# Radius of curvature

## Homework Statement

This is not exactly a question, but I am trying to understand the derivation of radius of curvature from a boof I'm reading. I would be extremely grateful if someone is able to help me.

## Homework Equations

Let u and n be the tangent and normal unit vectors respectively. If r(s) is the path function, we know that:

$$\textbf{u}=\frac{d\textbf{r}}{ds}$$

and

$$\frac{d\textbf{u}}{ds}=k\textbf{n}$$

where k is called the curvature.

We can expand r(s) around a certain point:

$$\textbf{\textbf{r(s)}}=\textbf{a}+s\frac{d\textbf{r}}{ds}+\frac{1}{2}s^{2}\frac{d^{2}\textbf{r}}{ds^{2}}+...$$

using the equations for u and n:

$$\textbf{r(s)}=\textbf{a}+s\textbf{u}+\frac{1}{2}s^{2}k\textbf{n}+...$$

which is the same as:

$$\textbf{r(s)}=\textbf{a}+\frac{sin(ks)}{k}\textbf{u}+\frac{1}{k}(1-cos(ks))\textbf{n}+...$$

And from here they conclude that this is the equation of a circle with radius $$\frac{1}{k}$$, which I don't quite understand. Tried to square it, extract sines and cosins, but still don't understand why this is a circle.

Thanks to advance to whoever is able to help.

## Answers and Replies

The parametric equation of a circle is ( a+ Rcosu, b+Rsinu).

## Homework Statement

This is not exactly a question, but I am trying to understand the derivation of radius of curvature from a boof I'm reading. I would be extremely grateful if someone is able to help me.

## Homework Equations

Let u and n be the tangent and normal unit vectors respectively. If r(s) is the path function, we know that:

$$\textbf{u}=\frac{d\textbf{r}}{ds}$$

and

$$\frac{d\textbf{u}}{ds}=k\textbf{n}$$

where k is called the curvature.

We can expand r(s) around a certain point:

$$\textbf{\textbf{r(s)}}=\textbf{a}+s\frac{d\textbf{r}}{ds}+\frac{1}{2}s^{2}\frac{d^{2}\textbf{r}}{ds^{2}}+...$$

using the equations for u and n:

$$\textbf{r(s)}=\textbf{a}+s\textbf{u}+\frac{1}{2}s^{2}k\textbf{n}+...$$

which is the same as:

$$\textbf{r(s)}=\textbf{a}+\frac{sin(ks)}{k}\textbf{u}+\frac{1}{k}(1-cos(ks))\textbf{n}+...$$

And from here they conclude that this is the equation of a circle with radius $$\frac{1}{k}$$, which I don't quite understand. Tried to square it, extract sines and cosins, but still don't understand why this is a circle.

Thanks to advance to whoever is able to help.

I'm wondering if your latter equation is well derived. Did you take it from your book?

Yes, it is from the book and I was able to derive it by myself.

@Eynstone, that accounts only for the sin term, not for the -cos term

Oh wait! it's not expressed differently, it's just rearranged so that you can manage to get the radius of curvature, check that first website I handled, and enjoy up to point (31). ;)