aa r X i v : . [ m a t h . G M ] J a n ON A NEW TYPE BERTRAND CURVE
C¸ ET˙IN CAMCI
Abstract.
In this paper, we define a new type Bertrand curve and thiscurves are said V -Bertrand curve, f -Bertrand curve and a -Bertrandcurve. In addition, we give charectarization of the V -Bertrand curveand we define a Bertrand surface. Introduction
Let γ : I −→ R ( s → γ ( s )) be a unit-speed curve. Formula of theSerret-Frenet equations are given by T ´ N ´ B ´ = κ − κ τ − τ TNB where { T, N, B, κ, τ } is the Serret-Frenet apparatus of the curves([3], [4]). In 3-Euclidean spaces, Saint-Venant proposed that the ruled surfacewhose based curve is the principal normals of a curve and another curvewhich have same principal normals can exist ([9]). Solution of the problemwas given by Bertrand([10]). Let M be a unit-speed curve with coordinateneighborhood ( I, γ ) of the curve. So we can define a curve by β ( s ) = γ ( s ) + λ ( s ) N ( s )where λ : I −→ R ( s → λ ( s )) is differentiable function. Let (cid:8) T , N , B (cid:9) beFrenet frame of the curve and κ and τ be curvatures of the curves K .If (cid:8) N, N (cid:9) is linear dependent (cid:0) N = ǫN, ǫ = ± (cid:1) , then it is said that M is Bertrand mate of K ([13],[14]). If there is Bertrand mate of the curve Mathematics Subject Classification.
Primary 53C15; Secondary 53C25.
Key words and phrases.
Sasakian Space, curve . M , then it is said that M is Bertrand curve ([13],[14]). For example, anyplanar curve and circular helix is Bertrand curve. It is well known theoremthat M is Bertrand curve if and only if there exist λ , µ ∈ R such that λκ + µτ = 1 where λ = 0 , µ = λ cot θ and θ is angle between T and T ([13],[14]). In fact, M is Bertrand or B − Bertrand curve if and only if thereare λ , µ ∈ R such that λκ + µτ = 1.If the curve is a unit-speed spherical curve, for all s ∈ I , γ ( s ) is perpen-dicular to T = γ ′ ( s ). So we have Sabban frame as (cid:8) γ ( s ), T ( s ) = γ ′ ( s ), Y ( s ) = γ ( s ) × T ( s ) (cid:9) ([5]). Serret-Frenet formula of the this curve is given by γ ′ T ′ Y ′ = − κ g − κ g γTY where κ g ( s ) = det ( γ, T, T ′ ) ([5]). Under the above notation, the spacecurve is definned by(1.1) e γ ( s ) = a Z γ ( s ) ds + a cot θ Z Y ( s ) ds + c where a, θ ∈ R and c is constant vector ([5]). Izumiya and Tkeuchi provethat all Bertrand curves can be constructed from equation (1.1) and theygave some Bertrand curve examples by using equation (1.1) ([5]). In thispaper, we give two method different from above method. Furthermore, wedefine a V -Bertrand curve, f -Bertrand curve and a -Bertrand curve whereBertrand curve is T − Bertrand curve and a ∈ R . Furthermore, we modifycharectarization of the Bertrand curve. In addition, we define a Bertrandsurface and we give an example Bertrand surface. Moreover, we define aequivalence relation on Bertrand surface set.2. Preliminaries
In 3-Euclidean spaces, let M be a unit-speed curve and { T, N, B, κ, τ } is Serret-Frenet apparatus of the curves. The curve lies on 2-sphere if and only if below equation is hold(2.1) (cid:18)(cid:18) κ (cid:19) ′ τ (cid:19) ′ + τκ = 0([6],[8]). From the solution of the equation (2.1), we have(2.2) 1 κ = R cos s Z τ ( u ) du + θ . where radius of the sphere is equal to R ( [6],[7], [11]). It is well known thatif M is spherical curve, then M and osculating circle lies on this sphere.From equation (2.2), we have1 κ = cos θ = R R . If R is equal to 1, then we have cos θ = R = κ ([12]).Choi and Kim defined a unit vector field V which is given by V ( s ) = u ( s ) T ( s ) + v ( s ) N ( s ) + w ( s ) B ( s )and integral curve of V is define by γ V ( s ) = Z V ( s ) ds where u, v, w are functions from I to R and u + v + w = 1 ([1]). Withoutloss of generality, we suppose that arc-lenght parameter of M and integralcurve of V are same ([1]). If u ( s ) = 1 , v ( s ) = w ( s ) = 0, for all s ∈ I , thenwe have γ T ( s ) = Z T ( s ) ds = γ ( s ) . If u ( s ) = w ( s ) = 0, v ( s ) = 1, for all s ∈ I , then we have γ N ( s ) = Z N ( s ) ds and if u ( s ) = v ( s ) = 0, w ( s ) = 1, for all s ∈ I , then we have γ B ( s ) = Z B ( s ) ds ([1]). ( γ N ) (resp.( γ B )) is called principal-direction curve (resp. binormal-direction curve) of γ ([1]). Let κ = p κ + τ , τ = κ κ + τ (cid:16) τκ (cid:17) ′ and T = N , N = N ′ k N ′ k , B = T × N ( resp. κ β = | τ | , τ β = κ and T β = B, N β = N, B β = − T ) be curvatures and Frenet vectors of principal-direction curve (resp. binormal-direction curve)([1], [2]). If u ( s ) = 0 , v ( s ) = − cos (cid:18)Z τ ( s ) ds (cid:19) = 0 , w ( s ) = sin (cid:18)Z τ ( s ) ds (cid:19) then γ pdo ( s ) = R V ( s ) ds is called principal-donor curve of γ , for all s ∈ I ([1]). So we have κ pdo = κ cos (cid:18)Z τ ( s ) ds (cid:19) , τ pdo = κ sin (cid:18)Z τ ( s ) ds (cid:19) and T pdo = V , N pdo = T , B pdo = T pdo × N pdo .3. V-Bertrand Curve
Let γ : I −→ R ( s → γ ( s )) be a unit-speed curve and { T, N, B, κ, τ } be Serret-Frenet apparatus of the curve. We can define a curve K by β ( s ) = Z V ( s ) ds + λ ( s ) N ( s )where λ : I −→ R ( s → λ ( s )) is differentiable function and V = uT + vN + wB is unit vector field. Let (cid:8) T , N , B (cid:9) be orthonormal frame of thecurve and κ and τ be curvatures of the curve K . So we can give followingdefinition. Definition 3.1. If (cid:8) N, N (cid:9) is linear dependent (cid:0) N = ǫN, ǫ = ± (cid:1) , then itis said that ( M, K ) is V -Bertrand mate. If V = T , then ( M, K ) is Bertrandmate. Theorem 3.1.
Let M be a unit-speed curve and { T, N, B, κ, τ } be Serret-Frenet apparatus of the curves. M is V -Bertrand curve if and only if it issatisfied that (3.1) λ ( κ tan θ + τ ) = u tan θ − w and (3.2) λ ( s ) = − Z v ( s ) ds where θ is constant angle between T and T .Proof. If M is V -Bertrand curve, then V -Bertrand mate of M is equal to(3.3) β ( s ) = Z V ( s ) ds + λ ( s ) N ( s )and (cid:8) N, N (cid:9) is linear dependent. If we derivative equation (3.3), we have dsds T = ( u − λκ ) T + (cid:0) λ ′ + v (cid:1) N + ( w + λτ ) B. Since (cid:8)
N, N (cid:9) is linear dependent, we have λ ( s ) = − Z v ( s ) ds and we get(3.4) T = dsds ( u − λκ ) T + dsds ( w + λτ ) B = cos θ ( s ) T + sin θ ( s ) B where cos θ ( s ) = dsds ( u − λκ ) and sin θ ( s ) = dsds ( w + λτ ). So we havetan θ = w + λτu − λκ or λ ( κ tan θ + τ ) = u tan θ − w .If we take the derivative of a equation (3.4), we have dsds κN = − θ ′ sin θT + ( κ cos θ − τ sin θ ) N + θ ′ cos θB So that (cid:8)
N, N (cid:9) is linear dependent, we have θ ′ = 0 or θ = const . Con-versely, we define a curve by(3.5) β ( s ) = Z V ( s ) ds + λ ( s ) N ( s )where λ : I −→ R ( s → λ ( s )) is differentiable function such that λ ( s ) = − R v ( s ) ds . If we derivative equation (3.5), we have dsds T = ( u − λκ ) T + (cid:0) λ ′ + v (cid:1) N + ( w + λτ ) B. From equation (3.2), we get(3.6) T = dsds ( u − λκ ) T + dsds ( w + λτ ) B = cos φ ( s ) T + sin φ ( s ) B where cos φ ( s ) = dsds ( u − λκ ) and sin φ ( s ) = dsds ( v + λτ ). From equation(3.1) and (3.6) we have tan φ ( s ) = v + λτu − λκ = tan θ. So φ = θ + nπ is angle between T and T where n ∈ Z . If we take thederivative of a equation (3.6), we see that (cid:8) N, N (cid:9) is linear dependent. (cid:3)
Definition 3.2.
Let M be a unit-speed non-planar curve and { T, N, B, κ, τ } is Serret-Frenet apparatus of the curves. If there is λ = 0 , θ ∈ R such that λκ + λ cot θτ = 1 , then we said that M is Bertrand curve (or T -Bertrand curve ). If there is λ = 0 , θ ∈ R such that λ tan θκ + λτ = − , then we said that M is B -Bertrand curve. Corollary 3.1. If u ( s ) = 1 , v ( s ) = w ( s ) = 0 , we have Bertrand curve.From equation (3.2), λ is constant. From equation (3.1), we have λκ + µτ =1 where λ = 0 , µ = λ cot θ and θ is constant angle between T and T .If v ( s ) = 1 , u ( s ) = w ( s ) = 0 , we have B -Bertrand curve. From equation(3.1), we have λκ + µτ = − where µ = 0 is constant, λ = µ tan θ and θ isconstant angle between T and T . Corollary 3.2.
If the curve is a Salkowki curve, then curvature of thecurve is equal to (or constant). If λ = 1 , µ = λ cot θ = 0 , then we have λκ + µτ = 1 . if µ = 0 , λ = µ cot θ = − , then we have λκ + µτ = − .But this is contradiction. So a Salkowski curve is a Bertrand curve, but itis not B -Bertrand curve. If the curve is anti-Salkowki curve, then torsionof the curve is equal to (or constant). If µ = − , λ = µ tan θ = 0 , thenwe have λκ + µτ = − . if λ = 0 , µ = λ cot θ = − , then we have λκ + µτ = − . But this is a contradiction. So a anti-Salkowski curve is a B -Bertrand curve, but it is not a Bertrand curve. Corollary 3.3.
If the curve is planar curve ( τ = 0 ), then the curve is notonly Bertrand but also B -Bertrand curve. Corollary 3.4. If v ( s ) = 1 , u ( s ) = w ( s ) = 0 , we have N -Bertrand curve.From equation (3.2),we have λ = c − s . From equation (3.1), we have τκ = − tan θ where θ is constant angle between T and T . As a result, M is N -Bertrand curve if and only if M is general helix. Corollary 3.5.
If the curve M is Bertrand curve, there is λ = 0 , µ = 0 ∈ R such that λκ + µτ = 1 . If we define λ = − λ and µ = − µ , then we have λ κ + µ τ = − . As a result, the curve M is B -Bertrand curve. Conversely,similarly, If the curve M is B -Bertrand curve where there exist λ = 0 , µ = 0 ∈ R such that λκ + µτ = − , then the curve M is Bertrand curve.In this case, it is said that M is proper Bertrand curve Corollary 3.6.
Let ( α ) be a unit-speed curve and { T, N, B, κ, τ } beSerret-Frenet apparatus of the curves. We suppose that v = 0 , then wecan see that λ is non zero constant . So we get u + w = 1 and w = ε √ − u where ε = ± . From equation (3.1), we have (3.7) u tan θ − ε p − u = f So, from (3.7), we have (3.8) u ± = f tan θ ± q θ ) − f θ ) .If u + = f tan θ + √ θ ) − f θ ) , then we have w ± = ± q − ( u + ) If u − = f tan θ − √ θ ) − f θ ) , then we have w ± = ± q − ( u − ) So, we have V +1 = u + T + w +1 B, V − = u + T + w − B,V +2 = u − T + w +2 B, V − = u − T + w − B and β +1 = Z V +1 ds + λN, β − = Z V − ds + λNβ +2 = Z V +2 ds + λN, β − = Z V − ds + λN In this case, ( α ) is V +1 , V − , V +2 and V − − Bertrand curve. So we can givefollowing definition.
Definition 3.3.
The curves β +1 , β − , β +2 , β − are called f − Bertrand curvemate of ( α ) and the curve ( α ) is called f − Bertrand curve. If f = r ( const. ) ,then ( α ) is called r − Bertrand curve and the curves β +1 , β − , β +2 , β − are called r − Bertrand curve mate of ( α ) Example 3.1. If α ( s ) = ( a cos sc , a sin sc , bsc ) , then we have T = ( − ac sin sc , ac cos sc , bc ) N = ( − cos sc , − sin sc , B = ( bc sin sc , − bc cos sc , ac ) and κ = ac and τ = bc where c = a + b . If λ = λ tan θ and µ = λ , thenwe have λκ + µτ = 1 = f where tan θ = c − λbλa . From (3.8), we have u ± = tan θ ± tan θ θ ) So we have u − = 0 and u + = λa ( c − λb ) c ( a +( λ − b ) ) . If u − = 0 , then we have w +2 = 1 and w − = − . If u + = λa ( c − λb ) c ( a +( λ − b ) ) , then we have w +1 = λ a + c (cid:0) λb − c (cid:1) c (cid:16) a + ( λ − b ) (cid:17) and w − = − λ a + c (cid:0) λb − c (cid:1) c (cid:16) a + ( λ − b ) (cid:17) So we have V +2 = B, V − = − B, V +1 = u + T + w +2 B, V − = u − T + w − B and β ± = (cid:16) ( λ ∓ b ) cos sc , ( λ ∓ b ) sin sc , ± asc (cid:17) β ± = (cid:18)(cid:0) w ± b − u a − λ (cid:1) cos sc , (cid:0) w ± b − u a − λ (cid:1) sin sc , w ± a + u bc s (cid:19) . So, ( α ) is − Bertrand curve and β +1 , β − , β +2 , β − curves are all − Bertrandcurve mate of ( α ) . Example 3.2.
From (3.8), if λκ + µτ = tan θ = f , then we obtain u + = 1 and u − = − cos 2 θ where tan θ = c − λbλa . If u + = 1 (resp. u − = − cos 2 θ ),then we have w ± = 0 (resp. w ± = ± sin 2 θ ). So we have V +1 = u + T + w +1 B, V − = u − T + w − B, V ± = T and β ± = (cid:18) ( λ − a ) cos sc , ( λ − a ) sin sc , bsc (cid:19) β ± = (cid:18)(cid:0) w ± b − u a − λ (cid:1) cos sc , (cid:0) w ± b − u a − λ (cid:1) sin sc , w ± a + u bc s (cid:19) . ( α ) is tan θ − Bertrand curve and the curves β +1 , β − , β +2 , β − are all tan θ − Bertrandcurve mate of ( α ) . Let ( γ ) be Bertrand curve and { T, N, B, κ, τ } be Serret-Frenet appa-ratus of the curves. We suppose that v = 0, then we can see that λ isnon-zero constant . We can obtain V ( s ) = uT ( s ) + wB ( s )where u + w = 1 and u, w ∈ R . Integral curve of V is defined as(3.9) γ V ( s ) = Z V ( s ) ds Let { T V , N V , B V , κ V , τ V } be Serret-Frenet apparatus of the curves ( γ V ).From equation (3.9), we have(3.10) T V = V ( s ) = uT ( s ) + wB ( s )From equation (3.10), we can see that the curves ( γ ) and ( γ V ) have a samearc-length parameter. From equation (3.10), we have κ V N V = ( uκ − wτ ) N So we obtain(3.11) N V = N and(3.12) κ V = uκ − wτ From equation (3.10) and (3.11), we get(3.13) B V = T V × N V = − wT ( s ) + uB ( s )From equation (3.13), we have(3.14) − τ V N V = − ( wκ + uτ ) N So we obtain(3.15) τ V = wκ + uτ From equation (3.12) and (3.15), we get(3.16) κ = uκ V + wτ V and(3.17) τ = − wκ V + uτ V In this case, we can obtain following theorem.
Theorem 3.2. ( γ ) is the Bertrand (resp. B -Bertrand ) curve if and onlyif ( γ V ) is the Bertrand (resp. B -Bertrand ) curve. Proof.
We suppose that ( γ ) is the Bertrand curve. There is λ, µ ∈ R suchthat λκ + µτ = u − w cot θ = 1where λ = 0 , µ = λ cot θ and θ is constant angle between T and T . Fromequation (3.12) and (3.15), we get(3.18) λκ + µτ = λκ V + µτ V = 1where(3.19) λ = λu − µw = λ = 0and(3.20) µ = λw + µu = λ cot( − θ )So, ( γ V ) is the Bertrand curve. Similarly, other cases can be proved by thesame method. (cid:3) Let ( α ) be Bertrand curve and { T, N, B, κ, τ } is Serret-Frenet appa-ratus of the curves. There is λ , µ ∈ R such that λκ + µτ = 1 where µ = λ tan θ . If λ = λa and µ = µa , then λκ + µτ = a . So we have(3.21) u ± ( a ) = a tan θ ± q θ ) − a θ ) and w ± ( a ) = ± q − ( u + ( a )) , w ± ( a ) = ± q − ( u − ( a )) . In this case,we can give following definition. Definition 3.4.
Let ( α ) be Bertrand curve. So, we can define a surfaceas ϕ +1 ( t, s ) = Z V +1 ds + λN, ϕ − ( t, s ) = Z V − ds + λNϕ +2 ( t, s ) = Z V +2 ds + λN, ϕ − ( t, s ) = Z V − ds + λN where V +1 ( t, s ) = u + ( t ) T ( s ) + w +1 ( t ) B ( s ) , V − ( t, s ) = u + ( t ) T ( s ) + w − ( t ) B ( s ) ,V +2 ( t, s ) = u − ( t ) T ( s ) + w +2 ( t ) B ( s ) , V − ( t, s ) = u − ( t ) T ( s ) + w − ( t ) B ( s ) , u ± ( t ) = t tan θ ± q θ ) − t θ ) , w ± ( t ) = ± q − ( u + ( t )) and w ± ( t ) = ± q − ( u − ( t )) . These surfaces are said Bertrand surface taken by ( α ) . Example 3.3. If α ( s ) = (cos s √ , sin s √ , s √ ) , then we have T = ( − √ s √ , √ s √ , √ N = ( − cos s √ , − sin s √ , B = ( 1 √ s √ , − √ s √ , √ and κ = and τ = . If λ = λ tan θ = 1 and µ = λ = 1 , then we have λκ + µτ = 1 = f and tan θ = 1 . From (3.21), we have u ( t ) = t + √ − t and w ( t ) = 1 √ q − t p − t So, we have a surface K ( α ) as ϕ ( t, s ) = u ( t ) Z T ds + w ( t ) Z Bds + λN = (cid:18) a ( t ) cos s √ , a ( t ) sin s √ , b ( t ) s √ (cid:19) where a ( t ) = u ( t ) − w ( t ) − and b ( t ) = u ( t ) + w ( t ) . The surface K ( α ) isBertrand surface taken by ( α ) . Definition 3.5.
Let (( α ) , ( β )) be Bertrand curve mate and K ( α ) and K ( β ) be Bertrand surface. ( K ( α ) , K ( β )) is said Bertrand surface matetaken by (( α ) , ( β )) Bertrand mate.
We can define a set as K = { K ( α ) / ( α ) is a Bertrand curve } .So we can define a equivalence relation on K such that K ( α ) ∼ K ( β ) ifand only if (( α ) , ( β )) is a Bertrand mate. In Example 3.3, if ( β ) is Bertrand mate of ( α ), then we have β ( s ) = α ( s ) + λN = (cos s √ , sin s √ , s √ − cos s √ , − sin s √ , , , s √ Example 3.4. If β ( s ) = (0 , , s √ ) , then we have T = (0 , , N = ( − cos sr , − sin sr , B = (sin sr , − cos sr , and κ = 0 and τ = r . If λ = λ tan θ = 0 and µ = λ = r , then we have λκ + µτ = 1 = f and tan θ = 0 . From (3.21), we have u ( t ) = √ − t and w ( t ) = t . So, we have a surface K ( α ) as ϕ ( t, s ) = u ( t ) Z T ds + w ( t ) Z Bds + λN = (cid:18) − (cid:18) r + tr (cid:19) cos s √ , − (cid:18) r + tr (cid:19) sin s √ , p − t s (cid:19) The surface K ( β ) is Bertrand surface taken by ( β ) . Furhermore, ( K ( α ) , K ( β )) is Bertrand surface mate taken by (( α ) , ( β )) . Construction of the Bertrand and the Principal-donorcurves
In 3-Euclidean spaces, let M be a regular spherical curve with coordinateneighborhood ( I, γ ). So we can define a curve K ( α : I → E ) such that(4.1) α ( s ) = Z S M ( s ) γ ( s ) ds where S M : I −→ R ( s → S M ( s )) is differentiable function. Lemma 4.1. ( [12] ) The curve K is spherical curve if and only if S M ( s ) = (cid:13)(cid:13) γ ′ ( s ) (cid:13)(cid:13) cos s Z det( γ ( s ) , γ ′ ( s ) , γ ′′ ( s )) k γ ′ ( s ) k ds + θ . Let M be a unit-speed curve with coordinate neighborhood ( I, γ ) and { T, N, B, κ, τ } be Serret-Frenet apparatus of the curves. Because of unit-speed curve, we have k γ ′ ( s ) k = 1. From lemma 4.1, there is S T : I −→ R differentiable function such that (cid:13)(cid:13)(cid:13)(cid:13)Z S T ( s ) γ ′ ( s ) ds (cid:13)(cid:13)(cid:13)(cid:13) = 1where(4.2) S T ( s ) = κ ( s ) cos s Z τ ( u ) du + θ If we define a curve K with coordinate neighborhood ( I, β ) such that β ′ ( s ) = Z S T ( s ) γ ′ ( s ) ds then we have(4.3) β ′′ ( s ) = S T ( s ) γ ′ ( s )Arc-lenght parameter of M and K are same. Let (cid:8) T , N , B, κ, τ (cid:9) be Serret-Frenet apparatus of the curves where(4.4) S T ( s ) = κ ( s ) = κ ( s ) cos s Z τ ( u ) du + θ and(4.5) τ ( s ) = κ ( s ) sin s Z τ ( u ) du + θ .From (4.3), we have(4.6) N ( s ) = εT ( s )where ε = ±
1. From equation (4.3) and (4.6), we can see that principalnormal of K and tangent of M is colinear. Theorem 4.1. K is principal-donor curve of M if and only if M is prin-cipal direction curve of K .Proof. From equation of β ′′ ( s ) = S T ( s ) γ ′ ( s ), we have N = T . So we canwrite(4.7) T = vN + wB where v + w = 1. If we take the derivative of equation (4.7), we have(4.8) κN = − κvT + (cid:0) v ′ − wτ (cid:1) N + (cid:0) w ′ + τ v (cid:1) B So we have(4.9) κ = − κv , v ′ − wτ = 0, w ′ + τ v = 0From equation (4.9), we have v = − cos s Z τ ( u ) du + θ and w = sin s Z τ ( u ) du + θ Thus, we have(4.10) T = − cos s Z τ ( u ) du + θ N + sin s Z τ ( u ) du + θ B From equation (4.10), K is principal-donor curve of M . Conversely, fromequation β ′′ ( s ) = κγ ′ ( s ), we have N = T . So we have γ ( s ) = s Z N ( u ) du . (cid:3) Theorem 4.2.
In 3-Euclidean spaces, let M , K be a unit-speed curves withunit coordinate neighborhoods ( I, γ ) and ( I, β ) , respectively, such that β ′′ ( s ) = S T ( s ) γ ′ ( s ) where (4.11) S T ( s ) = κ ( s ) cos s Z τ ( u ) du + θ .Following equations are equivalent,1) M is spherical curve2) K is Bertrand curve or B -Bertrand curve.Proof. If M is spherical curve, from (2.2) we have1 κ ( s ) = A cos s Z τ ( u ) du + θ where A and θ are constant. If θ = 0, then we have κ ( s ) = A ( cont. ). So K is Salkowski curve. If θ = − π , then we have τ ( s ) = A ( const. ). So, K is anti-Salkowski curve. If θ = 0 and θ = − π , then we have1 = A cos ( θ ) ǫκ ( s ) cos θ ( s ) − A sin ( θ − c ) ǫκ ( s ) sin θ ( s )where θ ( s ) = s Z τ ( u ) du . So we have λκ ( s ) + µτ ( s ) = 1 where λ = A cos ( θ ) = const. and µ = − A sin ( θ ) = const. Furthermore, K is proper BertranD curve. Conversely, if K is Bertrandcurve or B -Bertrand curve, there are λ , µ ∈ R such that λκ ( s ) + µτ ( s ) = 1.So we have1 = λκ ( s ) cos s Z τ ( u ) du + θ + µκ ( s ) sin s Z τ ( u ) du + θ and 1 κ ( s ) = A cos ( θ ) cos s Z τ ( u ) du + sin ( θ ) sin s Z τ ( u ) du where θ ∈ [0 , π ] , cos θ = λ √ λ + µ , sin θ = − µ √ λ + µ and A = p λ + µ .So we have κ ( s ) = A cos s Z τ ( u ) du + θ . (cid:3) Example 4.1.
In 3-Euclidean spaces, let M be a circle where γ ( s ) =( r cos ws, − r sin ws, and w = r . So we have κ = w and τ = 0 . From(4.4) and (4.5), we obtain κ ( s ) = ǫw cos c = A ( const. ) and τ ( s ) = ǫw sin c = B ( const. ) . So, from (4.3) , we get β ′′ ( s ) = A ( − sin ws, − cos ws, If we integrate of the above equation, we have a curve K as β ( s ) = ( Ar cos ( ǫws ) , Ar sin ( ǫws ) , bws + c ) Because K is unit spreed curve, we have b = r sin c . If a = Ar = ǫr cos c and c = 0 , we have β ( s ) = ( a cos ( ws ) , a sin ( ws ) , bws ) where r = √ a + b . References [1] Choi, J.H. and Kim Y.H.: Associated curves of Frenet curves and their applications;Applied Mathematics and Computation 218,9116-9124(2012)[2] Monterde, J.: Salkowski curves revisted: A family of curves with constant curvatureand non-constant torsion, Comput.Aided Geomet. Design 26(2009), 271-278[3] Frenet, F.: Sur les courbe `a double courbure, Jour. de Math. 17, 437-447 (1852)Extrait d’une These (Toulouse, 1847)[4] Serret, J.A.: Sur quelquees formules relatives `a la th´erie de courbes `a double cour-bure, J. De Math., 16(1851)[5] Izumiya, S and Tkeuchi N.: Generic properties of helices and Bertrand curves, J.geom., 74, 97-109, (2002).[6] Kreyszig, E.: Diferential Geometry, Mathematical Exposition, 11.Uni.of TorontoPress (1959).[7] Breuer, E. and Gottlied, D.: Exblicit charaterization of spherical curves, Proceedingsof the American Mathematical Society (1971)27, No.1.[8] Struik, D.J.: Lectures on Classical Differential Geometry, Dover, New-York, (1988). [9] Saint-Venant J.C.: Memoire sur les lignes courbas non planes, journal d’Ecoleplytechnique (1845) 1-76.[10] Bertrand J.: Memoire sur la theorie des courbes a double courbure, comptes Rendus36 (1850); journal de mathematiques press et Appliquees 15(1850) 332-350[11] Wong Y. C.: On an explicit characterization of spherical curves. Proc. Amer. Math.Soc., 34(1) 239–242, 1972.[12] Camcı, C¸ .: How can we construct a k -slant curve from a given spherical curve?,arXiv:1912.13392[math.DG] 23 December 2019.[13] do Carmo, M.P.: Diferential Geometry of Curves and Surfaces, Prentice-Hall, En-glewood Cliffs. NJ.(1976).[14] Hsiung, C.C.: A first course in Diferential Geometry, International Press, Cam-bridge.MA(1997). Department of Mathematics, Onsekiz Mart University, 17020 C¸ anakkale,Turkey
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