2

I've been trying to paramaterize

$$x^2+y^2+z^2=9,\ x^2-y^2=3$$

but haven't had any luck. I was thinking to let $x=\sqrt{3}\ \sec(t), y=\sqrt{3}\ \tan(t)$ to complete the identity on the right equation, but when going back to the left one and solving for $z$, I get a nonreal result.

HK Lee
  • 19,964
James Snyder
  • 307
  • 3
  • 12
  • This is the intersection of a sphere with a hyperbolic cylinder. The intersection will be two disjoint "loops" on the sphere. Perfectly good coordinates on a sphere are spherical coordinates ... – Eric Towers Jan 27 '14 at 03:02
  • Are you trying to parametrize each equation on its own or their intersection? – Mhenni Benghorbal Jan 27 '14 at 03:05
  • I am not as familiar with spherical coordinates. The overall problem is to find the equation of an oscillating plane, but I would like them in terms of parameter t. – James Snyder Jan 27 '14 at 03:06
  • And I'm trying to do it on their intersection I assume (so that the x and y satisfy both equations) – James Snyder Jan 27 '14 at 03:06
  • The intersection "curve" has two disjoint pieces, so you can't parameterize the whole thing. The solutions already given will allow you to parameterize either one piece or the other, and that's probably all you need. – bubba Jan 27 '14 at 06:43

1 Answers1

3

The sphere equation becomes

$$ z^2 \ = \ 9 \ - \ 3 \sec^2 t \ - \ 3 \tan^2 t $$

$$ = \ 9 \ - \ 3 \ (\tan^2 t \ + \ 1 ) - \ 3 \tan^2 t \ = \ 6 \ - \ 6 \tan^2 t \ . $$

Doesn't seem like there's anything problematic there, since $ \ |z| \ \le \ 3 \ $ ; the upper bound on $ \ z^2 \ $ of 6 represents the greatest "altitude" from the $ \ xy-$ plane where the surfaces intersect. There will be two square-roots, corresponding to locations on the two hemispheres about the $ \ xy-$ plane. (I'm adding a picture of the surfaces.)

enter image description here

colormegone
  • 10,842