Why are normal unit vectors in a different direction than acceleration vectors? We learned today in class about normal unit vectors of vector-valued functions. I don't understand why they aren't in the same direction as the acceleration vector.
The acceleration is the second derivative and isn't the normal unit vector just a double scaled version of the second derivative? The magnitude of $T(t)$ and $N(t)$ act as scalars when they are multiplied to make the vectors unit vectors, right? Therefore wouldn't the direction remain the same as the second derivative? What am I missing?
 A: Let's take a curve in 2D parametrized in $t$ as $\bf r(t) =(x(t),y(t))$.   
If the curve is traced at constant speed $v$
$$
v = \sqrt {x'(t)^{\,2}  + y'(t)^{\,2} } \quad  \to \quad x'(t)^{\,2}  + y'(t)^{\,2}  = v^{\,2} 
$$
then derivating the above we get
$$
0 = 2x'(t)x''(t) + 2y'(t)y''(t) = x'(t)x''(t) + y'(t)y''(t)
$$
i.e.
$$
{\bf r}'(t)\, \cdot \,{\bf r}''(t) = 0
$$
so acceleration is normal to speed. 
However, a given curve $y(x)$ can be parametrized in many different ways, that is it can be traced
by a point moving at speed not constant.
In that case the above clearly does not hold. the $dv/dt$ will add to the acceleration a component parallel
to ${\bf r}'(t)$ which disrupts the normality between the two vectors.  
Analitically, given the curve $y(x)$ we define the tangent and the normal unit vectors to it as
$$
\left\{ \matrix{
  ds = \sqrt {1 + \left( {{{dy} \over {dx}}} \right)^{\,2} } dx = \sqrt {1 + \left( {{{y'(t)} \over {x'(t)}}} \right)^{\,2} } x'(t)dt = \sqrt {x'(t)^{\,2}  + y'(t)^{\,2} } dt \hfill \cr 
  {\bf t}(s) = {d \over {ds}}{\bf r}(s)\quad {\bf n}(s) = {d \over {ds}}{\bf t}(s) = {{d^{\,2} } \over {ds^{\,2} }}{\bf r}(s) \hfill \cr}  \right.
$$
then for the speed and acceleration wrt to the parameter $t$, we get
$$
\eqalign{
  & {\bf v}(t) = {d \over {dt}}{\bf r}(t) = {d \over {ds}}{\bf r}(s){{ds(t)} \over {dt}} = v(t)\,{\bf t}(t)  \cr 
  & {\bf a}(t) = {d \over {dt}}{\bf v}(t) = {{d^{\,2} } \over {ds^{\,2} }}{\bf r}(s)\left( {{{ds(t)} \over {dt}}} \right)^{\,2}  + {d \over {ds}}{\bf r}(s){{d^{\,2} s(t)} \over {dt^{\,2} }} =   \cr 
  &  = v(t)^{\,2} {\bf n}(t) + a(t){\bf t}(t) \cr} 
$$
so that if $s$ is a linear function of  $t$, $a(t)$ is null
and the vectorial acceleration  will be only centripetal.
A: If the particle is speeding up or slowing down, there will be a component of acceleration in the direction of travel.
