Given the equation of positively oriented curve
$\mathbf{r}(s) = (x(s); y(s))$,
we obtain
$\mathbf{T} = ({dx \over ds}; {dy \over ds})$ (tangent vector)
$\mathbf{N} = ({-dy \over ds}; {dx \over ds})$ (normal vector).
I can't understand how to obtain these vectors, starting with these equations:
$\mathbf{T} = {\mathbf{r'}(t) \over ||\mathbf{r'}(t)||}$
$\mathbf{N} = {\mathbf{T'}(t) \over ||\mathbf{T'}(t)||}$,
and assuming that
$t = t(s)$
and
${dt \over ds} = {1 \over ||\mathbf{r'}(t)||}$
Can You help me ?
Thanks!
I think the parameter $s$ is the arc length of the curve $\textbf{r}(s)$.
The arc length $s$ of the curve $\textbf{r}(t)$ is defined by
$$\frac{\mathrm{d} s}{\mathrm{d} t}=\sqrt{\left(\frac{\mathrm{d} x}{\mathrm{d} t}\right)^{2}+\left(\frac{\mathrm{d} y}{\mathrm{d} t}\right)^{2}}={\left \| \mathbf{r}'(t) \right \|},$$
where $t$ is another parameter. Then,
$$\left \| \textbf{r}'(s) \right \|^{2}=\left(\frac{\mathrm{d} x}{\mathrm{d} s}\right)^{2}+\left(\frac{\mathrm{d} y}{\mathrm{d} s}\right)^{2}=\left(\frac{\mathrm{d} t}{\mathrm{d} s}\right)^{2}\left [ \left ( \frac{\mathrm{d} x}{\mathrm{d} t} \right )^{2}+\left ( \frac{\mathrm{d} y}{\mathrm{d} t}^{2} \right ) \right ]=1.$$
Hence we have
$$\mathbf{T}=\frac{\mathbf{r}'(t)}{\left \| \mathbf{r}'(t) \right \|}=\frac{\mathrm{d} t}{\mathrm{d} s}\mathbf{r}'(t)=\left(\frac{\mathrm{d} x}{\mathrm{d} s},\frac{\mathrm{d} y}{\mathrm{d} s}\right).$$
I want to derive the formula of $\mathbf{N}$ generally; that is, in three dimensions.
First,
$$\textbf{T}'(t)=\frac{\mathrm{d} }{\mathrm{d} t}\frac{\textbf{r}'(t)}{\left \| \textbf{r}'(t) \right \|}=\frac{\textbf{r}''(t)}{\left \| \textbf{r}'(t) \right \|}-\frac{\textbf{r}'(t)}{\left \| \textbf{r}'(t) \right \|^{2}}\frac{\mathrm{d} }{\mathrm{d} t}\left \| \textbf{r}'(t) \right \|$$ Note that,
$$\frac{\mathrm{d} }{\mathrm{d} t}\left \| \textbf{r}'(t) \right \|=\frac{x'(t)x''(t)+y'(t)y''(t)}{\left \| \textbf{r}'(t) \right \|}=\frac{\textbf{r}'(t)\cdot \textbf{r}''(t)}{\left \| \textbf{r}'(t) \right \|}$$ Then we have,
$$\textbf{T}'(t)=\frac{1}{\left \| \mathbf{r}'(t) \right \|^{3}}\left [ \left \| \textbf{r}'(t) \right \|^{2}\textbf{r}''(t)-\mathbf{r}'(t)\left ( \mathbf{r}'(t)\cdot \textbf{r}''(t) \right ) \right ]=\frac{\mathbf{r}'(t)\times \left ( \textbf{r}''(t)\times \textbf{r}'(t) \right )}{\left \| \textbf{r}'(t) \right \|^{3}}$$
Thus,
$$\left \| \textbf{T}'(t) \right \|=\frac{\left \| \textbf{r}''(t)\times \mathbf{r}'(t) \right \|}{\left \| \textbf{r}'(t) \right \|^{2}}$$
and,
$$\textbf{N}(t)=\frac{\textbf{r}'(t)\times \left [ \textbf{r}''(t)\times \mathbf{r}'(t) \right ]}{\left \| \textbf{r}'(t) \right \|\left \| \textbf{r}''(t)\times \textbf{r}'(t) \right \|}=\mathbf{T}(t)\times \frac{\left ( x''(t)y'(t)-x'(t)y''(t) \right )\mathbf{k}}{\left | x''(t)y'(t)-x'(t)y''(t) \right |}=\frac{\left ( x''(t)y'(t)-x'(t)y''(t) \right )}{\left | x''(t)y'(t)-x'(t)y''(t) \right |}(\frac{\mathrm{d} y}{\mathrm{d} s},-\frac{\mathrm{d} x}{\mathrm{d} s})$$
Since the curve is positive-oriented, the direction of $\mathbf{r}''(t)\times \mathbf{r}'(t)$ is downward;namely, $x''(t)y'(t)-x'(t)y''(t)<0$.
Finally we have,
$$\mathbf{N}(t)=-\left ( \frac{\mathrm{d} y}{\mathrm{d} s},-\frac{\mathrm{d} x}{\mathrm{d} s} \right )=\left ( -\frac{\mathrm{d} y}{\mathrm{d} s},\frac{\mathrm{d} x}{\mathrm{d} s} \right )$$