Mathematical Sciences And Applications E-Notes SOME

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Sep 15, 2015 - the Euclidean space E3. For a regular curve c : I ⊂ R → R3 with curvature and torsion, κ and τ, the following Frenet-Serret formulae are given in ...
Mathematical Sciences And Applications E-Notes c Volume 3 No. 2 pp. 64–73 (2015) MSAEN

SOME CHARACTERIZATIONS OF SLANT AND SPHERICAL HELICES DUE TO SABBAN FRAME ¨ BULENT ALTUNKAYA AND LEVENT KULA (Communicated by Yusuf YAYLI)

Abstract. In this paper, we are investigating that under which conditions of the geodesic curvature of unit speed curve γ that lies on the unit sphere, the curve c which is obtained by using γ, is a spherical helix or slant helix.

1. Introduction Izumiya and Takeuchi [4],[2] have defined helices, spherical helices, slant helices and conical geodesic curve and given a classification of special developable surfaces under the condition of the existence of such a special curve as a geodesic [1]. Encheva and Georgiev [3] have used a similar method and determined a F renet curve up to a direct similarity of R3 . In this paper we have used the the method in [2], [3], and [4] to construct spherical helices and slant helices. In section 2, we recall some basic concepts of differential geometry of space curves that we will use later. In the next section we will provide some new theorems and proofs about the construction of spherical helices and slant helices. 2. Basic Concepts We now recall some basic concepts on classical differential geometry of curves in the Euclidean space E 3 . For a regular curve c : I ⊂ R → R3 with curvature and torsion, κ and τ , the following Frenet-Serret formulae are given in [5] written in the matrix form for κ > 0  0    T 0 κν 0 T N 0  = −κν 0 τ ν  N  0 0 −τ ν 0 B B

Date: Received: June 26, 2015; Revised: September 15, 2015; Accepted: October 12, 2015. 2010 Mathematics Subject Classification. 53A04,53A05. Key words and phrases. Helices, Spherical Helices, Slant Helices, Sabban Frame. 64

SOME CHARACTERIZATIONS OF SLANT AND SPHERICAL HELICES...

65

where (2.1)

0 D 0 E 00 00 000

0 (t) × c (t) c (t) × c (t), c (t)

c

ν = ν(t) = c (t) , κ = κ(t) = , τ = τ (t) = 3 2 kc0 (t)k kc0 (t) × c00 (t)k and 0

(2.2)

0

00

c (t) c (t) × c (t) T = T (t) = 0 , B = B(t) = 0 , N = N (t) = B(t) × T (t) kc (t)k kc (t) × c00 (t)k

In the formulae above, we denote unit tangent vector with T , binormal unit vector with B, unit principal normal vector with N , cross product with ×, and inner product with . A regular curve c with κ > 0 is a cylindrical helix if and only if the ratio τ κ is constant [5]. A regular curve c with κ > 0 is a slant helix if and only if the geodesic curvature of the spherical image of the principal normal indicatrix of c !  τ 0 κ2 (t) (2.3) σ(t) = 3/2 κ ν (κ2 + τ 2 ) is constant [4]. A regular curve c with κ > 0 lies on the surface of a sphere which has a radius r if and only if    0 !2 1 1 1  (t) (2.4) r2 =  2 + κ ντ κ satisfies [6]. We can easily simplify this equation as follows  "   0 #0 τ 1 1 1 (2.5) +  (t) = 0. ν ντ κ κ Let γ : I → S 2 be a unit speed spherical curve with an arc length parameter s 0 0 and denote γ (s) = t (s) where γ (s) = dγ ds . If we set a vector p (s) = γ (s) × t (s), by definition we have an orthonormal frame {γ (s) , t (s) , p (s)} along γ. This frame is called the Sabban f rame of γ. Then we have the following F renet − Serret formulae of γ 0

(2.6)

γ (s) = t (s) t (s) = −γ (s) + kg (s) p (s) 0 p (s) = −kg (s) t (s) 0

where kg (s) is the geodesic curvature of the curve γ on S 2 which is   0 kg (s) = det γ (s) , t (s) , t (s)

¨ BULENT ALTUNKAYA AND LEVENT KULA

66

In [2] Izuyama and Takeuchi showed a way to construct all Bertrand curves by the following formula Z s Z s p (ϕ) dϕ + a γ (ϕ) dϕ + b cot θ (2.7) c (s) = b s0

s0

where b, θ are constant numbers, a is a constant vector, and γ is a unit speed curve on S 2 with the Sabban f rame above. Also they showed that the spherical curve γ is a circle if and only if the corresponding Bertrand curves are circular helices. In [3] Encheva and Georgiev showed a way to construct all F renet curves (κ > 0) by the following formula Z R (2.8) c (s) = b e k(s)ds γ (s)ds + a where b is a constant number, a is a constant vector, γ is a unit speed curve on S 2 with the Sabban f rame above, and k : I → R is a function of class C 1 . Also they showed that the spherical curve γ is a circle if and only if the corresponding F renet curves are cylindrical helices. If we use the equations in (2.1), (2.2), (2.6), (2.7), and (2.8) we can easily see the F renet f rame {T, N, B} of the curve c and the Sabban f rame {γ, t, p} of the curve γ coincides. Therefore we can say the tangent indicatrix of the curve c is γ. 3. Characterizations of Helices Now, we can deduce some results from the equations above. First, we want to show, under which circumstances the equation (2.8) is a spherical helix. As we know before, if γ is a circle, the geodesic curvature of it is constant. Therefore we can write the theorem below. Theorem 3.1. Let γ (s) be a unit speed spherical curve on S 2 with the geodesic curvature kg . If the curve γ is a circle, the curve c defined by (2.8) is a spherical helix if and only if the function k (s) = −kg tan [(kg ) (s − b1 )] where b1 ∈ R . Proof. For the curve Z c (s) = b

e

R

k(s)ds

γ (s)ds + a

If we calculate κ, τ , and ν of the curve c by using the equations at (2.1), we will find 1 κ (s) = beR k(s)ds k (s) τ (s) = beRR gk(s)ds . ν (s) = be k(s)ds

(3.1)

Now, by putting these equations in (2.5), we have  "   0 #0 τ 1 1 1 +  (s) = 0 ν ντ κ κ   

 1 be

R

kds

1 

R

be

kds

k Rg be kds

1 R1 be kds

!0 0  +

k Rg be kds R1 be kds

   (s) = 0

SOME CHARACTERIZATIONS OF SLANT AND SPHERICAL HELICES...



1 be



R

kds

1 kg e

R

kds

1  R kds  be kg

0

0

! + kg

(s) = 0

 h 0 R i R k e kds + k 2 e kds + kg (s) = 0

0

k (s) + k 2 (s) = −kg 2 . If we solve this differential equation, we will have k (s) = −kg tan [(kg ) (s − b1 )] Conversely, if we take k (s) = −kg tan [(kg ) (s − b1 )] in (2.8) then Z

Z −kg tan [(kg ) (s − b1 )] ds.

k (s) ds =

Let u = kg (s − b1 ) = kg s − kg b1 then kg ds = du, by using these equations Z

Z k (s) ds =

− tan udu

= ln cos u + ln b2 = ln [b2 cos (kg (s − b1 ))] we have Z c (s) = b

R

k(s)ds

R

−kg tan[(kg )(s−b1 )]ds

e Z

=b

e Z

=b

γ (s)ds + a γ (s)ds + a

eln [b2 cos (kg (s−b1 ))] γ (s)ds + a

Z =b

where b1 , b2 ∈ R.

b2 cos (kg (s − b1 ))γ (s)ds + a.

67

68

¨ BULENT ALTUNKAYA AND LEVENT KULA

Now, we must show that curve c is spherical. If we use (2.4) to do it, we will have    0 !!2 1 1 1  (s) r2 =  2 + κ ντ κ    !0 2 R 1 1    = b2 e2 kds +  R  (s) 1 k g be kds beR kds beR kds !   R 0  2 R 1 2 2 kds kds = b e + be (s) kg   R b2 k 2 2 R kds 2 2 kds (s) = b e + 2 e kg    R k2 (s) = b2 e2 kds 1 + 2 kg ! 2 (−kg tan [(kg ) (s − b1 )]) 2 2 2 = b b2 cos (kg (s − b1 )) 1 + kg2   1 = b2 b2 2 cos2 (kg (s − b1 )) cos2 (kg (s − b1 )) = b2 b2 2 . Therefore, we can say curve c lies on a sphere which has a radius |bb2 |    q     Example 3.1. Let’s take γ (s) = √13 cos 1/s√3 , √13 sin 1/s√3 , 23 , we ) ) ( ( √ 2 know that γ is a unit speed curve on S with the geodesic curvature 2. Then due to Theorem 3.1, k (s) = kg tan [(kg ) (s − b1 )] and

Z α (s) = b

b2 cos (kg (s − b1 ))γ (s)ds + a

where b, b1 , b2 ∈ R. If we take b = 2, b1 = 0, b2 = 1 then we have q √  √  √  √  α1 (s) = −2 23 cos 3s sin 2s + 2 cos 2s sin 3s √ √ √ √ √     α2 (s) = − 23 3 cos 2s cos 3s + 6 sin 2s sin 3s √ 2 sin ( 2s) √ α3 (s) = 3 where α (s) = (α1 (s) , α2 (s) , α3 (s)) and a = (0, 0, 0) Now, we can write a new thorem about (2.8) in which we are looking for the slant helix condition of the curve c. Theorem 3.2. Let γ (s) be a unit speed spherical curve on S 2 ; b, m, n be constant numbers; and a be a constant vector. The geodesic curvature of γ (s) satisfies 2

kg 2 (s) =

(ms + n)

2

1 − (ms + n)

SOME CHARACTERIZATIONS OF SLANT AND SPHERICAL HELICES...

69

Figure 1. Spherical Helice if and only if Z c (s) = b

e

R

k(s)ds

γ (s)ds + a

is a slant helix. Proof. Let, for γ kg 2 (s) =

(3.2)

(ms + n)

2 2.

1 − (ms + n)

For the curve c (s) we have 0

R

k(s)ds

γ (s)  0 c (s) = be k (s) γ (s) + γ (s)    R 000 0 0 00 c (s) = be k(s)ds k 2 (s) + k (s) γ (s) + 2k (s) γ (s) + γ (s) 1 κ (s) = beR k(s)ds k (s) g τ (s) = beRR k(s)ds ν (s) = be k(s)ds . 00

c (s) = be R

k(s)ds

The geodesic curvature of the spherical image of the principal normal indicatrix of c is as follows !  τ 0 κ2 σ(s) = (s) 3/2 κ ν (κ2 + τ 2 )   1

0  ν2 =  3/2 kg  (s) . 2 k ν ν12 + νg2

So we have 0

(3.3)

σ(s) =

kg (s) kg 2 (s) + 1

3/2

Now, let’s take u (s) = ms + n then we have (3.2) as follows (3.4)

kg 2 (s) =

u2 (s) . 1 − u2 (s)

70

¨ BULENT ALTUNKAYA AND LEVENT KULA

If we take the derivates of the both sides of (3.4) for s we have      0 0 2uu 1 − u2 − −2uu u2 0  (s) 2kg (s) k g (s) =  2 (1 − u2 ) ! 0 0 uu kg (s) k g (s) = (s) 2 (1 − u2 ) !

0

(3.5)

uu

0

kg (s) =

(1 − u2 )

r ε

2

1 − u2 u2

!! (s)

where ε = ±1. Putting (3.4) and (3.5) in (3.3), we have 0

σ(s) =

kg (s) kg 2 (s) + 1  √

 = ε 

3/2 0

1−u2 uu |u|(1−u2 )2

u2 1−u2



+1



 3/2  (s) !

0

1 − u2 uu

 2 3/2

2 1−u |u| (1 − u2 ) ! 2 1 − u2 u 0 ε u (s) 2 (1 − u2 ) |u|

=

ε

=

(s)

ms + n m |ms + n| = εm =ε

which is constant. Conversely, let c (s) be a slant helix, then the geodesic curvature of the spherical image of the principal normal indicatrix of c is a constant function. So we can take !  τ 0 κ2 σ(s) = (s) = m 3/2 κ ν (κ2 + τ 2 ) where m ∈ R. Therefore, from (3.3) m=

 τ 0

κ2 ν (κ2 + τ 2 )

3/2

κ

0

=

kg (s) kg 2 (s) + 1

3/2

If we solve this differential equation, we have kg (s) q

kg 2 (s) + 1

= ms + n

! (s)

SOME CHARACTERIZATIONS OF SLANT AND SPHERICAL HELICES...

71

where n ∈ R. Then, kg 2 (s) 2 = (ms + n) kg 2 (s) + 1 kg 2 (s) + 1 − 1 2 = (ms + n) kg 2 (s) + 1 1 2 1− 2 = (ms + n) kg (s) + 1 1 kg 2 (s) = 2 −1 1 − (ms + n) 2

kg 2 (s) =

(ms + n)

2.

1 − (ms + n)

 Furthermore, we can give a similar theorem for (2.7), Theorem 3.3. Let γ (s) be a unit speed spherical curve on S 2 ; b, m, n, θ be constant numbers; and a be a constant vector. The geodesic curvature of γ (s) satisfies 2

kg 2 (s) =

(ms + n)

2

1 − (ms + n)

if and only if s

Z c (s) = b

Z

s

γ (ϕ) dϕ + b cot θ s0

p (ϕ) dϕ + a s0

is a slant helix. Proof. Let, for γ kg 2 (s) =

(3.6)

(ms + n)

2 2.

1 − (ms + n)

For the curve c (s) we have 0

c (s) = b (γ (s) + cot θp (s)) c (s) = b (1 − cot θkg (s)) p (s) 0 000 c (s) = −b cot θkg (s) p (s) + b (1 − cot θkg (s)) (−γ (s) + kg (s) p (s)) sin2 θ(1−cot θkg (s)) κ (s) = ε b sin2 θ(kg (s)+cot θ) τ (s) = b ν (s) = εb csc θ. 00

where ε = ±1.

¨ BULENT ALTUNKAYA AND LEVENT KULA

72

The geodesic curvature of the spherical image of the principal normal indicatrix of c is as follows  τ 0

κ2

σ(s) =

3/2

! (s)

κ

ν (κ2 + τ 2 ) 

3

 =

ε sin θkg b3



(ε2 +cot2 θ−2(−1+ε2 ) cot θkg +(1+ε2 cot2 θ)kg 2 ) sin4 θ a2

 =

εkg 3

1 + cot2 θ

sin θ

 =

εkg sin3 θ

1 sin2 θ

 =



0

εkg

0

1 + kg 2

 3/2  (s)



0

3/2  (s) 1 + kg 2 



0

1 + kg 2 

3/2  (s)

3/2  (s)

So we have 0

(3.7)

εkg (s)

σ(s) =

kg 2 (s) + 1

3/2

Now, let’s take u (s) = ms + n then we have (3.6) as follows

kg 2 (s) =

(3.8)

u2 (s) . 1 − u2 (s)

If we take the derivates of the both sides of (3.8) for s we have     0 0 2uu 1 − u2 − −2uu u2  (s) 2kg (s) k g (s) =  2 (1 − u2 ) ! 0 0 uu kg (s) k g (s) = (s) 2 (1 − u2 ) 

0

!

0

(3.9)

0

kg (s) =

uu

(1 − u2 )

2

r ε

1 − u2 u2

!! (s) .

SOME CHARACTERIZATIONS OF SLANT AND SPHERICAL HELICES...

73

Putting (3.9) and (3.8) in (3.7), we have 0

σ(s) =

εkg (s) kg 2 (s) + 1  √

 = ε2  √ =

=

3/2 

0

1−u2 uu |u|(1−u2 )2

u2 1−u2

+1

 3/2  (s) !

0

1 − u2 uu

 2 3/2

2 1−u |u| (1 − u2 ) ! 2 1 − u2 u 0 u (s) 2 (1 − u2 ) |u|

(s)

ms + n m |ms + n| = εm

=

which is constant. Conversely, let c (s) be a slant helix, then the geodesic curvature of the spherical image of the principal normal indicatrix of c is a constant function. So we can take !  τ 0 κ2 σ(s) = (s) = m 3/2 κ ν (κ2 + τ 2 ) where m ∈ R. Therefore, from (3.7) m=

 τ 0

κ2 ν (κ2 + τ 2 )

3/2

!

κ

(s)

0

=

εkg (s) kg 2 (s) + 1

3/2

If we solve this differential equation, we have εkg (s) q

= ms + n

kg 2 (s) + 1

where n ∈ R. Then, kg 2 (s) 2 = (ms + n) kg 2 (s) + 1 kg 2 (s) + 1 − 1 2 = (ms + n) kg 2 (s) + 1 1 2 1− 2 = (ms + n) kg (s) + 1 2

kg 2 (s) =

(ms + n)

2.

1 − (ms + n)



74

¨ BULENT ALTUNKAYA AND LEVENT KULA

References [1] L. Kula, Y. Yaylı, On slant helix and its spherical indicatrix, Appl. Math. Comput., 169 (1), 600-607, 2005. [2] S. Izumiya, N. Takeuchi, Generic properties of helices and Bertrand curves , J. Geom. 74, 97-109, 2002. [3] R. Encheva, G. Georgiev, Shapes of space curves, Journal for Geometry and Graphics, Vol 7, No. 2, 145-155, 2003. [4] S. Izumiya, N. Takeuchi, New special curves and developable surfaces, Turk. J. Math. 28, 153-163, 2004. [5] B. O’Neill, Elementary Differential Geometry, Academic Press, 2006. [6] D. J. Struik, Lectures on Classical Differential Geometry, Dover, 1961. Department of Mathematics, Faculty of Education, University of Ahi Evran, Kırs¸ehir, Turkey E-mail address: [email protected] Department of Mathematics, Faculty of Science, University of Ahi Evran, Kırs¸ehir, Turkey E-mail address: [email protected]