Question

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?

Answer 1

the answer will be infinity

Answer 2

That's not what I asked

Answer 3

i know i am just working through it , i said so i ll know if u have the right answer or not

Answer 4

x = u , y = v, z = u^2 + v^2/9 -9
< u , v , u^2 + v^2/9 -9 >

Answer 5

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

Answer 6

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.. ! ) :)

Answer 7

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 \(\vec r (t) = <r \cos t, 3r \sin t , r^2 - 9>\)