Bézier Curves: where the 3 (or 1/3) constant comes from when moving from Hermite curves? I'm learning Bézier Curves, and I'm stuck at the reason why they use 3 (or 1/3) constant when moving from Hermite curves?
Like in this, this, and this source. 
e.g. why t0 = 3(q1 - q0) ?

or why v1 = v0 + 1/3 d0 in the first link:
x0         → v0
→ v1 = v0 + ⅓d0
→ v2 = v3 – ⅓d1
x1         → v3
 A: A cubic Bezier curve is represented as 
$$B(t)=(1-t)^3Q_0+3(1-t)^2tQ_1+3(1-t)t^2Q_2+t^3Q_3$$
and its first derivative is 
$$B'(t)=3(1-t)^2(Q_1-Q_0)+6t(1-t)(Q_2-Q_1)+3t^2(Q_3-Q_2)$$
For cubic Hermite curve, $P_0,T_0, P_1$ and $T_1$ are the position and first derivative vectors at curve's start ($t=0$) and end ($t=1$). So, we can evaluate the cubic Bezier curve's position vector and first derivative at $t=0$ and $t=1$ to obtain
$P_0=B(0)=Q_0$
$T_0=B'(0)=3(Q_1-Q_0)$
$P_1=B(1)=Q_3$
$T_1=B'(1)=3(Q_3-Q_2)$.
A: As the answer from @fang shows, the $\tfrac13$ arises from the use of the Bernstein polynomials (the blending functions you use to define Bezier curves). 
But there are other forms of the same curve that use different constants. The Timmer form of the curve uses a constant of $\tfrac14$, and the Ball form uses $\tfrac12$. Different blending functions will give you different constants. There's nothing magic about $\tfrac13$; in some ways $\tfrac14$ (the Timmer form) is a better choice.
There's a picture showing the Timmer control points in my answer to this question.
