Ways to convert Bezier curve to parametric equations. So my task was to generate Bezier curve by points and measure its length. The only way I found is to rewrite it as a parametric equation and then use numerical integration. But I have no idea how to convert it to a parametric equation. Any help would be appreciated!


Let us take the example of a Bezier curve defined by a moving point $M=M_t$ defined by 4 points $A,B,C,D$ in this order: starting at $M_0=A$, ending at $M_1=D$ and "influenced" by points $B$ and $C$. I will work in 2D (for 3D, add third coordinates):
This curve is defined by the synthetic expression: 
$$\tag{1}M=(1-t)^3A+3t(1-t)^2B+3t^2(1-t)C+t^3D,$$
(coefficients $v^3, 3v^2w, 3vw^2, w^3$ come from Newton's binomial expansion ; an essential property is that their sum is $(u+v)^3=1$ for all $t$).
where the meaning of (1) is a pair of parametric equations :
$$\binom{x}{y}=(1-t)^3\binom{x_A}{y_A}+3t(1-t)^2\binom{x_B}{y_B}+3t^2(1-t)\binom{x_C}{y_C}+t^3\binom{x_D}{y_D}$$
For example, if $A=\binom{0}{0}$, $B=\binom{1}{0}$, $C=\binom{1}{1}$, $D=\binom{0}{1}$, we will have:
$$\binom{x}{y}=\binom{3t(1-t)^2+3t^2(1-t)}{3t^2(1-t)+t^3}=\binom{3t(1-t)}{t^2(3-2t)} \ \text{with} \ 0 \leq t \leq 1.$$

