Let $P(t)$ be a Bézier curve of degree $N$ with control points $P_0, \ldots, P_N$ . Show that $P'(0) = N(P_1−P_0)$ and $P'(1) = N(P_N − P_{N−1})$

Question

I am trying to learn more about Bézier Curves and I am following an old professor's course online, and I can't seem to see how to go about this problem. I understand the format of finding $x(t)$ and $y(t)$ given control points. It's something along the lines of: \begin{align} x(t) &= x0(1-t)^3 + x1 * 3t(1-t)^2 + x2 * 3t^2 (1-t) + x3 * t^3 \\ y(t) &= y0(1-t)^3 + y1 * 3t(1-t)^2 + y2 * 3t^2 (1-t) + 3 * t^3 \end{align} assuming there are 4 control points of course.

Any help at all would be appreciated!

Answer 1

You seem to know already that the curve equation is $$ P(t) = (1-t)^3P_0 + 3t(1-t)^2P_1 + 3t^2(1-t)P_2 + t^3P_3 $$ Differentiate this with respect to $t$, and set $t=0$, and you will get $P'(0) = 3(P_1 - P_0)$.

Similarly, setting $t=1$ will give you $P'(1) = 3(P_3 - P_2)$.

To get the more general result (rather than the case $N=3$ that I outlined above), you have to start with the general equation of a Bézier curve of degree $N$.