# Normal and Osculating Planes ofΔ-Regular Curves

Authors: Sibel Paşalı Atmaca
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

We present the normal and osculating planes of the curves parameterized by a compact subinterval of a time scale.


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## Body

## 1. Introduction

Concept of calculus on time scales (or measure chains) was initiated by Hilger and Aulbach [1, 2] in order to unify discrete and continuous analyses.This theory is appealing because it provides a useful tool for modeling dynamical processes. Since a time-scale is a closed subset of the reals [3], curves may have scattered points in multidimensional time scale spaces. Therefore, Δ-differentiation plays a major role in investigation of curves parameterized by an arbitrary time scale.The results in this paper were motivated by geometric interpretation of the results presented in [4].In this paper, we consider planes whose normal isΔ-differentiable vector that is each component of the vector is Δ-differentiable (i.e., normal planes) and which contain first and second order Δ-differentiable vectors (i.e., osculating planes). In this study we present the normal and osculating planes of the curves parameterized by a compact subinterval of a time scale. Since we need vector valued functions to study Δ-differentiable vectors of curves, we first define the concept of vector valued functions on time scales in Section 2. In [5] Guseinov and Özylmaz introduced the tangent line for Δ-regular curves in 3-dimensional time scales; then in [4] Bohner and Guseinov obtained the equation of such tangent line. The tangent line can also be studied in the concept of partial Δ-differentiation. In Section 3, we obtain the equations of tangent vectors of planar curves by using partial Δ-differentiation. Then we derive the equation of the normal plane for a Δ-regular curve. In Section 4, we present the basic theorem to construct osculating plane of a curve and obtain the equation of this plane by using first- and-second order Δ-derivatives.We refer the reader to resources such as [3, 4, 6, 7] and [8, 9] for more detailed discussions on the calculus of time scales and on the differential geometry of curves, respectively.


## 2. Vector-Valued Functions on Time Scales

Letn be fixed. Let 𝕋i denote a time scale for each i∈{1,2,…,n}. Let us set
(2.1)Λn=𝕋1×𝕋2×⋯×𝕋n={t=(t1,…,tn):ti∈𝕋i∀i∈{1,2,…,n}}.
We call Λn an n-dimensional time scale. Λn is also a complete metric space with (2.2)d(x,y)=(∑in|xi-yi|2)1/2forx,y∈Λn.Let a time-scale parametert vary in an interval [a,b]. If to each value t∈[a,b] we assign a vector r(t), then we say that a vector-valued function r(t) with argument t∈[a,b] is given. Assume that coordinates x1,x2,…,xn are fixed; then the representation of vector-valued function r(t) is equivalent to the representation of scalar functions x1(t),x2(t),…,xn(t); that is, r(t)={x1(t),…,xn(t)}.Definition 2.1.
A vectorr0 is called the limit of the vector-valued function r(t) as t→t0 if the length of the vector r(t)-r(t0) tends to zero as t→t0. Here we write
(2.3)limt→t0r(t)=r(t0).
It is clear that the vector-valued function r(t) has a limit if and only if each one of the functions x1(t),…,xn(t) has a limit as t→t0.Definition 2.2.
Δ-Derivative of a vector-valued function can be obtained by Δ-differentiating components x1(t),…,xn(t) of r(t); that is,
(2.4)rΔ(t)={x1Δ(t),…,xnΔ(t)}.
Precisely, for the Δ-derivative rΔ(t) of the vector-valued function r(t), we call the limit
(2.5)lims→tr(σ(t))-r(s)σ(t)-s.
If this limit exists, then r(t) is called Δ-differentiable.Proposition 2.3.
Letr1(t) and r2(t)  be vector-valued functions. Then  (i)
(r1(t)+r2(t))Δ=r1Δ(t)+r2Δ(t),(ii)
(r1r2)Δ=r1Δr2+r1σr2Δ.TheΔ-differentiation of the inner products and vector products of vector-valued functions, is computed by the consecutive differentiation of the cofactors.Proposition 2.4.
Letr1(t) and r2(t) be vector-valued functions, let × be Euclidean vector product, · and let Euclidean inner product. Then  (i)
(r1·r2)Δ=r1Δ·r2+r1σ·r2Δ,(ii)
(r1×r2)Δ=r1Δ×r2+r1σ×r2Δ=r1Δ×r2σ+r1×r2Δ.Definition 2.5 (Taylor's expansion for vector-valued functions).
Assume thatn times Δ-derivative of the vector-valued function r(t) exist and are rd-continuous, then we can write Taylor's expansions for the components; x1(t),…,xn(t) as
(2.6)x1(t)=h0(t,t0)x1(t0)+h1(t,t0)x1Δ(t0)+h2(t,t0)x1Δ2(t0)+⋯+o1(gn(t,t0))⋮xn(t)=h0(t,t0)xn(t0)+h1(t,t0)xnΔ(t0)+h2(t,t0)xnΔ2(t0)+⋯+on(gn(t,t0)),
where h0(r,s)≡1, hk+1(r,s)=∫srhk(τ,s)Δτ for k∈ℕ0, and
(2.7)gn(t,t0)=∫t0ρn-1(t)hn-1(t,σ(τ))xiΔn(τ)Δτ
for i={1,…,n}.This system of three equations can be written as(2.8)r(t)=h0(t,t0)r(t0)+h1(t,t0)rΔ(t0)+h2(t,t0)rΔ2(t0)+⋯+o(gn(t,t0)),
where  o(gn(t,t0))  denotes a vector whose length is an infinitesimal since  limt→t0gn(t,t0)=0.Remark 2.6.
There exists one essential difference between Taylor's expansions of vector-valued function and scalar function. If we consider Taylor's expansion for a scalar functionf(t), then we have
(2.9)o(gn(t,t0))=fΔk+1(ξ)hk+1(t,t0),
where ξ is a point between ρn-1(t) and t0. For a vector-valued function we cannot write similar formula for the corresponding infinitesimal vector, because in general for different components of the vector o(gn(t,t0)) the corresponding points ξ are different. However, it is more important to note that the length of the vector o(gn(t,t0)) is an infinitesimal with respect to gn(t,t0).


## 3. Tangent Line to a Curve

Let𝕋 be a time scale.Definition 3.1.
AΔ-regular curve Γ is defined as a mapping
(3.1)x=f1(t),y=f2(t),z=f3(t),t∈[a,b]
of the segment [a,b]⊂𝕋, a<b, to the space ℝ3, where f1,f2,f3 are real-valued functions defined on [a,b] and Δ-differentiable on [a,b]κ with rd-continuous Δ-derivatives and
(3.2)|f1Δ(t)|2+|f2Δ(t)|2+|f3Δ(t)|2≠0.Definition 3.2.
A lineℒ0 passing through the point P0 is called the delta tangent line to the curve Γ at the point P0 if the following held.(i)
ℒ0 passes also through the point P0σ=(x(σ(t0)),y(σ(t0)),z(σ(t0))).(ii)
IfPo is not an isolated point of the curve Γ, then(3.3)limP→P0P≠P0d(P,ℒ0)d(P,P0),
where P is the moving point of the curve Γ, d(P,ℒ0) is the distance from the point P to the line ℒ0, and d(P,P0) is the distance from the point P to the point P0σ.Theorem 3.3.
For any pointP0 of the curve Γ there exists the tangent to Γ at P0 and the directing vector of the tangent is Δ-differential of its position vector function rΔ(t0), where r(t0)=P0 for t0∈𝕋.Proof.
This theorem can be proven as in [5], Theorem 3.3.Let three functionsx:𝕋→ℝ, y:𝕋→ℝ, and z:𝕋→ℝ be given. Let us set x(𝕋):=𝕋1, y(𝕋)=𝕋2, and z(𝕋)=𝕋3. We will assume that 𝕋1, 𝕋2, and 𝕋3 are time scales. Denote by σ1Δ1, σ2Δ2, σ3Δ3 the forward jump operators and delta operators for 𝕋1, 𝕋2, and 𝕋3, respectively.Under the above assumptions, let functionsϕ:𝕋×𝕋×𝕋→ℝ and φ:𝕋×𝕋×𝕋→ℝ be given.Consider a space curve given by two equations.(3.4)ϕ(x,y,z)=0,φ(x,y,z)=0.
If x=x(t), y=y(t), z=z(t) is the position vector of the considered curve, then, substituting these three functions into (3.4), we obtain two equalities:(3.5)ϕ(x(t),y(t),z(t))=0,φ(x(t),y(t),z(t))=0.
If the functions ϕ and φ are σ1-completely differentiable, then, Δ-differentiation of these two equalities leads(3.6)∂ϕΔ1xxΔ+∂ϕσ1Δ2yyΔ+∂ϕσ1Δ3zzΔ=0,∂φΔ1xxΔ+∂φσ1Δ2yyΔ+∂φσ1Δ3zzΔ=0.
If ϕ and φ are σ2-completely differentiable, then Δ-differentiation of (3.5) leads us to obtain the following two equations:(3.7)∂ϕσ2Δ1xxΔ+∂ϕΔ2yyΔ+∂ϕσ2Δ3zzΔ=0,∂φσ2Δ1xxΔ+∂φΔ2yyΔ+∂φσ2Δ3zzΔ=0.
If ϕ and φ are σ3-completely differentiable, then Δ-differentiation of (3.5) leads us to obtain the following two equations:(3.8)∂ϕσ3Δ1xxΔ+∂ϕσ3Δ2yyΔ+∂ϕΔ3zzΔ=0,∂φσ3Δ1xxΔ+∂φσ3Δ2yyΔ+∂φΔ3zzΔ=0.
Other combinations of σi-completely differentiability of ϕ and φ can be shown similarly. The components {xΔ,yΔ,zΔ} of the tangent vector satisfy the system consisting of two equations: (3.6), (3.7), and (3.8).Assume thatϕ is σ1-completely differentiable planar curve given by the equations ϕ(x,y)=0, z=0 satisfying the condition (∂ϕ/Δ1x)2+(∂ϕσ1/Δ2y)2≠0; then the components of the tangent vector rΔ={xΔ,yΔ} are the solution of the linear equation (3.9)∂ϕΔ1xxΔ+∂ϕσ1Δ2yyΔ=0.Therefore,{xΔ,yΔ}=μ{-∂ϕσ1/Δ2y,∂ϕ/Δ1x}, and the equation of tangent is (3.10)x̃-x0-∂ϕ(σ1(x0),y0)/Δ2y=ỹ-y0∂ϕ(x0,y0)/Δ1x.
If planar curve ϕ is σ2-completely differentiable, then equation of tangent plane becomes (3.11)(x̃-x0)-(∂ϕ(x0,y0)/Δ2y)=(ỹ-y0)∂ϕ(x0,σ2(y0))/Δ1x.Definition 3.4.
LetΓ be a smooth and completely differentiable space curve. The plane passing through points P0∈Γ and orthogonal to the vector tangent to Γ at P0 is called the plane normal to Γ at P0.Denote byr̂ the position vector of the normal plane. Since this plane is orthogonal to the vector rΔ and contains the point with position vector r̂-r(t0), the equation of the normal plane is (3.12)(r̂-r(t0))·rΔ(t0)=0.
The vectors orthogonal to the tangent are called the vectors normal to Γ.


## 4. Osculating Plane of a Curve

LetP0 be a point of a curve Γ. Take two points Q1,Q2∈Γ situated right side of P0σ. If the points Q1 and Q2 tend to P0σ, then the limit position of the plane containing P0,P0σ,Q1,Q2 is called the osculating plane of Γ at the point P0.Theorem 4.1.
LetΓ be a Δ-regular curve represented as r=r(t). Assume that the vectors rΔ and rΔ2 are not collinear at point P0. Then there exists the osculating plane of  Γ at P0 and it is spanned by the vectors rΔ and rΔ2.Proof.
IfP0=P0σ, that is, P0 is right-dense point of Γ, then this theorem can be proven as in differential geometry concept.
LetP0 be a right-scattered point of Γ. Then, the positions vector of →P0Q1 and →P0Q2 are a1=r(t0+τ1)-r(t0) and a2=r(t0+τ2)-r(t0), respectively. That is, these vectors, if linearly independent, span the plane E.
This plane is also spanned by the vectorsv(i)=ai/τi for i∈{1,2}  or by the vectors
(4.1)v(1),w=2(v(2)-v(1))τ2-τ1.
By the means of Taylor's formula, we have
(4.2)r(t0+τi)=h0(t,t0)r(t0)+h1(t,t0)rΔ(t0)+h2(t,t0)rΔ2(t0)+o(g2(t0)).
Hence, we obtain
(4.3)v(1)=rΔ(t0)+τ12rΔ2(t0)+o(τ1),w=rΔ2(t0)+o(1).
Consequently, if τi→0 for i∈{1,2},  then v(1)→rΔ(t0) and w→rΔ2(t0).
These vectors, if linearly independent, determine the limiting position of the planeE passing through the points P0,P0σ,Q1,Q2.Corollary 4.2.
If the vectorsrΔ(t0) and rΔ2(t0) are collinear, then the limit position of considering plane is not determined. For instance, take a straight line
(4.4)r(t)=a+bt,
where a,b are constant vectors and t∈𝕋. Then
(4.5)rΔ(t0)=b,rΔ2(t0)=0,
so the osculating plane of the straight line is not determined uniquely. If rΔ(t) and rΔ2(t) are collinear, then the corresponding point of Γ is called the straightening point of Γ.Theorem 4.3.
The osculating plane of a planar curve coincides with the plane containing this curve.Proof.
Let us consider the Taylor expansion of the position vectorr(t) at the neighborhood of P0:(4.6)r(t0+τ)=h0(t,t0)r(t0)+h1(t,t0)rΔ(t0)+h2(t,t0)rΔ2(t0)+o(g2(t,t0)).
The curve Γ¯, determined by the expantion,
(4.7)r¯=h0(t,t0)r(t0)+h1(t,t0)rΔ(t0)+h2(t,t0)rΔ2(t0)
is situated in the osculating plane of Γ at P0; the difference between the position vectors of Γ and Γ¯ is a sufficiently small vector
(4.8)r(t0+τ)-r¯(τ)=o(g2(t,t0)).
Hence a sufficiently small neighborhood of P0 on the space curve Γ is near to the planar curve Γ¯ situated in the osculating plane of Γ at P0.Now let us write the equation of the osculating plane ofΓ at P0. Let r̂ be the position vector of the osculating plane. Since rΔ and rΔ2 span the osculating plane, the vector product rΔ×rΔ2 is orthogonal to the osculating plane. The vector r̂-r(t0) belongs to the osculating plane; therefore, the inner product of these vectors is equal to zero: (4.9)(r̂-r(t0))·(rΔ×rΔ2)=0.With respect to coordinate functions, this equation has the following form:(4.10)det(x̂-x(t0)xΔxΔ2ŷ-y(t0)yΔyΔ2ẑ-z(t0)zΔzΔ2)=0.



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