Re: partial derivatives of a sphere

On Fri, 1 Sep 2006, vsgdp wrote:

x = r sin v cos u
y = r cos v
z = r sin v sin u

u in [0, 2pi]
v in [0, pi]

Now when I take dp/dv, the tangent vector will be tilted in the -y direction

How about attention to details? What's p? p = (x,y,z)?
What do you mean dp/dv? Do you mean the partial derivative p_v?
How about r? Is it a constant by any chance? Assuming maybe's:

p_u = (-r.sin v sin u, r cos v, r.sin v cos u)
p_v = (r.cos v cos u, -r.sin v, r.cos v sin u)

I would have been nice were you to present at least p_v.

because v (which comes from the spherical coordinates) is an angle measured
from +y down towards -y. So if we move a small increment dv, we are rolling
down the sphere so to speak.

However, I actually want the tangent vector to go the other way. For
example, if v = pi/2, I want dp/dv = (0, 1, 0). It seems to obtain this all
I need to do is negate dp/dv. Does this always work for every dp/dv on the
sphere?

p_v(u,pi/2) = (0, -r, 0)
p_v(u,-pi/2) = (0, r, 0)

I suppose another way to achieve my goal would be measure v from the -y up
towards +y and redo the spherical to rectangular coordinates based on this
to get a different parameterization of the sphere.

Are you clear about what you're thinking?
.

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