Why arc-length parametrized curves has unit tangent vector?

Question

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?

Answer 1

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 $q$?

1. $q(a) = 0$.

2. $q'(t) = \| \alpha'(t) \| > 0$ 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.

Answer 2

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.