The Differential Geometry of a Space Curve via a Constant Vector in ℝ³

Abstract

The differential geometry of space curves is a fascinating area of research for mathematicians and physicists, and this refers to its crucial applications in many areas. In this paper, a new method is derived to study the differential geometry of space curves. More specifically, the position vector of a constant vector in ${\mathbb{R}}^3$ is given in the Frenet apparatus of a space curve, and it is implemented to study the differential geometry of the given space curve. Easy and neat proofs of various well-known results are given using this new method. Also, new results and the properties of space curves are obtained in light of this new method. More specifically, the position vectors of helices are given in simple forms. Moreover, a new frame associated with a smooth curve is obtained, as well as new curvatures associated with the new frame. The new frame and its curvatures are investigated and used to give the position vector of slant helix in a simple and memorable form. Furthermore, some non-trivial examples are given to illustrate some of the results obtained in this article.

1. Introduction

Even though it is a more traditional topic in differential geometry, research on the differential geometry of curves and surfaces is still on going. This is due to the fact that it is used in numerous fields, including computer graphics, computer vision, physics, aerospace, and medical imaging.

In many engineering applications, as well as DNA structures, a helix is essential. In fact, the double helix can be used to describe a DNA molecule. Furthermore, it has been noted that two adjacent helices pointing in opposite directions are joined by hydrogen bonds in a molecular model of DNA [1]. A circular helix is a geometric curve with non-vanishing constant torsion $\mathfrak{T}$ and non-vanishing constant curvature $\mathcal{K}$, as seen from the perspective of differential geometry [2,3]. Helices can be used in the fields of computer-aided geometric design and computer graphics for various tasks, such as designing highways, simulating kinematic motion, and describing tool paths [4]. In Euclidean space ${\mathbb{R}}^3$, a curve with a constant slope, also known as a general helix, is defined by the tangents’ constant angle with the fixed straight line known as the general helix’s axis. According to a classical result first proposed by Lancret in 1802 and first demonstrated by de Saint Venant in 1845 [5], a curve must satisfy the following necessary and sufficient condition in order to be considered a general helix: the ratio ${\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}$ must remain constant along the curve. In the classical differential geometry of space curves, general helices or inclined curves are well-known [2,6,7,8,9]. The notion of a slant helix was first presented by Izumiya and Takeuchi [10], who stated that a fixed straight line and the normal lines form a constant angle if and only if the principal image of the principal normal indicatrix’s geodesic curvature ${\mathcal{K}}_g={\displaystyle \frac{{\mathcal{K}}^2{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}{{\left({\mathcal{K}}^2+{\mathfrak{T}}^2\right)}^{\frac{3}{2}}}}$ is a constant function. Joint kinematics are analyzed using the rectifying curves [11,12]. Salkowski curves are helpful in creating knotted curves, which are closed curves with continuous torsion and constant curvature [13]. An example of slant helices with constant curvatures are the Salkowski curves. Curves and their Serret–Frenet frames are essential for creating special surfaces in differential geometry [14,15,16,17,18,19]. At every point along a given curve, the Serret–Frenet frame associated with the given curve in ${\mathbb{R}}^3$, which is a moving frame along the curve, forming an orthonormal basis for ${\mathbb{R}}^3$. This makes it possible for geometers to examine a curve’s position vector as well as those of other curves. In [20,21], the terms “natural mate” and “conjugate” that are related to a smooth curve were presented and examined. For recent developments in the theory of curves and surfaces we direct the reader to Refs. [22,23].

2. Preliminaries

In this section, we summarize some of the fundamental ideas of the differential geometry of curves in ${\mathbb{R}}^3$; the reader can consult References [11,12,20,24,25,26,27] for further information. Initially, a parametrized smooth curve $\ddot{\mho }$ in ${\mathbb{R}}^3$ is a map $\ddot{\mho }:I\subseteq \mathbb{R}\to {\mathbb{R}}^3$ given by $\ddot{\mho }(t)=\left({\ddot{\mho }}_1(t),{\ddot{\mho }}_2(t),{\ddot{\mho }}_3(t)\right)$ such that ${\ddot{\mho }}_1(t),{\ddot{\mho }}_2(t)$ and ${\ddot{\mho }}_3(t)$ are smooth functions for all $t\in I$, where I is an open interval in $\mathbb{R}$. $\ddot{\mho }$ is called a regular parametrized curve if ${\ddot{\mho }}^{\prime }(t)\ne (0,0,0)$ for all $t\in I$. The curvature $\mathcal{K}$ and torsion $\mathfrak{T}$ formulas of $\ddot{\mho }$ are given by $\mathcal{K}={\displaystyle \frac{\parallel {\ddot{\mho }}^{\prime }(t)\times {\ddot{\mho }}^{″}(t)\parallel }{\parallel {\ddot{\mho }}^{\prime }{(t)\parallel }^3}}$ and $\mathfrak{T}={\displaystyle \frac{\left({\ddot{\mho }}^{\prime }(t)\times {\ddot{\mho }}^{″}(t)\right)·{\ddot{\mho }}^{‴}(t)}{\parallel {\ddot{\mho }}^{\prime }(t)\times {\ddot{\mho }}^{″}{(t)\parallel }^2}}$. The distance function of $\ddot{\mho }(t)$ is defined by $d = \parallel \ddot{\mho }(t)\parallel$. The unit tangent vector of a regular parametrized curve $\ddot{\mho }$ at t is given by $\mathbf{T}(\mathbf{t})={\displaystyle \frac{{\ddot{\mho }}^{\prime }(t)}{\parallel {\ddot{\mho }}^{\prime }(t)\parallel }}$. If $\mathcal{K}\ne 0$, then the unit principal normal and the unit binormal vectors of $\ddot{\mho }$ at t are given by $\mathbf{N}(\mathbf{t})={\displaystyle \frac{{\mathbf{T}}^{\prime }(\mathbf{t})}{\parallel {\mathbf{T}}^{\prime }(\mathbf{t})\parallel }}$ and $\mathbf{B}(\mathbf{t})=\mathbf{T}(\mathbf{t})\times \mathbf{N}(\mathbf{t})$, respectively. The ordered triple $\{\mathbf{T}(\mathbf{t}),\mathbf{N}(\mathbf{t}),\mathbf{B}(\mathbf{t})\}$ is called the Serret–Frenet frame of $\ddot{\mho }$ and the elements of this frame are related by $\mathbf{T}(\mathbf{t})=\mathbf{N}(\mathbf{t})\times \mathbf{B}(\mathbf{t})$, $\mathbf{N}(\mathbf{t})=\mathbf{B}(\mathbf{t})\times \mathbf{T}(\mathbf{t})$, and $\mathbf{B}(\mathbf{t})=\mathbf{T}(\mathbf{t})\times \mathbf{N}(\mathbf{t})$, respectively. The Serret–Frenet frame is an orthonormal basis for ${\mathbb{R}}^3$. The parametrized curve $\ddot{\mho }:I\to {\mathbb{R}}^3$ is called a unit speed curve if $\parallel {\ddot{\mho }}^{\prime }(t)\parallel = 1$ for all $t\in I$. Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve with non-vanishing curvature. The following equations are called Serret–Frenet equations.

(i)

${\mathbf{T}}^{\prime }=\mathcal{K}\mathbf{N}$;

(ii)

${\mathbf{N}}^{\prime }=\mathfrak{T}\mathbf{B}-\mathcal{K}\mathbf{T}$;

(iii)

${\mathbf{B}}^{\prime }=-\mathfrak{T}\mathbf{N}$.

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve with $\mathcal{K}>0$. Then, $\ddot{\mho }$ is called a rectifying curve if and only if

(i)

$\ddot{\mho }=(s+c_1)\mathbf{T}+c\mathbf{B}$, where $s\in I$ and $c_1$ and c are constants;

(ii)

$\ddot{\mho }·\mathbf{N}=0$,

(iii)

the distance function of the rectifying curve satisfies $\parallel \ddot{\mho }\parallel = \sqrt{s^2+c_{1}s+c_2}$, where $s\in I$ and $c_1$ and $c_2$ are constants;

(iv)

${\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}$ is a linear function in its arc length (see, for instance, [28,29,30,31,32]).

Theorem 1. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a regular parametrized curve with $\mathcal{K}\ne 0$ for all $t\in I$. Then, $\mathfrak{T} = 0$ for all $t\in I$ if and only if $\ddot{\mho }(I)$ lies in a plane.

Theorem 2. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a regular parametrized curve with $\mathcal{K}\ne 0$ and $\mathfrak{T}\ne 0$ for all $t\in I$. Then, $\ddot{\mho }$ is a Bertrand curve if and only if there exist constants a and b such that

$$a\mathcal{K}+b\mathfrak{T} = 1.$$

Definition 1. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a regular parametrized curve with $\mathcal{K}\ne 0$. Then, $\ddot{\mho }$ is a helix if its tangent makes a fixed angle with a fixed direction.

Definition 2. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a regular parametrized curve with Frenet–Serret apparatus $\{\mathcal{K},\mathfrak{T},\mathbf{T},\mathbf{N},\mathbf{B}\}$ and $\mathbf{U}$ be a constant unit vector. Then, $\ddot{\mho }$ is a proper slant helix if $\mathbf{U}·\mathbf{N} = c$, where $c\ne 0$ and c is a constant.

Theorem 3. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a regular parametrized curve with $\mathcal{K}\ne 0$ and $\mathfrak{T}\ne 0$ for all $t\in I$. Then, $\ddot{\mho }$ is a general helix if and only if there exists a constant c such that

$${\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}} = c.$$

Theorem 4. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a regular parametrized curve with non-vanishing curvature and torsion. Then, $\ddot{\mho }$ is a circular helix if and only if the curvature and torsion are constants.

Definition 3. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a regular parametrized curve. Then, $\ddot{\mho }$ is a slant helix if its normal vector makes a fixed angle with a fixed direction.

Theorem 5. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a regular parametrized curve with $\mathcal{K}\ne 0$ and $\mathfrak{T}\ne 0$ for all $t\in I$. Then, $\ddot{\mho }$ is a slant helix if and only if there exists a constant c such that

$${\displaystyle \frac{{\mathcal{K}}^2{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}{{\left({\mathcal{K}}^2+{\mathfrak{T}}^2\right)}^{\frac{3}{2}}}} = c.$$

Definition 4. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve with $\mathcal{K}\ne 0$ and $\mathfrak{T}\ne 0$. Then, the natural mate of $\ddot{\mho }$ is defined by

$$\ddot{{\rm Y}} = \int \mathbf{N}(\mathbf{s})ds.$$

Theorem 6. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve with $\mathcal{K}\ne 0$, $\mathfrak{T}\ne 0$ and $\ddot{{\rm Y}}$ be its natural mate. Then, $\ddot{\mho }$ is a slant helix if and only if $\ddot{{\rm Y}}$ is a general helix.

Theorem 7. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve with $\mathcal{K}\ne 0$, $\mathfrak{T}\ne 0$ and $\ddot{{\rm Y}}$ be its natural mate. Then, the curvature ${\mathcal{K}}_{\ddot{{\rm Y}}}$ and the torsion ${\mathfrak{T}}_{\ddot{{\rm Y}}}$ of $\ddot{{\rm Y}}$ are given by

$(i)$

${\mathcal{K}}_{\ddot{{\rm Y}}} = \sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}$;

$(ii)$

${\mathfrak{T}}_{\ddot{{\rm Y}}} = {\displaystyle \frac{{\mathcal{K}}^2{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}{{\mathcal{K}}^2+{\mathfrak{T}}^2}}$.

Definition 5. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve. The Darboux vector filed along $\ddot{\mho }$ is defined by

$$\mathbf{D} = \mathfrak{T}\mathbf{T}+\mathcal{K}\mathbf{B},$$

Theorem 8. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve and $\mathbf{D}$ be the Darboux vector filed along $\ddot{\mho }$. Then,

$${\mathbf{T}}^{\prime } = \mathbf{D}\times \mathbf{T},$$

$${\mathbf{N}}^{\prime } = \mathbf{D}\times \mathbf{N},$$

$${\mathbf{B}}^{\prime } = \mathbf{D}\times \mathbf{B}.$$

3. Position Vector of a Constant Vector Using Serret–Frenet Frame

In this section, we give the position vector formula for any constant vector using the Serret–Frenet frame. We use the position vector to study the differential geometry of space curves and many new results will be obtained.

Theorem 9. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve with $\mathcal{K}\ne 0$ and $\mathbf{U}$ be a constant unit vector in ${\mathbb{R}}^3.$ Then, $\mathbf{U}$ is given by

$$\mathbf{U} = \int \mathcal{K}cos\ddot{\mathbf{\Theta }}\mathbf{ds}\mathbf{T}+cos\ddot{\mathbf{\Theta }}\mathbf{N}-\int \mathfrak{T}cos\ddot{\mathbf{\Theta }}\mathbf{ds}\mathbf{B},$$

(1)

where $\ddot{\Theta }$ is the angle function between $\mathbf{U}$ and N.

Proof. 

Let $\mathbf{U}$ be a constant unit vector in ${\mathbb{R}}^3$, then, $\mathbf{U}$ can be written in the form

$$\mathbf{U} = {\alpha }_1\mathbf{T}+{\alpha }_2\mathbf{N}+{\alpha }_3\mathbf{B}.$$

(2)

Taking the dot product of Equation (2) with T, N, and B, we have

$$\mathbf{U}·\mathbf{T} = {\alpha }_1,$$

(3)

$$\mathbf{U}·\mathbf{N} = {\alpha }_2,$$

(4)

$$\mathbf{U}·\mathbf{B} = {\alpha }_3.$$

(5)

Since $\mathbf{U}·\mathbf{N} = \parallel \mathbf{U}\parallel \parallel \mathbf{N}\parallel cos\ddot{\mathbf{\Theta }}$, then, from Equation (4) we obtain

$$cos\ddot{\Theta } = {\alpha }_2.$$

(6)

From Equations (4) and (6), we obtain

$$\mathbf{U}·\mathbf{N} = cos\ddot{\mathbf{\Theta }}.$$

(7)

Multiplying Equation (7) by $\mathcal{K}$, we obtain

$$\begin{array}{ccc}\hfill \mathcal{K}\mathbf{U}·\mathbf{N}& = & \mathcal{K}cos\ddot{\Theta }\hfill \\ \\ \hfill \mathbf{U}·{\mathbf{T}}^{\prime }& = & \mathcal{K}cos\ddot{\Theta }\hfill \\ \\ \hfill {(\mathbf{U}·\mathbf{T})}^{\prime }& = & \mathcal{K}cos\ddot{\Theta }\hfill \\ \\ \hfill \int {(\mathbf{U}·\mathbf{T})}^{\prime }\mathbf{ds}& = & \int \mathcal{K}cos\ddot{\Theta }ds,\hfill \end{array}$$

which implies that

$$\mathbf{U}·\mathbf{T} = \int \mathcal{K}cos\ddot{\Theta }ds.$$

(8)

From Equations (3) and (8), we obtain

$$\int \mathcal{K}cos\ddot{\Theta }ds = {\alpha }_1.$$

(9)

Multiplying Equation (7) by $\mathfrak{T}$, we obtain

$$\begin{array}{ccc}\hfill \mathfrak{T}\mathbf{U}·\mathbf{N}& = & \mathfrak{T}cos\ddot{\Theta }\hfill \\ \\ \hfill -\mathbf{U}·{\mathbf{B}}^{\prime }& = & \mathfrak{T}cos\ddot{\Theta }\hfill \\ \\ \hfill {(\mathbf{U}·\mathbf{B})}^{\prime }& = & -\mathfrak{T}cos\ddot{\Theta }\hfill \\ \\ \hfill \int {(\mathbf{U}·\mathbf{B})}^{\prime }ds& = & -\int \mathfrak{T}cos\ddot{\Theta }ds.\hfill \end{array}$$

Which implies that

$$\mathbf{U}·\mathbf{B} = -\int \mathfrak{T}cos\ddot{\Theta }ds.$$

(10)

From Equations (5) and (10), we obtain

$${\alpha }_3 = -\int \mathfrak{T}cos\ddot{\Theta }ds.$$

(11)

Finally, using Equations (6), (9), and (11) in Equation (2), we conclude that

$$\mathbf{U} = \int \mathcal{K}cos\ddot{\mathbf{\Theta }}\mathbf{ds}\mathbf{T}+cos\ddot{\mathbf{\Theta }}\mathbf{N}-\int \mathfrak{T}cos\ddot{\mathbf{\Theta }}\mathbf{ds}\mathbf{B}.$$

□

As a consequence of Theorem 9, we have the following corollary.

Corollary 1. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve and $\mathbf{U}$ be a constant unit vector in ${\mathbb{R}}^3$, then,

$${\left(\int \mathcal{K}cos\ddot{\Theta }ds\right)}^2+{\left(\int \mathfrak{T}cos\ddot{\Theta }ds\right)}^2\le 1.$$

(12)

Proof. 

Let $\mathbf{U}$ be a constant unit vector in ${\mathbb{R}}^3$. By Theorem 9,

$$\begin{array}{ccc}\hfill \mathbf{U}& = & \int \mathcal{K}cos\ddot{\Theta }ds\mathbf{T}+cos\ddot{\Theta }\mathbf{N}-\int \mathfrak{T}cos\ddot{\Theta }ds\mathbf{B}\hfill \\ \hfill \mathbf{U}·\mathbf{U}& = & {\left(\int \mathcal{K}cos\ddot{\Theta }ds\right)}^2+{cos}^2\ddot{\Theta }+{\left(\int \mathfrak{T}cos\ddot{\Theta }ds\right)}^2\hfill \\ \hfill 1& = & {\left(\int \mathcal{K}cos\ddot{\Theta }ds\right)}^2+{cos}^2\ddot{\Theta }+{\left(\int \mathfrak{T}cos\ddot{\Theta }ds\right)}^2\hfill \\ \hfill 1-{cos}^2\ddot{\Theta }& = & {\left(\int \mathcal{K}cos\ddot{\Theta }ds\right)}^2+{\left(\int \mathfrak{T}cos\ddot{\Theta }ds\right)}^2\hfill \\ \hfill {sin}^2\ddot{\Theta }& = & {\left(\int \mathcal{K}cos\ddot{\Theta }ds\right)}^2+{\left(\int \mathfrak{T}cos\ddot{\Theta }ds\right)}^2\le 1.\hfill \end{array}$$

□

In view of the results from Theorem 9, we introduce the following applications.

4. Applications

In this section, we show the importance of Theorem 9 in proving some important results for various famous curves in differential geometry, including the plane curve; Bertrand curve; and circular helix, general helix, slant helix, and rectifying curves. Also, we give a new frame generated from the Serret–Frenet frame, and we obtain significant novel results, as well as new proofs, for some well-known results, using our proposed methods from the new frame.

4.1. Plane Curve

In this subsection, we derive the angle function between a constant vector and the unit normal vector for a plane curve in ${\mathbb{R}}^3.$

Theorem 10. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve and $\mathbf{U}$ be a constant unit vector. If $\ddot{\mho }$ is a plane curve, then,

$$\ddot{\Theta } = \int \mathcal{K}ds+c,$$

where c is a constant.

Proof. 

Let $\ddot{\mho }$ be a plane curve and U be a constant unit vector in ${\mathbb{R}}^3,$ then, by Theorem 9,

$$\begin{array}{ccc}\hfill \mathbf{U}& = & \int \mathcal{K}cos\ddot{\Theta }ds\mathbf{T}+cos\ddot{\Theta }\mathbf{N}\hfill \\ \hfill \mathbf{U}·\mathbf{U}& = & {\left(\int \mathcal{K}cos\ddot{\Theta }ds\right)}^2+{cos}^2\ddot{\Theta }\hfill \\ \hfill 1& = & {\left(\int \mathcal{K}cos\ddot{\Theta }ds\right)}^2+{cos}^2\ddot{\Theta }\hfill \\ \hfill 1-{cos}^2\ddot{\Theta }& = & {\left(\int \mathcal{K}cos\ddot{\Theta }ds\right)}^2\hfill \\ \hfill {sin}^2\ddot{\Theta }& = & {\left(\int \mathcal{K}cos\ddot{\Theta }ds\right)}^2.\hfill \end{array}$$

Thus,

$$sin\ddot{\Theta } = \int \mathcal{K}cos\ddot{\Theta }ds.$$

(13)

Differentiating Equation (13) with respect to s, we obtain

$$\begin{array}{ccc}\hfill {\ddot{\Theta }}^{\prime }cos\ddot{\Theta }& = & \mathcal{K}cos\ddot{\Theta }\hfill \\ \hfill {\ddot{\Theta }}^{\prime }& = & \mathcal{K}\hfill \\ \hfill \ddot{\Theta }& = & \int \mathcal{K}ds+c.\hfill \end{array}$$

□

4.2. Bertrand Curve

In this subsection, we use Theorem 9 to give an important new characteristic of the Bertrand curve, which is as follows:

Theorem 11. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve and $\mathbf{U}$ be a constant unit vector. Then, $\ddot{\mho }$ is a Bertrand curve if and only if

$$\int cos\ddot{\Theta }ds = acos\ddot{\Phi }-bcos\ddot{\Psi },$$

(14)

where a and b are constants, $cos\ddot{\Phi }\ne 0$, $cos\ddot{\Psi }\ne 0$, $cos\ddot{\Theta }\ne 0$, $\ddot{\Phi }$ is the angle between U and T, $\ddot{\Psi }$ is the angle between U and B, and $\ddot{\Theta }$ is the angle between U and N.

Proof. 

Let $\ddot{\mho }$ be a Bertrand curve and U be a constant unit vector in ${\mathbb{R}}^3$, then by Theorem 9,

$$\begin{array}{ccc}\hfill \mathbf{U}& = & \int \mathcal{K}cos\ddot{\Theta }ds\mathbf{T}+cos\ddot{\Theta }\mathbf{N}-\int \mathfrak{T}cos\ddot{\Theta }ds\mathbf{B}\hfill \\ \hfill \mathbf{U}·\mathbf{T}& = & \int \mathcal{K}cos\ddot{\Theta }ds\hfill \\ \hfill \parallel \mathbf{U}\parallel \parallel \mathbf{T}\parallel cos\ddot{\Phi }& = & \int \mathcal{K}cos\ddot{\Theta }ds,\hfill \end{array}$$

which implies that

$$cos\ddot{\Phi } = \int \mathcal{K}cos\ddot{\Theta }ds.$$

(15)

Differentiating Equation (15) with respect to s, we obtain

$$-{\ddot{\Phi }}^{\prime }sin\ddot{\Phi } = \mathcal{K}cos\ddot{\Theta }.$$

Since $cos\ddot{\Theta }\ne 0$, then,

$$\mathcal{K} = -{\displaystyle \frac{{\ddot{\Phi }}^{\prime }sin\ddot{\Phi }}{cos\ddot{\Theta }}}.$$

(16)

Similarly,

$$\begin{array}{ccc}\hfill \mathbf{U}·\mathbf{B}& = & -\int \mathfrak{T}cos\ddot{\Theta }ds\hfill \\ \hfill \parallel \mathbf{U}\parallel \parallel \mathbf{B}\parallel cos\ddot{\Psi }& = & -\int \mathfrak{T}cos\ddot{\Theta }ds,\hfill \end{array}$$

which implies that

$$cos\ddot{\Psi } = -\int \mathfrak{T}cos\ddot{\Theta }ds.$$

(17)

Differentiating Equation (17) with respect to s, we obtain

$$-{\ddot{\Psi }}^{\prime }sin\ddot{\Psi } = -\mathfrak{T}cos\ddot{\Theta }.$$

Therefore,

$$\mathfrak{T} = {\displaystyle \frac{{\ddot{\Psi }}^{\prime }sin\ddot{\Psi }}{cos\ddot{\Theta }}}.$$

(18)

Now, $\ddot{\mho }$ is a Bertrand curve, consequently,

$$a\mathcal{K}+b\mathfrak{T} = 1,$$

(19)

where a and b are constants. Substituting Equations (16) and (18) in Equation (19), we obtain

$$\begin{array}{ccc}\hfill -{\displaystyle \frac{a{\ddot{\Phi }}^{\prime }sin\ddot{\Phi }}{cos\ddot{\Theta }}}+{\displaystyle \frac{b{\ddot{\Psi }}^{\prime }sin\ddot{\Psi }}{cos\ddot{\Theta }}}& = & 1\hfill \\ \hfill b{\ddot{\Psi }}^{\prime }sin\ddot{\Psi }-a{\ddot{\Phi }}^{\prime }sin\ddot{\Phi }& = & cos\ddot{\Theta }\hfill \\ \hfill \int \left(b{\ddot{\Psi }}^{\prime }sin\ddot{\Psi }ds-a{\ddot{\Phi }}^{\prime }sin\ddot{\Phi }\right)ds& = & \int cos\ddot{\Theta }ds\hfill \\ \hfill b\int {\ddot{\Psi }}^{\prime }sin\ddot{\Psi }ds-a\int {\ddot{\Phi }}^{\prime }sin\ddot{\Phi }ds& = & \int cos\ddot{\Theta }ds\hfill \\ \hfill acos\ddot{\Phi }-bcos\ddot{\Psi }& = & \int cos\ddot{\Theta }ds.\hfill \end{array}$$

Conversely, assume that $\ddot{\mho }$ is a unit speed curve in ${\mathbb{R}}^3$ and

$$\int cos\ddot{\Theta }ds = acos\ddot{\Phi }-bcos\ddot{\Psi }.$$

(20)

Substituting Equations (15) and (17) in Equation (20), we obtain

$$\begin{array}{ccc}\hfill \int cos\ddot{\Theta }ds& = & \int \left(a\mathcal{K}+b\mathfrak{T}\right)cos\ddot{\Theta }ds\hfill \\ \hfill cos\ddot{\Theta }& = & \left(a\mathcal{K}+b\mathfrak{T}\right)cos\ddot{\Theta }\hfill \\ \hfill 1& = & a\mathcal{K}+b\mathfrak{T}.\hfill \end{array}$$

So that $\ddot{\mho }$ is a Bertrand curve. □

4.3. Circular Helix

In this subsection, we prove that the angle function between the normal vector of a circular helix and a constant unit vector is a linear function in the arc-length of a given circular helix. Also, we use the position vector of a constant vector to give the position vector of a circular helix.

Theorem 12. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve and $\mathbf{U}$ be a constant unit vector. If $\ddot{\mho }$ is a circular helix, then,

$$\ddot{\Theta } = \sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c,$$

(21)

where c is a constant and $\ddot{\Theta }$ is the angle between $\mathbf{U}$ and $\mathbf{N}$.

Proof. 

Let $\ddot{\mho }$ be a circular helix and $\mathbf{U}$ be a constant unit vector in ${\mathbb{R}}^3$, except for the axis of $\ddot{\mho }$, then, by Theorem 9,

$$\begin{array}{ccc}\hfill \mathcal{K}\int cos\ddot{\Theta }ds\mathbf{T}+cos\ddot{\Theta }\mathbf{N}-\mathfrak{T}\int cos\ddot{\Theta }ds\mathbf{B}& = & \mathbf{U}\hfill \\ \hfill {\left(\mathcal{K}\int cos\ddot{\Theta }ds\right)}^2+{cos}^2\ddot{\Theta }+{\left(\mathfrak{T}\int cos\ddot{\Theta }ds\right)}^2& = & \mathbf{U}·\mathbf{U}\hfill \\ \hfill {\left(\mathcal{K}\int cos\ddot{\Theta }ds\right)}^2+{cos}^2\ddot{\Theta }+{\left(\mathfrak{T}\int cos\ddot{\Theta }ds\right)}^2& = & 1\hfill \\ \hfill {\mathcal{K}}^2{\left(\int cos\ddot{\Theta }ds\right)}^2+{\mathfrak{T}}^2{\left(\int cos\ddot{\Theta }ds\right)}^2& = & 1-{cos}^2\ddot{\Theta }\hfill \\ \hfill \left({\mathcal{K}}^2+{\mathfrak{T}}^2\right){\left(\int cos\ddot{\Theta }ds\right)}^2& = & {sin}^2\ddot{\Theta }\hfill \\ \hfill \sqrt{\left({\mathcal{K}}^2+{\mathfrak{T}}^2\right){\left(\int cos\ddot{\Theta }ds\right)}^2}& = & sin\ddot{\Theta }.\hfill \end{array}$$

Thus,

$$sin\ddot{\Theta } = \sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}\int cos\ddot{\Theta }ds$$

(22)

Differentiating Equation (22) with respect to s, we obtain

$$\begin{array}{ccc}\hfill {\ddot{\Theta }}^{\prime }cos\ddot{\Theta }& = & \sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}cos\ddot{\Theta }\hfill \\ \hfill {\ddot{\Theta }}^{\prime }& = & \sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}\hfill \\ \hfill \ddot{\Theta }& = & \int \sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}ds = \sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c.\hfill \end{array}$$

□

Theorem 13. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve. If $\ddot{\mho }$ is a circular helix, then,

$$\ddot{\mho } = \left({\displaystyle \frac{\mathcal{K}}{{\mathcal{K}}^2+{\mathfrak{T}}^2}}cos\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s\right),{\displaystyle \frac{\mathcal{K}}{{\mathcal{K}}^2+{\mathfrak{T}}^2}}sin\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s\right),\pm {\displaystyle \frac{\mathfrak{T}}{\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}}}s\right).$$

Proof. 

Let $\ddot{\mho } = ({\ddot{\mho }}_1,{\ddot{\mho }}_2,{\ddot{\mho }}_3)$ be a circular helix and $\mathbf{U}$ be a constant unit vector. Then, by Theorems 9 and 12, we obtain

$$\begin{array}{ccc}\hfill \mathbf{U}& = & \int \mathcal{K}cos(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c)ds\mathbf{T}+cos(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c)\mathbf{N}-\int \mathfrak{T}cos(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c)ds\mathbf{B}.\hfill \end{array}$$

Without loss of generality, we may assume that the axis of the circular helix is ${\mathbf{e}}_{\mathbf{3}} = (0,0,1)$, then,

$$\begin{array}{ccc}\hfill {\mathbf{e}}_{\mathbf{3}}·\mathbf{N}& = & 0\hfill \\ \\ \hfill (0,0,1)·(n_1,n_2,n_3)& = & 0.\hfill \end{array}$$

Therefore,

$$n_3 = 0.$$

(23)

Since $\mathcal{K}\ne 0$, then, multiplying Equation (23) by $\mathcal{K},$ we obtain

$$\mathcal{K}{n}_3 = 0.$$

Since ${\mathbf{T}}^{\prime } = \mathcal{K}\mathbf{N}$, we have

$$\begin{array}{ccc}\hfill t_3^{\prime }& = & 0\hfill \\ \hfill \int t_3^{\prime }ds& = & \int 0ds,\hfill \end{array}$$

Thus,

$$t_3 = c,$$

(24)

where c is a constant. Since ${\parallel \mathbf{T}\parallel }^2 = 1$, then $t_1^2+t_2^2+t_3^2 = 1$, and from Equation (24), we have

$$t_1^2+t_2^2 = 1-c^2.$$

(25)

Since ${\parallel \mathbf{N}\parallel }^2 = 1$, then $n_1^2+n_2^2+n_3^2 = 1$, and from Equation (23), we obtain

$$n_1^2 = 1-n_2^2.$$

(26)

Since $n_1 = {\mathbf{e}}_{\mathbf{1}}·\mathbf{N}$, then,

$$n_1 = cos\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c\right).$$

(27)

We substitute Equation (27) in Equation (26), and we obtain

$$\begin{array}{ccc}\hfill n_2^2& = & 1-{cos}^2\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c\right) = {sin}^2\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c\right).\hfill \end{array}$$

(28)

Moreover,

$$n_2 = sin\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c\right).$$

(29)

Since $\mathcal{K}\ne 0$, then, by multiplying Equation (27) by $\mathcal{K}$, we obtain

$$\begin{array}{ccc}\hfill \mathcal{K}{n}_1& = & \mathcal{K}cos\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c\right)\hfill \\ \hfill t_1^{\prime }& = & \mathcal{K}cos\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c\right)\hfill \\ \hfill \int t_1^{\prime }ds& = & \int \mathcal{K}cos\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c\right)ds\hfill \end{array}$$

$$t_1 = {\displaystyle \frac{\mathcal{K}}{\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}}}sin\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c\right).$$

(30)

Similarly,

$$t_2 = -{\displaystyle \frac{\mathcal{K}}{\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}}}cos\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c\right).$$

(31)

We substitute Equations (30) and (31) in Equation (25), and we obtain

$$\begin{array}{ccc}\hfill {\displaystyle \frac{{\mathcal{K}}^2}{{\mathcal{K}}^2+{\mathfrak{T}}^2}}{sin}^2\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c\right)+{\displaystyle \frac{{\mathcal{K}}^2}{{\mathcal{K}}^2+{\mathfrak{T}}^2}}{cos}^2\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c\right)& = & 1-c^2\hfill \\ \\ \hfill {\displaystyle \frac{{\mathcal{K}}^2}{{\mathcal{K}}^2+{\mathfrak{T}}^2}}\left({sin}^2\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c\right)+{cos}^2\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c\right)\right)& = & 1-c^2.\hfill \end{array}$$

So that

$$c^2 = {\displaystyle \frac{{\mathfrak{T}}^2}{{\mathcal{K}}^2+{\mathfrak{T}}^2}},$$

(32)

which gives

$$c = \pm {\displaystyle \frac{\mathfrak{T}}{\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}}}.$$

(33)

Since ${\ddot{\mho }}_1 = \int t_{1}ds$, then, by Equation (30), we obtain

$${\ddot{\mho }}_1 = \int {\displaystyle \frac{\mathcal{K}}{\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}}}sin\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c\right)ds = -{\displaystyle \frac{\mathcal{K}}{{\mathcal{K}}^2+{\mathfrak{T}}^2}}cos\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c\right).$$

(34)

Similarly,

$${\ddot{\mho }}_2 = \int -{\displaystyle \frac{\mathcal{K}}{\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}}}cos\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c\right)ds = -{\displaystyle \frac{\mathcal{K}}{{\mathcal{K}}^2+{\mathfrak{T}}^2}}sin\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c\right).$$

(35)

From Equations (24) and (33), we have

$$\begin{array}{ccc}\hfill {\ddot{\mho }}_3& = & \int t_{3}ds = \int cds\hfill \\ \\ & = & \int \pm {\displaystyle \frac{\mathfrak{T}}{\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}}}ds.\hfill \end{array}$$

Thus,

$${\ddot{\mho }}_3 = \pm {\displaystyle \frac{\mathfrak{T}}{\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}}}s+b,$$

(36)

where b is a constant.

• We conclude from Equations (34)–(36) that

$$\ddot{\mho } = \left({\displaystyle \frac{-\mathcal{K}}{{\mathcal{K}}^2+{\mathfrak{T}}^2}}cos\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c\right),{\displaystyle \frac{-\mathcal{K}}{{\mathcal{K}}^2+{\mathfrak{T}}^2}}sin\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+c\right),{\displaystyle \frac{\pm \mathfrak{T}s}{\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}}}+b\right).$$

(37)

We can use $c = \pi$ and $b = 0$ from Equation (37) and obtain

$$\begin{array}{ccc}\hfill \ddot{\mho }& = & \left(-{\displaystyle \frac{\mathcal{K}}{{\mathcal{K}}^2+{\mathfrak{T}}^2}}cos\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+\pi \right),-{\displaystyle \frac{\mathcal{K}}{{\mathcal{K}}^2+{\mathfrak{T}}^2}}sin\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+\pi \right),\pm {\displaystyle \frac{\mathfrak{T}}{\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}}}s\right)\hfill \\ & = & \left({\displaystyle \frac{\mathcal{K}}{{\mathcal{K}}^2+{\mathfrak{T}}^2}}cos\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s\right),{\displaystyle \frac{\mathcal{K}}{{\mathcal{K}}^2+{\mathfrak{T}}^2}}sin\left(\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s\right),\pm {\displaystyle \frac{\mathfrak{T}}{\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}}}s\right).\hfill \end{array}$$

□

4.4. General Helix

In light of Theorem 9, we present a new characteristic of the general helix and find its position vector.

Theorem 14. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve with $\mathcal{K}\ne 0$. If $\ddot{\mho }$ is a general helix, then,

$$\ddot{\Theta } = \pm \sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds,$$

(38)

where $\ddot{\Theta }$ is the angle between $\mathbf{U}$ and $\mathbf{N}$.

Proof. 

Let $\ddot{\mho }$ be a general helix and U be a constant unit vector in ${\mathbb{R}}^3$, except for the axis of $\ddot{\mho }$, then, by Theorem 9,

$$\begin{array}{ccc}\hfill \int \mathcal{K}cos\ddot{\Theta }ds\mathbf{T}+cos\ddot{\Theta }\mathbf{N}-c\int \mathcal{K}cos\ddot{\Theta }ds\mathbf{B}& = & \mathbf{U}\hfill \\ \hfill {\left(\int \mathcal{K}cos\ddot{\Theta }ds\right)}^2+{cos}^2\ddot{\Theta }+c^2{\left(\int \mathcal{K}cos\ddot{\Theta }ds\right)}^2& = & \mathbf{U}·\mathbf{U}\hfill \\ \hfill {\left(\int \mathcal{K}cos\ddot{\Theta }ds\right)}^2+{cos}^2\ddot{\Theta }+c^2{\left(\int \mathcal{K}cos\ddot{\Theta }ds\right)}^2& = & 1\hfill \\ \hfill (1+c^2){\left(\int \mathcal{K}cos\ddot{\Theta }ds\right)}^2& = & 1-{cos}^2\ddot{\Theta }\hfill \\ \hfill (1+c^2){\left(\int \mathcal{K}cos\ddot{\Theta }ds\right)}^2& = & {sin}^2\ddot{\Theta }.\hfill \end{array}$$

Therefore,

$$\begin{array}{ccc}\hfill sin\ddot{\Theta }& = & \pm \sqrt{1+c^2}\int \mathcal{K}cos\ddot{\Theta }ds.\hfill \end{array}$$

(39)

Differentiating Equation (39) with respect to s, we obtain

$$\begin{array}{ccc}\hfill {\ddot{\Theta }}^{\prime }cos\ddot{\Theta }& = & \pm \sqrt{1+c^2}\mathcal{K}cos\ddot{\Theta }\hfill \\ \hfill {\ddot{\Theta }}^{\prime }& = & \pm \sqrt{1+c^2}\mathcal{K}\hfill \\ \hfill \int {\ddot{\Theta }}^{\prime }ds& = & \pm \int \sqrt{1+c^2}\mathcal{K}ds\hfill \\ \hfill \ddot{\Theta }& = & \pm \sqrt{1+c^2}\int \mathcal{K}ds.\hfill \end{array}$$

Since $\ddot{\mho }$ is a general helix, then, ${\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}} = c$ is a constant. Thus,

$$\begin{array}{ccc}\hfill \ddot{\Theta }& = & \pm \sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds.\hfill \end{array}$$

□

Theorem 15. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve and $\mathbf{U}$ be a constant unit vector. Then, $\ddot{\mho }$ is a general helix if and only if $cos\ddot{\Psi }+Ccos\ddot{\Phi }$ is a constant, where C is a constant, $cos\ddot{\Psi }\ne 0$, and $cos\ddot{\Phi }\ne 0$.

Proof. 

Let $\ddot{\mho }$ be a general helix and U be a constant unit vector in ${\mathbb{R}}^3$. If U is the axis of $\ddot{\mho }$, then,

$$\mathbf{U} = C_1\mathbf{T}+C_2\mathbf{B},$$

(40)

where $C_1$ and $C_2$ are constants.

Taking the inner product between $\mathbf{U}$ and $\mathbf{T}$, we have

$$\begin{array}{ccc}\hfill \mathbf{U}·\mathbf{T}& = & C_1\hfill \\ \hfill \parallel \mathbf{U}\parallel \parallel \mathbf{T}\parallel cos\ddot{\Phi }& = & C_1\hfill \\ \hfill cos\ddot{\Phi }& = & C_1.\hfill \end{array}$$

Similarly,

$$\begin{array}{ccc}\hfill \mathbf{U}·\mathbf{B}& = & C_2\hfill \\ \hfill \parallel \mathbf{U}\parallel \parallel \mathbf{T}\parallel cos\ddot{\Psi }& = & C_2\hfill \\ \hfill cos\ddot{\Psi }& = & C_2.\hfill \end{array}$$

Since $cos\ddot{\Psi } = C_2$ and $cos\ddot{\Phi } = C_1$, we conclude that

$$\begin{array}{ccc}\hfill {\displaystyle \frac{cos\ddot{\Psi }}{cos\ddot{\Phi }}}& = & {\displaystyle \frac{C_2}{C_1}}\hfill \\ \hfill cos\ddot{\Psi }& = & {\displaystyle \frac{C_2}{C_1}}cos\ddot{\Phi }\hfill \\ \hfill cos\ddot{\Psi }-{\displaystyle \frac{C_2}{C_1}}cos\ddot{\Phi }& = & 0\hfill \\ \hfill cos\ddot{\Psi }+Ccos\ddot{\Phi }& = & 0,\hfill \end{array}$$

where $C = -{\displaystyle \frac{C_2}{C_1}}$ is a constant. Assume that U is not the axis of $\ddot{\mho }$ and recall that

$$\mathbf{U} = \int \mathcal{K}cos\ddot{\Theta }ds\mathbf{T}+cos\ddot{\Theta }\mathbf{N}-\int \mathfrak{T}cos\ddot{\Theta }ds\mathbf{B},$$

(41)

Now, Taking the inner product between $\mathbf{U}$ and $\mathbf{T}$, we obtain

$$\begin{array}{ccc}\hfill \mathbf{U}·\mathbf{T}& = & \int \mathcal{K}cos\ddot{\Theta }ds\hfill \\ \hfill \parallel \mathbf{U}\parallel \parallel \mathbf{T}\parallel cos\ddot{\Phi }& = & \int \mathcal{K}cos\ddot{\Theta }ds.\hfill \end{array}$$

Thus,

$$cos\ddot{\Phi } = \int \mathcal{K}cos\ddot{\Theta }ds.$$

(42)

Differentiating Equation (42) with respect to s, we obtain

$$-{\ddot{\Phi }}^{\prime }sin\ddot{\Phi } = \mathcal{K}cos\ddot{\Theta }.$$

(43)

Similarly,

$$\begin{array}{ccc}\hfill \mathbf{U}·\mathbf{B}& = & -\int \mathfrak{T}cos\ddot{\Theta }ds\hfill \\ \hfill \parallel \mathbf{U}\parallel \parallel \mathbf{B}\parallel cos\ddot{\Psi }& = & -\int \mathfrak{T}cos\ddot{\Theta }ds.\hfill \end{array}$$

Therefore,

$$cos\ddot{\Psi } = -\int \mathfrak{T}cos\ddot{\Theta }ds.$$

(44)

Differentiating Equation (44) with respect to s, we obtain

$${\ddot{\Psi }}^{\prime }sin\ddot{\Psi } = \mathfrak{T}cos\ddot{\Theta }.$$

(45)

From Equations (45) and (43), we obtain

$${\displaystyle \frac{{\ddot{\Psi }}^{\prime }sin\ddot{\Psi }}{{\ddot{\Phi }}^{\prime }sin\ddot{\Phi }}} = -{\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}.$$

Since $\ddot{\mho }$ is a general helix, then, ${\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}} = C$ is a constant. Thus,

$$\begin{array}{ccc}\hfill {\displaystyle \frac{{\ddot{\Psi }}^{\prime }sin\ddot{\Psi }}{{\ddot{\Phi }}^{\prime }sin\ddot{\Phi }}}& = & -C\hfill \\ \hfill {\ddot{\Psi }}^{\prime }sin\ddot{\Psi }& = & -C{\ddot{\Phi }}^{\prime }sin\ddot{\Phi }\hfill \\ \hfill \int {\ddot{\Psi }}^{\prime }sin\ddot{\Psi }ds& = & -C\int {\ddot{\Phi }}^{\prime }sin\ddot{\Phi }ds\hfill \\ \hfill cos\ddot{\Psi }& = & -Ccos\ddot{\Phi }+a\hfill \\ \hfill cos\ddot{\Psi }+Ccos\ddot{\Phi }& = & a,\hfill \end{array}$$

where a is a constant.

Conversely, assume that $cos\ddot{\Psi }+Ccos\ddot{\Phi } = a$, where a and C are constants, $cos\ddot{\Psi }\ne 0$, and $cos\ddot{\Phi }\ne 0$. From Equations (44) and (42), we obtain

$$\begin{array}{ccc}\hfill -\int \mathfrak{T}cos\ddot{\Theta }ds+C\int \mathcal{K}cos\ddot{\Theta }ds& = & a\hfill \\ \hfill -{\displaystyle \frac{d}{ds}}\left(\int \mathfrak{T}cos\ddot{\Theta }ds\right)+C{\displaystyle \frac{d}{ds}}\left(\int \mathcal{K}cos\ddot{\Theta }ds\right)& = & {\displaystyle \frac{d}{ds}}(a)\hfill \\ \hfill -\mathfrak{T}cos\ddot{\Theta }& = & -C\mathcal{K}cos\ddot{\Theta }\hfill \\ \hfill \mathfrak{T}& = & C\mathcal{K}\hfill \\ \hfill {\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}& = & C.\hfill \end{array}$$

Therefore, $\ddot{\mho }$ is a general helix. □

Theorem 16. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve. If $\ddot{\mho }$ is a general helix, then

$$\ddot{\mho } = \left({\displaystyle \frac{1}{\sqrt{1+{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^2}}}\int sin\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds\right)ds,{\displaystyle \frac{-1}{\sqrt{1+{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^2}}}\int cos\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds\right)ds,{\displaystyle \frac{\pm \mathfrak{T}s}{\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}}}\right).$$

Proof. 

Let $\ddot{\mho } = ({\ddot{\mho }}_1,{\ddot{\mho }}_2,{\ddot{\mho }}_3)$ be a general helix and $\mathbf{U}$ be a constant unit vector given in Theorem 9. By Theorem 14, we obtain

$$\mathbf{U} = \int \mathcal{K}cos\left(\pm \sqrt{1+c^2}\int \mathcal{K}ds\right)ds\mathbf{T}+cos\left(\pm \sqrt{1+c^2}\int \mathcal{K}ds\right)\mathbf{N}-\int \mathfrak{T}cos\left(\pm \sqrt{1+c^2}\int \mathcal{K}ds\right)ds\mathbf{B},$$

where $c = {\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}$ is a constant. We may assume that the axis of the general helix is ${\mathbf{e}}_{\mathbf{3}} = (0,0,1)$ and $\mathbf{N} = (n_1,n_2,n_3)$, then

$$\begin{array}{ccc}\hfill {\mathbf{e}}_{\mathbf{3}}·\mathbf{N}& = & 0.\hfill \end{array}$$

Therefore,

$$n_3 = 0.$$

(46)

Since $\mathcal{K}\ne 0$, then, multiplying Equation (46) by $\mathcal{K}$, we obtain

$$\begin{array}{ccc}\hfill \mathcal{K}{n}_3& = & 0.\hfill \end{array}$$

Let $T^{\prime } = (t_1^{\prime },t_2^{\prime },t_3^{\prime })$, then, using the Serret–Frenet equations,

$$\begin{array}{ccc}\hfill t_3^{\prime }& = & 0\hfill \\ \hfill \int t_3^{\prime }ds& = & \int 0ds.\hfill \end{array}$$

This implies that

$$t_3 = c,$$

(47)

where c is a constant. Since ${\parallel \mathbf{T}\parallel }^2 = 1$, then, $t_1^2+t_2^2+t_3^2 = 1$, and therefore,

$$t_1^2+t_2^2 = 1-c^2.$$

(48)

Since ${\parallel \mathbf{N}\parallel }^2 = 1$, then, $n_1^2+n_2^2+n_3^2 = 1$. By using Equation (46), we obtain

$$n_1^2 = 1-n_2^2.$$

(49)

Since $n_1 = {\mathbf{e}}_{\mathbf{1}}·\mathbf{N}$, then,

$$n_1 = cos\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds\right).$$

(50)

Substituting Equation (50) in Equation (49), we have

$$\begin{array}{ccc}\hfill n_2^2& = & 1-{cos}^2\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds\right) = {sin}^2\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds\right).\hfill \end{array}$$

(51)

This implies that

$$n_2 = sin\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds.\right).$$

(52)

Since $\mathcal{K}\ne 0$, then, multiplying Equation (50) by $\mathcal{K}$, we obtain

$$\begin{array}{ccc}\hfill \mathcal{K}{n}_1& = & \mathcal{K}cos\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds\right)\hfill \\ \\ \hfill t_1^{\prime }& = & \mathcal{K}cos\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds\right)\hfill \\ \\ \hfill \int t_1^{\prime }ds& = & \int \mathcal{K}cos\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds\right)ds\hfill \end{array}$$

$$t_1 = {\displaystyle \frac{1}{\sqrt{1+{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^2}}}sin\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds\right).$$

(53)

Similarly,

$$t_2 = -{\displaystyle \frac{1}{\sqrt{1+{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^2}}}cos\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds\right).$$

(54)

Substituting Equations (53) and (54) in Equation (48), and obtain

$$\begin{array}{ccc}\hfill {\displaystyle \frac{1}{1+{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^2}}{sin}^2\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds\right)+{\displaystyle \frac{1}{1+{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^2}}{cos}^2\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds\right)& = & 1-c^2\hfill \\ \hfill {\displaystyle \frac{1}{1+{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^2}}\left({sin}^2\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds\right)+{cos}^2\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds\right)\right)& = & 1-c^2\hfill \end{array}$$

So that

$$c^2 = {\displaystyle \frac{{\mathfrak{T}}^2}{{\mathcal{K}}^2+{\mathfrak{T}}^2}}.$$

(55)

Thus,

$$c = \pm {\displaystyle \frac{\mathfrak{T}}{\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}}}.$$

(56)

Since ${\ddot{\mho }}_1 = \int t_{1}ds$, then, by Equation (53), we obtain

$${\ddot{\mho }}_1 = {\displaystyle \frac{1}{\sqrt{1+{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^2}}}\int sin\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds\right)ds.$$

(57)

Similarly,

$${\ddot{\mho }}_2 = -{\displaystyle \frac{1}{\sqrt{1+{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^2}}}\int cos\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds\right)ds.$$

(58)

From Equations (47) and (56), we have

$$\begin{array}{ccc}\hfill {\ddot{\mho }}_3& = & \int t_{3}ds = \int cds\hfill \\ & = & \int \pm {\displaystyle \frac{\mathfrak{T}}{\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}}}ds.\hfill \end{array}$$

So that

$${\ddot{\mho }}_3 = \pm {\displaystyle \frac{\mathfrak{T}}{\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}}}s+b,$$

(59)

where b is a constant. We conclude from Equations (57)–(59) that

$$\ddot{\mho } = \left({\displaystyle \frac{1}{\sqrt{1+{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^2}}}\int sin\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds\right)ds,-{\displaystyle \frac{1}{\sqrt{1+{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^2}}}\int cos\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds\right)ds,\pm {\displaystyle \frac{\mathfrak{T}}{\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}}}s+b\right).$$

We may take $b = 0$ in the above equation to obtain

$$\ddot{\mho } = \left({\displaystyle \frac{1}{\sqrt{1+{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^2}}}\int sin\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds\right)ds,-{\displaystyle \frac{1}{\sqrt{1+{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^2}}}\int cos\left(\sqrt{1+{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^2}\int \mathcal{K}ds\right)ds,\pm {\displaystyle \frac{\mathfrak{T}}{\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}}}s\right).$$

□

4.5. Slant Helix

In this subsection, some properties and characterizations of the slant helix and proper slant helix are derived and presented in light of Theorem 9.

Theorem 17. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve with $\mathcal{K}\ne 0$ and $\mathbf{U}$ be a constant unit vector. If $\ddot{\mho }$ is a slant helix with axis  U, then,  U  is given by

$$\mathbf{U} = c\int \mathcal{K}ds\mathbf{T}+c\mathbf{N}-c\int \mathfrak{T}ds\mathbf{B},$$

(60)

where $c\ne 0$ is a constant.

Proof. 

Let $\ddot{\mho }$ be a slant helix with axis U. By Definition 3, we obtain

$$\begin{array}{ccc}\hfill \mathbf{U}·\mathbf{N}& = & c\hfill \\ \hfill \parallel \mathbf{U}|\parallel \parallel \mathbf{N}\parallel cos\ddot{\Theta }& = & c,\hfill \end{array}$$

which implies that

$$cos\ddot{\Theta } = c,$$

(61)

where $c\ne 0$ is a constant. Substituting Equation (61) in Equation (1), we obtain

$$\mathbf{U} = c\int \mathcal{K}ds\mathbf{T}+c\mathbf{N}-c\int \mathfrak{T}ds\mathbf{B}.$$

□

Theorem 18. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve with $\mathcal{K}\ne 0$ and $\mathbf{U}$ be a constant unit vector. If $\ddot{\mho }$ is a slant helix with axis  U, then,

$${\left(\int \mathcal{K}ds\right)}^2+{\left(\int \mathfrak{T}ds\right)}^2\mathrm{is} \mathrm{a} \mathrm{constant}.$$

Proof. 

Let $\ddot{\mho }$ be a slant helix with axis U. Then, from Theorem 17, we have

$$\begin{array}{ccc}\hfill \mathbf{U}& = & c\int \mathcal{K}ds\mathbf{T}+c\mathbf{N}-c\int \mathfrak{T}ds\mathbf{B}\hfill \\ \hfill \mathbf{U}·\mathbf{U}& = & c^2{\left(\int \mathcal{K}ds\right)}^2+c^2+c^2{\left(\int \mathfrak{T}ds\right)}^2\hfill \\ \hfill 1& = & c^2\left[{\left(\int \mathcal{K}ds\right)}^2+1+{\left(\int \mathfrak{T}ds\right)}^2\right]\hfill \\ \hfill {\displaystyle \frac{1}{c^2}}& = & {\left(\int \mathcal{K}ds\right)}^2+1+{\left(\int \mathfrak{T}ds\right)}^2\hfill \\ \hfill {\displaystyle \frac{1}{c^2}}-1& = & {\left(\int \mathcal{K}ds\right)}^2+{\left(\int \mathfrak{T}ds\right)}^2.\hfill \end{array}$$

Thus, ${\left(\int \mathcal{K}ds\right)}^2+{\left(\int \mathfrak{T}ds\right)}^2$ is a constant. □

Theorem 19. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve with $\mathcal{K}\ne 0$ and $\mathbf{U}$ be a constant unit vector. If $\ddot{\mho }$ is a proper slant helix with axis  U  and $\ddot{{\rm Y}}$ is its natural mate with ${\mathfrak{T}}_{\ddot{{\rm Y}}}\ne 0$, then,

(i)

$\int \mathfrak{T}ds = -{\displaystyle \frac{\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}$,

(ii)

$\int \mathcal{K}ds = {\displaystyle \frac{\mathfrak{T}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}$,

(iii)

$\mathbf{U} = {\displaystyle \frac{c\mathfrak{T}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\mathbf{T}+c\mathbf{N}+{\displaystyle \frac{c\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\mathbf{B}$, where c is a constant.

Proof. 

Let $\ddot{\mho }$ be a proper slant helix with axis U. Then, by Theorem 17, we have

$$\mathbf{U} = c\int \mathcal{K}ds\mathbf{T}+c\mathbf{N}-c\int \mathfrak{T}ds\mathbf{B}.$$

(62)

Differentiating Equation (62) with respect to s and using the Serret–Frenet equations, we obtain

$$\begin{array}{ccc}\hfill 0& = & c\mathcal{K}\mathbf{T}+c\mathcal{K}\int \mathcal{K}ds\mathbf{N}+c\mathfrak{T}\mathbf{B}-c\mathcal{K}\mathbf{T}-c\mathfrak{T}\mathbf{B}+c\mathfrak{T}\int \mathfrak{T}ds\mathbf{N}\hfill \\ \hfill 0& = & c\left(\mathcal{K}\int \mathcal{K}ds+\mathfrak{T}\int \mathfrak{T}ds\right)\mathbf{N},\hfill \end{array}$$

which implies that

$$\mathcal{K}\int \mathcal{K}ds+\mathfrak{T}\int \mathfrak{T}ds = 0.$$

(63)

Therefore,

$$\int \mathcal{K}ds = -{\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\int \mathfrak{T}ds.$$

(64)

Differentiating Equation (63) again, with respect to s, we obtain

$$\begin{array}{ccc}\hfill {\mathcal{K}}^{\prime }\int \mathcal{K}ds+{\mathfrak{T}}^{\prime }\int \mathfrak{T}ds& = & -\left({\mathcal{K}}^2+{\mathfrak{T}}^2\right).\hfill \end{array}$$

(65)

Substituting Equation (64) in Equation (65), we obtain

$$\begin{array}{ccc}\hfill -{\displaystyle \frac{{\mathcal{K}}^{\prime }\mathfrak{T}}{\mathcal{K}}}\int \mathfrak{T}ds+{\mathfrak{T}}^{\prime }\int \mathfrak{T}ds& = & -\left({\mathcal{K}}^2+{\mathfrak{T}}^2\right)\hfill \\ \hfill \left(-{\displaystyle \frac{{\mathcal{K}}^{\prime }\mathfrak{T}}{\mathcal{K}}}+{\mathfrak{T}}^{\prime }\right)\int \mathfrak{T}ds& = & -\left({\mathcal{K}}^2+{\mathfrak{T}}^2\right)\hfill \\ \hfill {\displaystyle \frac{\mathcal{K}{\mathfrak{T}}^{\prime }-{\mathcal{K}}^{\prime }\mathfrak{T}}{\mathcal{K}}}\int \mathfrak{T}ds& = & -\left({\mathcal{K}}^2+{\mathfrak{T}}^2\right)\hfill \\ \hfill \mathcal{K}\left({\displaystyle \frac{\mathcal{K}{\mathfrak{T}}^{\prime }-{\mathcal{K}}^{\prime }\mathfrak{T}}{{\mathcal{K}}^2}}\right)\int \mathfrak{T}ds& = & -\left({\mathcal{K}}^2+{\mathfrak{T}}^2\right)\hfill \\ \hfill \mathcal{K}{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^{\prime }\int \mathfrak{T}ds& = & -\left({\mathcal{K}}^2+{\mathfrak{T}}^2\right).\hfill \end{array}$$

Therefore,

$$\int \mathfrak{T}ds = {\displaystyle \frac{-\left({\mathcal{K}}^2+{\mathfrak{T}}^2\right)}{\mathcal{K}{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}}.$$

(66)

Substituting Equation (66) in Equation (64), we obtain

$$\begin{array}{ccc}\hfill \int \mathcal{K}ds& = & {\displaystyle \frac{\mathfrak{T}\left({\mathcal{K}}^2+{\mathfrak{T}}^2\right)}{{\mathcal{K}}^2{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}}.\hfill \end{array}$$

By Theorem 7, we conclude that

$$\begin{array}{ccc}\hfill \int \mathcal{K}ds& = & {\displaystyle \frac{\mathfrak{T}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}and\int \mathfrak{T}ds = -{\displaystyle \frac{\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}.\hfill \end{array}$$

So that

$$\mathbf{U} = {\displaystyle \frac{c\mathfrak{T}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\mathbf{T}+c\mathbf{N}+{\displaystyle \frac{c\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\mathbf{B}.$$

(67)

□

Theorem 20. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve with $\mathcal{K}\ne 0$ and let $\mathbf{U}$ be a constant unit vector. If $\ddot{\mho }$ is a proper slant helix with axis  U  and $\ddot{{\rm Y}}$ is its natural mate with ${\mathfrak{T}}_{\ddot{{\rm Y}}}\ne 0$, then

$$\begin{array}{ccc}\hfill tan\ddot{\Theta }& = & \pm {\displaystyle \frac{\sqrt{{\left({\mathcal{K}}^2+{\mathfrak{T}}^2\right)}^3}}{{\mathcal{K}}^2{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}},\hfill \end{array}$$

(68)

where $\ddot{\Theta }$ is the angle between U and N.

Proof. 

Let $\ddot{\mho }$ be a proper slant helix with axis U and let $\ddot{{\rm Y}}$ be the natural mate of $\ddot{\mho }$, then, from Theorem 19, we have

$$\begin{array}{ccc}\hfill \mathbf{U}& = & {\displaystyle \frac{c\mathfrak{T}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\mathbf{T}+c\mathbf{N}+{\displaystyle \frac{c\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\mathbf{B}\hfill \\ \\ \hfill {\parallel \mathbf{U}\parallel }^2& = & {\displaystyle \frac{c^2{\mathfrak{T}}^2}{{\mathfrak{T}}_{\ddot{{\rm Y}}}^2}}+c^2+{\displaystyle \frac{c^2{\mathcal{K}}^2}{{\mathfrak{T}}_{\ddot{{\rm Y}}}^2}}\hfill \\ \\ \hfill 1& = & {\displaystyle \frac{c^2{\mathfrak{T}}^2}{{\mathfrak{T}}_{\ddot{{\rm Y}}}^2}}+c^2+{\displaystyle \frac{c^2{\mathcal{K}}^2}{{\mathfrak{T}}_{\ddot{{\rm Y}}}^2}}\hfill \\ \\ \hfill 1& = & c^2\left({\displaystyle \frac{{\mathfrak{T}}^2+{\mathcal{K}}^2}{{\mathfrak{T}}_{\ddot{{\rm Y}}}^2}}\right)+c^2\hfill \\ \\ \hfill {\displaystyle \frac{1-c^2}{c^2}}& = & {\displaystyle \frac{{\mathcal{K}}^2+{\mathfrak{T}}^2}{{\mathfrak{T}}_{\ddot{{\rm Y}}}^2}}.\hfill \end{array}$$

By using Equation (61), we obtain

$$\begin{array}{ccc}\hfill {\displaystyle \frac{1-{cos}^2\ddot{\Theta }}{{cos}^2\ddot{\Theta }}}& = & {\displaystyle \frac{{\mathcal{K}}^2+{\mathfrak{T}}^2}{{\mathfrak{T}}_{\ddot{{\rm Y}}}^2}}\hfill \\ \hfill {\displaystyle \frac{{sin}^2\ddot{\Theta }}{{cos}^2\ddot{\Theta }}}& = & {\displaystyle \frac{{\mathcal{K}}^2+{\mathfrak{T}}^2}{{\mathfrak{T}}_{\ddot{{\rm Y}}}^2}}\hfill \\ \hfill {tan}^2\ddot{\Theta }& = & {\displaystyle \frac{{\mathcal{K}}^2+{\mathfrak{T}}^2}{{\mathfrak{T}}_{\ddot{{\rm Y}}}^2}}\hfill \\ \hfill tan\ddot{\Theta }& = & \pm {\displaystyle \frac{\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}.\hfill \end{array}$$

By using Theorem 7, we obtain

$$\begin{array}{ccc}\hfill tan\ddot{\Theta }& = & \pm {\displaystyle \frac{\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}}{\frac{{\mathcal{K}}^2{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}{{\mathcal{K}}^2+{\mathfrak{T}}^2}}}\hfill \\ & = & \pm {\displaystyle \frac{\sqrt{{\left({\mathcal{K}}^2+{\mathfrak{T}}^2\right)}^3}}{{\mathcal{K}}^2{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}}.\hfill \end{array}$$

□

Theorem 21. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve with $\mathcal{K}\ne 0$. If $\ddot{\mho }$ is a proper slant helix, then, its axis makes a constant angle with the Darboux vector field associated with $\ddot{\mho }$.

Proof. 

Let $\ddot{\mho }$ be a proper slant helix with axis U and let $\ddot{{\rm Y}}$ be the natural mate of $\ddot{\mho }$. Then, from Theorem 19, we obtain

$$\begin{array}{ccc}\hfill \mathbf{U}& = & {\displaystyle \frac{c\mathfrak{T}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\mathbf{T}+c\mathbf{N}+{\displaystyle \frac{c\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\mathbf{B}.\hfill \end{array}$$

(69)

Let $\mathbf{D}$ be the Darboux vector field associated with $\ddot{\mho }$, then,

$$\begin{array}{ccc}\hfill \mathbf{D}& = & \mathfrak{T}\mathbf{T}+\mathcal{K}\mathbf{B}.\hfill \end{array}$$

(70)

Thus,

$$\begin{array}{ccc}\hfill \mathbf{U}·\mathbf{D}& = & \left({\displaystyle \frac{c\mathfrak{T}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\mathbf{T}+c\mathbf{N}+{\displaystyle \frac{c\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\mathbf{B}\right)·(\mathfrak{T}\mathbf{T}+\mathcal{K}\mathbf{B})\hfill \\ & = & {\displaystyle \frac{c{\mathfrak{T}}^2}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}+{\displaystyle \frac{c{\mathcal{K}}^2}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\hfill \\ & = & {\displaystyle \frac{c{\mathfrak{T}}^2+c{\mathcal{K}}^2}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\hfill \\ & = & {\displaystyle \frac{c({\mathfrak{T}}^2+{\mathcal{K}}^2)}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}.\hfill \end{array}$$

By using Theorem 7, we obtain

$$\begin{array}{ccc}\hfill \mathbf{U}·\mathbf{D}& = & {\displaystyle \frac{c{\mathcal{K}}_{\ddot{{\rm Y}}}^2}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\hfill \\ \hfill \parallel \mathbf{U}\parallel \parallel \mathbf{D}\parallel cos\ddot{\Phi }& = & {\displaystyle \frac{c{\mathcal{K}}_{\ddot{{\rm Y}}}^2}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\hfill \\ \hfill \sqrt{{\mathfrak{T}}^2+{\mathcal{K}}^2}cos\ddot{\Phi }& = & {\displaystyle \frac{c{\mathcal{K}}_{\ddot{{\rm Y}}}^2}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\hfill \\ \hfill {\mathcal{K}}_{\ddot{{\rm Y}}}cos\ddot{\Phi }& = & {\displaystyle \frac{c{\mathcal{K}}_{\ddot{{\rm Y}}}^2}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}},\hfill \end{array}$$

which implies that

$$cos\ddot{\Phi } = {\displaystyle \frac{c{\mathcal{K}}_{\ddot{{\rm Y}}}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}},$$

(71)

where $\ddot{\Phi }$ is the angle between $\mathbf{U}$ and $\mathbf{D}$. Since $\ddot{\mho }$ is a proper slant helix, then, ${\displaystyle \frac{{\mathfrak{T}}_{\ddot{{\rm Y}}}}{{\mathcal{K}}_{\ddot{{\rm Y}}}}}$ is a constant and hence $cos\ddot{\Phi }$ is constant. □

Theorem 22. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve with $\mathcal{K}\ne 0$ and let $\mathbf{U}$ and $\mathbf{W}$ be constants unit vectors such that $\mathbf{W}\ne \pm \mathbf{U}$. If $\ddot{\mho }$ is a proper slant helix with axis  U, then,

$$\begin{array}{ccc}\hfill {\displaystyle \frac{1}{c}}\mathbf{U}·\mathbf{W}& = & {\displaystyle \frac{1}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\mathbf{W}·\mathbf{D}+cos\ddot{\Theta },\hfill \end{array}$$

where ${\mathfrak{T}}_{\ddot{{\rm Y}}}\ne 0$, D is the Darboux vector field associated with $\ddot{\mho }$ and $\ddot{\Theta }$ is the angle between W and N.

Proof. 

Let $\ddot{\mho }$ be a proper slant helix with axis U. Then, by Theorem 19, we obtain

$$\mathbf{U} = {\displaystyle \frac{c\mathfrak{T}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\mathbf{T}+c\mathbf{N}+{\displaystyle \frac{c\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\mathbf{B}.$$

Let $\mathbf{W}$ be any constant unit vector such that $\mathbf{W}\ne \pm \mathbf{U}$, we assume

$$\begin{array}{ccc}\hfill \mathbf{W}& = & \int \mathcal{K}cos\ddot{\Theta }ds\mathbf{T}+cos\ddot{\Theta }\mathbf{N}-\int \mathfrak{T}cos\ddot{\Theta }ds\mathbf{B}.\hfill \end{array}$$

(72)

Since $\mathbf{D}$ is the Darboux vector field associated with $\ddot{\mho }$, then,

$$\begin{array}{ccc}\hfill \mathbf{D}& = & \mathfrak{T}\mathbf{T}+\mathcal{K}\mathbf{B}.\hfill \end{array}$$

(73)

Now,

$$\begin{array}{ccc}\hfill \mathbf{W}·\mathbf{D}& = & \mathfrak{T}\int \mathcal{K}cos\ddot{\Theta }ds-\mathcal{K}\int \mathfrak{T}cos\ddot{\Theta }ds.\hfill \end{array}$$

(74)

Moreover,

$$\begin{array}{ccc}\hfill \mathbf{U}·\mathbf{W}& = & {\displaystyle \frac{c\mathfrak{T}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\int \mathcal{K}cos\ddot{\Theta }ds+ccos\ddot{\Theta }-{\displaystyle \frac{c\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\int \mathfrak{T}cos\ddot{\Theta }ds\hfill \\ \\ & = & c\left({\displaystyle \frac{\mathfrak{T}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\int \mathcal{K}cos\ddot{\Theta }ds+cos\ddot{\Theta }-{\displaystyle \frac{\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\int \mathfrak{T}cos\ddot{\Theta }ds\right)\hfill \\ \\ \hfill {\displaystyle \frac{1}{c}}\mathbf{U}·\mathbf{W}& = & {\displaystyle \frac{\mathfrak{T}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\int \mathcal{K}cos\ddot{\Theta }ds+cos\ddot{\Theta }-{\displaystyle \frac{\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\int \mathfrak{T}cos\ddot{\Theta }ds,\hfill \end{array}$$

which implies that

$${\displaystyle \frac{1}{c}}\mathbf{U}·\mathbf{W} = {\displaystyle \frac{1}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\left(\mathfrak{T}\int \mathcal{K}cos\ddot{\Theta }ds-\mathcal{K}\int \mathfrak{T}cos\ddot{\Theta }ds\right)+cos\ddot{\Theta }.$$

(75)

Substituting Equation (74) in Equation (75), we obtain

$$\begin{array}{ccc}\hfill {\displaystyle \frac{1}{c}}\mathbf{U}·\mathbf{W}& = & {\displaystyle \frac{1}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\mathbf{W}·\mathbf{D}+cos\ddot{\Theta }.\hfill \end{array}$$

□

Theorem 23. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve with $\mathcal{K}\ne 0$ and let $\mathbf{U}$ and $\mathbf{W}$ be constants unit vectors such that $\mathbf{W}\ne \pm \mathbf{U}$. If $\ddot{\mho }$ is a proper slant helix with axis   U  and $\mathbf{W}\perp \mathbf{U}$, then,

$$cos\ddot{\Theta }+acos\ddot{\Phi } = 0,$$

where $\ddot{\Phi }$ is the angle between $\mathbf{W}$ and $\mathbf{D}$, $\ddot{\Theta }$ is the angle between $\mathbf{W}$ and $\mathbf{N}$, and a is a constant.

Proof. 

Assume that $\mathbf{W}\perp \mathbf{U}$, then, $\mathbf{W}·\mathbf{U} = 0$. Using Theorem 22, we obtain

$$\begin{array}{ccc}\hfill {\displaystyle \frac{1}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\mathbf{W}·\mathbf{D}+cos\ddot{\Theta }& = & 0\hfill \\ \hfill cos\ddot{\Theta }& = & -{\displaystyle \frac{1}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\mathbf{W}·\mathbf{D}\hfill \\ & = & -{\displaystyle \frac{1}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\parallel \mathbf{W}\parallel \parallel \mathbf{D}\parallel cos\ddot{\Phi }\hfill \\ & = & -{\displaystyle \frac{\sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}cos\ddot{\Phi }\hfill \\ & = & -{\displaystyle \frac{{\mathcal{K}}_{\ddot{{\rm Y}}}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}cos\ddot{\Phi }.\hfill \end{array}$$

Since $\ddot{\mho }$ is a propper slant helix, then, ${\displaystyle \frac{{\mathcal{K}}_{\ddot{{\rm Y}}}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}} = a$ is a constant.

$$\begin{array}{ccc}\hfill cos\ddot{\Theta }& = & -acos\ddot{\Phi }\hfill \\ \hfill cos\ddot{\Theta }+acos\ddot{\Phi }& = & 0.\hfill \end{array}$$

□

Theorem 24. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve with $\mathcal{K}\ne 0$ and let $\mathbf{U}$ be a constant unit vector. If $\ddot{\mho }$ is a proper slant helix with axis $\mathbf{U}$, then, $cos\ddot{\Phi } = sin\ddot{\Theta },$ where $\ddot{\Phi }$ is the angle between $\mathbf{U}$ and $\mathbf{D}$, $\ddot{\Theta }$ is the angle between $\mathbf{U}$ and $\mathbf{N}$.

Proof. 

Let $\ddot{\mho }$ be a proper slant helix, then, by Theorem 19, we obtain

$$\begin{array}{ccc}\hfill tan\ddot{\Theta }& = & \pm {\displaystyle \frac{\sqrt{{\left({\mathcal{K}}^2+{\mathfrak{T}}^2\right)}^3}}{{\mathcal{K}}^2{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}}\hfill \\ & = & \pm \sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}\left({\displaystyle \frac{{\mathcal{K}}^2+{\mathfrak{T}}^2}{{\mathcal{K}}^2{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}}\right).\hfill \end{array}$$

From Theorem 7, we obtain

$$tan\ddot{\Theta } = \pm {\displaystyle \frac{{\mathcal{K}}_{\ddot{{\rm Y}}}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}.$$

Using Equation (71), we obtain

$$\begin{array}{ccc}\hfill tan\ddot{\Theta }& = & \pm {\displaystyle \frac{1}{c}}cos\ddot{\Phi },\hfill \\ \hfill cos\ddot{\Phi }& = & ctan\ddot{\Theta }\hfill \\ & = & {\displaystyle \frac{csin\ddot{\Theta }}{cos\ddot{\Theta }}}.\hfill \end{array}$$

where c is a constant.

Since $\ddot{\mho }$ is a proper slant helix, then,

$$cos\ddot{\Phi } = {\displaystyle \frac{csin\ddot{\Theta }}{c}}.$$

Hence,

$$\begin{array}{c}\hfill cos\ddot{\Phi } = sin\ddot{\Theta }.\end{array}$$

□

4.6. Rectifying Curve

The following result provides the characterization of the rectifying curves in light of Theorem 9.

Theorem 25. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve with $\mathcal{K}\ne 0$, $\mathfrak{T}\ne 0$ and ${\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^{\prime }\ne 0$. If $\ddot{\mho }$ is a rectifying curve, then, $\ddot{\mho } = {\displaystyle \frac{c}{\mathcal{K}}}\mathbf{D}$, where $c = {\displaystyle \frac{1}{{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}}$ is a constant.

Proof. 

Let U be a unit constant vector in ${\mathbb{R}}^3$, then, by Theorem 9, we obtain

$$\mathbf{U} = \int \mathcal{K}cos\ddot{\Theta }ds\mathbf{T}+cos\ddot{\Theta }\mathbf{N}-\int \mathfrak{T}cos\ddot{\Theta }ds\mathbf{B}.$$

Assume that $\ddot{\mho }$ is a rectifying curve, then,

$$\ddot{\mho } = (s+c_1)\mathbf{T}+c\mathbf{B},$$

(76)

where $c_1$ and c are constants. Taking the dot product of Equation (76) with $\mathbf{U}$, we have

$$\begin{array}{ccc}\hfill \ddot{\mho }·\mathbf{U}& = & (s+c_1)\mathbf{T}·\mathbf{U}+c\mathbf{B}·\mathbf{U},\hfill \end{array}$$

which implies that

$$\ddot{\mho }·\mathbf{U} = (s+c_1)\int \mathcal{K}cos\ddot{\Theta }ds-c\int \mathfrak{T}cos\ddot{\Theta }ds.$$

(77)

Differentiating Equation (77) with respect to s, we obtain

$$\begin{array}{ccc}\hfill \mathbf{T}·\mathbf{U}& = & (s+c_1)\mathcal{K}cos\ddot{\Theta }+\int \mathcal{K}cos\ddot{\Theta }ds-c\mathfrak{T}cos\ddot{\Theta }\hfill \\ \hfill \int \mathcal{K}cos\ddot{\Theta }ds& = & (s+c_1)\mathcal{K}cos\ddot{\Theta }+\int \mathcal{K}cos\ddot{\Theta }ds-c\mathfrak{T}cos\ddot{\Theta }\hfill \\ \hfill 0& = & (s+c_1)\mathcal{K}cos\ddot{\Theta }-c\mathfrak{T}cos\ddot{\Theta }.\hfill \end{array}$$

Therefore,

$$\left[(s+c_1)\mathcal{K}-c\mathfrak{T}\right]cos\ddot{\Theta } = 0.$$

(78)

Since $cos\ddot{\Theta }\ne 0$, then,

$$\begin{array}{ccc}\hfill (s+c_1)\mathcal{K}-c\mathfrak{T}& = & 0\hfill \end{array}$$

(79)

$$\begin{array}{ccc}\hfill c\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)& = & s+c_1.\hfill \end{array}$$

(80)

Differentiating Equation (78) with respect to s, we obtain

$$\begin{array}{ccc}\hfill -{\ddot{\Theta }}^{\prime }sin\ddot{\Theta }\left[(s+c_1)\mathcal{K}-c\mathfrak{T}\right]+\left[\mathcal{K}+(s+c_1){\mathcal{K}}^{\prime }-c{\mathfrak{T}}^{\prime }\right]cos\ddot{\Theta }& = & 0.\hfill \end{array}$$

From Equation (79), we obtain

$$\begin{array}{ccc}\hfill \left[\mathcal{K}+(s+c_1){\mathcal{K}}^{\prime }-c{\mathfrak{T}}^{\prime }\right]cos\ddot{\Theta }& = & 0\hfill \end{array}$$

This implies that

$$\begin{array}{ccc}\hfill \mathcal{K}+(s+c_1){\mathcal{K}}^{\prime }-c{\mathfrak{T}}^{\prime }& = & 0.\hfill \end{array}$$

(81)

Substituting Equation (80) in Equation (81), we obtain

$$\begin{array}{ccc}\hfill \mathcal{K}+c\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right){\mathcal{K}}^{\prime }-c{\mathfrak{T}}^{\prime }& = & 0\hfill \\ \hfill \mathcal{K}-c\left[{\mathfrak{T}}^{\prime }-{\displaystyle \frac{\mathfrak{T}{\mathcal{K}}^{\prime }}{\mathcal{K}}}\right]& = & 0\hfill \\ \hfill {\mathcal{K}}^2-c\left[\mathcal{K}{\mathfrak{T}}^{\prime }-\mathfrak{T}{\mathcal{K}}^{\prime }\right]& = & 0\hfill \\ \hfill {\mathcal{K}}^2& = & c\left[\mathcal{K}{\mathfrak{T}}^{\prime }-\mathfrak{T}{\mathcal{K}}^{\prime }\right]\hfill \\ \hfill 1& = & c\left[{\displaystyle \frac{\mathcal{K}{\mathfrak{T}}^{\prime }-\mathfrak{T}{\mathcal{K}}^{\prime }}{{\mathcal{K}}^2}}\right]\hfill \\ \hfill 1& = & c{\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^{\prime },\hfill \end{array}$$

which gives

$$c = {\displaystyle \frac{1}{{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}}.$$

(82)

We substitute Equation (82) in Equation (80) and obtain

$$s+c_1 = {\displaystyle \frac{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}{{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}}.$$

(83)

By substituting Equations (82) and (83) in Equation (76), we obtain

$$\begin{array}{ccc}\hfill \ddot{\mho }& = & {\displaystyle \frac{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}{{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}}\mathbf{T}+{\displaystyle \frac{1}{{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}}\mathbf{B} = {\displaystyle \frac{1}{{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}}\left[{\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\mathbf{T}+\mathbf{B}\right]\hfill \\ & = & {\displaystyle \frac{1}{{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}}{\displaystyle \frac{1}{\mathcal{K}}}\left[\mathfrak{T}\mathbf{T}+\mathcal{K}\mathbf{B}\right] = {\displaystyle \frac{1}{{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}}{\displaystyle \frac{1}{\mathcal{K}}}\mathbf{D}\hfill \\ & = & {\displaystyle \frac{c}{\mathcal{K}}}\mathbf{D}.\hfill \end{array}$$

□

Theorem 26. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve with $\mathcal{K}\ne 0,\mathfrak{T}\ne 0$ and ${\left({\displaystyle \frac{\mathfrak{T}}{\mathcal{K}}}\right)}^{\prime }\ne 0$. If $\ddot{\mho }$ is a rectifying curve, then,

$$\begin{array}{ccc}\hfill {\mathcal{K}}_{\alpha }& = & {\displaystyle \frac{d}{\left|c\right|}}\mathrm{and}{\mathfrak{T}}_{\alpha } = {\displaystyle \frac{c}{\mathcal{K}{d}^2}},\hfill \end{array}$$

where ${\mathcal{K}}_{\alpha }$ and ${\mathfrak{T}}_{\alpha }$ are the curvature and the torsion of the spherical indicatrix of the tangent α, respectively; c is a constant; and d is the distance function of $\ddot{\mho }$.

Proof. 

Assume that $\ddot{\mho }$ is a rectifying curve and $\mathbf{D}$ is the Darboux vector field, then, by Theorem 25, we have

$$\begin{array}{ccc}\hfill \ddot{\mho }& = & {\displaystyle \frac{1}{{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}}{\displaystyle \frac{1}{\mathcal{K}}}D\hfill \\ & = & {\displaystyle \frac{1}{{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}}{\displaystyle \frac{1}{\mathcal{K}}}\left[\mathfrak{T}\mathbf{T}+\mathcal{K}\mathbf{B}\right]\hfill \\ \hfill \ddot{\mho }·\ddot{\mho }& = & {\displaystyle \frac{{\mathfrak{T}}^2+{\mathcal{K}}^2}{{\mathcal{K}}^2{\left[{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }\right]}^2}}.\hfill \end{array}$$

Using Equation (82), we obtain

$$\begin{array}{ccc}\hfill d^2& = & c^2\left({\displaystyle \frac{{\mathfrak{T}}^2+{\mathcal{K}}^2}{{\mathcal{K}}^2}}\right)\hfill \\ \\ & = & c^2{\mathcal{K}}_{\alpha }^2.\hfill \end{array}$$

Hence, ${\mathcal{K}}_{\alpha } = {\displaystyle \frac{d}{\left|c\right|}}$. Since $d^2 = {\displaystyle \frac{{\mathfrak{T}}^2+{\mathcal{K}}^2}{{\mathcal{K}}^2{\left[{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }\right]}^2}}$, then, $d^2 = {\displaystyle \frac{1}{{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}}{\displaystyle \frac{1}{\mathcal{K}}}{\displaystyle \frac{{\mathfrak{T}}^2+{\mathcal{K}}^2}{\mathcal{K}{\left(\frac{\mathfrak{T}}{\mathcal{K}}\right)}^{\prime }}}$. Therefore, $d^2 = {\displaystyle \frac{c}{\mathcal{K}{\mathfrak{T}}_{\alpha }}},$ which implies that ${\mathfrak{T}}_{\alpha } = {\displaystyle \frac{c}{\mathcal{K}{d}^2}}.$ □

Theorem 27. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a rectifying curve. Then, $\ddot{\mho }$ is a slant helix if and only if

$$\mathcal{K} = {\displaystyle \frac{c_2}{{\left(s^2+c_{3}s+c_4\right)}^{\frac{3}{2}}}},$$

where $c_2$, $c_3$, and $c_4$ are constants.

Proof. 

Assume that $\ddot{\mho }$ is a slant helix, then, the spherical indicatarise $\alpha = \mathbf{T}$ is a general helix. Thus, ${\displaystyle \frac{{\mathfrak{T}}_{\alpha }}{{\mathcal{K}}_{\alpha }}} = a$, where a is a constant. Since $\ddot{\mho }$ is a rectifying curve, then, by Theorem 26, we obtain

$$\begin{array}{ccc}\hfill \left({\displaystyle \frac{c}{\mathcal{K}{d}^2}}\right)\left({\displaystyle \frac{\left|c\right|}{d}}\right)& = & a\hfill \\ \hfill {\displaystyle \frac{c^2}{\mathcal{K}{d}^3}}& = & \left|a\right|\hfill \\ \hfill \mathcal{K}& = & {\displaystyle \frac{c^2}{\left|a\right|d^3}} = {\displaystyle \frac{c_2}{d^3}}\hfill \\ & = & {\displaystyle \frac{c_2}{{\left(s^2+c_{3}s+c_4\right)}^{\frac{3}{2}}}},\hfill \end{array}$$

where $c_2 = {\displaystyle \frac{c^2}{\left|a\right|}}$ is a constant. □

4.7. New Frame Generated from the Serret–Frenet Frame

We provide a new frame constructed from the Serret–Frenet frame, which can be used to derive extremely significant new results, as well as new proofs, for several well-known results.

Lemma 1. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve with {$\mathcal{K}$,$\mathfrak{T}$,T,N,B} as the Serret Frenet apparatus of $\ddot{\mho }$. Then,

$$\begin{array}{ccc}\hfill {\mathbf{e}}_{\mathbf{1}}& = & t_1\mathbf{T}+n_1\mathbf{N}+b_1\mathbf{B},\hfill \\ \hfill {\mathbf{e}}_{\mathbf{2}}& = & t_2\mathbf{T}+n_2\mathbf{N}+b_2\mathbf{B},\hfill \\ \hfill {\mathbf{e}}_{\mathbf{3}}& = & t_3\mathbf{T}+n_3\mathbf{N}+b_3\mathbf{B},\hfill \end{array}$$

where $\{{\mathbf{e}}_{\mathbf{1}},{\mathbf{e}}_{\mathbf{2}},{\mathbf{e}}_{\mathbf{3}}\}$ is the standard basis in ${\mathbb{R}}^3$ and

$$\begin{array}{ccc}\hfill \mathbf{T}& = & (t_1,t_2,t_3),\hfill \\ \hfill \mathbf{N}& = & (n_1,n_2,n_3),\hfill \\ \hfill \mathbf{B}& = & (b_1,b_2,b_3).\hfill \end{array}$$

Proof. 

Since ${\mathbf{e}}_{\mathbf{1}} = (1,0,0)$ is a constant unit vector in ${\mathbb{R}}^3$, then, by Theorem 9, ${\mathbf{e}}_{\mathbf{1}}$ is given by

$${\mathbf{e}}_{\mathbf{1}} = \int \mathcal{K}cos\ddot{\Theta }ds\mathbf{T}+cos\ddot{\Theta }\mathbf{N}-\int \mathfrak{T}cos\ddot{\Theta }ds\mathbf{B}.$$

(84)

Taking the dot product of Equation (84) with N, we have

$$\begin{array}{ccc}\hfill cos\ddot{\Theta }& = & {\mathbf{e}}_{\mathbf{1}}·\mathbf{N}\hfill \\ & = & (1,0,0)·(n_1,n_2,n_3)\hfill \\ & = & n_1.\hfill \end{array}$$

Taking the dot product of Equation (84) with T, we have

$$\begin{array}{ccc}\hfill \int \mathcal{K}cos\ddot{\Theta }ds& = & {\mathbf{e}}_{\mathbf{1}}·\mathbf{T}\hfill \\ & = & (1,0,0)·(t_1,t_2,t_3)\hfill \\ & = & t_1.\hfill \end{array}$$

Taking the dot product of Equation (84) with B, we have

$$\begin{array}{ccc}\hfill -\int \mathfrak{T}cos\ddot{\Theta }ds& = & {\mathbf{e}}_{\mathbf{1}}·\mathbf{B}\hfill \\ & = & (1,0,0)·(b_1,b_2,b_3)\hfill \\ & = & b_1.\hfill \end{array}$$

We substitute $cos\ddot{\Theta } = n_1$, $\int \mathcal{K}cos\ddot{\Theta }ds = t_1$, and $-\int \mathfrak{T}ds = b_1$ in Equation (84), and we obtain

$${\mathbf{e}}_{\mathbf{1}} = t_1\mathbf{T}+n_1\mathbf{N}+b_1\mathbf{B}.$$

Similarly,

$${\mathbf{e}}_{\mathbf{2}} = t_2\mathbf{T}+n_2\mathbf{N}+b_2\mathbf{B},$$

$${\mathbf{e}}_{\mathbf{3}} = t_3\mathbf{T}+n_3\mathbf{N}+b_3\mathbf{B}.$$

□

Definition 6. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a regular parametrized curve. We define a vector ${\mathbf{V}}_i = (t_i,n_i,b_i)$, where $i = 1,2,3$; and

$$\begin{array}{ccc}\hfill \mathbf{T}& = & (t_1,t_2,t_3),\hfill \\ \hfill \mathbf{N}& = & (n_1,n_2,n_3),\hfill \\ \hfill \mathbf{B}& = & (b_1,b_2,b_3).\hfill \end{array}$$

Theorem 28. 

$\{{\mathbf{V}}_{\mathbf{1}},{\mathbf{V}}_{\mathbf{2}},{\mathbf{V}}_{\mathbf{3}}\}$ is an orthonormal basis in ${\mathbb{R}}^3.$

Proof. 

Since ${\mathbf{e}}_{\mathbf{1}} = (1,0,0)$ is a constant unit vector in ${\mathbb{R}}^3$, then, by Lemma 1, we have

$$\begin{array}{ccc}\hfill {\mathbf{e}}_{\mathbf{1}}& = & t_1\mathbf{T}+n_1\mathbf{N}+b_1\mathbf{B}\hfill \\ \hfill \parallel {\mathbf{e}}_{\mathbf{1}}\parallel & = & t_1^2+n_1^2+b_1^2\hfill \\ \hfill 1& = & t_1^2+n_1^2+b_1^2\hfill \\ \hfill 1& = & \parallel {\mathbf{V}}_1\parallel .\hfill \end{array}$$

Similarly, $\parallel {\mathbf{V}}_2\parallel = 1$ and $\parallel {\mathbf{V}}_3\parallel = 1$. Since ${\mathbf{e}}_1·{\mathbf{e}}_2 = 0$, then,

$$\begin{array}{ccc}\hfill \left(t_1\mathbf{T}+n_1\mathbf{N}+b_1\mathbf{B}\right)·\left(t_2\mathbf{T}+n_2\mathbf{N}+b_2\mathbf{B}\right)& = & 0\hfill \\ \hfill t_1{t}_2+n_1{n}_2+b_1{b}_2& = & 0\hfill \\ \hfill (t_1,n_1,b_1)·(t_2,n_2,b_2)& = & 0\hfill \\ \hfill {\mathbf{V}}_1·{\mathbf{V}}_2& = & 0.\hfill \end{array}$$

Similarly, ${\mathbf{V}}_1·{\mathbf{V}}_3 = 0$ and ${\mathbf{V}}_2·{\mathbf{V}}_3 = 0.$ □

Theorem 29. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a regular parametrized curve. The vectors ${\mathbf{V}}_{\mathbf{1}}$,${\mathbf{V}}_{\mathbf{2}}\mathbf{0}$, and ${\mathbf{V}}_{\mathbf{3}}$ are related by ${\mathbf{V}}_{\mathbf{1}}\times {\mathbf{V}}_{\mathbf{2}} = {\mathbf{V}}_{\mathbf{3}}$, ${\mathbf{V}}_{\mathbf{2}}\times {\mathbf{V}}_{\mathbf{3}} = {\mathbf{V}}_{\mathbf{1}}$, and ${\mathbf{V}}_{\mathbf{3}}\times {\mathbf{V}}_{\mathbf{1}} = {\mathbf{V}}_{\mathbf{2}}$.

Proof. 

Since $\{{\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3\}$ is the standard basis in ${\mathbb{R}}^3$, then, by Lemma 1, we have

$$\begin{array}{cc}\hfill {\mathbf{e}}_{\mathbf{3}}& = {\mathbf{e}}_{\mathbf{1}}\times {\mathbf{e}}_{\mathbf{2}}\hfill \\ \hfill t_3\mathbf{T}+n_3\mathbf{N}+b_3\mathbf{B}& = \left(t_1\mathbf{T}+n_1\mathbf{N}+b_1\mathbf{B}\right)\times \left(t_2\mathbf{T}+n_2\mathbf{N}+b_2\mathbf{B}\right)\hfill \\ & = \left|\begin{array}{ccc}\mathbf{T}& \mathbf{N}& \mathbf{B}\\ t_1& n_1& b_1\\ t_2& n_2& b_2\end{array}\right|\hfill \\ & = \left|\begin{array}{cc}{n}_1& b_1\\ n_2& b_2\end{array}\right|\mathbf{T}-\left|\begin{array}{cc}{t}_1& b_1\\ t_2& b_2\end{array}\right|\mathbf{N}+\left|\begin{array}{cc}{t}_1& n_1\\ t_2& n_2\end{array}\right|\mathbf{B}\hfill \\ & = (n_1{b}_2-n_2{b}_1)\mathbf{T}-(t_1{b}_2-t_2{b}_1)\mathbf{N}+(t_1{n}_2-t_2{n}_1)\mathbf{B}.\hfill \end{array}$$

By equating the coefficients, we obtain

$$\left.\begin{array}{c}{\displaystyle t_3 = n_1{b}_2-n_2{b}_1}\hfill \\ {\displaystyle n_3 = t_2{b}_1-t_1{b}_2}\hfill \\ {\displaystyle b_3 = t_1{n}_2-t_2{n}_1.}\hfill \end{array}\right\}$$

(85)

Moreover,

$$\begin{array}{cc}\hfill {\mathbf{e}}_{\mathbf{2}}& = {\mathbf{e}}_{\mathbf{3}}\times {\mathbf{e}}_{\mathbf{1}}\hfill \\ \hfill t_2\mathbf{T}+n_2\mathbf{N}+b_2\mathbf{B}& = \left(t_3\mathbf{T}+n_3\mathbf{N}+b_3\mathbf{B}\right)\times \left(t_1\mathbf{T}+n_1\mathbf{N}+b_1\mathbf{B}\right)\hfill \\ & = \left|\begin{array}{ccc}\mathbf{T}& \mathbf{N}& \mathbf{B}\\ t_3& n_3& b_3\\ t_1& n_1& b_1\end{array}\right|\hfill \\ & = \left|\begin{array}{cc}{n}_3& b_3\\ n_1& b_1\end{array}\right|\mathbf{T}-\left|\begin{array}{cc}{t}_3& b_3\\ t_1& b_1\end{array}\right|\mathbf{N}+\left|\begin{array}{cc}{t}_3& n_3\\ t_1& n_1\end{array}\right|\mathbf{B}\hfill \\ & = (n_3{b}_1-n_1{b}_3)\mathbf{T}-(t_3{b}_1-t_1{b}_3)\mathbf{N}+(t_3{n}_1-t_1{n}_3)\mathbf{B}.\hfill \end{array}$$

By equating the coefficients, we obtain

$$\left.\begin{array}{c}{\displaystyle t_2 = n_3{b}_1-n_1{b}_3}\hfill \\ {\displaystyle n_2 = t_1{b}_3-t_3{b}_1}\hfill \\ {\displaystyle b_2 = t_3{n}_1-t_1{n}_3.}\hfill \end{array}\right\}$$

(86)

Similarly,

$$\begin{array}{cc}\hfill {\mathbf{e}}_{\mathbf{1}}& = {\mathbf{e}}_{\mathbf{2}}\times {\mathbf{e}}_{\mathbf{3}}\hfill \\ \hfill t_1\mathbf{T}+n_1\mathbf{N}+b_1\mathbf{B}& = \left(t_2\mathbf{T}+n_2\mathbf{N}+b_2\mathbf{B}\right)\times \left(t_3\mathbf{T}+n_3\mathbf{N}+b_3\mathbf{B}\right)\hfill \\ & = \left|\begin{array}{ccc}\mathbf{T}& \mathbf{N}& \mathbf{B}\\ t_2& n_2& b_2\\ t_3& n_3& b_3\end{array}\right|\hfill \\ & = \left|\begin{array}{cc}{n}_2& b_2\\ n_3& b_3\end{array}\right|\mathbf{T}-\left|\begin{array}{cc}{t}_2& b_2\\ t_3& b_3\end{array}\right|\mathbf{N}+\left|\begin{array}{cc}{t}_2& n_2\\ t_3& n_3\end{array}\right|\mathbf{B}\hfill \\ & = (n_2{b}_3-n_3{b}_2)\mathbf{T}-(t_2{b}_3-t_3{b}_2)\mathbf{N}+(t_2{n}_3-t_3{n}_2)\mathbf{B}.\hfill \end{array}$$

Thus,

$$\left.\begin{array}{c}{\displaystyle t_1 = n_2{b}_3-n_3{b}_2}\hfill \\ {\displaystyle n_1 = t_3{b}_2-t_2{b}_3}\hfill \\ {\displaystyle b_1 = t_2{n}_3-t_3{n}_2.}\hfill \end{array}\right\}$$

(87)

Now, we want to calculate ${\mathbf{V}}_{\mathbf{1}}\times {\mathbf{V}}_{\mathbf{2}}$.

$$\begin{array}{cc}\hfill {\mathbf{V}}_{\mathbf{1}}\times {\mathbf{V}}_{\mathbf{2}}& = \left(t_1,n_1,b_1\right)\times \left(t_2,n_2,b_2\right)\hfill \\ & = \left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ t_1& n_1& b_1\\ t_2& n_2& b_2\end{array}\right|\hfill \\ & = \left|\begin{array}{cc}{n}_1& b_1\\ n_2& b_2\end{array}\right|\mathbf{i}-\left|\begin{array}{cc}{t}_1& b_1\\ t_2& b_2\end{array}\right|\mathbf{j}+\left|\begin{array}{cc}{t}_1& n_1\\ t_2& n_2\end{array}\right|\mathbf{k},\hfill \end{array}$$

which implies that

$${\mathbf{V}}_{\mathbf{1}}\times {\mathbf{V}}_{\mathbf{2}} = (n_1{b}_2-n_2{b}_1)\mathbf{i}-(t_1{b}_2-t_2{b}_1)\mathbf{j}+(t_1{n}_2-t_2{n}_1)\mathbf{k}.$$

(88)

We substitute Equation (85) in Equation (88), and we obtain

$$\begin{array}{cc}\hfill {\mathbf{V}}_{\mathbf{1}}\times {\mathbf{V}}_{\mathbf{2}}& = t_3\mathbf{i}+n_3\mathbf{j}+b_3\mathbf{k}\hfill \\ & = (t_3,n_3,b_3)\hfill \\ & = {\mathbf{V}}_{\mathbf{3}}.\hfill \end{array}$$

In a similar manner, we obtain ${\mathbf{V}}_{\mathbf{3}}\times {\mathbf{V}}_{\mathbf{1}} = {\mathbf{V}}_{\mathbf{2}}$ and ${\mathbf{V}}_{\mathbf{2}}\times {\mathbf{V}}_{\mathbf{3}} = {\mathbf{V}}_{\mathbf{1}}$. □

Definition 7. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a regular parametrized curve and  D  = $(D_1,D_2,D_3)$ be the Darboux vector field associated with $\ddot{\mho }$. Then,

$$k_i = D_i,i = 1,2,3$$

(89)

where $k_i$ represents the new curvatures associated with the new frame $\{{\mathbf{V}}_{\mathbf{1}},{\mathbf{V}}_{\mathbf{2}},{\mathbf{V}}_{\mathbf{3}}\}$ of $\ddot{\mho }$.

Theorem 30. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve and  D =$(D_1,D_2,D_3)$ as the Darboux vector field associated with $\ddot{\mho }$. Then,

$$\sum _{i = 1}^3{k}_i^2 = {\mathcal{K}}^2+{\mathfrak{T}}^2.$$

(90)

Proof. 

Let $\mathbf{D}$ be the Darboux vector field associated with $\ddot{\mho }$, then,

$$\begin{array}{ccc}\hfill \mathbf{D}& = & \mathfrak{T}\mathbf{T}+\mathcal{K}\mathbf{B}\hfill \\ \hfill {\parallel \mathbf{D}\parallel }^2& = & {\mathfrak{T}}^2+{\mathcal{K}}^2\hfill \\ \hfill {D_1}^2+{D_2}^2+{D_3}^2& = & {\mathfrak{T}}^2+{\mathcal{K}}^2.\hfill \end{array}$$

Since $k_i = D_i$ and $i = 1,2,3$, then, we obtain

$$\begin{array}{ccc}\hfill \sum _{i = 1}^3{k}_i^2& = & {\mathfrak{T}}^2+{\mathcal{K}}^2.\hfill \end{array}$$

□

Theorem 31. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve and   D be the Darboux vector field associated with $\ddot{\mho }$. Then,

$$\left.\begin{array}{c}{\displaystyle {\mathbf{V}}_{\mathbf{1}}^{\prime } = -k_3{\mathbf{V}}_{\mathbf{2}}+k_2{\mathbf{V}}_{\mathbf{3}}}\hfill \\ {\displaystyle {\mathbf{V}}_{\mathbf{2}}^{\prime } = k_3{\mathbf{V}}_{\mathbf{1}}-k_1{\mathbf{V}}_{\mathbf{3}}}\hfill \\ {\displaystyle {\mathbf{V}}_{\mathbf{3}}^{\prime } = -k_2{\mathbf{V}}_{\mathbf{1}}+k_1{\mathbf{V}}_{\mathbf{2}}}\hfill \end{array}\right\}.$$

(91)

Proof. 

Let $\mathbf{D}$ be the Darboux vector field associated with $\ddot{\mho }$, then,

$$\begin{array}{ccc}\hfill \mathbf{D}& = & \mathfrak{T}\mathbf{T}+\mathcal{K}\mathbf{B}\hfill \\ \hfill (D_1,D_2,D_3)& = & \mathfrak{T}(t_1,t_2,t_3)+\mathcal{K}(b_1,b_2,b_3)\hfill \\ & = & (\mathfrak{T}{t}_1+\mathcal{K}{b}_1,\mathfrak{T}{t}_2+\mathcal{K}{b}_2,\mathfrak{T}{t}_3+\mathcal{K}{b}_3).\hfill \end{array}$$

By equating the coefficients, we obtain

$$\left.\begin{array}{c}{\displaystyle D_1 = \mathfrak{T}{t}_1+\mathcal{K}{b}_1}\hfill \\ {\displaystyle D_2 = \mathfrak{T}{t}_2+\mathcal{K}{b}_2}\hfill \\ {\displaystyle D_3 = \mathfrak{T}{t}_3+\mathcal{K}{b}_3}\hfill \end{array}\right\}.$$

(92)

Since $k_i = D_i$, $i = 1,2,3$, then, we substitute equations $k_1 = D_1$, $k_2 = D_2$, and $k_3 = D_3$ in Equation (92), and we obtain

$$\left.\begin{array}{c}{\displaystyle k_1 = \mathfrak{T}{t}_1+\mathcal{K}{b}_1}\hfill \\ {\displaystyle k_2 = \mathfrak{T}{t}_2+\mathcal{K}{b}_2}\hfill \\ {\displaystyle k_3 = \mathfrak{T}{t}_3+\mathcal{K}{b}_3}\hfill \end{array}\right\}.$$

(93)

To prove that ${\mathbf{V}}_{\mathbf{1}}^{\prime } = -k_3{\mathbf{V}}_{\mathbf{2}}+k_2{\mathbf{V}}_{\mathbf{3}}$, we use Equation (93), and we have

$$\begin{array}{ccc}\hfill -k_3{\mathbf{V}}_{\mathbf{2}}+k_2{\mathbf{V}}_{\mathbf{3}}& = & -k_3(t_2,n_2,b_2)+k_2(t_3,n_3,b_3)\hfill \\ & = & (-\mathfrak{T}{t}_3-\mathcal{K}{b}_3)(t_2,n_2,b_2)+(\mathfrak{T}{t}_2+\mathcal{K}{b}_2)(t_3,n_3,b_3)\hfill \\ & = & (-\mathcal{K}{b}_3{t}_2+\mathcal{K}{b}_2{t}_3,-\mathfrak{T}{t}_3{n}_2+\mathfrak{T}{t}_2{n}_3-\mathcal{K}{b}_3{n}_2+\mathcal{K}{b}_2{n}_3,-\mathfrak{T}{t}_3{b}_2+\mathfrak{T}{t}_2{b}_3).\hfill \end{array}$$

Moreover,

$$-k_3{\mathbf{V}}_{\mathbf{2}}+k_2{\mathbf{V}}_{\mathbf{3}} = (\mathcal{K}(b_2{t}_3-b_3{t}_2),\mathfrak{T}(t_2{n}_3-t_3{n}_2)-\mathcal{K}(b_3{n}_2-b_2{n}_3),-\mathfrak{T}(t_3{b}_2-t_2{b}_3)).$$

(94)

By substituting Equation (87) in Equation (94), we obtain

$$\begin{array}{ccc}\hfill -k_3{\mathbf{V}}_{\mathbf{2}}+k_2{\mathbf{V}}_{\mathbf{3}}& = & (\mathcal{K}{n}_1,\mathfrak{T}{b}_1-\mathcal{K}{t}_1,-\mathfrak{T}{n}_1)\hfill \\ & = & (t_1^{\prime },n_1^{\prime },b_1^{\prime })\hfill \\ & = & {\mathbf{V}}_{\mathbf{1}}^{\prime }.\hfill \end{array}$$

To prove that ${\mathbf{V}}_{\mathbf{2}}^{\prime } = k_3{\mathbf{V}}_{\mathbf{1}}-k_1{\mathbf{V}}_{\mathbf{3}}$, we use Equation (93), and have

$$\begin{array}{ccc}\hfill k_3{\mathbf{V}}_{\mathbf{1}}-k_1{\mathbf{V}}_{\mathbf{3}}& = & k_3(t_1,n_1,b_1)-k_1(t_3,n_3,b_3)\hfill \\ & = & (\mathfrak{T}{t}_3+\mathcal{K}{b}_3)(t_1,n_1,b_1)-(\mathfrak{T}{t}_1+\mathcal{K}{b}_1)(t_3,n_3,b_3)\hfill \\ & = & (\mathcal{K}{b}_3{t}_1-\mathcal{K}{b}_1{t}_3,\mathfrak{T}{t}_3{n}_1-\mathfrak{T}{t}_1{n}_3-\mathcal{K}{b}_1{n}_3+\mathcal{K}{b}_3{n}_1,\mathfrak{T}{t}_3{b}_1-\mathfrak{T}{t}_1{b}_3).\hfill \end{array}$$

Therefore,

$$k_3{\mathbf{V}}_{\mathbf{1}}-k_1{\mathbf{V}}_{\mathbf{3}} = (\mathcal{K}(b_3{t}_1-b_1{t}_3),\mathfrak{T}(t_3{n}_1-t_1{n}_3)-\mathcal{K}(b_1{n}_3-b_3{n}_1),-\mathfrak{T}(t_1{b}_3-t_3{b}_1)).$$

(95)

By substituting Equation (86) in Equation (95), we obtain

$$\begin{array}{ccc}\hfill k_3{\mathbf{V}}_{\mathbf{1}}-k_1{\mathbf{V}}_{\mathbf{3}}& = & (\mathcal{K}{n}_2,\mathfrak{T}{b}_2-\mathcal{K}{t}_2,-\mathfrak{T}{n}_2)\hfill \\ & = & (t_2^{\prime },n_2^{\prime },b_2^{\prime })\hfill \\ & = & {\mathbf{V}}_{\mathbf{2}}^{\prime }.\hfill \end{array}$$

To prove that ${\mathbf{V}}_{\mathbf{2}}^{\prime } = -k_2{\mathbf{V}}_{\mathbf{1}}+k_1{\mathbf{V}}_{\mathbf{2}}$, we use Equation (93), and have

$$\begin{array}{ccc}\hfill -k_2{\mathbf{V}}_{\mathbf{1}}+k_1{\mathbf{V}}_{\mathbf{2}}& = & -k_2(t_1,n_1,b_1)+k_1(t_2,n_2,b_2)\hfill \\ & = & (-\mathfrak{T}{t}_2-\mathcal{K}{b}_2)(t_1,n_1,b_1)+(\mathfrak{T}{t}_1+\mathcal{K}{b}_1)(t_2,n_2,b_2)\hfill \\ & = & (-\mathcal{K}{b}_2{t}_1+\mathcal{K}{b}_1{t}_2,-\mathfrak{T}{t}_2{n}_1+\mathfrak{T}{t}_1{n}_2-\mathcal{K}{b}_2{n}_1+\mathcal{K}{b}_1{n}_2,-\mathfrak{T}{t}_2{b}_1+\mathfrak{T}{t}_1{b}_2).\hfill \end{array}$$

Thus,

$$-k_2{\mathbf{V}}_{\mathbf{1}}+k_1{\mathbf{V}}_{\mathbf{2}} = (\mathcal{K}(b_1{t}_2-b_2{t}_1),\mathfrak{T}(t_1{n}_2-t_2{n}_1)-\mathcal{K}(b_2{n}_1-b_1{n}_2),-\mathfrak{T}(t_2{b}_1-t_1{b}_2)).$$

(96)

By substituting Equation (85) in Equation (96), we obtain

$$\begin{array}{ccc}\hfill -k_2{\mathbf{V}}_{\mathbf{1}}+k_1{\mathbf{V}}_{\mathbf{2}}& = & (\mathcal{K}{n}_3,\mathfrak{T}b3-\mathcal{K}{t}_3,-\mathfrak{T}{n}_3)\hfill \\ & = & (t_3^{\prime },n_3^{\prime },b_3^{\prime })\hfill \\ & = & {\mathbf{V}}_{\mathbf{3}}^{\prime }.\hfill \end{array}$$

□

Definition 8. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve. The quasi-Darboux vector filed along $\ddot{\mho }$ is defined by

$$\mathcal{W} = \sum _{i = 1}^3{k}_i{\mathbf{V}}_{\mathbf{i}},$$

(97)

where $k_1,k_2$, and $k_3$ are the new curvatures associated with the new frame $\{{\mathbf{V}}_{\mathbf{1}},{\mathbf{V}}_{\mathbf{2}},{\mathbf{V}}_{\mathbf{3}}\}$ of $\ddot{\mho }$.

Theorem 32. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve, then,

$$\left.\begin{array}{c}{\displaystyle {\mathbf{V}}_{\mathbf{1}}^{\prime } = {\mathbf{V}}_{\mathbf{1}}\times \mathcal{W}}\hfill \\ {\displaystyle {\mathbf{V}}_{\mathbf{2}}^{\prime } = {\mathbf{V}}_{\mathbf{2}}\times \mathcal{W}}\hfill \\ {\displaystyle {\mathbf{V}}_{\mathbf{3}}^{\prime } = {\mathbf{V}}_{\mathbf{3}}\times \mathcal{W}}\hfill \end{array}\right\},$$

(98)

where $\mathcal{W}$ is the quasi-Darboux vector field associated with $\ddot{\mho }$.

Proof. 

Let $\ddot{\mho }$ be a unit speed curve and $\mathcal{W}$ be the quasi-Draboux vector field associated with $\ddot{\mho }$, then, by Definition 8, we obtain

$$\begin{array}{ccc}\hfill \mathcal{W}& = & k_1{\mathbf{V}}_{\mathbf{1}}+k_2{\mathbf{V}}_{\mathbf{2}}+k_3{\mathbf{V}}_{\mathbf{3}}.\hfill \end{array}$$

(99)

Using Equation (99), Theorems 29 and 31, we obtain

$$\begin{array}{ccc}\hfill {\mathbf{V}}_{\mathbf{1}}\times \mathcal{W}& = & {\mathbf{V}}_{\mathbf{1}}\times (k_1{\mathbf{V}}_{\mathbf{1}}+k_2{\mathbf{V}}_{\mathbf{2}}+k_3{\mathbf{V}}_{\mathbf{3}})\hfill \\ & = & k_1({\mathbf{V}}_{\mathbf{1}}\times {\mathbf{V}}_{\mathbf{1}})+k_2({\mathbf{V}}_{\mathbf{1}}\times {\mathbf{V}}_{\mathbf{2}})+k_3({\mathbf{V}}_{\mathbf{1}}\times {\mathbf{V}}_{\mathbf{3}}).\hfill \\ & = & k_2{\mathbf{V}}_{\mathbf{3}}-k_3{\mathbf{V}}_{\mathbf{2}} = {\mathbf{V}}_{\mathbf{1}}^{\prime }\hfill \\ \hfill {\mathbf{V}}_{\mathbf{2}}\times \mathcal{W}& = & {\mathbf{V}}_{\mathbf{2}}\times (k_1{\mathbf{V}}_{\mathbf{1}}+k_2{\mathbf{V}}_{\mathbf{2}}+k_3{\mathbf{V}}_{\mathbf{3}})\hfill \\ & = & k_1({\mathbf{V}}_{\mathbf{2}}\times {\mathbf{V}}_{\mathbf{1}})+k_2({\mathbf{V}}_{\mathbf{2}}\times {\mathbf{V}}_{\mathbf{2}})+k_3({\mathbf{V}}_{\mathbf{2}}\times {\mathbf{V}}_{\mathbf{3}})\hfill \\ & = & -k_1{\mathbf{V}}_{\mathbf{3}}+k_3{\mathbf{V}}_{\mathbf{1}}\hfill \\ & = & {\mathbf{V}}_{\mathbf{2}}^{\prime }.\hfill \\ \hfill {\mathbf{V}}_{\mathbf{3}}\times \mathcal{W}& = & {\mathbf{V}}_{\mathbf{3}}\times (k_1{\mathbf{V}}_{\mathbf{1}}+k_2{\mathbf{V}}_{\mathbf{2}}+k_3{\mathbf{V}}_{\mathbf{3}})\hfill \\ & = & k_1({\mathbf{V}}_{\mathbf{3}}\times {\mathbf{V}}_{\mathbf{1}})+k_2({\mathbf{V}}_{\mathbf{3}}\times {\mathbf{V}}_{\mathbf{2}})+k_3({\mathbf{V}}_{\mathbf{3}}\times {\mathbf{V}}_{\mathbf{3}})\hfill \\ & = & k_1{\mathbf{V}}_{\mathbf{2}}-k_2{\mathbf{V}}_{\mathbf{1}}\hfill \\ & = & {\mathbf{V}}_{\mathbf{3}}^{\prime }.\hfill \end{array}$$

□

Theorem 33. 

Let $\ddot{\mho }(s) = \left(acos\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),asin\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),{\displaystyle \frac{bs}{\sqrt{a^2+b^2}}}\right)$ be a unit speed circular helix. Then, $k_1 = k_2 = 0$ but $k_3\ne 0$, where $a>0$ and $b\ne 0$ are constants.

Proof. 

Suppose that $\ddot{\mho }(s)$ is given by

$$\ddot{\mho }(s) = \left(acos\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),asin\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),{\displaystyle \frac{bs}{\sqrt{a^2+b^2}}}\right).$$

Then,

$$\begin{array}{ccc}\hfill {\ddot{\mho }}^{\prime }(s)& = & \left({\displaystyle \frac{-a}{\sqrt{a^2+b^2}}}sin\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),{\displaystyle \frac{a}{\sqrt{a^2+b^2}}}cos\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),{\displaystyle \frac{b}{\sqrt{a^2+b^2}}}\right),\hfill \\ \hfill {\ddot{\mho }}^{″}(s)& = & \left({\displaystyle \frac{-a}{a^2+b^2}}cos\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),{\displaystyle \frac{-a}{a^2+b^2}}sin\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),0\right),\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill {\ddot{\mho }}^{‴}(s)& = & \left({\displaystyle \frac{a}{(a^2+b^2)\sqrt{a^2+b^2}}}sin\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),{\displaystyle \frac{-a}{(a^2+b^2)\sqrt{a^2+b^2}}}cos\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),0\right).\hfill \end{array}$$

Since $\ddot{\mho }$ is a unit speed curve, then, $\parallel {\ddot{\mho }}^{\prime }(s)\parallel = \parallel \mathbf{T}\parallel = 1$ and ${\mathbf{T}}^{\prime } = {\ddot{\mho }}^{″}(s)$. Moreover,

$$\parallel {\mathbf{T}}^{\prime }\parallel = {\displaystyle \frac{a}{a^2+b^2}}.$$

Now,

$$\begin{array}{ccc}\hfill \mathbf{N}& = & {\displaystyle \frac{{\mathbf{T}}^{\prime }}{\parallel {\mathbf{T}}^{\prime }\parallel }} = \left(-cos\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),-sin\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),0\right),\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \mathbf{B}& = & \mathbf{T}\times \mathbf{N} = \left(bsin\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),-{\displaystyle \frac{b}{\sqrt{a^2+b^2}}}cos\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),-{\displaystyle \frac{a}{\sqrt{a^2+b^2}}}\right).\hfill \end{array}$$

Recall that

$$\begin{array}{ccc}\hfill {\mathbf{V}}_{\mathbf{1}}& = & \left(t_1,n_1,b_1\right)\hfill \\ & = & \left(-{\displaystyle \frac{a}{\sqrt{a^2+b^2}}}sin\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),-cos\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),bsin\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right)\right)\hfill \\ \hfill {\mathbf{V}}_{\mathbf{2}}& = & \left(t_2,n_2,b_2\right)\hfill \\ & = & \left({\displaystyle \frac{a}{\sqrt{a^2+b^2}}}cos\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),-sin\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),-{\displaystyle \frac{b}{\sqrt{a^2+b^2}}}cos\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right)\right)\hfill \\ \hfill {\mathbf{V}}_{\mathbf{3}}& = & \left(t_3,n_3,b_3\right)\hfill \\ & = & \left({\displaystyle \frac{b}{\sqrt{a^2+b^2}}},0,{\displaystyle \frac{-a}{\sqrt{a^2+b^2}}}\right).\hfill \end{array}$$

Since ${\mathbf{V}}_{\mathbf{3}}$ is a constant vector, then, ${\mathbf{V}}_{\mathbf{3}}^{\prime } = 0$. From Theorem 31 and ${\mathbf{V}}_{\mathbf{3}}^{\prime } = 0$, we obtain $-k_2{\mathbf{V}}_{\mathbf{1}}+k_1{\mathbf{V}}_{\mathbf{2}} = 0$. Since ${\mathbf{V}}_{\mathbf{1}}$ and ${\mathbf{V}}_{\mathbf{2}}$ are linearly independent, then, $k_1 = k_2 = 0$.

To prove that $k_3\ne 0$, we calculate ${\ddot{\mho }}^{\prime }\times {\ddot{\mho }}^{″}$, $\parallel {\ddot{\mho }}^{\prime }\times {\ddot{\mho }}^{″}{\parallel }^2$ and $({\ddot{\mho }}^{\prime }\times {\ddot{\mho }}^{″})·{\ddot{\mho }}^{‴}$, and we obtain

$$\begin{array}{cc}\hfill {\ddot{\mho }}^{\prime }\times {\ddot{\mho }}^{″}& = {\displaystyle \frac{ab}{{(a^2+b^2)}^{\frac{3}{2}}}}sin\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right)\mathbf{i}-{\displaystyle \frac{ab}{{(a^2+b^2)}^{\frac{3}{2}}}}cos\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right)\mathbf{j}+{\displaystyle \frac{a^2}{{(a^2+b^2)}^{\frac{3}{2}}}}\mathbf{k}\hfill \\ \\ \hfill \parallel {\ddot{\mho }}^{\prime }\times {\ddot{\mho }}^{″}{\parallel }^2& = {\displaystyle \frac{a^2{b}^2}{{(a^2+b^2)}^3}}+{\displaystyle \frac{a^4}{{(a^2+b^2)}^3}}\hfill \\ & = {\displaystyle \frac{a^2}{{(a^2+b^2)}^2}}\hfill \\ \hfill ({\ddot{\mho }}^{\prime }\times {\ddot{\mho }}^{″})·{\ddot{\mho }}^{‴}& = {\displaystyle \frac{a^{2}b}{{(a^2+b^2)}^3}}.\hfill \end{array}$$

Since $\mathfrak{T} = {\displaystyle \frac{({\ddot{\mho }}^{\prime }\times {\ddot{\mho }}^{″})·{\ddot{\mho }}^{‴}}{\parallel {\ddot{\mho }}^{\prime }\times {\ddot{\mho }}^{″}{\parallel }^2}}$ and $\mathcal{K} = {\displaystyle \frac{\parallel {\mathbf{T}}_{\ddot{\mho }}^{\prime }\parallel }{\parallel {\ddot{\mho }}^{\prime }\parallel }}$, then, $\mathfrak{T} = {\displaystyle \frac{b}{a^2+b^2}}$ and $\mathcal{K} = {\displaystyle \frac{a}{a^2+b^2}}$.

$k_1 = k_2 = 0$ and $k_1^2+k_2^2+k_3^2 = {\mathcal{K}}^2+{\mathfrak{T}}^2$, then,

$$\begin{array}{ccc}\hfill k_3^2& = & {\displaystyle \frac{a^2}{{(a^2+b^2)}^2}}+{\displaystyle \frac{b^2}{{(a^2+b^2)}^2}} = {\displaystyle \frac{a^2+b^2}{{(a^2+b^2)}^2}} = {\displaystyle \frac{1}{a^2+b^2}}\hfill \\ \hfill k_3& = & {\displaystyle \frac{1}{\sqrt{a^2+b^2}}}.\hfill \end{array}$$

Thus, $k_3$ is a constant and $k_3\ne 0.$ □

Theorem 34. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve. If $\ddot{\mho }$ is a plane curve, then, ${\mathbf{e}}_{\mathbf{3}}·{\mathbf{V}}_{\mathbf{i}}^{\prime } = \mathbf{0}$, where $\mathbf{i} = 1,2,3$.

Proof. 

Since $\ddot{\mho }$ is a plane curve and ${\mathbf{B}}^{\prime } = -\mathfrak{T}\mathbf{N}$, that means $B^{\prime } = (b_1^{\prime },b_2^{\prime },b_3^{\prime }) = (0,0,0)$, which implies that $b_1$, $b_2$, and $b_3$ are constants. Hence, $\mathbf{B}$ is constant vector. Now, we calculate ${\mathbf{e}}_{\mathbf{3}}·{\mathbf{V}}_{\mathbf{1}}^{\prime }$, and we obtain

$$\begin{array}{ccc}\hfill {\mathbf{e}}_{\mathbf{3}}·{\mathbf{V}}_{\mathbf{1}}^{\prime }& = & (0,0,1)·(t_1^{\prime },n_1^{\prime },b_1^{\prime })\hfill \\ & = & b_1^{\prime }\hfill \\ & = & 0\hfill \end{array}$$

Similar calculations imply ${\mathbf{e}}_{\mathbf{3}}·{\mathbf{V}}_{\mathbf{2}}^{\prime } = 0$ and ${\mathbf{e}}_{\mathbf{3}}·{\mathbf{V}}_{\mathbf{3}}^{\prime } = 0.$ □

Theorem 35. 

Let $\ddot{\mho }:I\to {\mathbb{R}}^3$ be a unit speed curve. If $\ddot{\mho }$ is a slant helix, then,

$$\ddot{\mho } = \left(-\int \left(\int {\displaystyle \frac{k_1\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}ds\right)ds,-\int \left(\int {\displaystyle \frac{k_2\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}ds\right)ds,\int \left(\int c\mathcal{K}ds\right)ds\right).$$

Proof. 

Let $\ddot{\mho } = ({\ddot{\mho }}_1,{\ddot{\mho }}_2,{\ddot{\mho }}_3)$ be a proper slant helix. Without loss of generality, we may assume that the axis of the proper slant helix is ${\mathbf{e}}_{\mathbf{3}} = (0,0,1)$, and we aim to compute the parametrization of $\ddot{\mho }$. We know that

$$\begin{array}{ccc}\hfill {\mathbf{e}}_{\mathbf{3}}·\mathbf{N}& = & c\hfill \\ \hfill (0,0,1)·(n_1,n_2,n_3)& = & c,\hfill \end{array}$$

where c is a constant. So that

$$n_3 = c.$$

(100)

Since $\mathcal{K}\ne 0$, multiplying Equation (100) by $\mathcal{K}$, we obtain

$$\begin{array}{ccc}\hfill \mathcal{K}{n}_3& = & c\mathcal{K}\hfill \\ \hfill t_3^{\prime }& = & c\mathcal{K}\hfill \\ \hfill \int t_3^{\prime }ds& = & \int c\mathcal{K}ds,\hfill \end{array}$$

which implies that

$$t_3 = \int c\mathcal{K}ds.$$

(101)

We substitute Equation (101) in ${\ddot{\mho }}_3 = \int t_{3}ds$ and obtain

$$\begin{array}{ccc}\hfill {\ddot{\mho }}_3& = & \int \left(\int c\mathcal{K}ds\right)ds.\hfill \end{array}$$

(102)

Since ${\mathbf{e}}_{\mathbf{1}}·{\mathbf{e}}_{\mathbf{3}} = 0$ and ${\mathbf{e}}_{\mathbf{1}}·\mathbf{N} = n_1$, then, from Theorem 22, we obtain

$$\begin{array}{ccc}\hfill n_1& = & {\mathbf{e}}_{\mathbf{1}}·\mathbf{N}\hfill \\ & = & -{\displaystyle \frac{1}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}{\mathbf{e}}_{\mathbf{1}}·\mathbf{D},\hfill \end{array}$$

which implies that

$$n_1 = -{\displaystyle \frac{k_1}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}.$$

(103)

Since $\mathcal{K}\ne 0$, then, multiplying Equation (103) by $\mathcal{K}$, we have

$$\begin{array}{ccc}\hfill \mathcal{K}{n}_1& = & -{\displaystyle \frac{k_1\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\hfill \\ \hfill t_1^{\prime }& = & -{\displaystyle \frac{k_1\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}\hfill \\ \hfill \int t_1^{\prime }ds& = & -\int {\displaystyle \frac{k_1\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}ds.\hfill \end{array}$$

Thus,

$$t_1 = -\int {\displaystyle \frac{k_1\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}ds.$$

(104)

Substituting Equation (104) in ${\ddot{\mho }}_1 = \int t_{1}ds$, we obtain

$$\begin{array}{ccc}\hfill {\ddot{\mho }}_1& = & -\int \left(\int {\displaystyle \frac{k_1\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}ds\right)ds.\hfill \end{array}$$

(105)

Similarly,

$$\begin{array}{ccc}\hfill {\ddot{\mho }}_2& = & -\int \left(\int {\displaystyle \frac{k_2\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}ds\right)ds.\hfill \end{array}$$

(106)

Therefore, from Equations (102), (105), and (106), we obtain

$$\ddot{\mho } = \left(-\int \left(\int {\displaystyle \frac{k_1\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}ds\right)ds,-\int \left(\int {\displaystyle \frac{k_2\mathcal{K}}{{\mathfrak{T}}_{\ddot{{\rm Y}}}}}ds\right)ds,\int \left(\int c\mathcal{K}ds\right)ds\right).$$

(107)

□

5. Examples

In this section, we give some examples to illustrate some of our results in this paper.

Example 1. 

In this example, we consider the circular helix $\ddot{\mho }(t) = \left(cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right),sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right),{\displaystyle \frac{t}{\sqrt{2}}}\right)$ to illustrate the results of Lemma 1 and Theorems 28 and 29.

• Now,

$${\ddot{\mho }}^{\prime }(t) = \left(-{\displaystyle \frac{1}{\sqrt{2}}}sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right),{\displaystyle \frac{1}{\sqrt{2}}}cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right),{\displaystyle \frac{1}{\sqrt{2}}}\right).$$

The norm of ${\ddot{\mho }}^{\prime }(t)$ is given by

$$\parallel {\ddot{\mho }}^{\prime }(t)\parallel = \sqrt{{\displaystyle \frac{1}{2}}\left({cos}^2\left({\displaystyle \frac{t}{\sqrt{2}}}\right)+{sin}^2\left({\displaystyle \frac{t}{\sqrt{2}}}\right)+1\right)} = 1.$$

The unit tangent vector of $\ddot{\mho }$ is given by

$$\mathbf{T}(\mathbf{t}) = {\displaystyle \frac{{\ddot{\mho }}^{\prime }(t)}{\parallel {\ddot{\mho }}^{\prime }(t)\parallel }} = {\ddot{\mho }}^{\prime }(t) = \left(-{\displaystyle \frac{1}{\sqrt{2}}}sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right),{\displaystyle \frac{1}{\sqrt{2}}}cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right),{\displaystyle \frac{1}{\sqrt{2}}}\right) = \left(t_1,t_2,t_3\right).$$

Moreover,

$${\ddot{\mho }}^{″}(t) = {\mathbf{T}}^{\prime }(\mathbf{t}) = \left(-{\displaystyle \frac{1}{2}}cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right),-{\displaystyle \frac{1}{2}}sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right),0\right)$$

$$\parallel {\mathbf{T}}^{\prime }(\mathbf{t})\parallel = \sqrt{{\displaystyle \frac{1}{4}}\left({cos}^2\left({\displaystyle \frac{t}{\sqrt{2}}}\right)+{sin}^2\left({\displaystyle \frac{t}{\sqrt{2}}}\right)\right)} = {\displaystyle \frac{1}{2}}.$$

The unit normal vector of $\ddot{\mho }(t)$ is given by

$$\mathbf{N}(\mathbf{t}) = {\displaystyle \frac{{\mathbf{T}}^{\prime }(\mathbf{t})}{\parallel {\mathbf{T}}^{\prime }(\mathbf{t})\parallel }} = \left(-cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right),-sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right),0\right) = \left(n_1,n_2,n_3\right).$$

The unit binormal vector of $\ddot{\mho }(t)$ is given by

$$\mathbf{B}(\mathbf{t}) = \mathbf{T}(\mathbf{t})\times \mathbf{N}(\mathbf{t}) = \left({\displaystyle \frac{1}{\sqrt{2}}}sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right),-{\displaystyle \frac{1}{\sqrt{2}}}cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right),{\displaystyle \frac{1}{\sqrt{2}}}\right) = \left(b_1,b_2,b_3\right)$$

Thus,

$${\mathbf{V}}_{\mathbf{1}} = \left(t_1,n_1,b_1\right) = \left(-{\displaystyle \frac{1}{\sqrt{2}}}sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right),-cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right),{\displaystyle \frac{1}{\sqrt{2}}}sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right)\right)$$

$$\parallel {\mathbf{V}}_{\mathbf{1}}\parallel = \sqrt{{\displaystyle \frac{1}{2}}{sin}^2\left({\displaystyle \frac{t}{\sqrt{2}}}\right)+{cos}^2\left({\displaystyle \frac{t}{\sqrt{2}}}\right)+{\displaystyle \frac{1}{2}}{sin}^2\left({\displaystyle \frac{t}{\sqrt{2}}}\right)} = \sqrt{{sin}^2\left({\displaystyle \frac{t}{\sqrt{2}}}\right)+{cos}^2\left({\displaystyle \frac{t}{\sqrt{2}}}\right)} = 1$$

$${\mathbf{V}}_{\mathbf{2}} = \left(t_2,n_2,b_2\right) = \left({\displaystyle \frac{1}{\sqrt{2}}}cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right),-sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right),-{\displaystyle \frac{1}{\sqrt{2}}}cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right)\right)$$

$$\parallel {\mathbf{V}}_{\mathbf{2}}\parallel = \sqrt{{\displaystyle \frac{1}{2}}{cos}^2\left({\displaystyle \frac{t}{\sqrt{2}}}\right)+{sin}^2\left({\displaystyle \frac{t}{\sqrt{2}}}\right)+{\displaystyle \frac{1}{2}}{cos}^2\left({\displaystyle \frac{t}{\sqrt{2}}}\right)} = \sqrt{{sin}^2\left({\displaystyle \frac{t}{\sqrt{2}}}\right)+{cos}^2\left({\displaystyle \frac{t}{\sqrt{2}}}\right)} = 1$$

• ${\mathbf{V}}_{\mathbf{3}} = \left(t_3,n_3,b_3\right) = \left({\displaystyle \frac{1}{\sqrt{2}}},0,{\displaystyle \frac{1}{\sqrt{2}}}\right)$
• $\parallel {\mathbf{V}}_{\mathbf{3}}\parallel = \sqrt{{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}} = 1.$

The dot product of ${\mathbf{V}}_{\mathbf{1}}$ and ${\mathbf{V}}_{\mathbf{2}}$ gives

$${\mathbf{V}}_{\mathbf{1}}·{\mathbf{V}}_{\mathbf{2}} = -{\displaystyle \frac{1}{2}}sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right)cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right)+sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right)cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right)-{\displaystyle \frac{1}{2}}sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right)cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right) = 0,$$

and similarly, ${\mathbf{V}}_{\mathbf{2}}·{\mathbf{V}}_{\mathbf{3}} = {\mathbf{V}}_{\mathbf{1}}·{\mathbf{V}}_{\mathbf{3}} = 0$.

• Now, we calculate ${\mathbf{V}}_{\mathbf{1}}\times {\mathbf{V}}_{\mathbf{2}}$ and obtain

$$\begin{array}{cc}\hfill {\mathbf{V}}_{\mathbf{1}}\times {\mathbf{V}}_{\mathbf{2}}& = \left({\displaystyle \frac{1}{\sqrt{2}}},0,{\displaystyle \frac{1}{\sqrt{2}}}\right) = {\mathbf{V}}_{\mathbf{3}}.\hfill \end{array}$$

Similarly, ${\mathbf{V}}_{\mathbf{2}}\times {\mathbf{V}}_{\mathbf{3}} = {\mathbf{V}}_{\mathbf{1}}$ and ${\mathbf{V}}_{\mathbf{3}}\times {\mathbf{V}}_{\mathbf{1}} = {\mathbf{V}}_{\mathbf{2}}$.

• Now, we calculate $t_3\mathbf{T}+n_3\mathbf{N}+b_3\mathbf{B}$ and obtain

$$\begin{array}{ccc}\hfill t_3\mathbf{T}+n_3\mathbf{N}+b_3\mathbf{B}& = & {\displaystyle \frac{1}{\sqrt{2}}}\mathbf{T}+0\mathbf{N}+{\displaystyle \frac{1}{\sqrt{2}}}\mathbf{B}\hfill \\ & = & {\displaystyle \frac{1}{\sqrt{2}}}\mathbf{T}+{\displaystyle \frac{1}{\sqrt{2}}}\mathbf{B}\hfill \\ & = & \left(-{\displaystyle \frac{1}{2}}sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right),{\displaystyle \frac{1}{2}}cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right),{\displaystyle \frac{1}{2}}\right)+\left({\displaystyle \frac{1}{2}}sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right),-{\displaystyle \frac{1}{2}}cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right),{\displaystyle \frac{1}{2}}\right)\hfill \\ & = & \left(0,0,1\right)\hfill \\ & = & {\mathbf{e}}_{\mathbf{3}}.\hfill \end{array}$$

Similarly, ${\mathbf{e}}_{\mathbf{2}} = t_2\mathbf{T}+n_2\mathbf{N}+b_2\mathbf{B}$ and ${\mathbf{e}}_{\mathbf{1}} = t_1\mathbf{T}+n_1\mathbf{N}+b_1\mathbf{B}$.

Example 2. 

In this example, we consider the unit speed circular helix

$$\ddot{\mho }(s) = \left(acos\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),asin\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),{\displaystyle \frac{bs}{\sqrt{a^2+b^2}}}\right)$$

to illustrate the result of Theorem 12. Then, $\mathbf{N}(\mathbf{t}) = \left(-sin\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),-cos\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),0\right)$. Since $\mathbf{U} = \left({\displaystyle \frac{1}{\sqrt{2}}},{\displaystyle \frac{1}{\sqrt{2}}},0\right)$ is a constant unit vector, then,

$$\begin{array}{ccc}\hfill \mathbf{U}·\mathbf{N}& = & \left({\displaystyle \frac{1}{\sqrt{2}}},{\displaystyle \frac{1}{\sqrt{2}}},0\right)·\left(-sin\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),-cos\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right),0\right)\hfill \\ \hfill \parallel \mathbf{U}\parallel \parallel \mathbf{N}\parallel cos\ddot{\Phi }& = & -{\displaystyle \frac{1}{\sqrt{2}}}sin\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right)-{\displaystyle \frac{1}{\sqrt{2}}}cos\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right)\hfill \\ \hfill cos\ddot{\Phi }& = & cos\left({\displaystyle \frac{3\pi }{4}}\right)cos\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right)-sin\left({\displaystyle \frac{3\pi }{4}}\right)cos\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}\right)\hfill \\ & = & cos\left({\displaystyle \frac{s}{\sqrt{a^2+b^2}}}+{\displaystyle \frac{3\pi }{4}}\right)\hfill \\ \hfill \ddot{\Phi }& = & {\displaystyle \frac{s}{\sqrt{a^2+b^2}}}+{\displaystyle \frac{3\pi }{4}}.\hfill \end{array}$$

Since $\mathcal{K} = {\displaystyle \frac{a}{a^2+b^2}}$ and $\mathfrak{T} = {\displaystyle \frac{b}{a^2+b^2}}$, then, ${\mathcal{K}}^2+{\mathfrak{T}}^2 = {\displaystyle \frac{1}{a^2+b^2}}$.

• Therefore, $\ddot{\Phi } = \sqrt{{\mathcal{K}}^2+{\mathfrak{T}}^2}s+{\displaystyle \frac{3\pi }{4}}$ is a linear function.

Example 3. 

In this example, we consider the circular helix $\ddot{\mho }(t) = \left(cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right),sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right),{\displaystyle \frac{t}{\sqrt{2}}}\right)$ to illustrate the result of Theorem 15. Then, $\mathbf{T}(\mathbf{t}) = \left(-{\displaystyle \frac{1}{\sqrt{2}}}sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right),{\displaystyle \frac{1}{\sqrt{2}}}cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right),{\displaystyle \frac{1}{\sqrt{2}}}\right)$ and $\mathbf{B}(\mathbf{t}) = \left({\displaystyle \frac{1}{\sqrt{2}}}sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right),-{\displaystyle \frac{1}{\sqrt{2}}}cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right),{\displaystyle \frac{1}{\sqrt{2}}}\right).$

• Since $\mathbf{U} = \left({\displaystyle \frac{1}{\sqrt{3}}},{\displaystyle \frac{1}{\sqrt{3}}},{\displaystyle \frac{1}{\sqrt{3}}}\right)$ is a constant unit vector, then,

$$\begin{array}{ccc}\hfill \mathbf{U}·\mathbf{T}& = & \left({\displaystyle \frac{1}{\sqrt{3}}},{\displaystyle \frac{1}{\sqrt{3}}},{\displaystyle \frac{1}{\sqrt{3}}}\right)·\left(-{\displaystyle \frac{1}{\sqrt{2}}}sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right),{\displaystyle \frac{1}{\sqrt{2}}}cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right),{\displaystyle \frac{1}{\sqrt{2}}}\right)\hfill \\ \\ \hfill \parallel \mathbf{U}\parallel \parallel \mathbf{T}\parallel cos\ddot{\Phi }& = & -{\displaystyle \frac{1}{\sqrt{6}}}sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right)+{\displaystyle \frac{1}{\sqrt{6}}}cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right)+{\displaystyle \frac{1}{\sqrt{6}}}.\hfill \end{array}$$

Therefore,

$$cos\ddot{\Phi } = -{\displaystyle \frac{1}{\sqrt{6}}}sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right)+{\displaystyle \frac{1}{\sqrt{6}}}cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right)+{\displaystyle \frac{1}{\sqrt{6}}},$$

(108)

where $\ddot{\Phi }$ is the angle between $\mathbf{U}$ and $\mathbf{T}$.

• Now,

$$\begin{array}{ccc}\hfill \mathbf{U}·\mathbf{B}& = & \left({\displaystyle \frac{1}{\sqrt{3}}},{\displaystyle \frac{1}{\sqrt{3}}},{\displaystyle \frac{1}{\sqrt{3}}}\right)·\left(-{\displaystyle \frac{1}{\sqrt{2}}}sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right),-{\displaystyle \frac{1}{\sqrt{2}}}cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right),{\displaystyle \frac{1}{\sqrt{2}}}\right)\hfill \\ \\ \hfill \parallel \mathbf{U}\parallel \parallel \mathbf{B}\parallel cos\ddot{\Psi }& = & {\displaystyle \frac{1}{\sqrt{6}}}sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right)-{\displaystyle \frac{1}{\sqrt{6}}}cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right)+{\displaystyle \frac{1}{\sqrt{6}}}.\hfill \end{array}$$

Therefore,

$$cos\ddot{\Psi } = {\displaystyle \frac{1}{\sqrt{6}}}sin\left({\displaystyle \frac{t}{\sqrt{2}}}\right)-{\displaystyle \frac{1}{\sqrt{6}}}cos\left({\displaystyle \frac{t}{\sqrt{2}}}\right)+{\displaystyle \frac{1}{\sqrt{6}}},$$

(109)

where $\ddot{\Psi }$ is the angle between $\mathbf{U}$ and $\mathbf{B}$. Adding Equations (108) and (109), we obtain

$$\begin{array}{ccc}\hfill cos\ddot{\Phi }+cos\ddot{\Psi }& = & {\displaystyle \frac{2}{\sqrt{6}}}.\hfill \end{array}$$