# (Curvature) Why the radius of the circle that gives the best approximation of a curve $\frac{\|T'\|}{\|\gamma '\|}$

Let $\alpha :I\to \mathbb R^2$ a curve. I want the curvature of $\alpha$ at $p=\alpha (u),$ $u\in I$. I therefore want to find the radius of the circle that gives the best approximation of the curve at $p$. I consider $$T(u)=\frac{\alpha (u)}{\|\alpha '(u)\|}.$$ I know that $$0=\underbrace{\left<T(u),T(u)\right>'}_{=1}=2\left<T(u),T'(u)\right>,$$ and thus $T\perp T'$. So, if $N$ is the unit normal of the curve then $$T'=\kappa N.$$ Now why the radius of the larger that approximate the curve at $p$ is the circle of radius $$\frac{\|T'\|}{\|\alpha '\|} \ \ ?$$

• The curvature is the reciprocal of the radius of curvature. "Optimal" here means approximating the curve to second order - not only must the first derivatives (tangent lines) of the curve and the circle agree at $p$, but their second derivatives (curvatures) must also. Dec 3, 2017 at 12:23
• @AnthonyCarapetis : I changed a little bit my question (or in fact adapt more the question and correct things).
– MSE
Dec 3, 2017 at 13:51

First, your notation is inconsistent. I presume that $\gamma'$ should be $\alpha'$. Your Frenet equation is correct only for an arclength-parametrized curve. Differentiating with respect to $u$, the chain rule tells you that you will have $T' = \kappa \|\alpha'\|N$, and so $\kappa = \|T'\|/\|\alpha'\|$. The osculating circle at $p$ should have the same curvature as the curve has at $p$, and so its radius will be $$\frac1\kappa = \frac{\|\alpha'\|}{\|T'\|}.$$