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Parametrization of a curve

  • Thread starter DougUTPhy
  • Start date
  • #1
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Homework Statement



I'm doing a line integral and can't seem to figure out the parametrization of this curve:
[itex]x^2+y^2+z=2\pi[/itex]



Homework Equations


Looking to get it to the form:
[itex]\textbf{c}(r,t)=(x(r,t),y(r,t),z(r,t))[/itex] (I don't even know if this is right though).


The Attempt at a Solution


Trying to use [itex]x=r \cos t[/itex] and [itex]y=r \sin t[/itex] but I still can't get anywhere.

I have a feeling I'm totally in the wrong direction.
The [itex]2\pi[/itex] is killing me too!
 
Last edited:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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Your basic problem is NOT the "[itex]2\pi[/itex]". It is that [itex]x^2+ y^2+ z= 2\pi[/itex] does NOT define a line (or curve or path) in three dimensions. It can be written as [itex]z= 2\pi- x^2- y^2[/itex] which is a surface (specifically, a paraboloid). Essentially, given any x and y you can solve for z so this is a two dimensional figure, not one dimension.

Please tell us what the entire problem really is.
 
  • #3
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I realized this after thinking about for a while, the real parametrizaion I can't figure out is a curve that is a corkscrew getting narrower as it goes up around the parabolioid, starting at [itex] (\sqrt{2\pi},0,0) [/itex] and ending at the top of the paraboloid, [itex] (0,0,2\pi) [/itex]
 

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