EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 13, No. 2, 2020, 216-226 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global On the Involute of the Cubic Bezier Curve By Using Matrix Representation in E3 Şeyda Kılıçoğlu1,∗, Süleyman Şenyurt2 1 Faculty of Education, Department of Mathematics, Başkent University, Ankara, Turkey 2 Faculty of Arts and Sciences, Department of Mathematics, Ordu Universty, Ordu, Turkey Abstract. In this study we have examined, involute of the cubic Bezier curve based on the control points with matrix form in E3. Frenet vector fields and also curvatures of involute of the cubic Bezier curve are examined based on the Frenet apparatus of the first cubic Bezier curve in E3. 2020 Mathematics Subject Classifications: 53A04,53A05 Key Words and Phrases: Bezier curves, Frenet vector fields, Cubic Bezier curve 1. Introduction and Preliminaries In 1962 Bézier curves was studied by the French engineer Pierre Bézier, who used them to design automobile bodies. But the study of these curves was first developed in 1959 by mathematician Paul de Casteljau using de Casteljau’s algorithm, a numerically stable method to evaluate Bézier curves. A Bézier curve is frequently used in computer graphics and related fields, in vector graphics, used in animation as a tool to control motion. For more datail using computer graphics see in [8]. In [2] some properties of Bezier curves are examined. To guarantee smoothness, the control point at which two curves meet must be on the line between the two control points on either side. In animation applications, such as Adobe Flash and Synfig, Bézier curves are used to outline, for example, movement. Users outline the wanted path in Bézier curves, and the application creates the needed frames for the object to move along the path. For 3D animation Bézier curves are often used to define 3D paths as well as 2D curves for keyframe interpolation. We have been motivated by the following studies. First Bezier-curves with curvature and torsion continuity has been examined in [5]. Also in [10] Bezier curves and surfaces has been given. In [3] planar Bezier curves and Bishop Frame of Bezier Curves are examined, respectively. Recently equivalence conditions of control points and application to planar Bezier curves have been examined in [6]. In this study we will define and work on Frenet apparatus of Bézier curves in E3. So we need the derivates of them. Recently Bezier-Like curves has been defined and cubic Bezier curves transitions have been studied in [7]. Also in [9] designing the ruled surface are examined as a new approach. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v13i2.3648 Email addresses: seyda@baskent.edu.tr (Ş. Kılıçoğlu), senyurtsuleyman52@gmail.com (S. Şenyurt) http://www.ejpam.com 216 c© 2020 EJPAM All rights reserved. Ş. Kılıçoğlu, S. Şenyurt / Eur. J. Pure Appl. Math, 13 (2) (2020), 216-226 217 Theorem 1. The set, whose elements are Frenet vector fields and the curvatures of a curve α (t) ⊂ IE3, is called Frenet apparatus of the curves. Let α(t) be the curve, with η = ‖α′ (t)‖ 6= 1 and Frenet apparatus are {T (t) , N (t) , B (t) , κ (t) , τ (t)}. Frenet vector fields are given for a non arc-lengthed curve T (t) = α′ (t) ‖α′ (t)‖ , N (t) = B (t) ΛT (t) , B (t) = α′ (t) Λα′′ (t) ‖α′ (t) Λα′′ (t)‖ where curvature functions are defined by κ (t) = ∥∥∥α′ (t) Λα ′′ (t) ∥∥∥ ‖α′ (t)‖3 , τ (t) = 〈 α′ (t) Λα ′′ (t) , α′′′(t) 〉 ‖α′ (t) Λα′′ (t)‖2 . Also Frenet formulae are well known as T ′ N ′ B′  =  0 ηκ 0 −ηκ 0 ητ 0 −ητ 0  T N B  , [4]. Theorem 2. The Frenet-Serret vectors fields of the involute α∗ = α (t)+λ (t)T (t) , which is not an arclengthed curve with ‖α′‖ = η 6= 1, based on the its evolute curve α are T ∗ = N,N∗ = −κT + τB (κ2 + τ2) 1 2 , B∗ = τT + κB (κ2 + τ2) 1 2 . The first and the second curvatures of involute α∗, are κ∗ = √ κ2 + τ2 (c− ηt)κ , τ∗ = −τ2 ( κ τ )′ (c− ηt)κ (κ2 + τ2) (1) respectively, where dt ds∗ = 1 c− ηt , [4]. Generaly Béziers curve can be defined by n + 1 control points P0, P1, ..., Pn with the parametrization B(t) = n∑ i=0 ( n i ) ti (1− t)n−i (t) [Pi] . (2) In this study we will define and work on cubic Bézier curves which are defined in E3. For more detail see [1] . Definition 1. A cubic Bézier curve is a special Bézier curve has only four points P0, P1, P2 and P3, with the parametrization B(t) = 3∑ i=0 ( 3 i ) ti (1− t)3−i (t) [Pi] , (3) B (t) = (1− t)3 P0 + 3t (1− t)2 P1 + 3t2 (1− t)P2 + t3P3. Ş. Kılıçoğlu, S. Şenyurt / Eur. J. Pure Appl. Math, 13 (2) (2020), 216-226 218 The matrix form of the cubic Bezier curve with control points P0, P1, P2, P3, is α (t) = [ t3 t2 t 1 ]  −1 3 −3 1 3 −6 3 0 −3 3 0 0 1 0 0 0   P0 P1 P2 P3  . Also using the derivatives of a cubic Bézier curve Frenet apparatus {T (t) , N (t) , B (t) , κ (t) , τ (t)} have already been given in [1] as in the following theorems by using matrix representation. For more detail see in [1]. Theorem 3. The first derivative of a cubic Bézier curve by using matrix representation is α′(t) = [ t2 t 1 ]  1 −2 1 −2 2 0 1 0 0  Q0 Q1 Q2  (4) with the control points Q0 = 3 (P1 − P0) = (x0, y0, z0) , Q1 = 3 (P2 − P1) = (x1, y1, z1)Q2 = 3 (P3 − P2) = (x2, y2, z2) . Theorem 4. The second derivative of a cubic Bézier curve by using matrix representation is α′′(t) = [ t 1 ] [ −1 1 1 0 ] [ R0 R1 ] (5) with the control points R0 = 6 (P2 − 2P1 + P0) = 6 (x1 − x0, y1 − y0, z1 − z0) , R1 = 6 (P3 − 2P2 + P1) = 6 (x2 − x1, y2 − y1, z2 − z1) . Theorem 5. The third derivative of a cubic Bézier curve by using matrix representation is α′′′(t) = [R0R1] (6) with the control points [R0R1] = R1 −R0 = 2 [Q1Q2]− 2 [Q0Q1] = 6 (P3 − 3P2 + 3P1 − P0) . 1.1. Frenet apparatus of a cubic Bezier curve Frenet apparatus {T (t) , N (t) , B (t) , κ (t) , τ (t)} of a cubic Bézier curve have already been given in [1] as in the following theorems by using the matrix representation. Theorem 6. Tangent vector field of a cubic Bezier curve by using the matrix representa- tion is T (t) = 1 η [ t2 t 1 ]  1 −2 1 −2 2 0 1 0 0  x0 y0 z0 x1 y1 z1 x2 y2 z2  where η = ‖α′‖ . Ş. Kılıçoğlu, S. Şenyurt / Eur. J. Pure Appl. Math, 13 (2) (2020), 216-226 219 Theorem 7. Binormal vector field of a cubic Bezier curve by using the matrix represen- tation is B (t) = 6 m [ t2 t 1 ]  b11 b12 b13 b21 b22 b23 b31 b32 b33  where ‖α′Λα′′‖ = m b11 = y0(z1 − z2) + y1(z2 − z0) + y2(z0 − z1) b12 = −x0 (z1 − z2)− x1 (z2 − z0)− x2 (z0 − z1) b13 = x0 (y1 − y2) + x1 (y2 − y0) + x2 (y0 − y1) b21 = 2y1z0 + y0z2 − 2y0z1 − y2z0 b22 = 2x0z1 − 2x1z0 − x0z2 + x2z0 b23 = 2x1y0 − 2x0y1 + x0y2 − x2y0 b31 = y0z1 − y1z0 b32 = x1z0 − x0z1 b33 = x0y1 − x1y0. Theorem 8. Normal vecror field of a cubic Bezier curve by using the matrix representation is N (t) = 6 ηm [ t4 t3 t2 t1 1 ]  n11 n12 n13 n21 n22 n23 n31 n32 n33 n41 n41 n43 n51 n51 n53  where n11 = b12d13 − b13d12 n21 = b12d23 − b13d22 + b22d13 − b23d12 n31 = b12d33 − b13d32 + b22d23 − b23d22 + b32d13 − b33d12 n41 = b22d33 − b23d32 + b32d23 − b33d22 n51 = b32d33 − b33d32 n12 = b11d13 − b13d11 n22 = −b11d23 − b21d13 + b13d21 + b23d11 n32 = b23d21 + b33d11 − b11d33 − b21d23 + b13d31 − b31d13 n42 = −b21d33 − b31d23 + b23d31 + b33d21 n52 = −b31d33 + b33d31 n13 = b11d12 − b12d11 n23 = b11d22 − b12d21 + b21d12 − b22d11 n33 = b11d32 − b12d31 + b21d22 − b22d21 + b31d12 − b32d11 n43 = b21d32 − b22d31 + b31d22 − b32d21 n53 = b31d32 − b32d31. Theorem 9. First and second curvatures of a cubic Bezier curve by using the matrix Ş. Kılıçoğlu, S. Şenyurt / Eur. J. Pure Appl. Math, 13 (2) (2020), 216-226 220 representation are κ (t) = 6 η3 [ t4 t3 t2 t 1 ]  b211 + b212 + b213 2b11b21 + 2b12b22 + 2b13b23 2b11b31 + 2b12b32 + 2b13b33 + b221 + b222 + b223 2b21b31 + 2b22b32 + 2b23b33 b231 + b232 + b233,  , τ (t) = x0y1z2 − x0y2z1 − x1y0z2 + x1y2z0 + x2y0z1 − x2y1z0 ‖α′ (t) Λα′′ (t)‖2 . 2. Involute of cubic Bezier curve Definition 2. If the curve α∗ which lies on the tangent surface intersect the tangent lines orthogonally is called an involute of α. If a curve α∗ is an involute of α, then by definition α which is not an arclengthed curve.is an evolute of α∗. Hence given α, its evolutes are the curves whose tangent lines intersect α orthogonally. Let the quantities {T ∗, N∗, B∗, κ∗, τ∗} be collectively Frenet-Serret apparatus of the curve α∗ which is not an arclengthed curve with ‖α′‖ = η 6= 1, [4]. The equation of involute of the curve α has the following parametrization; α∗ (t) = α (t) + λ (t)T (t) . (7) Also since λ = c− ηt it can be written as in the following parametrization α∗ (t) = α (t) + (c− ηt) η α′ (t) . Theorem 10. The involute of a cubic Bezier curve has the matrix form based on the control points P0, P1, P2 and P3 of any cubic Bezier curve α∗ (t) = [ t3 t2 t 1 ]  −1 3 −3 1 3− 3µ −6 + 9µ 3− 9µ 3µ −3 + 6µ 3− 12µ 6µ 0 1 + 3µ −3 0 0   P0 P1 P2 P3  with µ = c−ηt η . Proof. Lets µ = (c− ηt) η , since α∗ = α (t) + µα′ (t) α∗ = [ t3 t2 t 1 ]  −1 3 −3 1 3 −6 3 0 −3 3 0 0 1 0 0 0   P0 P1 P2 P3 +µ [ t2 t 1 ]  1 −2 1 −2 2 0 1 0 0  Q0 Q1 Q2  Ş. Kılıçoğlu, S. Şenyurt / Eur. J. Pure Appl. Math, 13 (2) (2020), 216-226 221 and Q0 = 3 (P1 − P0) , Q1 = 3 (P2 − P1) , Q2 = 3 (P3 − P2) it can be written in matrix form as in α∗ (t) = [ t3 t2 t 1 ]  −1 3 −3 1 (3− 3µ) (−6 + 9µ) (3− 9µ) 3µ (−3 + 6µ) (3− 12µ) 6µ 0 (1 + 3µ) −3 0 0   P0 P1 P2 P3  we have its matrix product form as α∗ (t) = [ t3 t2 t 1 ]  3P1 − P0 − 3P2 + P3 P1 (9µ− 6)− P2 (9µ− 3)− P0 (3µ− 3) + 3µP3 P0 (6µ− 3)− P1 (12µ− 3) + 6µP2 P0 (3µ+ 1)− 3P1  . (8) Theorem 11. The control points of the involute of any cubic Bezier curve with constant speed, based on the control points of cubic Bezier curve, as in the following way I0 = 3 c η P1 − P0 ( 3 c η − 1 ) , I1 = 3 c η P1 − P0 ( c η − 1 ) − 2 c η P2, I2 = ( 6 c η + 2 ) P1 − P2 ( 7 c η + 1 ) + c η P3, I3 = ( 3 c η − 2 ) P3 − P2 ( 15 c η − 3 ) + 12 c η P1. Proof. Let I0, I1, I2, and I3 be control points of involute α∗, so we can write α∗ (t) = [ t3 t2 t 1 ]  −1 3 −3 1 3 −6 3 0 −3 3 0 0 1 0 0 0   I0 I1 I2 I3  . (9) From the equality of the left sides of 8 and 9, we have −1 3 −3 1 3 −6 3 0 −3 3 0 0 1 0 0 0   I0 I1 I2 I3  =  2P0 − 6P1 + 6P2 − 2P3( −3− 3c η ) P0 + ( 6 + 9c η ) P1 − ( 3 + 9c η ) P2 + 3c η P3 6c η P0 − 6c η P2( 1− 3c η ) P0 + 3c η P1  =  2 −6 6 −2( −3− 3c η ) ( 6 + 9c η ) − ( 3 + 9c η ) +3c η 6c η 0 −6c η 0( 1− 3c η ) 3c η 0 0   P0 P1 P2 P3  Ş. Kılıçoğlu, S. Şenyurt / Eur. J. Pure Appl. Math, 13 (2) (2020), 216-226 222 using the inverse matrix I0 I1 I2 I3  =  −1 3 −3 1 3 −6 3 0 −3 3 0 0 1 0 0 0  −1  2 −6 6 −2( −3− 3c η ) ( 6 + 9c η ) − ( 3 + 9c η ) +3c η 6c η 0 −6c η 0( 1− 3c η ) 3c η 0 0   P0 P1 P2 P3  . (10) Theorem 12. The control points of the involute of any cubic Bezier curve, under the condition c η − t = µ = constant, can be given, I0 = 3µP1 − (3µ− 1)P0, I1 = 2µP2 − µP0 − P1 (µ− 1) , I2 = µP3 − 2µP1 + P2 (µ+ 1) , I3 = (3µ+ 1)P3 − 3µP2. Proof. If c η − t = µ is constant, α∗ (t) = [ t3 t2 t 1 ]  −1 3 −3 1 3 −6 3 0 −3 3 0 0 1 0 0 0   P0 P1 P2 P3  +µ [ t2 t 1 ]  1 −2 1 −2 2 0 1 0 0  3 (P1 − P0) 3 (P2 − P1) 3 (P3 − P2)  . Hence −1 3 −3 1 3 −6 3 0 −3 3 0 0 1 0 0 0   I0 I1 I2 I3  =  −P0 + 3P1 − 3P2 + P3 ((3− 3µ)P0 + (−6 + 9µ)P1 + (3− 9µ)P2 + 3µP3) ((−3 + 6µ)P0 + (3− 12µ)P1 + 6µP2) (1− 3µ)P0 + 3µP1  =  −1 3 −3 1 (3− 3µ) (−6 + 9µ) (3− 9µ) 3µ (−3 + 6µ) (3− 12µ) 6µ 0 (1− 3µ) 3µ 0 0   P0 P1 P2 P3  using the inverse matrix we can find the control points of the involute of any cubic Bezier curve with constant µ, based on the control points of cubic Bezier curve, as in the following way: I0 I1 I2 I3  =  −1 3 −3 1 3 −6 3 0 −3 3 0 0 1 0 0 0  −1  −1 3 −3 1 (3− 3µ) (−6 + 9µ) (3− 9µ) 3µ (−3 + 6µ) (3− 12µ) 6µ 0 (1− 3µ) 3µ 0 0   P0 P1 P2 P3  , Ş. Kılıçoğlu, S. Şenyurt / Eur. J. Pure Appl. Math, 13 (2) (2020), 216-226 223  I0 I1 I2 I3  =  0 0 0 1 0 0 1 3 1 0 1 3 2 3 1 1 1 1 1   −1 3 −3 1 (3− 3µ) (−6 + 9µ) (3− 9µ) 3µ (−3 + 6µ) (3− 12µ) 6µ 0 (1− 3µ) 3µ 0 0   P0 P1 P2 P3  . (11) 2.1. Frenet apparatus of the involute curve of any cubic Bezier curve in E3 Theorem 13. Tangent vector field of involute curve of any cubic Bezier curve is T ∗ = 6 mη [ t4 t3 t2 t1 1 ]  n11 n12 n13 n21 n22 n23 n31 n32 n33 n41 n41 n43 n51 n51 n53  . Proof. We have already known that tangent vector field of involute curve T ∗ is lineer dependent N, that is why T ∗ = N. Theorem 14. Normal vector field of involute α∗of any cubic Bezier curve in E3is N∗ = [ t2 t 1 ] η (κ2+τ2) 1 2 κ  2x1 − x0 − x2 + 6ητ mκ b11 2y1 − y0 − y2 + 6ητ mκ b12 2z1 − z0 − z2 + 6ητ mκ b13 2x0 − 2x1 + 6ητ mκ b21 2y0 − 2y1 + 6ητ mκ b22 2z0 − 2z1 + 6ητ mκ b23 6ητ mκ b31 − x0 6ητ mκ b32 − y0 6ητ mκ b33 − z0  . Proof. Since N∗ = −κT + τB (κ2 + τ2) 1 2 , we have N∗ = −κ η [ t2 t 1 ]  1 −2 1 −2 2 0 1 0 0  x0 y0 z0 x1 y1 z1 x2 y2 z2 + τ6 m [ t2 t 1 ]  b11 b12 b13 b21 b22 b23 b31 b32 b33  (κ2 + τ2) 1 2 N∗ = [ t2 t 1 ] (κ2 + τ2) 1 2 −κ η  1 −2 1 −2 2 0 1 0 0  x0 y0 z0 x1 y1 z1 x2 y2 z2 + 6τ m  b11 b12 b13 b21 b22 b23 b31 b32 b33  . Hence it is easy to calculate the following result N∗ = [ t2 t 1 ] η (κ2+τ2) 1 2 κ  2x1 − x0 − x2 + 6ητ mκ b11 2y1 − y0 − y2 + 6ητ mκ b12 2z1 − z0 − z2 + 6ητ mκ b13 2x0 − 2x1 + 6ητ mκ b21 2y0 − 2y1 + 6ητ mκ b22 2z0 − 2z1 + 6ητ mκ b23 6ητ mκ b31 − x0 6ητ mκ b32 − y0 6ητ mκ b33 − z0  . Ş. Kılıçoğlu, S. Şenyurt / Eur. J. Pure Appl. Math, 13 (2) (2020), 216-226 224 Theorem 15. Binormal vector field of involute α∗of any cubic Bezier curve in E3is B∗ = [ t2 t 1 ] (κ2+τ2) 1 2 τ η  x0 − 2x1 + x2 + 6ηκ mτ b11 y0 − 2y1 + y2 + 6ηκ mτ b12 z0 − 2z1 + z2 + 6ηκ mτ b13 2x1 − 2x0 + 6ηκ mτ b21 2y1 − 2y0 + 6ηκ mτ b22 2z1 − 2z0 + 6ηκ mτ b23 x0 + 6ηκ mτ b31 y0 + 6ηκ mτ b32 z0 + 6ηκ mτ b33  . Proof. Since B∗ = τT+κB (κ2+τ2) 1 2 , B∗ = τ η [ t2 t 1 ]  1 −2 1 −2 2 0 1 0 0  x0 y0 z0 x1 y1 z1 x2 y2 z2 + 6κ m [ t2 t 1 ]  b11 b12 b13 b21 b22 b23 b31 b32 b33  (κ2 + τ2) 1 2 , B∗ = [ t2 t 1 ] τη  1 −2 1 −2 2 0 1 0 0  x0 y0 z0 x1 y1 z1 x2 y2 z2 + 6κ m  b11 b12 b13 b21 b22 b23 b31 b32 b33  (κ2 + τ2) 1 2 . Hence it is easy to give the proof. 2.2. The first and second curvature of involute α∗ Theorem 16. The first curvature of involute α∗of any cubic Bezier curve in E3is κ∗ = 1 m2(c− ηt) √ m4 + η6 (x0y1z2 − x0y2z1 − x1y0z2 + x1y2z0 + x2y0z1 − x2y1z0)2 Proof. Since κ∗ = √ κ2 + τ2 (c− ηt)κ , and κ = m η3 κ∗ = √ m2 η6 + (x0y1z2−x0y2z1−x1y0z2+x1y2z0+x2y0z1−x2y1z0)2 m2 (c− ηt)m η3 (c− ηt)κ > 0, κ 6= 0. It is trivial. Theorem 17. The second curvature of involute α∗of any cubic Bezier curve in E3is τ∗ = η3 m3 ( κ− κ′ (x0y1z2 − x0y2z1 − x1y0z2 + x1y2z0 + x2y0z1 − x2y1z0) (c− ηt) (κ2 + τ2) ) . Proof. Considering the τ∗ = κτ ′ − κ′τ (c− ηt)κ (κ2 + τ2) equation, we have τ∗ = κ− κ′ (x0y1z2 − x0y2z1 − x1y0z2 + x1y2z0 + x2y0z1 − x2y1z0) m2(c− ηt)m η3 (κ2 + τ2) . REFERENCES 225 Example 1. Find the involute of the cubic Bezier curve with control points P0 = (1, 2, 3) , P1 = (1, 1, 1) , P2 = (2, 1, 3) , and P3 = (1,−1, 0) The cubic Bezier curve has the followow- ing matrix representation α (t) = [ t3 t2 t 1 ]  −1 3 −3 1 3 −6 3 0 −3 3 0 0 1 0 0 0   P0 P1 P2 P3  . (12) The involute α∗ of the cubic Bezier curve α, has the followowing matrix representation α∗ (t) = [ t3 t2 t 1 ]  −1 3 −3 1 3− 3µ −6 + 9µ 3− 9µ 3µ −3 + 6µ 3− 12µ 6µ 0 1 + 3µ −3 0 0   P0 P1 P2 P3  . (13) Using control points as α∗ (t) = [ t3 t2 t 1 ]  −1 3 −3 1 3− 3µ −6 + 9µ 3− 9µ 3µ −3 + 6µ 3− 12µ 6µ 0 1 + 3µ −3 0 0   1 2 3 1 1 1 2 1 3 1 −1 0  we have α∗ (t) = ( 3µ+ 6tµ− 9t2µ+ 3t2 − 3t3 − 2, 6µ− 3t+ 6tµ− 9t2µ+ 3t2 − 3t3, 24tµ− 19µ− 6t− 27t2µ+ 12t2 − 9t3 ) . Also under the condition constant c = η = ‖α′‖ and µ = (1− t) , we can find the special involute of α as in the following way α∗ (t) = [ t3 t2 t 1 ]  −1 3 −3 1 3− 3 (1− t) −6 + 9 (1− t) 3− 9 (1− t) 3 (1− t) −3 + 6 (1− t) 3− 12 (1− t) 6 (1− t) 0 1 + 3 (1− t) −3 0 0   1 2 3 1 1 1 2 1 3 1 −1 0  α∗ (t) = [ t3 t2 t 1 ]  −1 3 −3 1 3t 3− 9t 9t− 6 3− 3t 3− 6t 12t− 9 6− 6t 0 4− 3t −3 0 0   P0 P1 P2 P3  . References [1] Ş Kılıçoğlu and S Şenyurt. 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