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 $\frac{r''(t)}{||r''(t)||}$?
On
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 $$\frac{r'(t)}{\|r'(t)\|}.$$
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 $T(t)$.