Bézier curve

Remarks

The coefficients $\binom{n}{k}(1-t)^{n-k} t^k$ are the Bernstein basis polynomials of degree $n$; they are nonnegative on $[0, 1]$ and sum to $1$. So $B(t)$ lies in the convex hull of the control points, $B(0) = P_0$, $B(1) = P_n$, $B'(0) = n(P_1 - P_0)$ and $B'(1) = n(P_n - P_{n-1})$. Example: $n = 2$ with $P_0 = (0, 0)$, $P_1 = (1, 2)$, $P_2 = (2, 0)$ gives $B(t) = (2t, 4t - 4t^2)$, an arc of the parabola $y = 2x - x^2$. Non-example: no arc of a circle of positive length is a Bézier curve. Indeed, a polynomial map $t \mapsto (x(t), y(t))$ with $x^2 + y^2 = 1$ must be constant. If $x$ and $y$ had top degree $m \ge 1$, the coefficient of $t^{2m}$ in $x^2 + y^2$ would be a positive sum of squares. A rational Bézier curve adds positive weights, $B(t) = \sum_k w_k b_{k,n}(t) P_k / \sum_k w_k b_{k,n}(t)$, and does represent circular arcs.