Show that $\dot{n_s}=-\kappa_s t$

Question

I found the question in a differential geometry textbook while studying. This question seems so intesting to me. So please help me solving it.

Show that, if $\gamma$ is a unit-speed plane curve, $$\dot{\mathbf{n}}_s=-\kappa_s\mathbf t.$$

I know that

$$\dot t =\kappa_s n_s$$ and $$\kappa =|\kappa_s|$$

Answer 1

Check this out:

$\langle \mathbf n_s, \mathbf t_s \rangle = 0,\tag {1}$

whence

$\frac{d}{ds}\langle \mathbf n_s, \mathbf t_s \rangle = 0, \tag{2}$

so that

$\langle \dot{\mathbf n}_s, \mathbf t_s \rangle + \langle \mathbf n_s, \dot{\mathbf t}_s \rangle = 0, \tag{3}$

and using

$\dot {\mathbf t}_s =\kappa_s \mathbf n_s \tag{4}$

we get from (3)

$\langle \dot{\mathbf n}_s, \mathbf t_s \rangle = -\kappa_s \tag{5}$

since $\langle \mathbf n_s, \mathbf n_s \rangle = 1$. And since $\langle \mathbf n_s, \dot{\mathbf n}_s \rangle = 0$ and we are operating in $\Bbb R^2$, we can conclude that

$\dot {\mathbf n}_s = -\kappa \mathbf t_s. \tag{6}$

QED!

Hope this helps! Cheerio,

and as always,

Fiat Lux!!!