Abstract

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

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, -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 is -differentiable vector that is each component of the vector is -differentiable (i.e., normal planes) and which contain first and second order -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 -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 -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 -differentiation. In Section 3, we obtain the equations of tangent vectors of planar curves by using partial -differentiation. Then we derive the equation of the normal plane for a -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 -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

Let be fixed. Let denote a time scale for each . Let us set We call an -dimensional time scale. is also a complete metric space with

Let a time-scale parameter vary in an interval . If to each value we assign a vector , then we say that a vector-valued function with argument is given. Assume that coordinates are fixed; then the representation of vector-valued function is equivalent to the representation of scalar functions ; that is, .

Definition 2.1. A vector is called the limit of the vector-valued function as if the length of the vector tends to zero as . Here we write It is clear that the vector-valued function has a limit if and only if each one of the functions has a limit as .

Definition 2.2. -Derivative of a vector-valued function can be obtained by -differentiating components of ; that is, Precisely, for the -derivative of the vector-valued function , we call the limit If this limit exists, then is called -differentiable.

Proposition 2.3. Let and be vector-valued functions. Then (i) (ii)

The -differentiation of the inner products and vector products of vector-valued functions, is computed by the consecutive differentiation of the cofactors.

Proposition 2.4. Let and be vector-valued functions, let be Euclidean vector product, and let Euclidean inner product. Then (i) (ii)

Definition 2.5 (Taylor's expansion for vector-valued functions). Assume that times -derivative of the vector-valued function exist and are -continuous, then we can write Taylor's expansions for the components; as where , for , and for .

This system of three equations can be written as where denotes a vector whose length is an infinitesimal since .

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 function , then we have where is a point between and . For a vector-valued function we cannot write similar formula for the corresponding infinitesimal vector, because in general for different components of the vector the corresponding points are different. However, it is more important to note that the length of the vector is an infinitesimal with respect to .

3. Tangent Line to a Curve

Let be a time scale.

Definition 3.1. A -regular curve is defined as a mapping of the segment , , to the space , where are real-valued functions defined on and -differentiable on with -continuous -derivatives and

Definition 3.2. A line passing through the point is called the delta tangent line to the curve at the point if the following held.(i) passes also through the point (ii)If is not an isolated point of the curve , then where is the moving point of the curve , is the distance from the point to the line , and is the distance from the point to the point .

Theorem 3.3. For any point of the curve there exists the tangent to at and the directing vector of the tangent is -differential of its position vector function , where for .

Proof. This theorem can be proven as in [5], Theorem 3.3.

Let three functions , , and be given. Let us set , , and . We will assume that , and are time scales. Denote by , , the forward jump operators and delta operators for , and , respectively.

Under the above assumptions, let functions and be given.

Consider a space curve given by two equations. If , , is the position vector of the considered curve, then, substituting these three functions into (3.4), we obtain two equalities: If the functions and are -completely differentiable, then, -differentiation of these two equalities leads If and are -completely differentiable, then -differentiation of (3.5) leads us to obtain the following two equations: If and are -completely differentiable, then -differentiation of (3.5) leads us to obtain the following two equations: Other combinations of -completely differentiability of and can be shown similarly. The components of the tangent vector satisfy the system consisting of two equations: (3.6), (3.7), and (3.8).

Assume that is -completely differentiable planar curve given by the equations , satisfying the condition ; then the components of the tangent vector are the solution of the linear equation

Therefore, , and the equation of tangent is If planar curve is -completely differentiable, then equation of tangent plane becomes

Definition 3.4. Let be a smooth and completely differentiable space curve. The plane passing through points and orthogonal to the vector tangent to at is called the plane normal to at .

Denote by the position vector of the normal plane. Since this plane is orthogonal to the vector and contains the point with position vector , the equation of the normal plane is The vectors orthogonal to the tangent are called the vectors normal to .

4. Osculating Plane of a Curve

Let be a point of a curve . Take two points situated right side of . If the points and tend to , then the limit position of the plane containing is called the osculating plane of at the point .

Theorem 4.1. Let be a -regular curve represented as . Assume that the vectors and are not collinear at point . Then there exists the osculating plane of at and it is spanned by the vectors and .

Proof. If , that is, is right-dense point of , then this theorem can be proven as in differential geometry concept.
Let be a right-scattered point of . Then, the positions vector of and are and , respectively. That is, these vectors, if linearly independent, span the plane .
This plane is also spanned by the vectors for or by the vectors By the means of Taylor's formula, we have Hence, we obtain Consequently, if for , then and .
These vectors, if linearly independent, determine the limiting position of the plane passing through the points .

Corollary 4.2. If the vectors and are collinear, then the limit position of considering plane is not determined. For instance, take a straight line where are constant vectors and . Then so the osculating plane of the straight line is not determined uniquely. If and are collinear, then the corresponding point of is called the straightening point of .

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 vector at the neighborhood of : The curve , determined by the expantion, is situated in the osculating plane of at ; the difference between the position vectors of and is a sufficiently small vector Hence a sufficiently small neighborhood of on the space curve is near to the planar curve situated in the osculating plane of at .

Now let us write the equation of the osculating plane of at . Let be the position vector of the osculating plane. Since and span the osculating plane, the vector product is orthogonal to the osculating plane. The vector belongs to the osculating plane; therefore, the inner product of these vectors is equal to zero:

With respect to coordinate functions, this equation has the following form: