How did we get from this to this? (TNB Frames and Calculus Question) Edit Fixed a typo: I am looking at a teachers notes and I see the following:
$$N = x'' – (x''\cdot T)T = x'' – \frac12 \frac{\mathrm{d}}{\mathrm{d}t} (T•T)T = x’'$$
I don't see how $(x'' \cdot T)T$ becomes
$$\frac12 \frac{\mathrm{d}}{\mathrm{d}t}(T•T)T$$
Can someone help me understand this? Thanks in advance
I understand that the $\frac12 \frac{\mathrm{d}}{\mathrm{d}t}$ will go to $0$ which is why we just get an answer of $x''$ but I don't see how we get the middle step.
 A: I found this to be a rather odd question, but one nevertheless oddly engaging.  In an effort to understand curves satisfying the OPs hypothesis (see (7) below), I tried to develop a few of its implications.  It is my hope that others with curiosity similar to mine will find the following "analysis" helpful.
Assuming $x(t)$ is a regular curve, by definition,
$\dfrac{ds}{dt} = \vert x' \vert \ne 0, \tag 1$
where $s$ is the arc-length along $x(t)$; then
$x' = \dfrac{ds}{dt}T, \tag 2$
$T$ being the unit tangent field to $x(t)$; differentiating (2) with respect to $t$,
$x'' = \dfrac{d^2s}{dt^2} T + \dfrac{ds}{dt} \dfrac{dT}{dt}; \tag 3$
in the usual manner the chain rule yields
$\dfrac{dT}{dt} = \dfrac{ds}{dt} \dfrac{dT}{ds}; \tag 4$
the first Frenet-Serret equation,
$\dfrac{dT}{ds} = \kappa N, \tag 5$
is valid provided $\kappa$, the curvature of $x(t)$, does not vanish, and serves to define $N$, the unit normal field to $x(t)$; via (4) and (5), (3) becomes
$x'' = \dfrac{d^2s}{dt^2}T + \left (\dfrac{ds}{dt} \right )^2 \kappa N. \tag 6$
We oberve that (1)-(4) apply to any regular curve $x(t)$, and (5) and (6) as well when $\kappa$ does not vanish, which in accord with (1), (4) and (5) is true provided that
$\dfrac{dT}{dt} \ne 0; \tag{6.5}$
if in addition $x(t)$ satisfies the given hypothesis
$N = x'' - (x'' \cdot T)T = x'' - \dfrac{1}{2}\dfrac{d}{dt}(T \cdot T)T, \tag 7$
then
$(x'' \cdot T)T = \dfrac{1}{2}\dfrac{d}{dt}(T \cdot T)T, \tag 8$
whence
$x'' \cdot T = \dfrac{1}{2}\dfrac{d}{dt}(T \cdot  T) = \dfrac{1}{2}\dfrac{d}{dt}(1) = 0; \tag 9$
in light of (6) we find
$\dfrac{d^2s}{dt^2} = \dfrac{d^2s}{dt^2}T \cdot T + \left (\dfrac{ds}{dt} \right )^2 \kappa N \cdot T = x'' \cdot T = 0; \tag{10}$
thus
$\dfrac{ds}{dt} = v, \text{a constant}, \tag{11}$
that is, $x(t)$ is a constant-speed curve; hence we may write
$s(t) = v(t - t_0) + s_0, \tag{12}$
where
$s_0 = s(t_0); \tag{13}$
furthermore, via (9), equation (7) becomes
$N = x''; \tag{14}$
combining this with (6) and (10) we find
$N = \left (\dfrac{ds}{dt} \right )^2 \kappa N, \tag{15}$
from which we infer
$\left (\dfrac{ds}{dt} \right )^2 \kappa = 1, \tag{16}$
that is,
$\kappa = \left(\dfrac{ds}{dt} \right )^{-2} = v^{-2} = \dfrac{1}{v^2}; \tag{17}$
note that (1) and (11) imply
$v \ne 0. \tag{18}$
We may also take this line one or two steps further and develop an expression for the torsion $\tau$ of $x(t)$ in terms of $v$ and $x'(t)$, $x''(t)$, and $x'''(t)$.  Said torsion may be defined via the third Frenet-Serret equation
$\dfrac{dB}{ds} = -\tau N, \tag{19}$
where the unit binormal vector $B$ is defined through
$B = T \times N; \tag{20}$
from (2) and (11) we have
$T = \dfrac{dt}{ds} x' = v^{-1}x'; \tag{21}$
thus, substituting this and (14) into (20),
$B = v^{-1}x' \times x''; \tag{22}$
thus
$\dfrac{dB}{ds} = \dfrac{dt}{ds}\dfrac{dB}{dt} = v^{-1}\dfrac{d}{dt}(v^{-1}x' \times x'')$
$= v^{-2}(x'' \times x'' + x' \times x''') = v^{-2}(x' \times x'''); \tag{23}$
in light of this equation and (14), (19) becomes
$v^{-2}(x' \times x''') = -\tau x'', \tag{24}$
or
$\tau x'' = -v^{-2}(x' \times x'''); \tag{25}$
now (14) implies $x''$ is a unit vector, whence
$\tau = \tau x'' \cdot x'' = -v^{-2}(x' \times x''') \cdot x''$
$= v^{-2}(x''' \times x') \cdot x'' = v^{-2}(x' \times x'') \cdot x''', \tag{26}$
where we have used [standard vector algebra][1] to effect these re-arrangements.  We may re-express this result via (14) and (21):
$\tau =  v^{-2}(x' \times x'') \cdot x''' = v^{-1}(v^{-1}x' \times x'') \cdot x'''$
$= v^{-1}(T \times N) \cdot x''' = v^{-1} B \cdot x'''. \tag{27}$
Our OP Nono4271 posed the specific question, "how $(x'' \cdot T)T$ becomes $\frac{1}{2}\frac{d}{dt}(T \cdot T)T$?" it is evident that
$(x'' \cdot T)T = \dfrac{1}{2}\dfrac{d}{dt}(T \cdot T)T \tag{28}$
if and only if
$x'' \cdot T = \dfrac{1}{2}\dfrac{d}{dt}(T \cdot T) = T' \cdot T, \tag{29}$
which is evidently a consequence of
$x'' = T'; \tag{30}$
inspection of (3) reveals that this holds when
$\dfrac{ds}{dt} = 1, \; \dfrac{d^2s}{dt^2} = 0, \tag{31}$
that is, when $ds/dt$ is contant, and in fact
$s = t + c \tag{32}$
for some constant $c$; thus $t$ is essentially the arc-length along $x(t)$.  Indeed, (32) implies
$\dfrac{ds}{dt} = 1, \; \dfrac{dt}{ds} = 1, \tag{33}$
whence
$x'(t) = \dfrac{dx}{dt} = \dfrac{dt}{ds}\dfrac{dx}{dt} = \dfrac{dx}{ds} = T(t), \tag{34}$
and
$x''(t) = T'(t), \tag{35}$
as required above ca. (30).
Returning momentarily to (7), suppose for the moment that $V$ is any unit vector field along the curve $x(t)$, and set
$M =  x'' - (x'' \cdot V)V; \tag {36}$
then $M$ is a vector field along $x(t)$, orthogonal to $V$, for
$M \cdot V = x'' \cdot V - (x'' \cdot V)V \cdot V = x'' \cdot V - x'' \cdot V = 0, \tag{37}$
since
$V \cdot V = 1; \tag{38}$
if in addition we choose
$V = T, \tag{39}$
we find that $M$ is normal to the curve $x(t)$; but it is not necessarily the normal $N$ given by (5); as such, the mere existence of such $M$ is not in and of itself a rich source of information concerning the curve $x(t)$; the specific hypothesis that $M = N$, however, allows specific inferences concerning the nature of $x(t)$ to be drawn, since it imposes significant restrictions on both $T$ and $N$.
It may very well be possible to derive further properties of these curves $x(t)$, but we have in fact found their curvature (17) and torsion (27), so perhaps this is a good enough place to leave off.
