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 ) &gt; 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.

Question

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  )  &amp;gt;  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&#039;s formula, but I wasn&#039;t able to solve this problem.

Steven Bates · Accepted Answer

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  )