Why arc-length parametrized curves has unit tangent vector? I'm studying that if we have a smooth  parametrized curve $r(t)$, we can reparametrize it according to its arc-length so that the derivative will always have module $1$. Is there a proof? 
 A: Yes, provided your curve has nonzero tangent vector at all points. 
Suppose your curve is $\alpha: [a, b] \to \Bbb R^2$. For $t \in [a, b]$, define
$$
q(t) = \int_a^t \| a'(s) \| ds.
$$
You can see that $q(t)$ represents "how long is $\alpha$ from $a$ up to $t$".
What can you say about the function $q$? 


*

*$q(a) = 0$.

*$q'(t) = \| \alpha'(t) \| > 0$ for every $t \in (a, b)$, by the fundamental theorem of calculus. 
Define $L = q(b)$ to be the length of the whole curve. 
Now: $q: [a, b] \to [0, L]$ is an increasing continuous function onto its codomain; hence it has an inverse function $u: [0, L] \to [a, b]$. We may not be able to easily write down the inverse, but it's there. And the derivative of $u$ at a point is (by the inverse function theorem) given by:
$$
u'(t) = \frac{1}{q'(q^{-1}(t))} = \frac{1}{q'(u(t))} = \frac{1}{\|a'(u(t))\|}.
$$
Hold that thought. 
Now let
$$
\beta: [0, L] \to \Bbb R^2 : t \mapsto \alpha(u(t)).
$$
Clearly $\beta$ traverses the same path as $\alpha$. But what's $\beta'(t)$? It is, by the chain rule, 
\begin{align}
\beta'(t) 
&= \alpha'(u(t)) \cdot u'(t)\\
&= \alpha'(u(t)) \cdot \frac{1}{\| \alpha'(u(t))\|},
\end{align}
which is a unit vector. QED.
A: we can decomponse the position vector $\vec{r}(s)$, parameterized by arc length (s) into its components in Cartesian coordinates:
$\vec{r}(s) = x(s)\hat{\text{i}} + y(s)\hat{\text{j}} + z(s)\hat{\text{k}}$
now differentiate both sides with respect to the arc length s.
$\frac{d\vec{r}(s)}{ds} = \frac{dx(s)}{ds}\hat{\text{i}} + \frac{dy(s)}{ds}\hat{\text{j}} + \frac{dz(s)}{ds}\hat{\text{k}}$
now find the magnitude:
$\begin{Vmatrix}\frac{d\vec{r}(s)}{ds}\end{Vmatrix}^2 = \bigg(\frac{dx(s)}{ds}\bigg)^2 + \bigg(\frac{dy(s)}{ds}\bigg)^2 + \bigg(\frac{dz(s)}{ds}\bigg)^2$
$\begin{Vmatrix}\frac{d\vec{r}(s)}{ds}\end{Vmatrix}^2 = \frac{1}{ds^2}\bigg[dx^2 + dy^2 +  dz^2\bigg]$
since s is the arc length we can substitute the formula for arc length differential into the equation ($ds^2 = dx^2 + dy^2 + dz^2$), thus:
$\begin{Vmatrix}\frac{d\vec{r}(s)}{ds}\end{Vmatrix}^2 = \frac{1}{dx^2 + dy^2 +  dz^2}\bigg[dx^2 + dy^2 +  dz^2\bigg]$
$\begin{Vmatrix}\frac{d\vec{r}(s)}{ds}\end{Vmatrix}^2 = 1$
$\begin{Vmatrix}\frac{d\vec{r}(s)}{ds}\end{Vmatrix} = 1$
we conclude from this that if the function is parameterized using the arclength of the curve then the derivative is a unit vector.  futher, all derivatives are tangent vectors, thus, its a unit tangent vector.
A: Hint: if $r:[a,b]\longrightarrow\Bbb R^n$ and
$$s(t) = \int_a^t\|r'\|$$
then $r\circ s^{-1}$ is parametrized by arclengh. Now, apply the chain rule.
