- #1
Grand
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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:
[tex]\textbf{u}=\frac{d\textbf{r}}{ds}[/tex]
and
[tex]\frac{d\textbf{u}}{ds}=k\textbf{n}[/tex]
where k is called the curvature.
We can expand r(s) around a certain point:
[tex]\textbf{\textbf{r(s)}}=\textbf{a}+s\frac{d\textbf{r}}{ds}+\frac{1}{2}s^{2}\frac{d^{2}\textbf{r}}{ds^{2}}+...[/tex]
using the equations for u and n:
[tex]\textbf{r(s)}=\textbf{a}+s\textbf{u}+\frac{1}{2}s^{2}k\textbf{n}+...[/tex]
which is the same as:
[tex]\textbf{r(s)}=\textbf{a}+\frac{sin(ks)}{k}\textbf{u}+\frac{1}{k}(1-cos(ks))\textbf{n}+...[/tex]
And from here they conclude that this is the equation of a circle with radius [tex]\frac{1}{k}[/tex], 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.