Expressing a unit tangent vector in terms of r(t) Is there a simple way to express $N(t)$, the unit normal vector of a vector curve, in terms of $r(t)$?  I know that $T(t)$=$\frac{r'(t)}{||r'(t)}||$ and that $N(t)$=$\frac{T'(t)}{||T'(t)||}$.  Is it possible to simplify the definition of $N(t)$, or is the simplest version [$\frac{r'(t)}{||r'(t)||}$]'?
Why is $N(t)$ not defined as just $\frac{r''(t)}{||r''(t)||}$?
 A: Note that the quotient rule means that the two expressions are not equal.
$$
\left(\frac{r'(t)}{\|r'(t)\|} \right)' =
\frac{\|r'(t)\|r''(t) - \|r'(t)\|'r'(t)}{\|r'(t)\|^2} \neq \frac{r''(t)}{\|r''(t)\|}
$$
Why use the more complicated-looking one? Well, we want $N(t)$ to be perpendicular to $T(t)$.
A: Because $N$ must be perpendicular to $T$, but
$$r'(t)\cdot r''(t)=0$$ has no reason to hold (this is equivalent to $\|r'(t)\|=Cst$).
$N$ also belongs to the plane defined by $r'(t)$ and $r''(t)$, so it must be parallel to $$(r''(t)\times r'(t))\times r'(t),$$or by the expulsion formula
$$r''(t)r'^2(t)-r'(t)(r''(t)\cdot r'(t)).$$
Dividing by $\|r'(t)\|^3$, you find
$$\frac{r''(t)}{\|r'(t)\|}-\frac{r'(t)(r''(t)\cdot r'(t)}{\|r'(t)\|^3},$$
precisely the derivative of
$$\frac{r'(t)}{\|r'(t)\|}.$$
A: It is possible to define it as $\frac{r''(t)}{||r''(t)||}$ if the curve is naturally parameterized ("constant velocity"), so that the first derivative is already normalized.
Physically, this is easy to understand. If $r(t)$ is trajectory, then $r'(t)$ is velocity, and $T(t)$ is the direction of velocity. $r''(t)$ is acceleration. If your velocity is constant -- $||r'(t)||=\text{const}$ (by scaling the parameter $t$ you can make it equal to $1$), then your only acceleration is the centripetal acceleration (locally equal to rotation). If your velocity is changing, you also have tangential acceleration, so you must normalize the velocity before taking the second derivative.
