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Vector equation of a spherical curve

  1. Feb 10, 2008 #1
    Given a curve described by the following function:
    r(t) = (cos^2(t), sin(t), sin(t)*cos(t)), 0 ≤ t ≤ 2*Pi

    How can I prove it describes a spherical shape? I know that the parametric representation is the following, but I'm not sure how to reconcile that with the expression of a sphere.

    x = cos^2(t)
    y = sin(t)
    z = sin(t)*cos(t)

    I'd greatly appreciate any insight, thanks.
  2. jcsd
  3. Feb 10, 2008 #2
    Well, for a sphere the x^2+y^2+z^2=r^2
    I presume in your case r is equal to one. Plug in the values for x, y,z above and apply trigonometric identities. That said, I don't think a curve can fill an area without appealing to the axiom of choice. Thus perhaps your curve is on the surface of a sphere but I don't think it is a sphere.
  4. Feb 10, 2008 #3
    Oh, right, I just didn't think of assuming a unit sphere centered at the origin - much easier now. And yeah, the curve just follows a spherical surface.

  5. Jan 10, 2009 #4
    You can construct space-filling curves without the use of Choice. For example, see the construction given at http://en.wikipedia.org/wiki/Space-filling_curve .

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