C om mun.Fac.Sci.U niv.A nk.Series A 1 Volum e 66, N umb er 2, Pages 44–52 (2017) D O I: 10.1501/C om mua1_ 0000000799 ISSN 1303–5991
http://com munications.science.ankara.edu.tr/index.php?series= A 1
CURVES OF CONSTANT BREADTH ACCORDING TO DARBOUX FRAME
BÜLENT ALTUNKAYA AND FERDAG KAHRAMAN AKSOYAK
Abstract. In this paper, we investigate constant breadth curves on a surface according to Darboux frame and give some characterizations of these curves.
1. Introduction
Since the …rst introduction of constant breadth curves in the plane by L. Euler in 1778 [3], many researchers focused on this subject and found out a lot of interesting properties about constant breadth curves in the plane [9], [2], [12]. Fujiwara [4] has introduced constant breadth curves, by taking a closed curve whose normal plane at a point P has only one more point Q in common with the curve and for which the distance d (P; Q) is constant.
After the development of cam design, researchers have been shown a strong interest to this subject again and many interesting properties have been discovered. For example Köse has de…ned a new concept called space curve pair of constant breadth in [7], a pair of unit speed space curves of class C3 with non-vanishing
curvature and torsion in E3; which have parallel tangents in opposite directions at
corresponding points, and the distance between these points is always constant by using the Frenet frame.
Many authors have been studied spaccelike and timelike curves of constant breadth in Minkowski 3-space [5, 10, 13]. Furthermore Akdo¼gan and Ma¼gden gen-eralized in n dimensional Euclidean space [1].
The characterizations of Köse’s paper [6] on constant breadth curves in the space has led us to investigate this topic according to Darboux frame on a surface.
2. Basic Concepts
Now, we introduce some basic concepts about our study. Let M be an oriented surface and be a unit speed curve of class C3on M: As we know, has a natural
Received by the editors: June 17, 2016; Accepted: November 02, 2016. 2010 Mathematics Subject Classi…cation. 53A04.
Key words and phrases. Darboux frame, constant breadth curve, Euclidean space.
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frame called Frenet frame fT; N; Bg with properties below: T0= {N;
N0= {T + B;
B0= N;
(1) where { is the curvature, is the torsion, T is the unit tangent vector …eld, N is the principal normal vector …eld and B is the binormal vector …eld of the curve : By using the unit tangent vector …eld of the curve and the unit normal vector …eld of the surface M on the curve we de…ne unit vector …eld g as g = (n ) T; where cross product. We will have a new frame called Darboux frame fT; g; n g: The relations between these two frames can be given as follows:
2 4 Tg n 3 5 = 2 4 10 cos0 sin0 0 sin cos 3 5 2 4 NT B 3 5 (2)
where is the angle between the vector …elds n and B: If we take the derivatives of T; g; n with respect to s, we will have
2 4 T 0 g0 (n )0 3 5 = 2 4 0kg k0g ktng kn tg 0 3 5 2 4 Tg n 3 5 (3)
where kg; kn and tg are called the geodesic curvature, the normal curvature and
the geodesic torsion respectively. Then, we will have following relations [11].
kg = { cos ; (4)
kn = { sin ;
tg = 0:
In the di¤erential geometry of surfaces, For a curve (s) lying on a surface, there are following cases:
i) is a geodesic curve if and only if kg= 0:
ii) is an asymptotic line if and only if kn= 0:
iii) is a principal line if and only if tg= 0:
3. Curves of Constant Breadth According to Darboux Frame Let (s) and (s ) be a pair of unit speed curves of class C3with non-vanishing curvature and torsion in E3 which have parallel tangents in opposite directions at corresponding points and the distance between these points is always constant.
If lies on a surface, it has Darboux frame in addition to Frenet frame with properties (1), (2), (3) and (4). So we may write for
If we di¤erentiate this equation with respect to s and use (3), we will have ( )0 = d ds ds ds = (1 + m 0 1 m2kg m3kn) T + (m1kg+ m02 m3tg) g (5) + (m1kn+ m2tg+ m03) (n ) and d ds = d ds ds ds = T ds ds As we know hT; T i = 1: Then ds ds = 1 + m 0 1 m2kg m3kn: So we …nd from (5), m01 = m2kg+ m3kn 1 ds ds; (6) m02 = m3tg m1kg; m03 = m1kn m2tg:
Let us denote the angle between the tangents at the points (s) and (s + 4s) with 4 : If we denote the vector T (s + 4s) T (s) with 4T; we know lim4s!04T4s =
lim4s!044s = dds = {: We called the angle of contingency to the angle 4 [12]. Let us denote the di¤erentiation with respect to with " ": By using the equation
d
ds = {, we can write (6) as follows:
_ m1 = (m2kg+ m3kn) f ( ); (7) _ m2 = (m3tg m1kg) ; _ m3 = ( m1kn m2tg) ; where = 1 {, = 1 { and + = f ( ):
Now we investigate curves of constant breadth according to Darboux frame for some special cases:
3.1. Case (For geodesic curves). Let be non straight line geodesic curve on a surface. Then kg = { cos = 0 and { 6= 0, we get cos = 0. So it implies that
kn= {; tg= : By using (7), we have following di¤erential equation system
_ m1 = m3 f ( ); (8) _ m2 = m3'; _ m3 = m1 m2'; where ' =
{: By using (8), we obtain a di¤erential equation as follows:
... m1+ •f d' d 1 ' m•1+ m1+ _f + 1 + ' 2 m_ 1+ '2f = 0: (9)
We assume that ( ; ) a pair of curve is constant breadth , then k k2= m21+ m22+ m23= constant which implies that
m1m_1+ m2m_2+ m3m_3= 0: (10)
By combining (8) and (10) then we get
m1f ( ) = 0:
3.1.1. Case f ( ) = 0:. We assume that f ( ) = 0: By using (9), we get ... m1 d' d 1 '( •m1+ m1) + 1 + ' 2 m_ 1= 0: (11)
If is a helix curve then ' = '0=constant. From (11), we have ...
m1+ 1 + '20 m_1= 0
whose the solution is m1= 1 p 1 + '2 0 c1sin q 1 + '2 0 c2cos q 1 + '2 0 + c3;
where c1; c2 and c3 are real constants. By using (8), we can …nd as m2 = 1
'0(m1+ •m1) and m3= _m1. In that case we can compute m2and m3as follows:
m2= '0 p 1 + '2 0 c1sin q 1 + '2 0 c2cos q 1 + '2 0 + '0c3 and m3= c1cos q 1 + '2 0 + c2sin q 1 + '2 0 :
If f ( ) = 0 then from (8), it can easily seen that the vector d = m1(s)T (s) +
m2(s)g(s) + m3(s) (n ) (s) is a constant vector. In that case the curve is the
translation of the curve along the vector d:
3.1.2. Case m1= 0:. We assume that m1= 0; then by using (9), we get
• f d' d 1 'f + '_ 2f = 0: (12)
If is a helix curve, then ' = '0=constant. From (12) we obtain
•
f + '20f = 0
whose the solution is
f ( ) = c1cos ('0 ) + c2sin ('0 ) :
Since m1 = 0; (8) implies that m3 = f ( ) and m2 = _ f ( )
'0 : In that case we can
compute m2and m3as follows:
and
m3= c1cos ('0 ) + c2sin ('0 ) :
Theorem 1. Let be a geodesic curve and a helix curve. Let ( ; ) be a pair of constant breadth curve. In that case can be expressed as one of the following cases: i) (s ) = (s) + m1(s)T (s) 1 '0 (m1(s) + •m1(s)) g(s) + _m1(s) (n ) (s); where m1= p1 1+'2 0 c1sin p 1 + '2 0 c2cos p 1 + '2 0 + c3. ii) (s ) = (s) f ( )_ '0 g(s) + f ( ) (n ) (s); where f ( ) = c1cos ('0 ) + c2sin ('0 ) :
3.2. Case (For asymptotic lines). Let be non straight line asymptotic line on a surface. Then kn= { sin = 0 and { 6= 0, we have sin = 0. So we get kg = "{;
tg = ; where " = 1: By using (7), we have following di¤erential equation system
_ m1 = "m2 f ( ); (13) _ m2 = m3' "m1; _ m3 = m2'; where ' =
{: By using (13), we obtain a di¤erential equation as follows:
... m1+ •f d' d 1 ' m•1+ m1+ _f + 1 + ' 2 m_ 1+ '2f = 0: (14)
We assume that ( ; ) a pair of curve is constant breadth then k k2= m21+ m22+ m23= constant which implies that
m1m_1+ m2m_2+ m3m_3= 0: (15)
By combining (13) and (15) then we get m1f ( ) = 0:
3.2.1. Case f ( ) = 0. We assume that f ( ) = 0: By using (14), we get ... m1 d' d 1 '( •m1+ m1) + 1 + ' 2 m_ 1= 0: (16)
If is a helix curve then ' = '0=constant. From (16), we have ...
whose the solution is m1= 1 p 1 + '2 0 c1sin q 1 + '2 0 c2cos q 1 + '2 0 + c3;
where c1; c2 and c3 are real constants. By using (13), we can …nd as m2 = " _m1
and m3= "'1
0 (m1+ •m1). In that case we can compute m2and m3as follows:
m2= " c1cos q 1 + '2 0 + c2sin q 1 + '2 0 and m3= "'0 p 1 + '2 0 c1sin q 1 + '2 0 c2cos q 1 + '2 0 "'0c3:
If f ( ) = 0 then from (13), it can easily seen that the vector d = m1(s)T (s) +
m2(s)g(s) + m3(s) (n ) (s) is a constant vector. In that case the curve is the
translation of the curve along the vector d:
3.2.2. Case m1= 0. We assume that m1= 0:Then by using (14), we get
• f d' d 1 ' _ f + '2f = 0 (17)
If is a helix curve, then ' = '0=constant. From (17) we obtain •
f + '20f = 0
whose the solution is
f ( ) = c1cos ('0 ) + c2sin ('0 )
Since m1= 0; (13) implies that m2 = "f ( ) and m3 = " _ f ( )
'0 : In that case we can
compute m2and m3as follows:
m2= " (c1cos ('0 ) + c2sin ('0 ))
and
m3= " ( c1sin ('0 ) + c2cos ('0 )) :
Theorem 2. Let be an asymptotic line and a helix curve. Let ( ; ) be a pair of constant breadth curve. In that case can be expressed as one of the following cases: i) (s ) = (s) + m1(s)T (s) + " _m1(s) g(s) + " '0(m1(s) + •m1(s)) (n ) (s); where m1= p1 1+'2 0 c1sin p 1 + '2 0 c2cos p 1 + '2 0 + c3. ii) (s ) = (s) + "f ( )g(s) + "f ( )_ '0 (n ) (s);
where f ( ) = c1cos ('0 ) + c2sin ('0 ) :
3.3. Case (For principal line). We assume that is a principal line. Then we have tg= 0 and it implies that = 0. By using (7), we get
_ m1 = m2cos + m3sin f ( ); (18) _ m2 = m1cos ; _ m3 = m1sin :
By using (18), we obtain following di¤erential equation (m...1+ _m1) + ( _m1+ f ) _2 sin Z m1cos d cos Z m1sin d • + •f = 0: (19) Since tg= 0 we obtain _ = {. We assume that ( ; ) a pair of curve is constant
breadth. In that case
k k2= m21+ m22+ m23= constant which implies that
m1m_1+ m2m_2+ m3m_3= 0: (20)
By combining (18) and (20) then we get m1f ( ) = 0:
3.3.1. Case f ( ) = 0. We assume that f ( ) = 0: By using (19), we get (m...1+ _m1) + _m1_2 sin
Z
m1cos d cos
Z
m1sin d • = 0: (21)
If is a helix curve then _ = { =constant. From (21), we have ... m1+ 1 + _2 m_1= 0: Then we get m1(s) = 1 p 1 + _2 c1sin p 1 + _2 c2cos p 1 + _2 + c3;
where c1, c2 and c3 are real constants. By using (18) we obtain
m2 = Z m1cos d ; m3 = Z m1sin d ; where =R ds:
If f ( ) = 0 then from (18), it can easily seen that the vector d = m1(s)T (s) +
m2(s)g(s) + m3(s) (n ) (s) is a constant vector. In that case the curve is the
Theorem 3. Let be a principal line and a helix curve. Let ( ; ) be a pair of constant breadth curve such that h ; T i = m1 6= 0: In that case can be
expressed as: = + m1(s)T (s) Z m1(s) cos d g(s) Z m1(s) sin d (n ) (s); where m1(s) = p 1 1+ _2 c1sin p 1 + _2 c2cos p 1 + _2 + c3:
3.3.2. Case m1= 0. If m1= 0; then from (19)
•
f + _2f = 0; (22)
where _ =
{: On the other hand since m1= 0 from (18) we have m2= c2=constant,
m3= c3=constant and
f = c2cos + c3sin : (23)
By combining (22) and (23)
• ( c2sin + c3cos ) = 0:
In that case, if • = 0 then we obtain that _ =
{ = constant. becomes a helix
curve. If c2sin + c3cos = 0 then we have = constant. This means that is
a planar curve.
Theorem 4. Let be a principal line. Let ( ; ) be a pair of constant breadth curve such that h ; T i = m1 = 0: In that case is a helix curve or a planar
curve and can be expressed as:
= + c2g(s) + c3(n ) (s)
References
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Current address : Ahi Evran University, Division of Elementary Mathematics Education, K¬r¸ se-hir, TURKEY
E-mail address, B. Altunkaya: bulent_altunkaya@hotmail.com E-mail address, F. Kahraman Aksoyak: ferda.kahraman@yahoo.com
