# Homework Help: Radius of curvature

1. Mar 29, 2010

### Grand

1. The problem statement, all variables and given/known data
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.

2. Relevant 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.

2. Mar 29, 2010

### Eynstone

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

3. Mar 29, 2010

### Redsummers

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

4. Mar 29, 2010

### Grand

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

5. Mar 29, 2010

### Redsummers

6. Mar 29, 2010

### Redsummers

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). ;)