In every curve $\alpha$ parametrized by arc length, $\alpha''(s) \cdot \alpha'(s) = 0$

Question

In my book, in the first step of the proof they say the following, where $\cdot$ means dot product:

Since $\alpha$ is parametrized by arc length, we have $$1 = |\alpha'(t)|^2 = (\alpha'(s) \cdot\alpha'(s))$$ for all t

This is not at all clear to me. I don't see why that statement is true and have no idea how to prove it, and am confused that there are two different variables...

Answer 1

I think the $t$ in your statement should be $s$. As written it's confusing.

In any parameterization the vector $\alpha '(t)$ is the tangent vector to the curve at point $\alpha(t)$. In a short time interval the position changes by $\alpha '(t) dt$; you integrate $|\alpha '(t)| dt$ with respect to $t$ to compute the arclength up to $t$.

If $t$ is the parameterization by arclength then the arclength up to $t$ is just $t$ so the integrand $|\alpha '(t)| $ must be identically $1$.

The last step is just the general fact that for any vector $v$ $$ |v|^2 = v \cdot v . $$

Answer 2

I'll address the question expressed in the comment. If

$$ s(t) = \int_{a}^t | \alpha'(u) | \, du $$

is the arclength of the curve $\alpha$ (from $\alpha(a)$ to $\alpha(t)$) then by the fundamental theorem of calculus and the chain rule, we have

$$ \frac{ds}{dt} = |\alpha'(t)|, \frac{dt}{ds} = \frac{1}{\frac{ds}{dt}} = \frac{1}{|\alpha'(t(s))|}. $$

If we consider the curve $\alpha(t(s))$ which is parametrized by arclength and take the derivative using the chain rule, we get

$$ \frac{d}{ds}(\alpha(t(s))) = \alpha'(t(s)) \frac{dt}{ds} = \frac{\alpha'(t(s))}{|\alpha'(t(s))|} $$

which has unit length so the derivative of $\alpha$ with respect to $s$ has unit length.