Are you asking for a point, the equation of the particular circle generated, or something else?
     
  13. Aug 24, 2009 #12

    tiny-tim

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    I'm asking for the x,y,z equation(s) of the circle generated by the initial point (x0,0,z0). :smile:
     
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