# On the curvature extrema of special cubic Bézier curves

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

 C(t)=3∑i=0Pi(3i)ti(1−t)3−i (1)

with the special control point configuration

 P0 =Q0, P1 =(1−a)Q0+aQ1, P2 =aQ1+(1−a)Q2, P3 =Q2 (2)

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,

 Q0 =(−1,0), Q1 =(b,h), Q2 =(1,0), (3)

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

 κ=x′y′′−x′′y′(x′2+y′2)32, (4)

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

 (κ2)′=2κκ′ =2(x′y′′−x′′y′)(x′y′′′−x′′′y)(x′2+y′2)3(x′2+y′2)6 −(x′y′′−x′′y′)23(x′2+y′2)2(2x′x′′+2y′y′′)(x′2+y′2)6 (5)

so

 (x′2+y′2)4κκ′ =(x′y′′−x′′y′)[(x′y′′′−x′′′y)(x′2+y′2) −3(x′y′′−x′′y′)(x′x′′+y′y′′)]. (6)

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

 (x′y′′′−x′′′y)(x′2+y′2)−3(x′y′′−x′′y′)(x′x′′+y′y′′)=0 (7)

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

 N(0,a)=−324a2h[(1+b)(12+3a2(5+b)−4a(7+b))+a(−4+3a)h2], (8)

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

 df0(a)da=6(b2+6b+5+h2)a−4(b2+8b+7+h2), (9)

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

 N(0,a)>0. (10)

The derivative of with respect to is given as

 ∂N(t,a)∂t=1296ah⋅f1(t,a)⋅f(t,a), (11)

where

 f1(t,a) =2t2−2t+1−(3t2−3t+1)a, (12) f(t,a) =−3a2(b2+b(10−20t)+h2+60(t−1)t+13) +4a(5b(1−2t)+60(t−1)t+11)+80(1−t)t−12. (13)

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

 f(0,a)<0, (14)

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: b≤3−2a

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

 t0=ab+3a−22(3a−2). (15)

When , this will be in the interval, and

 f(t0,a) =8−16a+6a2−3a2h2+2a2b2 (16) <8−16a+6a2−3a2h2+2a2(3−2a)2 (17) =(24−3h2)a2−40a+16=:f3(a). (18)

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: b>3−2a

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

 N(1,a) <0 when f(1,a)>0, (19) ∂f(0,a)∂t >0, (20) ∂2f(t,a)∂t2 <0. (21)

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)

Since

 f(1,a)=−3a2(15a−103a−b)2−3a2h2+36(a−23)(a−89), (22)

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

 f(1,a)|b<1

Consequently, the assumption implies .

Restructuring

 N(1,a) =−324a2h[12(b−1)+4a(7+(b−8)b+h2)−3a2(5+(b−6)b+h2)] (24) =−324a3(4−3a)h⎡⎣(b−−9a2+16a−6(4−3a)a)2−36(a−1)4(4−3a)2a2+h2⎤⎦ (25)

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

 ∂f(0,a)∂t=20[a(3a−2)b+3a(3a−4)+4], (26)

and for any , Eq. (20) is satisfied.

#### 4.2.3 Proof of Eq. (21)

Derivating twice,

 ∂2f(t,a)∂t2=40(12a−9a2−4), (27)

which is negative for all .

## References

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