/compile/output.dvi EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 8, No. 2, 2015, 255-270 ISSN 1307-5543 – www.ejpam.com Frenet Apparatus of the Curves and Some Special Curves in the Euclidean 5-Space E5 Melek Masal, A. Zeynep Azak∗ Department of Elementary Education, Faculty of Education, Sakarya University, Sakarya, Turkey Abstract. In this study, initially the geometric meanings of the curvatures of the curves parametrized with the arc length are given in E5. This is followed by the calculation of the Frenet vectors and curvatures of any curve. After these, some results have been given for the state of evolute curve X being a W-curve and the Frenet vectors and curvatures of involute curve Y have been calculated in terms of Frenet vectors and curvatures of the curve X. At last, the differential equation of the spherical curves, the equation of the radius and the center of the osculating hyperspheres have been achieved in E5. 2010 Mathematics Subject Classifications: 53A04, 14H50 Key Words and Phrases: Euclidean 5-space, involute-evolute curves, curvatures, Frenet apparatus, spherical curves 1. Introduction The involute-evolute curves and helices can often be seen in our daily lives. For example, the idea of a string involute is due to C. Huygens, who is also known for his works in optics. He discovered involutes while trying to build a more accurate clock [2, 7]. In addition to this, standard screws, bolts and a double-stranded molecule of DNA are the most common examples for helices in the nature and structures [11]. A. R. Forsyth (1930) [4] has took the hypothesis of curves and surfaces in the four-dimensional Euclidean space E4 [4], while H. Gluck (1966) [5] examined the curvatures of the curve in the n-dimensional Euclidean space En. Lately the studies in the four and five-dimensional spaces have been accelerated. For example, some characterizations for the spherical curves and helices have been obtained in the four-dimensional Euclidean space E4, [8, 10, 13]. Also some characterizations related to the inclined curves have been defined in the 5-dimensional Euclidean space E5 and 5-dimensional Lorentzian space L5, [1, 11]. The Frenet vectors of any curve and involute-evolute curves in E4 and E4 1 have been given by [12, 15]. In addition ∗Corresponding author. Email addresses: mmasal@sakarya.edu.tr (M. Masal), apirdal@sakarya.edu.tr (A. Azak) http://www.ejpam.com 255 c© 2015 EJPAM All rights reserved. M. Masal, A. Azak / Eur. J. Pure Appl. Math, 8 (2015), 255-270 256 to this, the curvatures and Frenet vectors of the curves parametrized with the arc length in E5 and L5 have been determined, [14, 16]. At last, Bertrand curves in E5 and L5 have been defined, [3, 8]. In this study, initially we have given the geometrical meanings of the curvatures of curves parametrized with arc length in the Euclidean 5-Space. Afterwards, we have calculated the Frenet vectors and curvatures of an arbitrary curve in E5. Moreover, we have given the Frenet vectors and curvatures of the involute curve Y in the state of the evolute curve X as the W- curve. Finally, we have defined the differential equation of the spherical curves, the equation of the center of osculating hyperspheres and the equation of their radius in E5. 2. Preliminaries In this section, we recall some basic concepts on classical differential geometry of space curve in the Euclidean 5-space and the definitions of special curves. Let X : I ⊂ R → E5 be an arbitrary curve in the Euclidean 5-space. We call the curve X as unit speed curve if 〈X ′(s), X ′(s)〉= 1, where 〈, 〉 is the standard scalar product of E5 given by 〈a, b〉= a1 b1 + a2 b2 + a3 b3 + a4 b4 + a5 b5, for each vectors a = (a1, a2, a3, a4, a5) and b = (b1, b2, b3, b4, b5) of E5, [6]. The norm of a vector a of E5 is given by ‖a‖= p 〈a, a〉, [6]. Let a = (a1, a2, a3, a4, a5), b = (b1, b2, b3, b4, b5), c = (c1, c2, c3, c4, c5) and d = (d1, d2, d3, d4, d5) be vectors in E5. The vectorial product of these vectors is defined by the determinant,[6] a ∧ b ∧ c ∧ d = � � � � � � � � � � e1 e2 e3 e4 e5 a1 a2 a3 a4 a5 b1 b2 b3 b4 b5 c1 c2 c3 c4 c5 d1 d2 d3 d4 d5 � � � � � � � � � � where ei for 1≤ i ≤ 5 are the standard basis vectors of E5 which satisfies e1∧ e2∧ e3∧ e4 = e5, e2 ∧ e3 ∧ e4 ∧ e5 = e1, e3 ∧ e4 ∧ e5 ∧ e1 = e2, e4 ∧ e5 ∧ e1 ∧ e2 = e3, e5 ∧ e1 ∧ e2 ∧ e3 = e4. Let � V1, V2, V3, V4, V5 denotes the moving Frenet Frame of the unit speed curve X . Then the Frenet formulas are given by       V ′1 V ′2 V ′3 V ′4 V ′5       =       0 k1 0 0 0 −k1 0 k2 0 0 0 −k2 0 k3 0 0 0 −k3 0 k4 0 0 0 −k4 0             V1 V2 V3 V4 V5       where Vi, i = 1,2,3,4,5 are called the i th Frenet vectors of the curve X and the functions ki , i = 1,2,3,4 are called the i th curvatures of the curve X ,[6]. The set, whose elements are frame vectors and curvatures of a curve, is called Frenet apparatus of the curve. A regular curve is M. Masal, A. Azak / Eur. J. Pure Appl. Math, 8 (2015), 255-270 257 called a W-curve if it has constant Frenet curvatures. A unit speed curve X is called inclined curve in E5 if its tangent vector V1 makes a constant angle with a unit fixed direction U . Let X and Y be unit speed curves in E5. Y is an involute of X if the tangent line V1 at X (s) and the tangent line V ∗1 at Y (s) are perpendicular for each s. X is an evolute of Y if Y is an involute of X . This curve couple is defined by [12] Y = X +µV1. The Euclidean hypersphere with the center C and radius r ∈ R+ in Euclidean 5-space E5 is defined by [6] S4 = � X ∈ E5|〈X − C , X − C〉= r2 If X ⊂ S4 is a regular curve in E5, then the curve X is called as a spherical curve in E5. The hypersphere is called as osculating hypersphere if it has six common points with the curve X at the point X (s),[6] 3. Geometric Meanings of the Curvatures in Euclidean 5-Space Let X = X (s) be a unit speed curve in Euclidean 5-space. The Frenet vectors and curvatures of X , are given by V1 =X ′, V2 = X ′′ k1 , V3 = X ′′ 2 (X ′′′ + X ′′ 2 X ′)− 〈X ′′, X ′′′〉X ′′ ‖X ′′‖2 (X ′′′ + ‖X ′′‖2 X ′)− 〈X ′′, X ′′′〉X ′′ , V4 =ηV3 ∧ V2 ∧ V1 ∧ V5, V5 =η V1 ∧ V2 ∧ X ′′′ ∧ X (4) V1 ∧ V2 ∧ X ′′′ ∧ X (4) , k1 = X ′′ , k2 = 〈X ′′′, V3〉 ‖X ′′‖ , k3 = V1 ∧ V2 ∧ X ′′′ ∧ X (4) � 〈X ′′′, V3〉 �2 , k4 = 〈X (4), V5〉 � 〈X ′′′, V3〉 �2 V1 ∧ V2 ∧ X ′′′ ∧ X (4) . where V1, V2, V3 , V4, V5 and k1, k2, k3, k4 denote the Frenet vectors and Frenet curvatures of the curve X , respectively. Also, η number is selected as+1 or -1, in order to make the determinant of � V1, V2, V3, V4, V5 � matrix +1. Thus, the Frenet frame will be directed positively, [16]. M. Masal, A. Azak / Eur. J. Pure Appl. Math, 8 (2015), 255-270 258 The geometric meanings of the curvatures at the initial point X (0) of the curve X can be given with respect to the Taylor expansion of the curve X at this point in the Euclidean 5-space E5 as if in the Euclidean 3-space E3. Firstly, let us write Taylor expansion about the point X (0) up to fifth order and take the terms including the lowest powers of s in every component. Thus the Taylor expansion can be given by X (s)∼= X (0) + s X ′(0) + s2 2 X ′′(0) + s3 6 X ′′′(0) + s4 4! X (4)(0) + s5 5! X (5)(0). and considering the Frenet formulas, we obtain X (s)∼=X (0) + sV1(0) + s2 2 k1(0)V2(0) + s3 3! k1(0)k2(0)V3(0) + s4 4! k1(0)k2(0)k3(0)V4(0) + s5 5! k1(0)k2(0)k3(0)k4(0)V5(0). (1) The first two terms of the equation (1) X1 (s) = X (0) + sV1(0) gives us a tangent line which is the best linear approach of the curve X in the neighborhood of X (0). The first three terms of the equation (1) X2 (s) = X (0) + sV1(0) + s2 2 k1(0)V2(0) is a parabola which is the best quadratic approach of the curve X in the neighborhood of X (0). Thus parabola lies on the plane spanned by the vectors V1 and V2. thus the curvature k1(0) indicates how much V2 changes in the direction that is tangent to the curve. The first four terms of the equation (1) X3 (s) = X (0) + sV1(0) + s2 2 k1(0)V2(0) + s3 3! k1(0)k2(0)V3(0) is cubic which is the best cubic approach of the curve X in the neighborhood of X (0). This curve lies on Sp � V1, V2, V3 -subspace. The torsion k2(0) indicates how much V3 changes in the direction orthogonal to the V1, V2-plane of the curve. If k2(0) is zero, then the curve X lies on the Sp � V1, V2 -plane. The first five terms of the equation (1) X4 (s) = X (0) + sV1(0) + s2 2 k1(0)V2(0) + s3 3! k1(0)k2(0)V3(0) + s4 4! k1(0)k2(0)k3(0)V4(0) is a curve which is the best quartic approach of the curve X in the neighborhood of X (0). This curve lies on the Sp � V1, V2, V3, V4 -subspace. The curvature k3(0) is the scale of the curve M. Masal, A. Azak / Eur. J. Pure Appl. Math, 8 (2015), 255-270 259 X separating from the Sp � V1, V2, V3 -subspace. If k3(0) is zero, then the curve X lies on the Sp � V1, V2, V3 -subspace. The first six terms of the equation (1) X5 (s) =X (0) + sV1(0) + s2 2 k1(0)V2(0) + s3 3! k1(0)k2(0)V3(0) + s4 4! k1(0)k2(0)k3(0)V4(0) + s5 5! k1(0)k2(0)k3(0)k4(0)V5(0) is a curve which is the best quintic approach of the curve X in the neighborhood of X (0). This curve lies on the Sp � V1, V2, V3, V4, V5 -subspace. The curvature k4(0) is the scale of the curve X separating from the Sp � V1, V2, V3, V4 -subspace. If k4(0) is zero, then the curve X lies on the Sp � V1, V2, V3, V4 -subspace. Therefore, the following theorem can be given. Theorem 1. (i) A unit speed curve is a line if and only if the first curvature is zero. (ii) A unit speed curve is a quadratic (to be on the Sp � V1, V2 -plane) if and only if the second curvature is zero. (iii) A unit speed curve is a cubic (to be on the Sp � V1, V2, V3 -subspace) if and only if the third curvature is zero. (iv) A unit speed curve is a quartic (to be on the Sp � V1, V2, V3, V4 -subspace) if and only if the fourth curvature is zero. (v) A unit speed curve is a quintic (to be on the Sp � V1, V2, V3, V4, V5 -subspace) if and only if the all curvatures are different from zero. 4. Calculation of the Frenet Apparatus of the curves in the Euclidean 5-Space The Frenet apparatus of a curve with respect to any parameter in the Euclidean 5-space can be calculated via the same method in the Euclidean 3-space. Let X be an arbitrary curve and a function is of class C5 in E5. If the derivatives of the curve X up to the fifth order are calculated with respect to parameter t in terms of the parameter s, the following equations are obtained Ẋ =v V1, v = ds d t 6= 0 (2) Ẍ =v̇ V1 + v2 k1 V2 (3) ... X =(v̈ − v3k2 1)V1 + (3vv̇k1 + v2k̇1)V2 + (v 3k1k2)V3 (4) X (4) =( ... v − 6v2 v̇k2 1 − 3v3k1k̇1)V1 + (4vv̈k1 − v4k3 1 + 3v̇2k1 + 5vv̇k̇1 + v2k̈1 − v4k1k2 2)V2 + (6v2 v̇k1k2 + 2v3k̇1k2 + v3k1k̇2)V3 + (v 4k1k2k3)V4 (5) M. Masal, A. Azak / Eur. J. Pure Appl. Math, 8 (2015), 255-270 260 X (5) =(v(4) − 15vv̇2k2 1 − 10v2 v̈k2 1 − 26v2 v̇k1k̇1 − 3v3k̇2 1 − 4v3k1k̈1 + v5k4 1 + v5k2 1k2 2)V1 + (5 ... v vk1 − 10v3 v̇k3 1 − 6v4k2 1 k̇1 + 10v̇ v̈k1 + 9vv̈k̇1 + 8v̇2k̇1 + 7vv̇k̈1 + v2 ... k1 − 10v3 v̇k1k2 2 − 3v4k̇1k2 2 − 3v4k1k2k̇2)V2 + (10v2 v̈k1k2 − v5k3 1k2 + 15vv̇2k1k2 + 17v2 v̇ k̇1k2 + 3v3k̈1k2 − v5k1k3 2 + 9v2 v̇k1k̇2 + 3v3k̇1k̇2 + v3k1k̈2 − v5k1k2k2 3)V3 + (10v3 v̇k1k2k3 + 3v4k̇1k2k3 + 2v4k1k̇2k3 + v4k1k2k̇3)V4 + (v5k1k2k3k4)V5 (6) where "·" denotes the derivative with respect to t. From the equation (2), we find v = Ẋ (7) and V1 = Ẋ Ẋ . (8) Since v2 = Ẋ , Ẋ � , if the derivative of this term is taken consecutively, we have v̇ = Ẋ , Ẍ � Ẋ (9) and v̈ = Ẍ 2 Ẋ 2 + Ẋ , ... X � Ẋ 2 − Ẋ , Ẍ �2 Ẋ 3 (10) If the first curvature is calculated from the equation (3), the following equations are obtained k1 = Ẋ 2 Ẍ − 〈Ẋ , Ẍ 〉Ẋ Ẋ 4 (11) and k2 1 = Ẍ 2 Ẋ 2 − 〈Ẋ , Ẍ 〉2 Ẋ 6 (12) If we take the derivative of both sides of the equation (11) with respect to t, we get k̇1 = Ẋ 4 〈Ẍ , ... X 〉+ 3〈Ẋ , Ẍ 〉3 − Ẋ 2 〈Ẋ , Ẍ 〉〈Ẋ , ... X 〉 − 3 Ẋ 2 Ẍ 2 〈Ẋ , Ẍ 〉 Ẋ 4 Ẋ 2 Ẍ − 〈Ẋ , Ẍ 〉Ẋ (13) In addition to this, the second Frenet vector from the equation (3) is V2 = Ẍ Ẋ 2 − 〈Ẋ , Ẍ 〉Ẋ Ẋ 2 Ẍ − 〈Ẋ , Ẍ 〉Ẋ . (14) M. Masal, A. Azak / Eur. J. Pure Appl. Math, 8 (2015), 255-270 261 By using the equation (4), we can write 〈 ... X , V3〉= v3k1k2. Substituting the equations (7) and (11) in the above equation, we obtain the second curvature k2 as follows k2 = 〈 ... X , V3〉 Ẋ Ẋ 2 Ẍ − 〈Ẋ , Ẍ 〉Ẋ . (15) Again, considering the equation (4), the third Frenet vector V3 of X is given by V3 = ... X − aV1 − bV2 ... X − aV1 − bV2 such that a =v̈ − v3k2 1 b =3vv̇k1 + v2k̇1 . (16) Substituting the equations (7), (9), (10), (11) and (13) in the equation (16), we have a = 〈Ẋ , ... X 〉 Ẋ . and b = Ẋ 2 〈Ẍ , ... X 〉 − 〈Ẋ , Ẍ 〉〈Ẋ , ... X 〉 Ẋ 2 Ẍ − 〈Ẋ , Ẍ 〉Ẋ . Now, we can compute the vector form V1 ∧ V2 ∧ ... X ∧ X (4) as the follows; V1 ∧ V2 ∧ ... X ∧ X (4) = v7k2 1k2 2k3V5. (17) then from the above equation V5 = η V1 ∧ V2 ∧ ... X ∧ X (4) V1 ∧ V2 ∧ ... X ∧ X (4) . (18) and η is taken ±1 to make det(V1, V2, V3, V4, V5) = +1. Substituting the equations (7), (12) and (15) in the equation (17), the third curvature is found k3 = V1 ∧ V2 ∧ ... X ∧ X (4) 〈 ... X , V3〉2 Ẋ . (19) The inner product 〈X (5), V5〉 gives us the fourth curvature k4 as k4 = 〈X (5), V5〉 v5k1k2k3 . (20) M. Masal, A. Azak / Eur. J. Pure Appl. Math, 8 (2015), 255-270 262 Then, if we substitute the equations (7), (11), (15) and (19) in the above equation, we imme- diately arrive to k4 = 〈X (5), V5〉〈 ... X , V3〉 V1 ∧ V2 ∧ ... X ∧ X (4) Ẋ . (21) Finally, the fourth Frenet vector is V3 ∧ V2 ∧ V1 ∧ V5 = V4 (22) Therefore, the following theorem can be given. Theorem 2. Let X be an arbitrary curve of class C5 in the Euclidean 5-space E5. In this regard, the Frenet vectors and curvatures of the curve X are V1 = Ẋ Ẋ , V2 = Ẍ Ẋ 2 − Ẋ Ẋ , Ẍ � Ẋ 2 Ẍ − Ẋ Ẋ , Ẍ � , V3 = ... X − aV1 − bV2 ... X − aV1 − bV2 , a = Ẋ , ... X 〉 Ẋ , b = Ẋ 2 Ẍ , ... X 〉 − Ẋ , Ẍ � Ẋ , ... X 〉 Ẋ 2 Ẍ − Ẋ , Ẍ � Ẋ , V4 =ηV3 ∧ V2 ∧ V1 ∧ V5, V5 =η V1 ∧ V2 ∧ ... X ∧ X (4) V1 ∧ V2 ∧ ... X ∧ X (4) , k1 = Ẋ 2 Ẍ − Ẋ , Ẍ � Ẋ Ẋ 4 , k2 = ... X , V3 � Ẋ Ẋ 2 Ẍ − Ẋ , Ẍ � Ẋ , k3 = V1 ∧ V2 ∧ ... X ∧ X (4) ... X , V3 � Ẋ , k4 = X (5), V5 � ... X , V3 � V1 ∧ V2 ∧ ... X ∧ X (4) Ẋ , respectively. 5. Involute-Evolute Curve Couples in the Euclidean 5-Space Let X be a W-curve and Y be the involute of X in E5. While the Frenet apparatus of X is � V1, V2, V3, V4, V5, k1, k2, k3, k4 , we will denote the Frenet apparatus of Y with M. Masal, A. Azak / Eur. J. Pure Appl. Math, 8 (2015), 255-270 263 � V Y 1 , V Y 2 , V Y 3 , V Y 4 , V Y 5 , kY 1 , kY 2 , kY 3 , kY 4 . So, from the definition of involute-evolute curve, we may express Y = X +µV1 (23) where s and sY denote the arc-parameters of the curves X and Y , respectively. Differentiating the both sides of the equation (23) with respect to s, one can obtain dY dsY dsY ds = dX ds + dµ ds V1 +µk1V2. (24) Since the tangent vector V1 of the curve X orthogonal to the tangent vector V Y 1 of the curve Y , it is easily seen that 1+ dµ ds = 0. (25) We know that µ= c − s from the equation (25). So, we can write Y = X + (c − s)V1 (26) and V Y 1 dsY ds = (c − s)k1V2. (27) Also the equation (27) yields Ẏ = (c − s)k1V2. (28) If we take the norm of Ẏ , we have Ẏ = (c − s)k1. (29) where the subscript dot "·" denotes the derivative of Y with respect to s. Moreover, the derivatives of the curve Y up to the fifth order are given by Ÿ =− (c − s)k2 1V1 − k1V2 + (c − s)k1k2V3, (30) ... Y =2k2 1V1 − (c − s)k1(k 2 1 + k2 2)V2 − 2k1k2V3 + (c − s)k1k2k3V4, (31) Y (4) =(c − s)k2 1(k 2 1 + k2 2)V1 + 3k1(k 2 1 + k2 2)V2 − (c − s)k1k2(k 2 1 + k2 2 + k2 3)V3 − 3k1k2k3V4 + (c − s)k1k2k3k4V5, (32) Y (5) =− 4k2 1(k 2 1 + k2 2)V1 + (c − s)k1 � k2 1(k 2 1 + k2 2) + k2 2(k 2 1 + k2 2 + k2 3) � V2 + 4k1k2(k 2 1 + k2 2 + k2 3)V3 − (c − s)k1k2k3(k 2 1 + k2 2 + k2 3 + k2 4)V4 − 4k1k2k3k4V5. (33) From the equation (8), the first Frenet vector of the curve Y can be written as V Y 1 = Ẏ Ẏ . M. Masal, A. Azak / Eur. J. Pure Appl. Math, 8 (2015), 255-270 264 Considering the equations (28) and (29), we find V Y 1 = V2. (34) From the equations (28) and (30), we get Ẏ 2 Ÿ − Ẏ , Ÿ � Ẏ = −(c − s)3k4 1V1 + (c − s)3k3 1k2V3 (35) and Ẏ 2 Ÿ − Ẏ , Ÿ � Ẏ = (c − s)3k3 1 q k2 1 + k2 2. (36) If we use the equations (11) and (14), then we will obtain the second Frenet vector and the first curvature of curve Y as follows; V Y 2 = − k1 q k2 1 + k2 2 V1 + k2 q k2 1 + k2 2 V3 (37) and kY 1 = q k2 1 + k2 2 (c − s)k1 (38) respectively. Besides, considering the equations (28), (29), (30), (31) and (36), one can calculate ... Y − Ẏ , ... Y 〉 Ẏ V Y 1 − Ẏ 2 Ÿ , ... Y 〉 − Ẏ , Ÿ � Ẏ , ... Y 〉 Ẏ 2 Ÿ − Ẏ , Ÿ � Ẏ V Y 2 = (c − s)k1k2k3V4. (39) The third Frenet vector of Y is obtained from the equation (16) as V Y 3 = V4. (40) If the equations (31) and (40) are taken into consideration, we get ... Y , V Y 3 � = (c − s)k1k2k3. (41) From the equations (15), (36) and (41), the second curvature of Y is found as kY 2 = k2k3 (c − s)k1 q k2 1 + k2 2 . (42) Moreover, from the equations (31), (32), (34) and (37), we have V Y 1 ∧ V Y 2 ∧ ... Y ∧ Y (4) = (c − s)2k2 1k3 2k2 3k4 q k2 1 + k2 2 V1 + (c − s)2k3 1k2 2k2 3k4 q k2 1 + k2 2 V3 + (c − s)2k3 1k2 2k3 3 q k2 1 + k2 2 V5 (43) and V Y 1 ∧ V Y 2 ∧ ... Y ∧ Y (4) = (c − s)2k2 1k2 2k2 3 q k2 1 + k2 2 q k2 1k2 3 + k2 2k2 4 + k2 1k2 4. (44) M. Masal, A. Azak / Eur. J. Pure Appl. Math, 8 (2015), 255-270 265 Thus, if we take the equations (18), (43) and (44), the fifth Frenet vector of the curve Y is V Y 5 =k2k4 q k2 1k2 3 + k2 2k2 4 + k2 1k2 4V1 + k1k4 q k2 1k2 3 + k2 2k2 4 + k2 1k2 4V3 + k1k3 q k2 1k2 3 + k2 2k2 4 + k2 1k2 4V5. (45) By (41) and (44), we obtain V Y 1 ∧ V Y 2 ∧ ... Y ∧ Y (4) � ... Y , V Y 3 ��2 Ẏ = q k2 1k2 3 + k2 2k2 4 + k2 1k2 4 (c − s)k1 q k2 1 + k2 2 . Therefore, from the equation (19) the third curvature of the curve Y can be found as follows kY 3 = q k2 1k2 3 + k2 2k2 4 + k2 1k2 4 (c − s)k1 q k2 1 + k2 2 . (46) If the vectorial product V Y 3 ∧ V Y 2 ∧ V Y 1 ∧ V Y 5 is calculated by using the equations (34), (37), (40) and (45), we find the fourth Frenet vector of the curve Y , considering the equation (22) as follows: V Y 4 = −k1k2k3 q k2 1 + k2 2 q k2 1k2 3 + k2 2k2 4 + k2 1k2 4 V1 − k2 1k3 q k2 1 + k2 2 q k2 1k2 3 + k2 2k2 4 + k2 1k2 4 V3 + k4(k 2 1 + k2 2) q k2 1 + k2 2 q k2 1k2 3 + k2 2k2 4 + k2 1k2 4 V5 (47) From the equations (33) and (45), the inner product of the vectors Y (5) and V Y 5 is Y (5), V Y 5 � = 4k2 1k2 2k4(1− k2) q k2 1k2 3 + k2 2k2 4 + k2 1k2 4 . (48) Finally, considering the equations (21), (41), (44) and (48) the fourth curvature of Y is found as kY 4 = 4k2k4(1− k2) q k2 1 + k2 2 k3(c − s)2(k2 1k2 3 + k2 2k2 4 + k2 1k2 4) . (49) Therefore, the following theorem and results can be given Theorem 3. Let X be a W-curve and Y be the involute of X in E5. � V1, V2, V3, V4, V5, k1, k2, k3, k4 and � V Y 1 , V Y 2 , V Y 3 , V Y 4 , V Y 5 , kY 1 , kY 2 , kY 3 , kY 4 denote the Frenet apparatus of the curves X and Y , respectively. The relation can be expressed as V Y 1 =V2, M. Masal, A. Azak / Eur. J. Pure Appl. Math, 8 (2015), 255-270 266 V Y 2 =− k1 q k2 1 + k2 2 V1 + k2 q k2 1 + k2 2 V3, V Y 3 =V4, V Y 4 =− k1k2k3 q k2 1 + k2 2 q k2 1k2 3 + k2 2k2 4 + k2 1k2 4 V1 − k2 1k3 q k2 1 + k2 2 q k2 1k2 3 + k2 2k2 4 + k2 1k2 4 V3 + k4 � k2 1 + k2 2 � q k2 1 + k2 2 q k2 1k2 3 + k2 2k2 4 + k2 1k2 4 V5, V Y 5 =k2k4 q k2 1k2 3 + k2 2k2 4 + k2 1k2 4V1 + k1k4 q k2 1k2 3 + k2 2k2 4 + k2 1k2 4V3 + k1k3 q k2 1k2 3 + k2 2k2 4 + k2 1k2 4V5, kY 1 = q k2 1 + k2 2 (c − s) k1 , kY 2 = k2k3 (c − s) k1 q k2 1 + k2 2 , kY 3 = q k2 1k2 3 + k2 2k2 4 + k2 1k2 4 (c − s) k1 q k2 1 + k2 2 , kY 4 = 4k2k4 � 1− k2 � q k2 1 + k2 2 (c − s)2 k3 � k2 1k2 3 + k2 2k2 4 + k2 1k2 4 � . Corollary 1. � V Y 1 , V Y 2 , V Y 3 , V Y 4 , V Y 5 is an orthonormal frame in E5. Corollary 2. While X is a W-curve, Y can not be a W-curve. Corollary 3. The involute curve Y can’t be an inclined curve. 6. The Spherical Curves in Euclidean 5-Space Let X ⊂ R5 curve be given with coordinate neighborhood (I , X ) and s ∈ I be arc-length parameter of X . Also, assume that S4 is a hypersphere which has six common coalescent points with the curve X . If X (s) is a point on this hypersphere, C is the center of this hypersphere and r is the radius of it, then the equation of the hypersphere S4 is 〈X (s)− C , X (s)− C〉 = r2. (50) On the other hand, for the base � V1, V2, V3, V4, V5 and mi(s) ∈ R C − X (s) = m1(s)V1(s) +m2(s)V2(s) +m3(s)V3(s) +m4(s)V4(s) +m5(s)V5(s), (51) can be written. Hence, mi(s) = C − X (s), Vi(s) � , 1≤ i ≤ 5 (52) M. Masal, A. Azak / Eur. J. Pure Appl. Math, 8 (2015), 255-270 267 In accordance with this, let us consider f :I → R s→ f (s) = 〈X − C , X − C〉 − r2. If we have the following equations f (s) = f ′(s) = f ′′ (s) = f ′′′ (s) = f (4) (s) = f (5)(s) = 0 then we say that the hypersphere touches to X at the fifth order to the curve at X (s). Therefore, f (s) = 〈X − C , X − C〉 − r2 = 0, (53) f ′(s) = V1, X − C � = 0⇒ m1 = 0, (54) f ′′(s) =0⇒ V2, X − C � = 1 k1 ⇒ m2 = 1 k1 , (55) f ′′′(s) =0⇒ V3, X − C � = m′2 k2 ⇒ m3 = m′2 k2 , (56) f (4)(s) =0⇒ V4, X − C � = m′3 + k2m2 k3 ⇒ m4 = m′3 + k2m2 k3 , (57) f (5)(s) =0⇒ V5, X − C � = m′4 + k3m3 k4 ⇒ m5 = m′4 + k3m3 k4 . (58) are obtained. Thus, the center of the hypersphere is C = X +m2V2 + � m′2 k2 � V3 +    � m′2 k2 �′ + k2m2 k3   V4 +        � m′2 k2 �′ + k2m2 k3    ′ + k3 � m′2 k2 �     1 k4 V5 (59) and for the square of the radius r2 = m2 2 + � m′2 k2 �2 +    � m′2 k2 �′ + k2m2 k3    2 +        � m′2 k2 �′ + k2m2 k3 + k3 � m′2 k2 �    ′     2 1 k2 4 (60) is attained. Differentiating the equation (59), we obtain the derivative of the center as follows C ′ = (m4k4 +m′5)V5. (61) Considering the equality (61), it can be said that the centers of the osculating hyperspheres of a spherical curve are in the direction of V5. In addition, all spherical curves satisfy the following M. Masal, A. Azak / Eur. J. Pure Appl. Math, 8 (2015), 255-270 268 differential equation: m2 2 + � m′2 k2 �2 + � � m′2 k2 �′ + k2m2 �2 1 k2 3 +        � m′2 k2 �′ + k2m2 k3 + k3 � m′2 k2 �    ′     2 1 k2 4 = a2. (62) If the curve is spherical, then the hypersphere is also an osculating hypersphere. Here a will be the radius of the hypersphere. Conversely, if the equation (62) is provided, the radius of the osculating hypersphere is constant. If the derivative of the equation (62) is taken, m5(m4k4 +m′5) = 0 (63) is found. Therefore, if we consider the equation (63) with (61), then C ′ = 0. This means that the center of the osculating hypersphere is constant. From the equation (63), all of the differential equations of the spherical curves are m4k4 +m′5 = 0 (64) or � � m′2 k2 �′ + k2m2 � k4 k3 +            � m′2 k2 �′ + k2m2 k3    ′ + k3 � m′2 k2 �     1 k4     ′ = 0. However, the following theorem can be given: Theorem 4. Let X be a unit speed curve in E5. (i) The curve X is a spherical curve if and only if the differential equation m4k4 +m′5 = 0 is satisfied. (ii) If X is a spherical curve, then the center of the hypersphere is C = X +m2V2 +m3V3 +m4V4 +m5V5 and the radius is r = q m2 2 +m2 3 +m2 4 +m2 5 such that m2 = 1 k1 , m3 = m′2 k2 , m4 = m′3 + k2m2 k3 , m5 = m′4 + k3m3 k4 . (iii) The radius of the osculating hypersphere is constant at the point X (s) if and only if the centers of the osculating hyperspheres are the same[9]. REFERENCES 269 ACKNOWLEDGEMENTS The authors thank to referees. References [1] A.T. Ali. Inclined curves in the euclidean 5-space e5. 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