Why is there no $B$ component of acceleration in my Multivariable Calculus class?

Solution 1:

First, a note to your note: the way the binormal vector $\mathbf{B}$ 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, $\mathbf{B}=\mathbf{T}\times\mathbf{N}$ and $\|\mathbf{B}\|=\|\mathbf{T}\|\|\mathbf{N}\|\sin(\pi/2)=1\cdot1\cdot1=1$.

Regarding your first question: in fact, we can say that $\mathbf{a}=a_T\mathbf{T}+a_N\mathbf{N}+a_B\mathbf{B}$, because the three vectors $\mathbf{T}$, $\mathbf{N}$, and $\mathbf{B}$ (when they are well-defined) form an orthonormal basis. But it turns out that $a_B=0$ always, hence the standard expression $\mathbf{a}=a_T\mathbf{T}+a_N\mathbf{N}$.

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 $\mathbf{v}=\mathbf{r}'(t)$ and $\mathbf{a}=\mathbf{r}''(t)$ 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 $\mathbf{T}$ and $\mathbf{N}$ do just that — they form an orthonormal basis for this plane. We will need a third basis vector $\mathbf{B}$ for any vectors sticking out of this plane, such as $\mathbf{r}'''(t)$, for example, which may or may not be in the same plane.