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anonymous

  • one year ago

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?

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  1. anonymous
    • one year ago
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    the answer will be infinity

  2. anonymous
    • one year ago
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    That's not what I asked

  3. anonymous
    • one year ago
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    i know i am just working through it , i said so i ll know if u have the right answer or not

  4. anonymous
    • one year ago
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    x = u , y = v, z = u^2 + v^2/9 -9 < u , v , u^2 + v^2/9 -9 >

  5. IrishBoy123
    • one year ago
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    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

  6. anonymous
    • one year ago
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    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.. ! ) :)

  7. IrishBoy123
    • one year ago
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    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 \(\vec r (t) = <r \cos t, 3r \sin t , r^2 - 9>\)

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