Why the magnitude of normal is always 1 when parameterised by arclength

Question

Why is it that whenever we parameterize a vector equation by its arclength, its tangent and normal always have a magnitude of 1?

Answer 1

The arc-length parameter $g(s) = s^{-1}(t)$ i.e;

$$g'(s) = \frac{1}{s'(g(s))} \ \ \ \ \ \ s(t)= \int_{a}^t \|\gamma'(u)\| \ du$$

The arc-length parametrization is $\gamma(g(s))$ and so;

$$\|(\gamma\circ g)'(s))\| = |\gamma'(g(s)) \cdot g'(s)\| = \left\|\gamma'(g(s)) \cdot \frac{1}{s'(g(s))}\right\|$$

Recall by F.T.C $s'(t) = \|\gamma'(t)\|$ i.e;

$$\left\|\gamma'(g(s)) \cdot \frac{1}{s'(g(s))}\right\| = \left\|s'(g(s)) \cdot \frac{1}{s'(g(s))}\right\| = 1$$

Now look at the definitions of $\vec{T}(t)$ and $\vec{N}(t)$ for $\vec{\gamma}(g(s))$ and your questions should be immediately answered.

Answer 2

By definition a curve $c$ is parameterized by arc length if the length of $c$ between $s$ and $t$ equals $t-s$: $$\int_s^t\|\dot c(x)\|\,dx=t-s.$$ So by the fundamental theorem of calculus we have $\|\dot c(t)\|=1$, i.e., the tangent vector has length $1$.

Answer 3

A parametric representation $$t\mapsto{\bf z}(t)=\bigl(x(t),y(t)\bigr)\qquad(0\leq t\leq T)$$ of a curve $\gamma\subset{\mathbb R}^2$ is regular if it is $C^1$, and if ${\bf z}'(t)=\bigl(x'(t),y'(t)\bigr)\ne{\bf 0}$ for all $t\in[0,T]$. If we are given such a curve $\gamma$ then we have to, and can, prove that the arc length $s$ along $\gamma$ is given by $$s(t)=\int_0^t|{\bf z}'(t)|\>dt=\int_0^t\sqrt{x'^2(t)+ y'^2(t)}\>dt\ .$$ The chosen parameter variable $t$ ("time") happens to be arc length if $s(t)\equiv t$, hence $s'(t)\equiv1$, which means that $$|{\bf z}'(t)|=\sqrt{x'^2(t)+ y'^2(t)}=1\qquad(0\leq t\leq T)\ .$$ Therefore the tangent vector ${\bf z}'(t)=\bigl(x'(t),y'(t)\bigr)$ is automatically a unit vector for all $t$ in this case, and so is the normal vector ${\bf n}(t)=\bigl(-y'(t),x'(t)\bigr)$ obtained by rotating ${\bf z}'(t)$ ninety degrees counterclockwise.