How are the tangential and normal components of the acceleration vector derived? Let acceleration be $$\overrightarrow a= a_T \overrightarrow T + a_N \overrightarrow N$$ where $\overrightarrow T$ is the unit tangent component and $\overrightarrow N$ is the unit normal component.
If we define $$\ v=\lVert \overrightarrow v(t) \rVert$$
Then the tangential and normal components of acceleration are given by,
$$\ a_T = v'= \frac{\overrightarrow r\ '(t)\ \cdot \overrightarrow r\ ''(t)}{\lVert\overrightarrow r\ '(t)\rVert} \ \ \ \ and\ \ \ \ \ \ a_N=kv^2 = \frac{\lVert \overrightarrow r\ '(t)\ \times \overrightarrow r\ ''(t)\rVert}{\lVert \overrightarrow r\ '(t)\rVert }$$
Where k is the curvature and $\overrightarrow T(t)$ is unit tangent vector given by $$\ k= \frac{\lVert \overrightarrow T\ '(t)\rVert}{\lVert \overrightarrow r\ '(t)\rVert}\ \ \ \ \ \ \overrightarrow T(t)=\frac{\overrightarrow r\ '(t)}{\lVert\overrightarrow r\ '(t)\rVert}$$  
Where do these equations come from? Why are they defined in terms of the speed $\ v=\lVert\overrightarrow v(t)\rVert$? is there a geometric interpretation of this definition?
edit: I'm still a little confused, but this is what I have gathered thus far.
The acceleration vector $\overrightarrow a$ is a non-zero vector so it can be represented as a linear combination of $\overrightarrow T$ and $\overrightarrow N$ such that $\overrightarrow a$ lies in the plane formed by $\overrightarrow T$ and $\overrightarrow N$. Therefore, $\overrightarrow a$ can also be represented as the sum of the projection of $\overrightarrow a$ onto $\overrightarrow T$ and the projection of $\overrightarrow a $ onto $\overrightarrow N$.
$$\overrightarrow a(t) = Proj_{\overrightarrow T}\overrightarrow a + Proj_{\overrightarrow N}\overrightarrow a$$
$$\overrightarrow a(t)= \frac{\overrightarrow a\cdot\overrightarrow T}{(\lVert \overrightarrow T\rVert)^2}\overrightarrow T + \frac{\overrightarrow a\cdot\overrightarrow N}{(\lVert \overrightarrow N\rVert)^2}\overrightarrow N$$
$$\overrightarrow a(t)=(\overrightarrow a\cdot\overrightarrow T)\overrightarrow T + (\overrightarrow a\cdot\overrightarrow N)\overrightarrow N$$
$$\overrightarrow a(t)= a_{T}\overrightarrow T + a_{N}\overrightarrow N$$
(From this point on, I'm dropping the overhead arrows to represent vectors.) 
I understand how the tangential component $\ a_{T}$ can be derived from the dot product of $\ a\cdot\ T$ but I am still unclear about the Normal component. 
How do you prove:$$\ a\cdot N = \frac{\lVert v\times a\rVert}{\lVert v\rVert}=\sqrt{\lVert a \rVert^2-a_{T}^2}$$
 A: This really isn't a definition, but rather a computation, decomposing the acceleration vector into its tangential and normal components. The unit tangent vector, curvature, and normal vector should not change when we reparametrize the curve; indeed, they are usually defined assuming the particle moves at constant speed $1$. The curvature tells us the rate at which the unit tangent vector changes (turns) when we move at speed $1$, and the unit normal vector $\vec N$ gives the direction of that change. That is, using $s$ to give arclength along the curve,
$$\frac{d\vec T}{ds} = \kappa\vec N.$$
Note, also, that $v = ds/dt$ (why?).
Now, the unit tangent vector is given by the equation
$$\vec v(t) = \|\vec v(t)\| \vec T(t),$$
so, differentiating, and using the chain rule,
\begin{align*}
\vec a(t) = \vec v'(t) &= \frac{d\vec v}{dt} = v'(t) \vec T(t) + v(t) \frac{d\vec T}{dt} = v'(t)\vec T(t) + v(t) \frac{d\vec T}{ds}\,\frac{ds}{dt}\\ &= v'\vec T + kv^2 \vec N.
\end{align*}
You can now see that $a_T = \vec a\cdot \vec T = \vec r''\cdot\left(\dfrac{\vec r'}{\|\vec r'\|}\right)$ and $|a_N| = \|\vec a\times \vec T\| = \left\|\vec r''\times\left(\dfrac{\vec r'}{\|\vec r'\|}\right)\right\|$, so $a_N = kv^2 = \dfrac{\|\vec r''\times \vec r'\|}{\|\vec r'\|}$. (Note that $kv^2\ge 0$, so we know that in fact $a_N\ge 0$ and the absolute value is unnecessary.)
A: Here's how I intuitively see it:
$(t,t\text{ sin(}t))$
(The dashed blue vector is the position vector), (the blue vector represents the derivative of the position vector, the tangent of the red curve), (and the orange one is the rate of change of the blue vector). We can think of the blue vector as the drawing the function while the orange one can be seen as pulling on the end on the blue vector. When we take $\vec{r}'(t)\text{ }⋅\text{ }\vec{r}''(t)$, it can be thought of as the length of the projection of the acceleration (orange) vector onto the velocity (blue) vector, scaled by the length of the velocity vector. Projection If you divide by the length of the velocity vector, this then tells us how much of the acceleration is in the same direction as the velocity, which is how quick the tangent vector grows and shrinks.
For example, in curvature, when we take the derivative of the unit tangent vector, we find how quick the unit tangent vector (or direction vector as I call it) changes direction, but since the vector never grows and shrinks at 1, the derivative is always perpendicular.
With the normal component, the cross product represents the perpendicular vector with the length of the parallelogram of width $r'(t)$ and slant described by $r''(t)$. Since we just want the length (magnitude), we just have the parallelogram's area. Parallelogram Since the area is base times height, we can divide the base to get the height. However, the "height" is just the part of the acceleration vector perpendicular to the tangent (and thus the slope). Here's the parallelogram in $(t,t^{-1})$. Because the component is perpendicular, the normal component of acceleration gives us the rate of change in direction.
A: Long time ago I did some computations here when I was learning differential geometry of curves: (GD)Let $\alpha: I \to \mathbb R^n (n = 2, 3)$ be a regular curve of any parameter $r \in I$...
For 2 dimonsional case, I wanted to note this non-trivial computation. It is interesting for me:
$$\vec a=v'\vec T+\kappa v^2\vec N=\frac{x'x''+y'y''}{v}\frac{\langle x',y'\rangle}{v}+\frac{x'y''-y'y''}{v^3}v^2\frac{\langle -y',x'\rangle}{v}=\langle x'',y'' \rangle$$
This is not general. But, I think, it is helpfull to learn this separation.
