The Curvature of a Curve in Euclidean 3-space $\mathbb{R}^3$

The quantity curvature is intended to be a measurement of the bending or turning of a curve. Let $\alpha: I\longrightarrow\mathbb{R}^3$ be a regular curve (i.e. a smooth curve whose derivative never vanishes). If $\alpha$ were to have the unit speed, i.e.
\begin{equation}
\label{eq:unitspped}
||\dot\alpha(t)||^2=\alpha(t)\cdot\alpha(t)=1.
\end{equation}
Differentiating \eqref{eq:unitspped}, we see that $\dot\alpha(t)\cdot\ddot\alpha(t)=0$, i.e. the acceleration is normal to the velocity which is tangent to $\alpha$. Hence, measuring the acceleration is measuring the curvature. So, if we denote the curvature by $\kappa$, then
\begin{equation}
\label{eq:curvature}
\kappa=||\ddot\alpha(t)||.
\end{equation}
Remember that the definition of curvature \eqref{eq:curvature} requires the curve $\alpha$ to be a unit speed curve, but it is not necessarily always the case. What we know is that we can always reparametrize a curve and reparametrization does not change the curve itself but only changes its speed. There is one particular parametrization that we are interested in as it results a unit speed curve. It is called paramtrization by arc-length. This time let us assume that $\alpha$ is not a unit speed curve and define
\begin{equation}
\label{eq:arclength}
s(t)=\int_a^t||\dot\alpha(u)||du,
\end{equation}
where $a\in I$. Since $\frac{ds}{dt}>0$, $s(t)$ is an increasing function and so it is one-to-one. This means that we can solve \eqref{eq:arclength} for $t$ and this allows us to reparametrize $\alpha(t)$ by the arc-length parameter $s$.

Example. Let $\alpha: (-\infty,\infty)\longrightarrow\mathbb{R}^3$ be given by
$$\alpha(t)=(a\cos t,a\sin t,bt)$$
where $a>0$, $b\ne 0$. $\alpha$ is a right circular helix. Its speed is
$$||\dot\alpha(t)||=\sqrt{a^2+b^2}\ne 1.$$
$s(t)=\sqrt{a^2+b^2}t$, so $t=\frac{s}{\sqrt{a^2+b^2}}$. The reparametrization of $\alpha(t)$ by $s$ is given by
$$\alpha(s)=\left(a\cos\frac{s}{\sqrt{a^2+b^2}},b\sin\frac{s}{\sqrt{a^2+b^2}},\frac{bs}{\sqrt{a^2+b^2}}\right).$$
Hence the curvature $\kappa$ is
$$\kappa=\frac{a}{a^2+b^2}.$$

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