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In our class, we're learning that you can split up the acceleration, $mathbfa$, of a particle into two convenient components, like so:

$$mathbfa = a_TmathbfT + a_NmathbfN$$

Where $a_T$ is the "tangential component" of acceleration, $a_N$ is the "normal component", and $mathbfT$ and $mathbfN$ are the unit tangent and unit normal vectors to the curve $mathbfr(t)$, respectively.

But we also learned earlier about a third kind of vector, $mathbfB$ - the "binormal vector" - which is orthogonal to both $mathbfN$ and $mathbfT$.

Why isn't the formula thus

$$mathbfa = a_TmathbfT + a_NmathbfN + a_BmathbfB?$$

Note: I know that the binormal vector $mathbfB$ is not generally defined as a unit vector for the purposes of Multivariable Calculus classes. But in this instance, just assume $mathbfB$ represents the unit binormal vector, and that $a_B$ represents the "binormal component" of acceleration.

I have a sneaking suspicion that the jounce, $mathbfj = mathbfr^(3)(t)$, of the particle moving along $mathbfr(t)$ is, in fact, defined by

$$mathbfj = j_TmathbfT + j_NmathbfN + j_BmathbfB.$$

...since, well,

$$mathbfv = VertmathbfvVertmathbfT = v_TmathbfT$$

and

$$mathbfa = a_TmathbfT + a_NmathbfN;$$

It just seems like each new order of derivative taken of $mathbfr(t)$ adds to the equation a new, orthogonal component of motion. If that's the case, why??

First, a note to your note: the way the binormal vector $mathbfB$ is defined, it is automatically a unit vector (whenever it's well-defined); but for some historical reason (which I don't know) the word "unit" isn't used in its name. More specifically, $mathbfB=mathbfTtimesmathbfN$ and $|mathbfB|=|mathbfT||mathbfN|cos(pi/2)=1cdot1cdot1=1$.

Regarding your first question: in fact, we can say that $mathbfa=a_TmathbfT+a_NmathbfN+a_BmathbfB$, because the three vectors $mathbfT$, $mathbfN$, and $mathbfB$ (when they are well-defined) form an orthonormal basis. But it turns out that $a_B=0$ always, hence the standard expression $mathbfa=a_TmathbfT+a_NmathbfN$.

Why is $a_B=0$? I presume you've read calculations leading to this formula, so instead I'll try to describe my personal way of understanding this intuitively. The two vectors $mathbfv=mathbfr'(t)$ and $mathbfa=mathbfr''(t)$ typically span a plane (unless they are collinear or one is zero), called the osculating plane. So to describe any vector in that plane we only need two basis vectors. And $mathbfT$ and $mathbfN$ do just that — they form an orthonormal basis for this plane. We will need a third basis vector $mathbfB$ for any vectors sticking out of this plane, such as $mathbfr'''(t)$, for example, which may or may not be in the same plane.