A parametrization of a circle by arc length may be written as γ ( t...

Awainaideannagi

Awainaideannagi

Answered

2022-07-18

A parametrization of a circle by arc length may be written as
γ ( t ) = c + r cos ( t / r ) e 1 + r sin ( t / r ) e 2 .
Suppose β is an unit speed curve such that its curvature κ satisfies κ ( 0 ) > 0.
How to show there is one, and only one, circle which approximates β in near t = 0 in the sense
γ ( 0 ) = β ( 0 ) , γ ( 0 ) = β ( 0 ) and γ ( 0 ) = β ( 0 ) .
I suppose we must use Taylor's formula, but I wasn't able to solve this problem.

Answer & Explanation

Steven Bates

Steven Bates

Expert

2022-07-19Added 15 answers

The parametrization of circle you stated assumes that γ ( t ) is parallel to e 2 at time t = 0. So, if β ( 0 ) is not parallel to e 2 , you have a problem. This nuisance can be avoided by writing
(1) γ ( t ) = c + r cos ( ( t + a ) / r ) e 1 + r sin ( ( t + a ) / r ) e 2
with constant a R . The Taylor formula says
γ ( t ) = ( c + r cos ( a / r ) e 1 + r sin ( a / r ) e 2 ) + ( sin ( a / r ) e 1 + cos ( a / r ) e 2 ) t + ( cos ( a / r ) e 1 sin ( a / r ) e 2 ) r 1 t 2 + O ( t 3 )
Comparing the above to
β ( t ) = β ( 0 ) + β ( 0 ) t + 1 2 β ( 0 ) t 2 + O ( t 3 )
we observe that
r 1 = 1 2 | β ( 0 ) | , so r is determined. (Unless β ( 0 ) = 0, when the radius of curvature is infinite and there is no circle you are looking for.)
Since β ( 0 ) is a unit vector, there exists a unique a (up to an integer multiple of 2 π r) such that β ( 0 ) = sin ( a / r ) e 1 + cos ( a / r ) e 2 .
c is uniquely determined from ( c + r cos ( a / r ) e 1 + r sin ( a / r ) e 2 ) = β ( 0 )

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