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Topic: tangent plane to sphere - how to prove this?
Replies: 3   Last Post: May 22, 2014 1:01 AM

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Roland Franzius

Posts: 433
Registered: 12/7/04
Re: tangent plane to sphere - how to prove this?
Posted: May 22, 2014 1:01 AM
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Am 22.05.2014 00:05, schrieb namenobodywants@gmail.com:
> On Wednesday, May 21, 2014 2:03:54 PM UTC-7, Ken Pledger wrote:
>> In article
>> <853956ce-23f8-4285-91f4-ee89f26d651d@googlegroups.com>,
>>
>> namenobodywants@gmail.com wrote:
>>
>>
>>

>>> ....
>>
>>> intuitively something like this seems to be the case: if plane T
>>> is tangent

>>
>>> to sphere S at point P then "T approximates S arbitrarily well in
>>> a

>>
>>> sufficiently small neighborhood of P". apparently what is needed
>>> here is a

>>
>>> one-one function f that maps points of S close to P to points of
>>> T close to

>>
>>> P, together with a proof that the ratio (distance on S from A to
>>> B) /

>>
>>> (distance on T from f(A) to f(B)) approaches unity for any A,B
>>> sufficiently

>>
>>> close to P. the two obvious choices for f(A) are (1) the
>>> intersection with T

>>
>>> with the line through A perpendicular to T, and (2) the
>>> intersection of T

>>
>>> with the line through A and the center of S....
>>
>>
>>
>>
>>
>> Let the sphere have centre O and radius r. Consider any point X
>> on
>>
>> the sphere, close to P. Cut the figure by the plane OPX to get a
>> great
>>
>> circle and its tangent at P. If the small angle POX is theta, then
>> the
>>
>> arc PX on the circle is r(theta).
>>
>>
>>
>> Can you see how your two proposed constructions give lengths
>>
>> r.sin(theta) and r.tan(theta) along the tangent? The ratios you
>> want
>>
>> are then (sin(theta))/theta and (tan(theta))/theta, each of
>> which
>>
>> tends to 1 as theta tends to zero.
>>
>>
>>
>> Ken Pledger.

>
>
> that's great, but the idea is to prove that the ratio of distances
> (straight line segment over great circle arc) between ANY two points
> sufficiently close to P is arbitrarily close to unity; your argument
> only addresses the case where one of the two points is P itself
>
>



Linearisation is linear. This means that retaining only linear terms in
the Taylor expansion at a point generates vector space of linear
functionals independent of the coordinates choosen.

If one approximates the points of a smooth 2-dimensional surface

p(u,v) = {x(u,v),y(uv))

then the differential of the point p gives the form

dp = {dx(u,v),dy(u,v)} = { dx/du + dx/dv dv , dy/du du + dy/dv dv }
= {dx/du, dy/du} du + { dx/dv, dy/dv } dv

= (d(xy)/d(uv)(P) . { du , dv}(P)

This primitive application of the multidimensional chainrule (linear is
linear is linear) shows that the space of differentials at the point p
is a linear vectorspace, that transforms linearly by the Jacobi matrix
(d(xy)/d(uv).

The linear combination of 1-forms alpha du + beta dv project any
tangent vector onto the component of the vector space tangent to point P
to the surface.

The difference of a point in the surface to its projection onto the
tangent plane by linearisation of the coordinate functions x(u,v),
y(u,v) is consequently an at least quadratic form over the tangent space
tangent to the point of expansion.

The quadratic form is an invariant geometric object describing a
quadratic surface as the next approxiamation after the linear tangent at
the point of approximations.

By duality, the tangents d/du, d/dv with

du(d/du) = 1, du(d/dv) = 0
dv(d/du) = 0, dv(d/dv) = 1

etc, make up a base of the tangent vector space consisting of the
velocity vectors of all curves tangent to the surface at the expansion
point.

--

Roland Franzius



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