On the Inclined Curves in Galilean 4-Space

Transcription

1 Applie Mathematical Sciences Vol no HIKARI Lt On the Incline Curves in Galilean 4-Space Dae Won Yoon Department of Mathematics Eucation an RINS Gyeongsang National University Jinju South Korea wyoon@gnu.ac.kr Copyright c 2013 Dae Won Yoon. This is an open access article istribute uner the Creative Commons Attribution License which permits unrestricte use istribution an reprouction in any meium provie the original work is properly cite. Abstract In this paper we give some characterizations for a incline curve by means of curvatures of a curve in a 4-imensional Galilean space. Mathematics Subject Classification: 53A04 Keywor: Galilean space Incline curve Curvature Frenet formula 1 Introuction In Eucliean 3-space R 3 a incline curve or general helix is a curve where the tangent lines make a constant angle with a fixe irection. A incline curve is characterize by the fact that the ratio τ/κ is constant along the curve where κ an τ enote the curvature an the torsion respectively. In Minkowski 3- space R1 3 an Galilean 3-space G3 one efines a incline curve in R1 3 an G3 by a similar way. Although ifferent expansions of the Frenet formula appear epening on the form of the Frenet vectors in many cases a incline curve is characterize by the constancy of the function τ/κ again [2][5]) Recently the stuy of a incline curve in 4-imensional space R 4 R1 4 an Q 4 quaternionic space examine by [3] [4 6] an [8] respectively. In this paper we investigate the properties of a incline curve in terms of the curvatures of a curve in Galilean 4-space G 4.

2 2194 D. W. Yoon 2 Preliminary Notes Let α : I R G 4 be an arbitrary curve in a 4-imensional Galilean space G 4 efine by αt) =xt)yt)zt)wt)) where xt)yt)zt)wt) are smooth functions. For any vectors x =x 1 y 1 z 1 w 1 ) an y =x 2 y 2 z 2 w 2 ) in Galilean space G 4 the Galilean scalar prouct of x an y is efine by { x1 x x y = 2 if x 1 0orx 2 0 y 1 y 2 + z 1 z 2 + w 1 w 2 if x 1 = 0 an x 2 =0. From this the Galilean norm of a vector x in G 4 is given by { x1 if x 1 0 x = y1 2 + z1 2 + w1 2 if x 1 =0. Let x = x 1 y 1 z 1 w 1 ) y = x 2 y 2 z 2 w 2 ) an z = x 3 y 3 z 3 w 3 ) be the vectors in G 4. Then the Galilean cross prouct on G 4 efine by x y z = 0 e 2 e 3 e 4 x 1 y 1 z 1 w 1 x 2 y 2 z 2 w 2 x 3 y 3 z 3 w 3 where e i are the stanar basis vectors. On the other han a curve α in G 4 parameterize by arc-length t = s given in coorinate form It follows that the tangent vector of α is given by αs) =s ys)zs)ws)). 1) t = α s) =1y s)z s)w s)). From this we obtain the first curvature k 1 as follows: k 1 s) = t s) = y ) 2 +z ) 2 +w ) 2. By the similar arguments as those of in the Eucliean ifferential geometry we have the following the Frenet vectors {t n 1 n 2 n 3 } of αs) ing 4 : ts) =1y s)z s)w s)) n 1 s) = 1 k 1 s) 0y s)z s)w s)) n 2 s) = 1 s) n 3 s) =εt n 1 n k 1 s) y s) ) 1 k 1 s) z s) ) 1 k 1 s) w s) ) )

3 On the incline curves 2195 where ε is taken ±1 to make +1 the eterminant tn 1 n 2 n 3 an s) = n 1 s) it is calle the secon curvature of αs). For their erivatives the following Frenet formula satisfies[cf. 7]) t s) =k 1 s)n 1 s) n 1 s) =s)n 2 s) n 2s) = s)n 1 s)+k 3 s)n 3 s) n 3 s) = k 3s)n 2 s). 2) Here k 3 = n 2 n 3 it sai to be the thir curvature of αs). 3 Characterizations of incline curves in G 4 Theorem 3.1 Let α = αs) be a unit spee curve in G 4 with non-zero curvatures k 1 s) s)) an k 3 s). Then α is a incline curve in G 4 if an only if the function ) 2 3) is a constant. 3 Proof. Let αs) be a incline curve in G 4 an the axis of the curve αs) be the unit vector u. Then we have t u = constant 4) along the curve α. By ifferentiating 4) with respect to s an using the Frenet formula 2) we have k 1 n 1 u =0 which implies that the unit vector u is in the subspace span{t n 2 n 3 } an can be written as follows where u = a 1 s)ts)+a 2 s)n 2 s)+a 3 s)n 3 s) 5) a 1 s) = t u = constant a 2 s) = n 2 u a 3 s) = n 3 u a a2 2 + a2 3 =1. The ifferentiation of 5) gives a 1 k 1 a 2 )n 1 +a 2 a 3 k 3 )n 2 +a 3 + a 2 k 3 )n 3 =0. Since the vectors n 1 n 2 n 3 are linearly inepenent we yiel a 1 k 1 a 2 =0a 2 a 3 k 3 =0a 3 + a 2 k 3 =0 that is a 2 = k 1 a 1 = 1 k 3 a 3 a 2 = a 3k 3. 6)

4 2196 D. W. Yoon Therefore from 6) we have the ODE for a 3 as follows a 3 k 3 a 3 + k k 3a 2 3 =0. 7) 3 If we change variables in 7) as t = s 0 k 3 then 7) becomes 2 a 3 t + a 2 3 =0. 8) Solving this ifferential equation we get the solution a 3 = A cos ts)+b sin ts) 9) for some constants A an B. From the first equation of 6) an 9) we fin a 2 = k 1a 1 = A sin ts) B cos ts) a 3 = 1 a 1 = A cos ts)+bsin ts). k 3 From the above equations we obtain ) k A = a 1 1 sin ts)+ 1 k 3 cos ts) ) 1 B = a 1 k 3 sin ts) k 1 cos ts) which imply Thus we have ) 2 A 2 + B 2 = a ) 2. k3 2 ) ) ) 2 = constant. 10) 3 Conversely if 4) hol then we can always fin a constant unit vector u satisfying t u =constant. We consier the unit vector efine by u = t + k 1 n k 3 ) n 3. Differentiation of u with the help of 10) gives u = 0 this mean that u is a constant vector. Consequently the curve αs) is a incline curve in G 4. Theorem 3.2 A unit spee curve αs) in G 4 is a incline curve if an only if there exists a C 2 -function f such that k 3 fs) = fs) = k 3. 11)

5 On the incline curves 2197 Proof. We assume that γ is a incline curve. Differentiation of 10) gives or equivalently Therefore we have ) If we efine f = fs) by then 13) becomes ) k 3 From 13) it can be written k k 3 fs) = = k ) =0 k 3 ) 1 k 3 k 2 1 k3 k 3 fs) = 1 k 3 By combining 14) an 15) we fin k 2 1 )) k3 ) fs) = k 3 ) =0. 12) )). 13). 14) = k 3. 15). 16) Conversely if 11) hol we efine a unit constant vector u by u = t + k 1 n 2 + fs)n 3. It follows t u = 1. Thus α is a incline curve. Theorem 3.3 Let α be a unit spee curve in G 4. curve if an only if the following conition hol; Then α is a incline where C 1 C 2 are constants an ts) = s 0 k 3. k 1 = C 1 cos t + C 2 sin t 17)

6 2198 D. W. Yoon Proof. Suppose that α is a incline curve. By using Theorem 3.1 let efine the C 2 -function ts) an the C 1 -functions ms) an ns) by ts) = s 0 k 3 18) ms) = k 1 cos t fs) sin t ns) = k 19) 1 sin t + fs) cos t. If we ifferentiate equation 19) with respect to s an take account of 18) 14) an 16) we have m = 0 an n = 0. Therefore m = C 1 an n = C 2 are constants. Thus from 19) we obtain k 1 = C 1 cos t + C 2 sin t. Conversely if the equation 17) hol. Then from 19) we have f = C 1 cos t + C 2 sin t it satisfies the conition of Theorem 3.3. Thus α is a incline curve in G 4. References [1] A. C. Coken an A. Tuna On the quaternionic incline curves in the semi- Eucliean space E 4 2 Applie Math. Computation ) [2] M. Barros A. Ferranez P. Lucas ans M. A. Merono Genernal helices in the three-imensional Lorentz space forms Rocky Mountain J. Math ) [3] A. Magen On the curves of constant slope YYU Fen Bilimleri Dergisi ) [4] H. Kocayigit an M. Oner Time-like curves of constant slope in Miknkowski space E 4 1 BU/JST 12007) [5] A. O. Ogrenmis M. Ergut an M. Bertas On the helices in the Galilean space G 3 Iranian J. Sci. Tech ) [6] M. Oner H. Kocayigit an M. Kazaz Space-like helices in Miknkowski 4-space E1 4 Ann Univ Ferrara ) [7] S. Yilmaz Construction of the Frenet-Serret frame of a curve in 4D Galilean space an some applications Inter. J. Phy. Sci )

7 On the incline curves 2199 [8] D. W. Yoon On the quaternionic general helices in Eucliean 4-space submitte for publication. Receive: February