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FibonacciMariachi Group Title

Find a parametrization of the surface Σ. Σ is the part of a cylinder x^2+y^2=1 that lies between the planes z=-1 and z=1.

  • 2 years ago
  • 2 years ago

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  1. henpen Group Title
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    |dw:1352924458586:dw|

    • 2 years ago
  2. henpen Group Title
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    I'm not sure what you mean by parametrise

    • 2 years ago
  3. FibonacciMariachi Group Title
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    Meaning put it into parametric form. <x,y,z>

    • 2 years ago
  4. henpen Group Title
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    OK. Forget about the z for a minute \[f(t)=cost \hat{x}+sint \hat{y} \] Is the circle

    • 2 years ago
  5. henpen Group Title
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    \[0 <t<2 \pi\]

    • 2 years ago
  6. FibonacciMariachi Group Title
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    Gotcha. Im mostly fine with finding. x=rcos(theta) y=rsin(theta). Im not sure how to find z though.

    • 2 years ago
  7. henpen Group Title
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    Now, to include the z co-ordinate:\[cost \hat{x}+sint \hat{x}+(\frac{t}{\pi}-1)\hat{z}\] \[ 0<t<2 \pi \]I basically twisted the z thing to get something that linearly went from z=-1 at t=0 to z=1 at t=2 pi

    • 2 years ago
  8. FibonacciMariachi Group Title
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    Hmmm, not 100% sure how you go to that point. Could you break it down for me?

    • 2 years ago
  9. FibonacciMariachi Group Title
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    Mainly, where did t/pi-1 come from? And what goes into t? The z coordinate says the answer is simply 'zk'

    • 2 years ago
  10. henpen Group Title
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    Because \[t=0, (\frac{0}{\pi}-1))=-1\] \[t=0, (\frac{2 \pi}{\pi}-1))=1\]

    • 2 years ago
  11. FibonacciMariachi Group Title
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    Hmmm I think I get it a little more now. :/ Still really murky on it though. the way I learned it was to just plug your parametrized x and y back into an equation with z in it and you find your z that way. Ehhh.

    • 2 years ago
  12. FibonacciMariachi Group Title
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    I really don't think I even understand the basic concept of this section to be honest.

    • 2 years ago
  13. henpen Group Title
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    It certainty doesn't help that I was incorrect. What I actually coded for was|dw:1352927439044:dw|

    • 2 years ago
  14. henpen Group Title
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    That is, a spiral rather than a full curve

    • 2 years ago
  15. henpen Group Title
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    http://tutorial.math.lamar.edu/Classes/CalcIII/ParametricSurfaces.aspx There's cylinder in there. I'll explain if more help needed, sorry for mistake earlier

    • 2 years ago
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