Dan
Goodman
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| Posted on Tuesday, 18
February, 2003 - 07:28 pm: |
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Yes it is, assuming you have a function f:R3® R with
the surface defined to be the set of points with f(x)=0. The proof is
not completely easy and uses the higher dimensional chain rule for
differentiation, but here goes.
Suppose you have a vector v in the tangent plane at a point
x on the surface. Draw a curve on the surface j(t) going
through x at t=0 with tangent vector v at
x, i.e. j ' (0)=v. Now, because j(t) is
a curve in the surface, you have that f(j(t))=0. So if we let
g(t)=f(j(t)) then g ' (t)=0. Using the chain rule we have that
0=g ' (0)=(Ñ f)(x).j ' (0)=(Ñ f)(x).v, which is what we wanted.
So, I've used the higher dimensional chain rule which says that if
f:R3® R and g:R® R3 and h:R® R defined by h(t)=f(g(t))
then h ' (t)=Ñf.g ' (t).
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David
Loeffler
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| Posted on Tuesday, 18
February, 2003 - 07:36 pm: |
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Yes. When you have a 2-D surface in 3-D space defined
by f(x,y,z)=0 where f is some (sufficiently smooth) function, then
the vector Ñf(x,y,z) is orthogonal to the surface at the point
(x,y,z).
There are various ways of looking at this. One is to use Taylor's theorem.
Let (x0,y0,z0) be a point on the surface, so f(x0,y0,z0)=0.
Then the higher-dimensional form of Taylor's theorem states that
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f(x0+r,y0+s,z0+t)=f(x0,y0,z0)+r |
¶f
¶x
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(x0, y0, z0)+s |
¶f
¶y
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(x0, y0, z0)+t |
¶ f
¶z
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(x0, y0, z0) |
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plus some terms of the order of r, s and t squared.
You can see that if (x0, y0, z0) is on the surface f(x,y,z)=0,
then the first term is zero. The second term you can recognise as
What does this mean? It means that locally the surface 'looks like'
the surface (x,y,z) . Ñf = constant. This is just a
plane with normal Ñf. So the normal at that one point
(x0, y0, z0) is parallel to Ñf at that point.
(This stuff is fairly hard, so if there's anything you don't
understand, please let me know and I'll explain it more
carefully.)
David
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