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Slant Helices Generated by Plane Curves in Euclidean 3-Space Mesut Altınok and Levent Kula

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Slant Helices Generated by Plane Curves in Euclidean 3-Space Mesut Altınok and Levent Kula Palestine Journal of Mathematics Vol. 5(2) (2016) , 164–174 © Palestine Polytechnic University-PPU 2016 Slant Helices Generated by Plane Curves in Euclidean 3-space Mesut Altınok and Levent Kula Communicated by Ayman Badawi MSC 2010 Classifications: Primary 53A04 ; Secondary 14H50. Keywords and phrases: Slant helix, plane curve and principal normal indicatrix. Abstract In this paper, we investigate the relationship between the plane curves and slant helices in R3. Moreover, we show how could be obtained to a slant helix from a plane curve. Finally, we give some slant helix examples generated by plane curves in Euclidean 3-space. 1 Introduction In [3], A slant helix in Euclidean space R3 was defined by the property that the principal normal makes a constant angle with a fixed direction. Moreover, Izumiya and Takeuchi showed that γ is a slant helix in R3 if and only if the geodesic curvature of the principal normal of a space curve γ is a constant function. In [5], Kula and Yayli have studied spherical images of tangent indicatrix and binormal indi- catrix of a slant helix and they showed that the spherical images are spherical helix. In [4], Kula, Ekmekci, Yayli and Ilarslan have studied the relationship between the plane curves and slant helices in R3. They obtained that the differential equations which are character- izations of a slant helix. In this paper we consider the relationship between the plane curves and slant helices in R3. Moreover, we get slant helix from plane curve. Also, we give some slant helix examples in Euclidean 3-space. 2 Preliminaries We now recall some basic concept on classical geometry of space curves and the definition of slant helix in R3. A curve γ˜ : I ⊂ R ! R3, with unit speed, is a space curve. T (s) = γ˜ 0(s) is a unit tangent vector of γ˜ at s. We define the curvature of γ˜ by κ = kγ˜ 00k. If κ(s) =6 0, then the unit principal normal vector N(s) of the curve γ˜ at s is given by γ˜ 00(s) = κ(s)N(s). The unit vector B(s) = T (s) ^ N(s) is called the unit binormal vector of γ˜ at s. For the derivatives of the Frenet frame the Serret-Frenet formula hold: T 0(s) = κ(s)N(s); N 0(s) = −κ(s)T (s) + τ(s)B(s); (2.1) B0(s) = −τ(s)N(s); where τ(s) is the torsion of the curve γ˜ at s. Definition 2.1. A curve γ˜ with κ(s) =6 0 is called a slant helix if the principal normal vector line of γ˜ make a constant angle with a fixed direction [3]. Theorem 2.2. γ˜ is a slant helix if and only if the geodesic curvature of the spherical image of the principal normal indicatrix (N) of γ˜ ! ( ) κ2 τ 0 κg(s) = 3 (s) (2.2) ν(κ2 + τ 2) 2 κ is a constant function and ν = kγ˜ 0k. Also, axis of the slant helix is τ sin θ κ sin θ ~a = 1 T + cos θN 1 B (2.3) (κ2 + τ 2) 2 (κ2 + τ 2) 2 Slant Helices Generated by Plane Curves in Euclidean 3-space 165 [3]. Theorem 2.3. Let γ˜ : I ! R3 be a space curve and β : I ! R3 be a space curve such that 0 γ0 β = kγ0k . γ˜ is a slant helix if and only if β is a slant helix. Proof. If we use that 0 kγ kκγ = κβ; 0 kγ kτγ = τβ: and by simple calculation, we show that β is a slant helix and thus the proof of theorem 2.3 is completed. Now we can give the following theorem. Theorem 2.4. Curves γ˜ and β have the same axis. Proof. Let γ˜ be a slant helix with Frenet frame fTγ˜ ;Nγ˜ ;Bγ˜ g, curvature κγ˜ and torsion τγ˜ . β is a slant helix with Frenet frame fTβ;Nβ;Bβg, curvature κβ and torsion τβ. By simple calculation, we get τβ sin θ κβ sin θ ~aβ = Tβ + cos θNβ Bβ 2 2 1 2 2 1 (κβ + τβ) 2 (κβ + τβ) 2 = ~aγ˜ : Thus the proof of theorem 2.4 is completed. Definition 2.5. The epitrochoid is traced by a point P attached at a distance h to the center of a circle of radius r2 rolling along another circle of radius r1 and its equation is r1 + r2 r1 + r2 γ(t) = ((r1 + r2) sin t − h sin(( )t); −(r1 + r2) cos t + h cos(( )t)) (2.4) r2 r2 [6]. Definition 2.6. The epicycloid is traced by a point P attached at a distance r2 to the center of a circle of radius r2 rolling along another circle of radius r1 and its equation is r1 + r2 r1 + r2 γ(t) = ((r1 + r2) sin t − r2 sin(( )t); −(r1 + r2) cos t + r2 cos(( )t)) (2.5) r2 r2 [6]. Definition 2.7. In definition 2.5 if we choose r1 = r2, then we get special case of epitrochoid curve. So we call this curve as limacon curve. γ(t) = ((2r1 + 2h cos t) cos t; (2r1 + 2h cos t) sin t) (2.6) [6]. Definition 2.8. If we choose r1 = r2 in equation (2.5), then the epicycloid curve is called a cardioid curve with equation γ(t) = (2r1 cos t − r1 cos(2t); 2r1 sin t − r1 sin(2t)): (2.7) Definition 2.9. If we choose r1 = 2r2 in equation (2.5), then the epicycloid curve is called a nephroid curve with equation γ(t) = (3r2 cos t − r2 cos(3t); 3r2 sin t − r2 sin(3t)): (2.8) 166 Mesut Altınok and Levent Kula 3 Plane curves and slant helix For a space curve, if the torsion always vanishes, then the curve is contained in a plane. In this case we denote the curve γ instead of γ˜ and the curvature κp instead of κ. Let γ˜ be a space curve with Frenet frame fT; N; Bg and with curvatures κ, τ in R3. The curve γ is given by γ = γe − hγ;~ae i~a; (3.1) where ~a is a constant vector. Theorem 3.1. Under the above notation, γ is a plane curve. Moreover, Frenet frame of the curve γ are 1 ~t = p (T − hT;~ai~a) 1 − hT;~ai2 1 ~n = p (−⟨N;~aiB + hB;~aiN) (3.2) 1 − hT;~ai2 and curvature of the curve γ is p κ 1 − hT;~ai2 − hN;~ai2 κp = p : (3.3) 1 − hT;~ai2 Proof. Differentiating the equation (3.1), we get γ0 = νT − νhT;~ai~a γ00 = ν0T + ν2κN − ν0hT;~ai~a − ν2κhN;~ai~a: (3.4) and kγ0k2 = ν2(1 − hT;~ai2): (3.5) Where ν = kγ˜ 0k. Thus tangent vector of the curve γ is found as 1 ~t = p (T − hT;~ai~a): (3.6) 1 − hT;~ai2 We can calculate that γ0 ^ γ00 = ν3κ(B − hN;~ai(T ^ ~a) + hT;~ai(N ^ ~a)) (3.7) and q kγ0 ^ γ00k = ν3κ 1 − hT;~ai2 − hN;~ai2: (3.8) By using (3.7) and (3.8), we obtain that principal binormal vector of the curve γ is 1 ~b = (B − hN;~ai(T ^ ~a) + hT;~ai(N ^ ~a)): 1 − hN;~ai2 − hT;~ai2 Differentiating the equation (3.4), we get γ000 = (ν00 − ν3κ2)T + (3ν0νκ + ν2κ0)N + ν2κτB (ν00hT;~ai + 3ν0νκhN;~ai + ν2κ0hN;~ai − ν3κ2hT;~ai + ν3κτhB;~ai)~a: Also det(γ0; γ00; γ000) = ν6κ2τ[1 − hT;~ai2 − hN;~ai2 − hB;~ai2] = 0: Then we know that ~a = hT;~aiT + hN;~aiN + hB;~aiB: (3.9) Slant Helices Generated by Plane Curves in Euclidean 3-space 167 By using (3.9), we get T ^ ~a = hN;~aiB − hB;~aiN N ^ ~a = −⟨T;~aiB + hB;~aiT (3.10) B ^ ~a = hT;~aiN − hN;~aiT: Also by using (3.9) and (3.11), we obtain that ~b = ~a: If we calculate ~n = ~b ^ ~t, The principal normal vector is found as 1 ~n = ~b ^ ~t = p (−⟨N;~aiB + hB;~aiN): 1 − hT;~ai2 Finally kγ0 ^ γ00k κ = p kγ0k3 p κ 1 − hT;~ai2 − hN;~ai2 = p 1 − hT;~ai2 and det(γ0; γ00; γ000) τ = = 0; p kγ0 ^ γ00k2 which means that γ is a plane curve. Corollary 3.2. If space curve is a unit speed curves in theorem 3.1, kγ0k ≤ 1: Corollary 3.3. Principal normal vector of the plane curve γ, defined by (3.1), is perpendicular to tangent vector of the space curve γ˜, i.e. h~n;T i = 0: Let γ : I ⊂ R ! R3 be a plane curve. Z t h 0 00i − γ ; γ γ˜ = γ (tan θ k 00k du)~a + ~c (3.11) t0 γ is a space curve. Where θ is a constant and ~a, ~c are constant vectors with hγ0(u);~ai = 0 and k~ak = 1. p 00 0 − kγ k 1−∥γ k2 k 0k Theorem 3.4. Under the above notation, if hγ0,γ00i = tan θ and γ < 1, then γ˜ is a unit speed slant helix. Proof. Suppose that γ is a plane curve with Frenet frame f~t; ~ng and with curvature κp. In this case, we will show that hN;~ai = cos θ. Differentiating the equation (3.11), we get hγ0; γ00i γ˜ 0 = kγ0k~t − tan θ( )~a kγ00k k 0kk 0k0 00 0 0 0 2 γ γ γ˜ = kγ k ~t + kγ k κp~n − p ~a: (3.12) 1 − kγ0k2 Since p kγ00k 1 − kγ0k2 tan θ = − ; (3.13) hγ0; γ00i 168 Mesut Altınok and Levent Kula we can write q tan θhγ0; γ00i 1 − kγ0k2 = − : (3.14) kγ00k By using equation (3.14) in equation (3.12), we obtain 00 0 0 0 2 00 γ˜ = kγ k ~t + kγ k κp~n + cot θkγ k~a: Then s hγ0; γ00i2 kγ˜ 0k = kγ0k2 + tan2 θ kγ00k2 s kγ00k2(1 − kγ0k2) hγ0; γ00i2 = kγ0k2 + hγ0; γ00i2 kγ00k2 = 1 and q T = kγ0k~t + 1 − kγ0k2~a: (3.15) Also q k 00k k 0k0 2 k 0k4 2 2 k 00k2 γ˜ = ( γ ) + γ κp + cot θ γ kγ00k = : sin θ So sin θ N = (kγ0k0~t + kγ0k2κ ~n + cot θkγ00k~a) (3.16) kγ00k p and q q sin θ B = (−∥γ0k2 1 − kγ0k2κ ~t+(kγ0k0 1 − kγ0k2 −cot θkγ0kkγ00k)~n+kγ0k3κ ~a): (3.17) kγ00k p p Moreover kγ00k κ = psin θ 1 − kγ0k2 τ = cos θ and κg = cot θ: (3.18) Consequently sin θ hN;~ai = cot θkγ00k = cos θ; (3.19) kγ00k which means that γ˜ is a slant helix.
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