© Wuhan University 2024

0 Introduction

We begin with the definition of the regular curve. Throughout this paper, we use ${\mathbb{R}}^n$ to denote the $n$-dimensional Euclidean space with Euclidean metric $|•|$. Let $I\subseteq \mathbb{R}$ be an interval. A continuous vector function $\mathit{r}:I\to {\mathbb{R}}^n,\mathit{r}=\mathit{r}(t)$ is called a regular curve if it is differentiable and ${\mathit{r}}^{\text{'}}(t)\ne \mathrm{0}$ for all $t\in I$, where $t$ is called its parameter. For example, given two linearly independent constant vectors $\mathit{a},\mathit{b}$, the curve $\mathit{r}(t)=\mathit{a}\mathrm{c}\mathrm{o}\mathrm{s}t+\mathit{b}\mathrm{s}\mathrm{i}\mathrm{n}t$ with $t\in (\mathrm{0,2}\mathrm{\pi })$ is regular. In particular, if $|{\mathit{r}}^{\text{'}}(t)|=\mathrm{1}$, $t$ is called arc length parameter^[1,2].

In differential geometry, one of the primary goals is to describe the regular curves using curvature and torsion. The well-known fundamental theorem of curves^[2] claims that the curvature and the torsion are their totally geometric invariants, that is, if two curves share the same curvature and the torsion at every $t\in I$, then these two arcs are congruent. However, it may be not enough to detect geometric information about the curves in practice using this method. For instance, it is necessary to give more precise descriptions of the curves' shape in the study of envelopes^[3], computerized design^[4,5], and so forth. Object shape comparability is a challenging problem in the field of pattern recognition and computer vision^[6]. The method based on geometric or algebraic invariants is the typical method in these applications.

Given an equation of a curve, it can comprehensively determine the shape of the curve. Hyperbolic functions are a class of important functions similar to common trigonometric functions in mathematics and other fields^[7-9]. The basic hyperbolic functions are hyperbolic sine, hyperbolic cosine, which is denoted by "sh, ch", respectively. However, there is few results about the curves defined by hyperbolic functions. So in this paper, we will investigate a class of curves' shape, which are defined by hyperbolic functions $\mathrm{c}\mathrm{h}t$ and $\mathrm{s}\mathrm{h}t$, namely, $\mathit{r}(t)=\mathit{a}\mathrm{c}\mathrm{h}t+\mathit{b}\mathrm{s}\mathrm{h}t+\mathit{c}$ where $\mathit{a},\mathit{b},\mathit{c}$ are three fixed vectors. Our main result is the following theorem.

Theorem 1   Let $\mathit{r}(t)=\mathit{a}\mathrm{c}\mathrm{h}t+\mathit{b}\mathrm{s}\mathrm{h}t+\mathit{c}$ be a regular curve. Then $\mathit{r}(t)$ is planar. Furthermore, when $\mathit{a}//\mathit{b}$, it is a straight line or line segment passing through the fixed point $\mathit{c}$; when $\mathit{a},\mathit{b}$ is linearly independent, it is a hyperbola.

The paper is organized as follows. In Section 1, we give some basic definitions and recall some known facts. The main result, Theorem 1, is proved in Section 2. Section 3 investigates the shape of some generalization of the curve in Theorem 1.

1 Preliminaries

We first recall some definitions and results of analytic geometry and differential geometry, more details see Refs. [1010,1,2].

Let $a{x}^{\mathrm{2}}+\mathrm{2}bxy+c{y}^{\mathrm{2}}+\mathrm{2}dx+\mathrm{2}ey+f=\mathrm{0}$ be a conic, where $a,b,c,d,e,f$ are arbitrary constants, but $a,b,c$ are not all equal to 0. We define $I_{\mathrm{1}},I_{\mathrm{2}},I_{\mathrm{3}}$ as follows:

$I_{\mathrm{1}}=a+c,\hspace{1em}{I}_{\mathrm{2}}=\left|\begin{array}{cc}a& b\\ b& c\end{array}\right|,\hspace{1em}{I}_{\mathrm{3}}=\left|\begin{array}{ccc}a& b& d\\ b& c& e\\ d& e& f\end{array}\right|$

and call them the invariants of the conic.

Proposition 1^[10] Suppose $a{x}^{\mathrm{2}}+\mathrm{2}bxy+c{y}^{\mathrm{2}}+\mathrm{2}dx+\mathrm{2}ey+f=\mathrm{0}$ is a conic. Then the shape of a conic depends on the invariants: if $I_{\mathrm{2}}\mathrm{0},$$I_{\mathrm{3}}\mathrm{0}$, it is an ellipse; if $I_{\mathrm{2}}\mathrm{0},$$I_{\mathrm{3}}\ne \mathrm{0}$, it is a hyperbola; if $I_{\mathrm{2}}=\mathrm{0},$$I_{\mathrm{3}}\ne \mathrm{0}$, it is a parabola.

According to Ref. [7] ,$\mathrm{s}\mathrm{h}t=\frac{{\mathrm{e}}^t-{\mathrm{e}}^{-t}}{\mathrm{2}},\mathrm{c}\mathrm{h}t=\frac{{\mathrm{e}}^t+{\mathrm{e}}^{-t}}{\mathrm{2}}$(See Fig.1, Fig.2).

Fig. 1 The function $\mathbf{s}\mathbf{h}\mathit{t}$

Fig. 2 The function $\mathbf{c}\mathbf{h}\mathit{t}$

Proposition 2^[7] 1) $\mathrm{s}{\mathrm{h}}^{\mathrm{2}}t+\mathrm{1}=\mathrm{c}{\mathrm{h}}^{\mathrm{2}}t$;

2) $\mathrm{s}\mathrm{h}(u+v)=\mathrm{s}\mathrm{h}u\mathrm{c}\mathrm{h}v+\mathrm{c}\mathrm{h}u\mathrm{s}\mathrm{h}v,\mathrm{c}\mathrm{h}(u+v)=\mathrm{c}\mathrm{h}u\mathrm{c}\mathrm{h}v+$

$\mathrm{s}\mathrm{h}u\mathrm{s}\mathrm{h}v;$

3) ${(\mathrm{s}\mathrm{h}t)}^{\text{'}}=\mathrm{c}\mathrm{h}t,{(\mathrm{c}\mathrm{h}t)}^{\text{'}}=\mathrm{s}\mathrm{h}t$.

Definition 1^[1,2] (Frenet Trihedron) Let $\mathit{r}=\mathit{r}(s)$ be a regular curve with arc-length parameter $s$. Then the ordered orthonormal basis $\mathit{\alpha },\mathit{\beta },\mathit{\gamma }$ are called the Frenet Trihedron if $\mathit{\alpha }=\frac{\mathrm{d}\mathit{r}}{\mathrm{d}s}=\dot{\mathit{r}}(s)$, $\mathit{\beta }=\frac{\dot{\mathit{\alpha }}(s)}{|\dot{\mathit{\alpha }}(s)|}$, $\mathit{\gamma }=\mathit{\alpha }\times \mathit{\beta }$.

Definition 2^[1,2] (Torsion of curve) Let $\mathit{r}=\mathit{r}(s)$ be a regular curve with arc-length parameter $s$. Then the real valued function $\tau$ satisfying $\tau =\mathit{\gamma }\cdot \dot{\mathit{\beta }}$ is called the torsion of $\mathit{r}=\mathit{r}(s)$ .

Definition 3^[1,2] (Spiral) Let $\mathit{r}=\mathit{r}(t)$ be a regular curve with $\kappa$ its curvature,$\tau$ its torsion. Then it is called a spiral if $\frac{\kappa }{\tau }\equiv \mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}t$ for all $t$ .

Theorem 2^[1,2] Let $\mathit{r}=\mathit{r}(t)$ be a regular curve and $\tau$ its torsion. Then $\mathit{r}=\mathit{r}(t)$ is a plane curve if and only if $\tau =\frac{({\mathit{r}}^{\text{'}},{\mathit{r}}^{″},{\mathit{r}}^{‴})}{({\mathit{r}}^{\text{'}}\times {\mathit{r}}^{″}{)}^{\mathrm{2}}}=\mathrm{0}$, where $({\mathit{r}}^{\text{'}},{\mathit{r}}^{″},{\mathit{r}}^{‴})$ denotes the mixed product of vectors. In particular, if its curvature $\kappa =\mathrm{0}$, it is a straight line or line segment.

Theorem 3^[1,2] Let $\mathit{r}=\mathit{r}(t)$ be a regular curve and $\kappa$ its curvature. Then $\kappa =\frac{|{\mathit{r}}^{\text{'}}\times {\mathit{r}}^{″}|}{|{\mathit{r}}^{\text{'}}{|}^{\mathrm{3}}}$.

Combining Theorem 2 with Theorem 3, we obtain the following lemma, which is important to complete the proof of Theorem 1.

Lemma 1   Let $\lambda =\lambda (t)$ be a differentiable function of $t$, $\mathit{m}\ne \mathrm{0}$ be a fixed vector. Then the curve satisfying $\mathit{r}(t)=\mathit{m}\lambda (t)+\mathit{n}$ is a straight line or line segment.

Proof   By an easy calculation, we can get that

${\mathit{r}}^{\text{'}}(t)=\mathit{m}{\lambda }^{\text{'}}(t),{\mathit{r}}^{″}(t)=\mathit{m}{\lambda }^{″}(t),{\mathit{r}}^{‴}(t)=\mathit{m}{\lambda }^{‴}(t)$

Obviously, its torsion is zero.

By Theorem 3, $\kappa =\frac{|{\mathit{r}}^{\text{'}}\times {\mathit{r}}^{″}|}{|{\mathit{r}}^{\text{'}}{|}^{\mathrm{3}}}=\frac{|{\lambda }^{\text{'}}\mathit{m}\times {\lambda }^{″}\mathit{m}|}{|{\mathit{r}}^{\text{'}}{|}^{\mathrm{3}}}=\mathrm{0}$.

So by Theorem 2, the curve $\mathit{r}(t)=\mathit{m}\lambda (t)+\mathit{n}$ is a straight line or line segment passing through the fixed point $\mathit{c}$.

2 Proof of Theorem 1

We still use all the notations in Section 1.To exclude the trivial case, we suppose that $\mathit{a},\mathit{b},\mathit{c}$ are not all equal to 0. We divide the proof of Theorem 1 into three steps.

Step 1   Prove the curve $\mathit{r}(t)=\mathit{a}\mathrm{c}\mathrm{h}t+\mathit{b}\mathrm{s}\mathrm{h}t+\mathit{c}$ is planar.

According to Proposition 2, we get

${\mathit{r}}^{\text{'}}(t)=\mathit{a}\mathrm{s}\mathrm{h}t+\mathit{b}\mathrm{c}\mathrm{h}t,{\mathit{r}}^{″}(t)=\mathit{a}\mathrm{c}\mathrm{h}t+\mathit{b}\mathrm{s}\mathrm{h}t,{\mathit{r}}^{‴}(t)=\mathit{a}\mathrm{s}\mathrm{h}t+\mathit{b}\mathrm{c}\mathrm{h}t.$

Obviously, $({\mathit{r}}^{\text{'}},{\mathit{r}}^{″},{\mathit{r}}^{‴})=\mathrm{0}$. So its torsion is zero.

Therefore, the curve $\mathit{r}(t)=\mathit{a}\mathrm{c}\mathrm{h}t+\mathit{b}\mathrm{s}\mathrm{h}t+\mathit{c}$ is planar.

Step 2   Prove that the curve is a straight line or a line segment if $\mathit{a}//\mathit{b}$.

Proof   Because the curve is regular, ${\mathit{r}}^{\text{'}}(t)=\mathit{a}\mathrm{s}\mathrm{h}t+\mathit{b}\mathrm{c}\mathrm{h}t\ne \mathrm{0}$ for every $t$.That is to say, $\mathit{a},\mathit{b}$ cannot be zero vectors at the same time.

1) $\mathit{a}=\mathrm{0},\mathit{b}\ne \mathrm{0},$

$\mathit{r}(t)=\mathit{b}\mathrm{s}\mathrm{h}t+\mathit{c}.$

According to Lemma 1, it is easy to see that the curve is a straight line or a line segment passing through the fixed point $\mathit{c}$.

2) $\mathit{a}\ne \mathrm{0},\mathit{b}=\mathrm{0},$

$\mathit{r}(t)=\mathit{a}\mathrm{c}\mathrm{h}t+\mathit{c}.$

We can obtain the same result in the similar way by Lemma 1.

3) $\mathit{a}\ne \mathrm{0},\mathit{b}\ne \mathrm{0}$.

Let $\mathit{a}=\mu \mathit{b}\ne \mathrm{0}$.$\mathit{r}(t)=\mathit{b}(\mu \mathrm{c}\mathrm{h}t+\mathrm{s}\mathrm{h}t)+\mathit{c}$.Set $\lambda =\lambda (t)=\mu \mathrm{c}\mathrm{h}t+\mathrm{s}\mathrm{h}t$, it cannot be zero and it is differentiable. So we get the same result by Lemma 1.

Step 3   Prove that the curve is a hyperbola if $\mathit{a},\mathit{b}$ is linearly independent.

Proof   It is easy to see that the fixed vector $\mathit{c}$ has no effect on the image, so we judge the shape of $\tilde{\mathit{r}}(t)=\mathit{a}\mathrm{c}\mathrm{h}t+\mathit{b}\mathrm{s}\mathrm{h}t$.

Since $\mathit{a},\mathit{b}$ are linearly independent, without loss of generality, let the coordinates of the vectors $\mathit{a},$$\mathit{b},\tilde{\mathit{r}}(t)$ be $\left(a,\mathrm{0}\right)$, $\left(b_{\mathrm{1}},b_{\mathrm{2}}\right)$, $(x,y)$, respectively. Then we obtain that

$\{\begin{array}{c}x=a\mathrm{c}\mathrm{h}t+b_{\mathrm{1}}\mathrm{s}\mathrm{h}t\\ y=b_{\mathrm{2}}\mathrm{s}\mathrm{h}t\end{array}.$(1)

By an easy calculation, we can get the equation as follows from (1):

$b_{\mathrm{2}}^{\mathrm{2}}{x}^{\mathrm{2}}-\mathrm{2}{b}_{\mathrm{1}}{b}_{\mathrm{2}}xy+\left(b_{\mathrm{1}}^{\mathrm{2}}-a^{\mathrm{2}}\right)y^{\mathrm{2}}-a^{\mathrm{2}}{b}_{\mathrm{2}}^{\mathrm{2}}=\mathrm{0}.$

namely, $\left(\begin{array}{ccc}x& y& \mathrm{1}\end{array}\right)\left(\begin{array}{ccc}{b}_{\mathrm{2}}^{\mathrm{2}}& -b_{\mathrm{1}}{b}_{\mathrm{2}}& \mathrm{0}\\ -b_{\mathrm{1}}{b}_{\mathrm{2}}& b_{\mathrm{1}}^{\mathrm{2}}-a^{\mathrm{2}}& \mathrm{0}\\ \mathrm{0}& \mathrm{0}& -a^{\mathrm{2}}{b}_{\mathrm{2}}^{\mathrm{2}}\end{array}\right)\left(\begin{array}{c}x\\ y\\ \mathrm{1}\end{array}\right)=\mathrm{0}.$

Obviously, $I_{\mathrm{2}}=\left|\begin{array}{cc}{b}_{\mathrm{2}}^{\mathrm{2}}& -b_{\mathrm{1}}{b}_{\mathrm{2}}\\ -b_{\mathrm{1}}{b}_{\mathrm{2}}& b_{\mathrm{1}}^{\mathrm{2}}-a^{\mathrm{2}}\end{array}\right|=-b_{\mathrm{2}}^{\mathrm{2}}{a}^{\mathrm{2}}\mathrm{0},I_{\mathrm{3}}\ne \mathrm{0}.$

By Proposition 1, we get that $\mathit{r}\left(t\right)$ is a hyperbola.

3 A Generalization of the Curve in Theorem 1

In this section, we investigate the shapes of the regular curve $\mathit{r}(t)=\mathit{a}\mathrm{c}\mathrm{h}t+\mathit{b}\mathrm{s}\mathrm{h}t+\mathit{c}t$, which is regarded as the generalization of the regular curve $\mathit{r}(t)=\mathit{a}\mathrm{c}\mathrm{h}t+\mathit{b}\mathrm{s}\mathrm{h}t+\mathit{c}$.

Theorem 4   Let $\mathit{a},\mathit{b},\mathit{c}$ be fixed vectors and $\mathit{r}(t)=\mathit{a}\mathrm{c}\mathrm{h}t+\mathit{b}\mathrm{s}\mathrm{h}t+\mathit{c}t$ a regular curve. Then

1) If $(\mathit{a},\mathit{b},\mathit{c})=\mathrm{0}$, it is planar. In particular, if the rank of vectors $\mathit{a},\mathit{b},\mathit{c}$ is equal to 1, it is a straight line or a line segment.

2) If $(\mathit{a},\mathit{b},\mathit{c})\ne \mathrm{0}$, it is spatial. In particular, if $\mathit{a},\mathit{b},\mathit{c}$ are pairwise perpendicular to each other and $|\mathit{a}|=|\mathit{b}|=|\mathit{c}|$, it is a spatial spiral.

Proof   We divide the proof into three steps.

Step 1   Give the relationship between the mixed product and the torsion.

Since $\mathit{r}(t)=\mathit{a}\mathrm{c}\mathrm{h}t+\mathit{b}\mathrm{s}\mathrm{h}t+\mathit{c}t$ is regular,${\mathit{r}}^{\text{'}}(t)=\mathit{a}\mathrm{s}\mathrm{h}t+\mathit{b}\mathrm{c}\mathrm{h}t+\mathit{c}\ne \mathrm{0}$. Thus $\mathit{a},\mathit{b},\mathit{c}$ cannot be null vector at the same time.

By an easy calculation, we can get

${\mathit{r}}^{″}(t)=\mathit{a}\mathrm{c}\mathrm{h}t+\mathit{b}\mathrm{s}\mathrm{h}t,{\mathit{r}}^{‴}(t)=\mathit{a}\mathrm{s}\mathrm{h}t+\mathit{b}\mathrm{c}\mathrm{h}t,$

$({\mathit{r}}^{\text{'}},{\mathit{r}}^{″},{\mathit{r}}^{‴})=[(\mathit{a}\mathrm{s}\mathrm{h}t+\mathit{b}\mathrm{c}\mathrm{h}t+\mathit{c})\cdot (\mathit{a}\mathrm{c}\mathrm{h}t+\mathit{b}\mathrm{s}\mathrm{h}t)]\cdot (\mathit{a}\mathrm{s}\mathrm{h}t+\mathit{b}\mathrm{c}\mathrm{h}t)=(\mathit{a},\mathit{b},\mathit{c}).$

Combing the formula of torsion, we know that the mixed product of vectors of $\mathit{a},\mathit{b},\mathit{c}$ controls the shape of the curve.

Step 2   The provement of Theorem 4 1).

From $(\mathit{a},\mathit{b},\mathit{c})=\mathrm{0}$,we can get that the torsion of the curve is zero. By Theorem 2, it is planar.

When the rank of vectors $\mathit{a},\mathit{b},\mathit{c}$ is equal to 1, it means that $\mathit{a},\mathit{b}$ and $\mathit{c}$ are parallel to each other. By Lemma 1, it is a straight line or line segment.

Step 3   The provement of Theorem 4 2).

From $(\mathit{a},\mathit{b},\mathit{c})\ne \mathrm{0}$,we can get that the torsion of the curve is not zero. So it is spatial.

Since $\mathit{a},\mathit{b},\mathit{c}$ is linearly independent and perpendicular to each other, without loss of generality, let the coordinates of the vectors $\mathit{a},$$\mathit{b},\mathit{c},\mathit{r}(t)$ be $\left(a,\mathrm{0,0}\right)$, $\left(\mathrm{0},b,\mathrm{0}\right),(\mathrm{0,0},c)$, $(x,y,z)$, respectively. Then we have

$\{\begin{array}{c}x=a\mathrm{c}\mathrm{h}t\\ y=b\mathrm{s}\mathrm{h}t\\ z=ct \end{array}.$

By an easy calculation, combining with $|\mathit{a}|=|\mathit{b}|=|\mathit{c}|$,we get

$\left|{\mathit{r}}^{\text{'}}\right|=\sqrt[]{a^{\mathrm{2}}\mathrm{s}{\mathrm{h}}^{\mathrm{2}}t+b^{\mathrm{2}}\mathrm{c}{\mathrm{h}}^{\mathrm{2}}t+c^{\mathrm{2}}}=\sqrt[]{\mathrm{2}}|a|\mathrm{c}\mathrm{h}t,$

$\left|{\mathit{r}}^{\text{'}}\times {\mathit{r}}^{″}\right|=\sqrt[]{b^{\mathrm{2}}{c}^{\mathrm{2}}\mathrm{s}{\mathrm{h}}^{\mathrm{2}}t+c^{\mathrm{2}}{a}^{\mathrm{2}}\mathrm{c}{\mathrm{h}}^{\mathrm{2}}t+a^{\mathrm{2}}{b}^{\mathrm{2}}}=\sqrt[]{\mathrm{2}}{a}^{\mathrm{2}}\mathrm{c}\mathrm{h}t,$

$\left({\mathit{r}}^{\text{'}},{\mathit{r}}^{″},{\mathit{r}}^{‴}\right)=\mathrm{d}\mathrm{e}\mathrm{t}\left|\begin{array}{ccc}a\mathrm{s}\mathrm{h}t& a\mathrm{c}\mathrm{h}t& a\mathrm{s}\mathrm{h}t\\ b\mathrm{c}\mathrm{h}t& b\mathrm{s}\mathrm{h}t& b\mathrm{c}\mathrm{h}t\\ c& \mathrm{0}& \mathrm{0}\end{array}\right|=abc.$

Thus we have

$k=\frac{\left|{\mathit{r}}^{\mathit{\text{'}}}\times {\mathit{r}}^{″}\right|}{{\left|{\mathit{r}}^{\text{'}}\right|}^{\mathrm{3}}}=\frac{\mathrm{1}}{\mathrm{2}|a|\mathrm{c}{\mathrm{h}}^{\mathrm{2}}t},$

$\tau =\frac{\left({\mathit{r}}^{\mathit{\text{'}}},{\mathit{r}}^{″},{\mathit{r}}^{‴}\right)}{{\left({\mathit{r}}^{\text{'}}\times {\mathit{r}}^{″}\right)}^{\mathrm{2}}}=\frac{abc}{\mathrm{2}{a}^{\mathrm{4}}\mathrm{c}{\mathrm{h}}^{\mathrm{2}}t},$

Obviously, $\frac{\kappa }{\tau }\equiv \mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}$.Therefore, the curve is a spatial spiral from Definition 3.

References

All Figures

	Fig. 1 The function $\mathbf{s}\mathbf{h}\mathit{t}$
In the text

	Fig. 2 The function $\mathbf{c}\mathbf{h}\mathit{t}$
In the text