Imagine a curve \(C\) made by tracing out a \(\textbf{r}(t)\), where \(\textbf{r}(t)\) represents the position of an object and is twice-differentiable.
Let \(\textbf{T}(t)\) be the unit tangent vector to \(C\):
$$ \textbf{T}(t) = \frac{ \textbf{r}'(t) }{ \Vert \textbf{r}'(t) \Vert} $$
let \(\textbf{N}(t)\) be the principle unit normal vector to \(C\):
$$ \textbf{N}(t) = \frac{ \textbf{T}'(t) }{ \Vert \textbf{T}'(t) \Vert} $$
Let \(\textbf{a}(t)\) be the acceleration vector that lies in the plane formed by \(\textbf{T}(t)\) and \(\textbf{N}(t)\).
Let \(\textbf{v}(t)=\textbf{r}'(t)\) be the velocity vector. This means:
$$ \textbf{T}(t) = \frac{ \textbf{r}'(t) }{ \Vert \textbf{r}'(t) \Vert} \implies \begin{align} \textbf{v}(t) &= \Vert \textbf{r}'(t) \Vert \textbf{T}(t) \\ \textbf{v}(t) &= \Vert \textbf{v}(t) \Vert \textbf{T}(t) \end{align}$$
Let \(v(t)=\Vert \textbf{v}(t) \Vert \) be the speed, which is a scalar value. If we differentiate the above:
$$\begin{align} \textbf{a}(t)=\textbf{v}'(t)=\frac{d}{dt} (v(t) \textbf{T}(t)) &= v '(t) \textbf{T}(t) + v(t) \textbf{T}'(t) \\ &= v '(t) \textbf{T}(t) + v(t) \Vert \textbf{T}'(t) \Vert \textbf{N}(t)\end{align}$$
Let \(k\) be the curvature:
$$k = \frac{\Vert\textbf{T}'(t) \Vert}{\Vert\textbf{r}'(t) \Vert} \implies \Vert\textbf{T}'(t) \Vert = k \Vert\textbf{r}'(t) \Vert = k \ v(t)$$
Therefore:
$$ \textbf{a}(t)= v '(t) \textbf{T}(t) + [v(t)]^2 \ k \ \textbf{N}(t)$$
The coefficients of \(\textbf{T}(t)\) and \(\textbf{N}(t)\) are referred to as the tangential component of acceleration and the normal component of acceleration. We write \(a_\textbf{T}\) to denote the tangential component and \(a_\textbf{N}\) to denote the normal component.
$$\begin{align} a_\textbf{T} &= v '(t) \\ a_\textbf{N} &= [v(t)]^2 \ k \\ \textbf{a}(t) &= a_\textbf{T} \textbf{T}(t) + a_\textbf{N} \textbf{N}(t) \end{align}$$
If we take the dot product of both sides with \(\textbf{T}(t)\):
$$\begin{align} \textbf{a}(t) \cdot \textbf{T}(t) &=a_\textbf{T} \textbf{T}(t) \cdot \textbf{T}(t) + a_\textbf{N} \textbf{N}(t) \cdot \textbf{T}(t) \\ \textbf{a}(t) \cdot \textbf{T}(t) &=a_\textbf{T} (\textbf{T}(t) \cdot \textbf{T}(t)) + 0 \\ \textbf{a}(t) \cdot \textbf{T}(t) &=a_\textbf{T} (1) \\ \textbf{a}(t) \cdot \textbf{T}(t) &=a_\textbf{T} \end{align}$$
This means:
$$ a_\textbf{T} = \textbf{a}(t) \cdot \textbf{T}(t) = \frac{\textbf{v}(t) \cdot \textbf{a}(t)}{\Vert \textbf{v}(t) \Vert }$$
If we do \(\textbf{v}(t) \times \textbf{a}\):
$$\begin{align} \textbf{v}(t) \times \textbf{a}(t) &= \textbf{v}(t) \times (a_\textbf{T} \textbf{T}(t) + a_\textbf{N} \textbf{N}(t)) \\ &= \textbf{v}(t) \times \left(a_\textbf{T} \frac{\textbf{v}(t)}{\Vert \textbf{v}(t) \Vert} + a_\textbf{N} \textbf{N}(t) \right) \\ &= \left( \textbf{v}(t) \times a_\textbf{T} \frac{\textbf{v}(t)}{\Vert \textbf{v}(t) \Vert}\right) + \left( \textbf{v}(t) \times a_\textbf{N} \textbf{N}(t) \right) \end{align}$$
Since \(\textbf{v}(t) \times \textbf{v}(t) = 0\):
$$\begin{align} \textbf{v}(t) \times \textbf{a}(t) &= 0 + \textbf{v}(t) \times a_\textbf{N} \textbf{N}(t) \\ &= \Vert \textbf{v}(t) \Vert \textbf{T}(t) \times a_\textbf{N} \textbf{N}(t) \end{align}$$
We can use "multiplication by a constant" property here:
$$\begin{align} \textbf{v}(t) \times \textbf{a}(t) &= \Vert \textbf{v}(t) \Vert (\textbf{T}(t) \times a_\textbf{N} \textbf{N}(t)) \\ &= \Vert \textbf{v}(t) \Vert a_\textbf{N} (\textbf{T}(t) \times \textbf{N}(t)) \\ \Vert \textbf{v}(t) \times \textbf{a}(t) \Vert &= \Vert \textbf{v}(t) \Vert a_\textbf{N} \ \Vert \textbf{T}(t) \times \textbf{N}(t) \Vert \end{align}$$
Since \(\textbf{T}(t)\) and \(\textbf{N}(t)\) are perpendicular unit vector, \(\Vert \textbf{T}(t) \times \textbf{N}(t) \Vert = 1\):
$$\begin{gather} \Vert \textbf{v}(t) \times \textbf{a}(t) \Vert = \Vert \textbf{v}(t) \Vert a_\textbf{N} \\ a_\textbf{N} = \frac{\Vert \textbf{v}(t) \times \textbf{a}(t) \Vert}{\Vert \textbf{v}(t) \Vert} \end{gather}$$
So we can conclude:
$$\begin{align} a_\textbf{T} &= \frac{\textbf{v}(t) \cdot \textbf{a}(t)}{\Vert \textbf{v}(t) \Vert } \\ a_\textbf{N} &= \frac{\Vert \textbf{v}(t) \times \textbf{a}(t) \Vert}{\Vert \textbf{v}(t) \Vert} \end{align}$$
There is also another expression involving \(a_\textbf{N}\) and \(a_\textbf{T}\). First let's start \(\Vert \textbf{a}(t) \Vert ^2\):
$$\begin{align} \Vert \textbf{a}(t) \Vert^2 &= \textbf{a}(t) \cdot \textbf{a}(t)= (a_\textbf{T} \textbf{T}(t) + a_\textbf{N} \textbf{N}(t)) \cdot (a_\textbf{T} \textbf{T}(t) + a_\textbf{N} \textbf{N}(t)) \\ &= a_\textbf{T}^2 (\textbf{T}(t) \cdot \textbf{T}(t)) + 2a_\textbf{T}a_\textbf{N} (\textbf{T}(t) \cdot \textbf{N}(t)) + a_\textbf{N}^2 (\textbf{N}(t) \cdot \textbf{N}(t)) \\ &= a_\textbf{T}^2 (1) + 0 + a_\textbf{N}^2 (1) \end{align}$$
Therefore:
$$\begin{align} a_\textbf{N}^2 = \sqrt{\Vert \textbf{a}(t) \Vert^2 - a_\textbf{T}^2 } \\ a_\textbf{T}^2 = \sqrt{\Vert \textbf{a}(t) \Vert^2 - a_\textbf{N}^2 } \end{align} $$
