Awainaideannagi

Answered

2022-07-18

A parametrization of a circle by arc length may be written as

$\gamma (t)=c+r\mathrm{cos}(t/r){e}_{1}+r\mathrm{sin}(t/r){e}_{2}.$

Suppose $\beta $ is an unit speed curve such that its curvature $\kappa $ satisfies $\kappa (0)>0$.

How to show there is one, and only one, circle which approximates $\beta $ in near $t=0$ in the sense

$\gamma (0)=\beta (0),{\gamma}^{{}^{\prime}}(0)={\beta}^{{}^{\prime}}(0)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\gamma}^{{}^{\u2033}}(0)={\beta}^{{}^{\u2033}}(0).$

I suppose we must use Taylor's formula, but I wasn't able to solve this problem.

$\gamma (t)=c+r\mathrm{cos}(t/r){e}_{1}+r\mathrm{sin}(t/r){e}_{2}.$

Suppose $\beta $ is an unit speed curve such that its curvature $\kappa $ satisfies $\kappa (0)>0$.

How to show there is one, and only one, circle which approximates $\beta $ in near $t=0$ in the sense

$\gamma (0)=\beta (0),{\gamma}^{{}^{\prime}}(0)={\beta}^{{}^{\prime}}(0)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\gamma}^{{}^{\u2033}}(0)={\beta}^{{}^{\u2033}}(0).$

I suppose we must use Taylor's formula, but I wasn't able to solve this problem.

Answer & Explanation

Steven Bates

Expert

2022-07-19Added 15 answers

The parametrization of circle you stated assumes that ${\gamma}^{\prime}(t)$ is parallel to ${e}_{2}$ at time $t=0$. So, if ${\beta}^{\prime}(0)$ is not parallel to ${e}_{2}$, you have a problem. This nuisance can be avoided by writing

$\begin{array}{}\text{(1)}& \gamma (t)=c+r\mathrm{cos}((t+a)/r){e}_{1}+r\mathrm{sin}((t+a)/r){e}_{2}\end{array}$

with constant $a\in \mathbb{R}$. The Taylor formula says

$\gamma (t)={\textstyle (}c+r\mathrm{cos}(a/r){e}_{1}+r\mathrm{sin}(a/r){e}_{2}{\textstyle )}+{\textstyle (}-\mathrm{sin}(a/r){e}_{1}+\mathrm{cos}(a/r){e}_{2}{\textstyle )}t+{\textstyle (}-\mathrm{cos}(a/r){e}_{1}-\mathrm{sin}(a/r){e}_{2}{\textstyle )}{r}^{-1}{t}^{2}+O({t}^{3})$

Comparing the above to

$\beta (t)=\beta (0)+{\beta}^{\prime}(0)t+\frac{1}{2}{\beta}^{\u2033}(0){t}^{2}+O({t}^{3})$

we observe that

${r}^{-1}=\frac{1}{2}|{\beta}^{\u2033}(0)|$, so $r$ is determined. (Unless ${\beta}^{\u2033}(0)=0$, when the radius of curvature is infinite and there is no circle you are looking for.)

Since ${\beta}^{\prime}(0)$ is a unit vector, there exists a unique $a$ (up to an integer multiple of $2\pi r$) such that ${\beta}^{\prime}(0)=-\mathrm{sin}(a/r){e}_{1}+\mathrm{cos}(a/r){e}_{2}$.

c is uniquely determined from $(}c+r\mathrm{cos}(a/r){e}_{1}+r\mathrm{sin}(a/r){e}_{2}{\textstyle )}=\beta (0)$

$\begin{array}{}\text{(1)}& \gamma (t)=c+r\mathrm{cos}((t+a)/r){e}_{1}+r\mathrm{sin}((t+a)/r){e}_{2}\end{array}$

with constant $a\in \mathbb{R}$. The Taylor formula says

$\gamma (t)={\textstyle (}c+r\mathrm{cos}(a/r){e}_{1}+r\mathrm{sin}(a/r){e}_{2}{\textstyle )}+{\textstyle (}-\mathrm{sin}(a/r){e}_{1}+\mathrm{cos}(a/r){e}_{2}{\textstyle )}t+{\textstyle (}-\mathrm{cos}(a/r){e}_{1}-\mathrm{sin}(a/r){e}_{2}{\textstyle )}{r}^{-1}{t}^{2}+O({t}^{3})$

Comparing the above to

$\beta (t)=\beta (0)+{\beta}^{\prime}(0)t+\frac{1}{2}{\beta}^{\u2033}(0){t}^{2}+O({t}^{3})$

we observe that

${r}^{-1}=\frac{1}{2}|{\beta}^{\u2033}(0)|$, so $r$ is determined. (Unless ${\beta}^{\u2033}(0)=0$, when the radius of curvature is infinite and there is no circle you are looking for.)

Since ${\beta}^{\prime}(0)$ is a unit vector, there exists a unique $a$ (up to an integer multiple of $2\pi r$) such that ${\beta}^{\prime}(0)=-\mathrm{sin}(a/r){e}_{1}+\mathrm{cos}(a/r){e}_{2}$.

c is uniquely determined from $(}c+r\mathrm{cos}(a/r){e}_{1}+r\mathrm{sin}(a/r){e}_{2}{\textstyle )}=\beta (0)$

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