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anonymous
 3 years ago
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.
anonymous
 3 years ago
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.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352924458586:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm not sure what you mean by parametrise

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Meaning put it into parametric form. <x,y,z>

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0OK. Forget about the z for a minute \[f(t)=cost \hat{x}+sint \hat{y} \] Is the circle

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Gotcha. Im mostly fine with finding. x=rcos(theta) y=rsin(theta). Im not sure how to find z though.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now, to include the z coordinate:\[cost \hat{x}+sint \hat{x}+(\frac{t}{\pi}1)\hat{z}\] \[ 0<t<2 \pi \]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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hmmm, not 100% sure how you go to that point. Could you break it down for me?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Mainly, where did t/pi1 come from? And what goes into t? The z coordinate says the answer is simply 'zk'

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Because \[t=0, (\frac{0}{\pi}1))=1\] \[t=0, (\frac{2 \pi}{\pi}1))=1\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hmmm 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I really don't think I even understand the basic concept of this section to be honest.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It certainty doesn't help that I was incorrect. What I actually coded for wasdw:1352927439044:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That is, a spiral rather than a full curve

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://tutorial.math.lamar.edu/Classes/CalcIII/ParametricSurfaces.aspx There's cylinder in there. I'll explain if more help needed, sorry for mistake earlier
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