Awainaideannagi

2022-07-18

A parametrization of a circle by arc length may be written as
$\gamma \left(t\right)=c+r\mathrm{cos}\left(t/r\right){e}_{1}+r\mathrm{sin}\left(t/r\right){e}_{2}.$
Suppose $\beta$ is an unit speed curve such that its curvature $\kappa$ satisfies $\kappa \left(0\right)>0$.
How to show there is one, and only one, circle which approximates $\beta$ in near $t=0$ in the sense
$\gamma \left(0\right)=\beta \left(0\right),{\gamma }^{{}^{\prime }}\left(0\right)={\beta }^{{}^{\prime }}\left(0\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\gamma }^{{}^{″}}\left(0\right)={\beta }^{{}^{″}}\left(0\right).$
I suppose we must use Taylor's formula, but I wasn't able to solve this problem.

Steven Bates

Expert

The parametrization of circle you stated assumes that ${\gamma }^{\prime }\left(t\right)$ is parallel to ${e}_{2}$ at time $t=0$. So, if ${\beta }^{\prime }\left(0\right)$ is not parallel to ${e}_{2}$, you have a problem. This nuisance can be avoided by writing
$\begin{array}{}\text{(1)}& \gamma \left(t\right)=c+r\mathrm{cos}\left(\left(t+a\right)/r\right){e}_{1}+r\mathrm{sin}\left(\left(t+a\right)/r\right){e}_{2}\end{array}$
with constant $a\in \mathbb{R}$. The Taylor formula says
$\gamma \left(t\right)=\left(c+r\mathrm{cos}\left(a/r\right){e}_{1}+r\mathrm{sin}\left(a/r\right){e}_{2}\right)+\left(-\mathrm{sin}\left(a/r\right){e}_{1}+\mathrm{cos}\left(a/r\right){e}_{2}\right)t+\left(-\mathrm{cos}\left(a/r\right){e}_{1}-\mathrm{sin}\left(a/r\right){e}_{2}\right){r}^{-1}{t}^{2}+O\left({t}^{3}\right)$
Comparing the above to
$\beta \left(t\right)=\beta \left(0\right)+{\beta }^{\prime }\left(0\right)t+\frac{1}{2}{\beta }^{″}\left(0\right){t}^{2}+O\left({t}^{3}\right)$
we observe that
${r}^{-1}=\frac{1}{2}|{\beta }^{″}\left(0\right)|$, so $r$ is determined. (Unless ${\beta }^{″}\left(0\right)=0$, when the radius of curvature is infinite and there is no circle you are looking for.)
Since ${\beta }^{\prime }\left(0\right)$ is a unit vector, there exists a unique $a$ (up to an integer multiple of $2\pi r$) such that ${\beta }^{\prime }\left(0\right)=-\mathrm{sin}\left(a/r\right){e}_{1}+\mathrm{cos}\left(a/r\right){e}_{2}$.
c is uniquely determined from $\left(c+r\mathrm{cos}\left(a/r\right){e}_{1}+r\mathrm{sin}\left(a/r\right){e}_{2}\right)=\beta \left(0\right)$

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