anonymous
  • anonymous
Parametrize paraboloid? z = x^2 + (y^2)/9 - 9 How to I progress parametrizing it? I know how to parametrize 2D curves, but this is 3D; What's different?
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
the answer will be infinity
anonymous
  • anonymous
That's not what I asked
anonymous
  • anonymous
i know i am just working through it , i said so i ll know if u have the right answer or not

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anonymous
  • anonymous
x = u , y = v, z = u^2 + v^2/9 -9 < u , v , u^2 + v^2/9 -9 >
IrishBoy123
  • IrishBoy123
why are you param'ing this? because, if is this is calculus with surface integrals and the like in mind, then you might wish to consider upping your game and going for polar-cylindrical what you have here is: \(z = x^2 + \frac{1}{9}y^2 - 9\) or: \(z + 9= x^2 + \frac{1}{9}y^2\) that bit on the RHS is an ellipse. let's say we set z = 0, then we have \(1= \frac{x^2}{3^2} + \frac{y^2}{9^2}\) compared to standard ellipse formulation: \(\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = 1 \) i will stop here, unless you think this is relevant :p
anonymous
  • anonymous
The way I understand this, I am supposed to find this; r(t) = [x(t), y(t), z(t)] (still working on this - understanding this takes time.. ! ) :)
IrishBoy123
  • IrishBoy123
i'd start with the ellipse part of \[z = x^2 + \frac{y^2}{9} - 9 \implies z +9= x^2 + (\frac{y}{3})^2 \] and look at this bit \(x^2 + \frac{y^2}{9}\) where you can say \(x = r \cos t, y = 3 r \sin t\) so \(z+9 = r^2, z = r^2 - 9\) from that you can say \(\vec r (t) = \)

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