A bezier curve B(t) has 4 control points P0 = (0,0,0) , P1 = (1,0,0), P2 = (1,1,1) , P3 = (0,1,1). Which of the following are correct:

a) B(-2) is not defined
b) B(-1) = (-1,-1,-1)
c) B(t) , t in [0,1] is in the plane that P0,P1,P2 define.

I don't know if t can have values in a range not in [0,1]. The Bezier curves are parametric relations in general. We can define parametric equations that describe curves and t not in [0,1], but i never have read about Bezier Curves with t not in [0,1] so i am confused. For c) i believe that the Bezier Curve is in the plain that ALL control points define and not in some of them.

Again we have a Bezier curve B(t) with 4 control points but now in 2-dimensions.

The P0 = (1,0) and the P3 = (0,1) The curve must be tangent at P0 in y-axis and tangent at P3 at x-axis [*]. When t = 0.5 must pass from the point A(4,4).

Find control points P1 and P2.

In the equation that describe the cubic Bezier curve we set t =0.5. Then we create two relations one for the x and one for the y. But if P1(x1,y1) and P2 = (x2,y2) we have 4 unknowns and 2 relations. So we need another 2. Here i am confused about the *. Especially * means that derivative of the equation that describe the cubic bezier curve must be 0 in x coordinate, for t = 0 ? How i can create the other two equations that we help me to solve the problem ?

  • 1
    $\begingroup$ RE: "For c) i believe that the Bezier Curve is in the plain that ALL control points define and not a some of them." A quadratic Bezier lies in the plane of the 3 control points. A cubic allows the curve not be constrained to a plane (assuming the control points are non-planar). $\endgroup$
    – Simon F
    Jul 21, 2021 at 9:51

1 Answer 1


Regarding the first question: a bezier curve is defined as linear combinations of control points and basis function (Bernstein polynomials). Although usually one considers a unit interval as parametrization you can also evaluate the polynomials with e.g. negative numbers. So i guess b) is right (i did not do the math, but the other two answers cannot be right).

Regarding the second question: you are right that you are looking for 4 unknowns. The tangent constraints constitute 2 equations (one for P0 one for P3). The other constraint that the curve needs to pass A make up 2 more equations,so the problem should be well defined.

One property of Bezier curves is that at its end points the curve is tangent to the control polygon. So in your example you have the additional conditions P1 - P0 = (0,a) and P3 - P2 = (b, 0).

  • $\begingroup$ In the second problem. I am confused about how i can create the four equations that i need. The first 2 are made by the replace of value t =0.5 in the equation that describe the cubric bezier curve. How i can make the other two equations that i need ? $\endgroup$ Jul 20, 2021 at 5:12
  • $\begingroup$ Do i need set dy/dt |(t=0 ) = 0 to be in x axis and dx/dt |(t=0) = 0 for y-axis or with this way find something other from tangent ? $\endgroup$ Jul 20, 2021 at 5:22

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