On the curvature extrema of special cubic Bézier curves

01/19/2021 ∙ by Kenjiro T. Miura, et al. ∙ 0

It is proved that special cubic Bézier curves, generated from quadratic curves by the use of a scalar parameter, have at most one local curvature extremum in the (0,1) parameter interval.



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1 Introduction

In this document we are going to prove that a cubic Bézier curve


with the special control point configuration


has at most one local extremum of curvature when and .

2 Special cases

We need to treat two special cases first. When , the curve degenerates to a line segment, which has a kink (a point with infinite curvature) at , so it clearly has exactly one local curvature extremum.

With this handled, let us set, without loss of generality,


where . The second special case is when . Once again, the curve degenerates to a line segment. When , its curvature is always , otherwise it has a single kink.

In the following we will assume .

3 Curvature extrema

The signed curvature of a planar polynomial curve is given [1] as


where is the first derivative of with respect to etc. The derivative of its square is




Here means that the curvature is equal to , and it corresponds to inflection points. Consequently,


corresponds to curvature extrema.

4 Main proof

Let denote the left-hand side of Eq. (7), applied to our curve, with control points given as in (3). We need to show that has at most one 0-crossing when and .

First note that


which is positive for . Let us denote the expression in brackets with . We have , and


from which we can see that is negative at , and increases with . We can conclude that as goes from to , first decreases, and then it may increase, but since , it always remains negative, so


The derivative of with respect to is given as




It is easy to see that , since its derivative with respect to is negative, and is positive. Consequently, the signs of and are the same.

As for , note that it is a quadratic function of , and the coefficient of is , which is negative for . We can also prove


by ascertaining that , and , as this means that decreases as goes from to , starting from a negative value, and thus itself is always negative, as well.

4.1 Case I:

The maximum of for a fixed value is found by solving , giving


When , this will be in the interval, and


Since is quadratic in , and also , and , we can see that as goes from to , the value of decreases, starting from a negative value, and while it may start to increase, it remains negative.

In summary, we have shown that in this case the maximum of is negative, so , i.e., the curvature decreases monotonically.

4.2 Case II:

In this case takes its maximum over the interval at . We can also state the following (these will be proved below):


Equation (21) shows that decreases monotonically as goes from to , while by Eq. (20) it starts from a positive value, so it may have at most one 0-crossing. Consequently first increases, starting from a negative value (Eq. 14), and then it may decrease.

When , since this is its maximal value, it means that is always negative, so the curvature decreases monotonically.

Otherwise has exactly one 0-crossing, so first decreases, starting from a positive value, and then increases, as goes from to . Since in this case (Eq. 19), there is exactly one curvature extremum.

4.2.1 Proof of Eq. (19)



it can be seen that can only be positive when . Under these constraints it holds that , so


Consequently, the assumption implies .



shows that is negative when is outside the circle with center and radius . But will be outside for any , which proves Eq. (19).

4.2.2 Proof of Eq. (20)

We have


and for any , Eq. (20) is satisfied.

4.2.3 Proof of Eq. (21)

Derivating twice,


which is negative for all .


  • [1] M. P. doCarmo, Differential geometry of curves and surfaces. Prentice-Hall, 1976.