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FibonacciMariachi
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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.
 one year ago
 one year ago
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.
 one year ago
 one year ago

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henpen Group TitleBest ResponseYou've already chosen the best response.1
dw:1352924458586:dw
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
I'm not sure what you mean by parametrise
 one year ago

FibonacciMariachi Group TitleBest ResponseYou've already chosen the best response.0
Meaning put it into parametric form. <x,y,z>
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
OK. Forget about the z for a minute \[f(t)=cost \hat{x}+sint \hat{y} \] Is the circle
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
\[0 <t<2 \pi\]
 one year ago

FibonacciMariachi Group TitleBest ResponseYou've already chosen the best response.0
Gotcha. Im mostly fine with finding. x=rcos(theta) y=rsin(theta). Im not sure how to find z though.
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Now, 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
 one year ago

FibonacciMariachi Group TitleBest ResponseYou've already chosen the best response.0
Hmmm, not 100% sure how you go to that point. Could you break it down for me?
 one year ago

FibonacciMariachi Group TitleBest ResponseYou've already chosen the best response.0
Mainly, where did t/pi1 come from? And what goes into t? The z coordinate says the answer is simply 'zk'
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Because \[t=0, (\frac{0}{\pi}1))=1\] \[t=0, (\frac{2 \pi}{\pi}1))=1\]
 one year ago

FibonacciMariachi Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

FibonacciMariachi Group TitleBest ResponseYou've already chosen the best response.0
I really don't think I even understand the basic concept of this section to be honest.
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
It certainty doesn't help that I was incorrect. What I actually coded for wasdw:1352927439044:dw
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
That is, a spiral rather than a full curve
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago
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