Understanding why $X''(s).X'(s)=0$ where $X$ is parametrized with respect to arc length. Say we have the parametrized curve $X(t)=(x_1(t),x_2(t),x_3(t))$. Then it is not necessary that $X''(t).X'(t)=0$. 
However, if we parametrize the same curve with respect to the arc length $s$, then it is necessary that $X''(s).X'(s)=0$. 
I know the proof to this. But I can't seem to develop a "feel for it". 
Say we have a point on the curve $(1,1,1)$. Evidently, the values of $s$ and $t$ for which this point is defined may be different. Let $s=s_0$ and $t=t_0$, where $s_0\neq t_0$. Won't the tangents $X'(s_0)$ and $X'(t_0)$ be in the same direction? If not, why so?
Thanks in advance!
 A: Well, this is rather rough, but here is a way to think about it.
When you parametrize by arc length, you are essentially setting the speed to a constant, that is, $\|\dot{x}(t)\| = \sigma$ for some $\sigma>0$. 
Since the 'forward' speed (that is the projection of the velocity in the direction of travel) is constant, the only way the velocity can change is in a direction perpendicular to the direction of travel. Hence the acceleration is perpendicular to the direction of travel.
More rigorously, if we differentiate $ {1 \over 2} \|\dot{x}(t)\|^2 = {1 \over 2} \sigma^2$, we obtain $\langle \ddot{x}(t), \dot{x}(t) \rangle = 0$.
Regarding your last question, just think about a single dimensional curve $x(t) = t^2$ on $[0,1]$. The parametrization is straightforward, it is just $\lambda(t) = \sqrt{t}$, so we obtain $y(t) = x(\lambda(t)) = t$.
In the first case, the acceleration is $2$ and the second it is $0$. The point here is that the component of acceleration in the 'direction' of travel is (in some sense) the difference between the curves.
To elaborate a little more, consider travel on an ellipse with very different minor/major axes. The constant speed parametrization will always have the acceleration perpendicular to travel (otherwise it would not be constant speed), but the constant angular frequency parametrization must speed up and slow down and so the acceleration has a non-zero component along the direction of travel (except at 4 points). It is this component of acceleration along the direction of travel that prevents the two tangents from lining up.
A: It's not so much that $X'(t_0)$ and $X'(s_0)$ don't point in the same direction (they always do) as it is that $X''(t_0)$ and $X''(s_0)$ don't point in the same direction, in general, unless the arc-length parameter $s$ is linearly related to $t$, i.e. $s = ct + b$. I will expand upon these remarks in what follows.
Consider the original curve $X(t)$ and its arc-length parametrized version $X(s)$.  If we assume we are initially given $X(t)$, we can compute it's arc-length function $s$ in the well-known manner  via the integral formula
$s(t) = \int_{t_0}^t \sqrt{X'(u) \cdot X'(u)} du + s_0, \tag{1}$
which fixes $s(t_0) = s_0$.  Having $s(t)$, we can re-parametrize $X(t)$ with $s$, obtaining $X(s)$, the arc-length parametrized version of $X(t)$; it should be noted at this point that, as long as the curve $X(t)$ is regular, meaning $X'(t) \ne 0$, that
$s'(t) = \dfrac{ds}{dt} = \sqrt{X'(t) \cdot X'(t)} > 0, \tag{2}$
showing both that $s(t)$ is monotonically increasing and invertible, at least locally, so that there is also a function $t(s))$ with $t(s(t)) = t$; $(ds / dt)(dt / ds) = 1$ as well.  Note that we have $X(t_0) = X(s_0)$, by virtue of equation (1), though $s_0$ is an arbitrary constant and hence the values of $t_0, s_0$ are essentially independent of one another; for any choice of $t_0$ in the domain of $X(t)$, any value of $s_0$ may be taken in (1).  It doesn't matter whether $s_0 = t_0$ or $s_0 \ne t_0$, just as it doesn't matter whether $X(t)$, $X(s)$ pass through $(1, 1, 1)$ or any other point, which may easily be arranged by translating $X(t)$ and/or $X(s)$ by a fixed vector $Z$:  $X(t) \to X(t) + Z$, $X(s) \to X(s) + Z$; it is the tangent vectors $X'(t)$, $X'(s)$ which matter here, not any specific point through which the curves may pass.  Indeed, $(X(t) + Z)' = X'(t)$, $(X(s) + Z)' = X'(s)$, so these tangents remain invariant under translations by (constant) vectors $Z$.
Having hopefully clarified these points, we proceed by examining in detail the relationship between $X'(t)$ and $X'(s)$.  Since $s = s(t)$ is a function of $t$, we can relate the derivatives of $X(t), X(s)$ to one another using the chain rule:
$X'(t) = \dfrac{dX(s(t))}{dt} = \dfrac{ds}{dt} \dfrac{dX(s)}{ds} = \dfrac{ds}{dt} X'(s),\tag{3}$
which, since $ds / dt > 0$ by (2), shows that $X'(t)$ and $X'(s)$ point in the same direction, always.  However, if we take things a step further and look at the second derivative of $X(t)$, which, if we interpret $t$ as a sort of time coordinate, would correspond the the acceleration of the curve $X(t)$, we see from (3) that
$X''(t) = \dfrac{d^2 s}{dt^2} X'(s) + (\dfrac{ds}{dt})^2 X''(s), \tag{4}$
the square of $ds/dt$ in the second term arising from the fact that
$\dfrac{X'(s)}{dt} = \dfrac{ds}{dt}X''(s), \tag{5}$
again making use of the chain rule.  Inspection of (4) reveals that, though $X'(s) \cdot X''(s) = 0$ by virtue of the fact that $X'(s) \cdot X'(s) = 1$, $X'(t) \cdot X''(t) \ne 0$ in general.  Indeed, it follows from (3) and (4) that
$X'(t) \cdot X''(t) =  \dfrac{ds}{dt}\dfrac{d^2 s}{dt^2} X'(s) \cdot X'(s) + (\dfrac{ds}{dt})^3 X''(s) \cdot X'(s) =  \dfrac{ds}{dt}\dfrac{d^2 s}{dt^2}; \tag{6}$
since $ds / dt > 0$, this equation shows that $X'(t) \cdot X''(t) = 0$ if and only if $d^2s/ dt^2 = 0$, that is, if and only if $s = ct + b$ for suitably chosen constants $c$ and $b$.  In the event that this is the case, we have $ds/dt = c$, whence from (2) we see that
$X'(t) \cdot X'(t) = c^2, \tag{7}$
so that $X'(t) \cdot X'(t)$ is constant as well.  Then differentiating (7) yields
$X''(t) \cdot X'(t) = 0, \tag{8}$
which shows that $X''(t)$ is normal to $X'(t)$ when $s$ is a linear function of $t$, that is, when $d^2s / dt^2 = 0$.  Furthermore, we then have by (3) that 
$X'(s) = c^{-1} X'(t), \tag{9}$
so that $X'(s)$ is merely a re-scaling of $X'(t)$ by the constant $c^{-1} = dt / ds$.  If (7) is inserted into (1) we find
$s = c(t - t_0) + s_0 = \Vert X'(t_0) \Vert (t - t_0) + s_0 = \Vert X'(t_0) \Vert t + (s_0 - \Vert X'(t_0) \Vert t_0), \tag{10}$
so that $c = \Vert X'(t_0) \Vert$ and $b = s_0 - \Vert X'(t_0) \Vert t_0$.  Equation (6) thus shows that, though $X'(t)$ and $X'(s)$ always point in the same direction, 
$X''(t)$ and $X''(s)$ generally do not, since $X''(t)$ may have a component, $d^2s/dt^2$, in the $X'(s)$ direction, whereas $X''(s)$ is always orthogonal to $X'(s)$.
The above discussion has some noteworthy interpretations in terms of the basic physics of curvilinear motion.  If, again, $t$ is taken to be time, then equation (4) for example tells us that the acceleration of the curve $X(t)$ has in general components both along $X'(s) = T$, the unit tangent vector field to $X(s)$, and $X''(s) = T'(s) = \kappa N(s)$, which as we see points in the direction of the unit normal field $N(s)$; $\kappa$ is of course the curvature of $X(t)$, $X(s)$.  (The equations $X'(s) = T$ and $T'(s) = \kappa N$ are indeed part of the Frenet-Serret apparatus of the curve $X(s)$.)  If we write out (4) in terms of $T(s)$ and $N(s)$ we find that
$X''(t) = \dfrac{d^2 s}{dt^2} T(s) + (\dfrac{ds}{dt})^2 \kappa N(s) = \dfrac{d}{dt}(\dfrac{ds}{dt})T(s) + (\dfrac{ds}{dt})^2 \kappa N(s), \tag{11}$
which indicates that the component of the acceleration vector $X''(t)$ along $X(t)$, $X(s)$ is simply the rate of change of the speed $ds/dt$, as in the case of one-dimensional motion along a line, whereas the component in the direction of the normal $N(s)$ is $(ds/dt)^2 \kappa$.  The term $(\dfrac{ds}{dt})^2 \kappa N(s)$ is in fact a generalization of the notion of centripetal force familiar from the dynamics of circular motion; this may be seen by noting that, for a circle of radius $r$, $\kappa = 1/ r$, so that $(ds/dt)^2 \kappa = (ds/dt)^2 /r$, the magnitude, per unit mass, of a centrally directed force required to maintain a trajectory in the round.  Taking things a step further, from (6) we see that
$X'(t) \cdot X''(t) =  \dfrac{ds}{dt}\dfrac{d^2 s}{dt^2} = \dfrac{d}{dt}(\dfrac{1}{2}(\dfrac{ds}{dt})^2), \tag{12}$
which is itself a generalization of another fact from basic dynamics:  the rate of change of kinetic energy, or power, per unit mass a particle undergoes is given by the velocity times the acceleration.  We see that all changes in the kinetic energy ($1/2(ds/dt)^2$) arise from the component of acceleration along $X(t)$; acceleration normal to $X(t)$, $X(s)$ alters the direction, but not the kinetic energy, of a particle.
It's probably worth observing that there is nothing essential about three dimensions in the preceding discussion.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
