This section discusses the resolution of $\mathbf{R}″\left(p\right)$ into components along the moving basis vectors T and N, a result captured in the decomposition equation

$\mathbf{R}″\left(p\right)\=\mathrm{\ρ}\prime \mathbf{T}\+\mathrm{\κ} {\mathrm{\ρ}}^2\mathbf{N}$

Moreover, these are the only two components in the vector $\mathbf{R}″\left(p\right)$, so that this vector always remains in the osculating plane, the plane determined by the vectors T and N.

In the special case where the parameter $p$ is the time $t$, the derivative $\mathbf{R}\prime \left(t\right)$ is V, the velocity vector; $\mathbf{R}″\left(t\right)$ is a, the acceleration vector, and $v\=∥\mathbf{R}\prime \left(t\right)∥$$$is the speed, the magnitude of the velocity vector.  In this event, the decomposition of the acceleration vector into components along the tangent and normal vectors is written

$\mathbf{a}\=\stackrel{\.}{v} \mathbf{T}\+\mathrm{\κ} v^2 \mathbf{N}$ 

where the overdot denotes differentiation with respect to $t$, so $\stackrel{\.}{v}\=\frac{\mathrm{dv}}{\mathrm{dt}}$ and $\mathbf{a}\=\stackrel{..}{\mathbf{R}}\left(t\right)$. Thus, the formula for the acceleration vector reads aloud with an alliterative "vee-dot tee", which helps this author remember which component goes with the vector T.

The term $\stackrel{\.}{v}$, called the rate of change of the speed, is not the "scalar acceleration $a$."  The reader is cautioned

 there is no scalar associated with acceleration

The  length of the velocity vector V is the scalar speed $v$, but the length of the acceleration vector a is not a scalar of any dynamic significance.  It is an egregious error to believe $\stackrel{\.}{v}$ is $a$, the length of the acceleration vector.  In fact, using the overdot to denote differentiation with respect to $t$, the acceleration vector is

$\mathbf{a}\=\stackrel{..}{x} \mathbf{i}\+\stackrel{..}{y} \mathbf{j}\+\stackrel{..}{z} \mathbf{k}$ 

and its length is $∥\mathbf{a}∥$ = $\sqrt{\stackrel{..}{x}{}^2\+\stackrel{..}{y}{}^2\+\stackrel{..}{z}{}^2}$, whereas the speed is $v\=\sqrt{\stackrel{\.}{x}{}^2\+\stackrel{\.}{y}{}^2\+\stackrel{\.}{z}{}^2}$; so the rate of change of the speed is given by the derivative

$\stackrel{\.}{v}\=\frac{\mathrm{dv}}{\mathrm{dt}}\=\frac{\stackrel{\.}{x}\stackrel{..}{x}\+\stackrel{\.}{y}\stackrel{..}{y}\+\stackrel{\.}{z}\stackrel{..}{z}}{\sqrt{\stackrel{\.}{x}{}^2\+\stackrel{\.}{y}{}^2\+\stackrel{\.}{z}{}^2}}$

It should be clear that $\stackrel{\.}{v}$ and $∥\mathbf{a}∥$ are completely unrelated.

A derivation of the decomposition formula $\mathbf{R}″\left(p\right)\=\mathrm{\rho }\prime \mathbf{T}\+\mathrm{\kappa } {\mathrm{\rho }}^2\mathbf{N}$ is given in Example 2.8.9.