# 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:$$\ a\cdot N = \frac{\lVert v\times a\rVert}{\lVert v\rVert}=\sqrt{\lVert a \rVert^2-a_{T}^2}$$

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.)
• by $\overrightarrow N$ do you mean the normal vector or the unit normal vector where, $$\overrightarrow N = \frac{\overrightarrow T\ '(t)}{\lVert \overrightarrow T\ '(t)\rVert}$$ – Nicholas Cousar Mar 8 '18 at 21:14
• Right, I mean the unit vector in the direction of $d\vec T/ds$, which is, after all, just a scalar multiple of $d\vec T/dt$. – Ted Shifrin Mar 8 '18 at 21:17
• A couple of criticisms: I would write $\hat T$ and $\hat N$ rather than $\vec T$ and $\vec N$. Also $\vec a\times\hat T=-a_N\hat B$ where $\hat B$ is the unit binormal vector. But you answer is OK otherwise. – user5713492 Mar 10 '18 at 6:16
• I was unclear on how you derived $\ a_{N}$ but I think I get it now. – Nicholas Cousar Mar 10 '18 at 7:50