# Vector equation of a spherical curve

1. Feb 10, 2008

### BilalX

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. Feb 10, 2008

### John Creighto

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.

3. Feb 10, 2008

### BilalX

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.

Thanks-

4. Jan 10, 2009

### Gumshoe

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 .