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Curves

Parametrized curves

Definition of differentiable curve

A parametrized differentiable curve is an infinitely differentiable map \(\alpha: I\rightarrow \mathbb{R}^3\) of an open interval into \(\mathbb{R}^3\).

The variable \(t\) is called the parameter of the curve. We do not exclude \(a=-\infty\) and \(b=\infty\).

The vector

\[ \alpha'(t)=(x'(t), y'(t), z'(t))\in\mathbb{R}^3 \]

is called the tangent vector of the curve \(\alpha\) at \(t\).

For the study of the differential geometry of a curve, it is essential to assume that there exsits such a tangent lint at every point.

Singular point, regular curve

(i) A point \(t\in I\) is called a singular point of \(\alpha\), if \(\alpha'(t)=0\).

(ii) A parametrized differentiable curve \(\alpha:I\rightarrow \mathbb{R}^3\) is said to be regular if \(\alpha'(t)\neq 0\) for all \(t\in I\).

A parameter called arc length is usually useful in further analysis.

Definition of Arc length

Given \(t_0\in I\), the arc length of a regular parametrized curve \(\alpha:I\rightarrow \mathbb{R}^3\) from the point \(t_0\) is defined by

\[ s=\int_{t_0}^t |\alpha'(\tau)|d\tau, \]

where \(|\alpha'(t)|=\sqrt{x'(t)^2+y'(t)^2+z'(t)^2}\).

Since \(|\alpha'(t)|>0\) for regular curve.

Now we talk about some invariance under reparametrization.

Vector product on E3

Equivalence: orientation

Suppose \(\{e_n\}\), \(\{f_n\}\) are two basis of \(n\)-dimensional space. They are said to have the same orientation, denoted by \(e\sim f\), if the matrix of change of basis has positive determinant.

Easy to show that orientation satisfies the equivalent relationship.

The vector product of \(u\) and \(v\) (in that order) is the unique vector \(u\times v\in \mathbb{R}^3\) such that

\[ (u\times v)\cdot w=det(u,v,w),\quad \forall w\in\mathbb{R}^3. \]

Easy to show that \((u\times v)\cdot u=0\) and \((u\times v)\cdot v=0\), so \(u\times v\neq 0\) is orthogonal to a plane generated by \(u\) and \(v\). To give a geometric interpretation of its norm and its direction, we proceed as follows.

Deduction

(i) Observe that \((u\times v)\cdot (u\times v)=|u\times v|^2>0\), so the determinant of \((u,v,u\times v)\) is positive and it could be a basis.

(ii) Prove that

\[ (u\times v)\cdot(x\times y)=\left|\begin{array}{cc}u\cdot x& v\cdot x\\u\cdot y& v\cdot y\end{array}\right| \]

where \(u,v,x,y\) are arbitrary vectors. Check for basis.

and show that

\[ |u\times v|^2=|u|^2|v|^2(1-\cos ^2\theta)=A^2. \]

(iii) The vector product is not associative. Because

\[ (u\times v)\times w = (u\cdot w)v - (v\cdot w)u, \]

and check for all basis.

The local theory of curves

Curvature

Let \(s\) be the arc length, and \(\alpha\) be parametrized by \(s\). \(\alpha'(s)\) has unit length, and \(|\alpha''(s)|\) measures how rapid the curve pulls away from the tangent line \(\alpha'(s)\), in the neighborhood of \(s\).

Curvature of a curve

Let \(\alpha:I\rightarrow \mathbb{E}^3\) be a curve parametrized be arc length \(s\in I\). The number \(\kappa(s):=|\alpha''(s)|\) is called the curvature of \(\alpha\) at \(s\).

Easy to show that for straight line, \(\alpha=us+v\), if and only if \(k(s)=0\). If we have \(k(s)\neq 0\) unless for \(s=s_0\), we could still find its curvature by leveraging the limit. If for its neighborhood, \(k(s)=0\), then it is a straight line.

When change the direction, we have tangent vector changes but the curvature does not. This is because, let \(\beta(s)=\alpha(-s)\), then

\[ \frac{d\beta(s)}{ds}=\frac{d\alpha(-s)}{ds}=(-1)\frac{d\alpha(-s)}{(-s)}. \]

If \(k(s)\neq 0\), then we have a unit normal vector \(n\) well defined by \(\alpha''(s)=k(s)n(s)\). The plane composed by \(t\) and \(n\) are called the osculating plane at \(s\).

Example. For \(k(s)=0\), check the following example.

\[ \alpha(t)=\begin{cases}(t,0,e^{-1/t^2}), \quad &t>0,\\ (0,0,0),\quad &t=0,\\ (t,e^{-1/t^2},0), \quad &t<0.\\\end{cases} \]

To proceed with the local analysis of curves, we assume \(\alpha''(s)\neq 0\) (the singular point of order \(1\), and \(\alpha'(s)=0\) is called the singular point of order \(0\)).

For a plane curve, we have the following description.

Example. Assume that \(\alpha(I) \subset \mathbb{E}^2\) and give \(k\) a sign as in the text.

Transport the vectors \(t(s)\) parallel to themselves in such a way that the origins of \(t(s)\) agree with the origin of \(\mathbb{E}^2\); the end points of \(t(s)\) then describe a parametrized curve \(s \rightarrow t(s)\) called the indicatrix of tangents of \(\alpha\).

Let \(\theta(s)\) be the angle from \(e_1\) to \(t(s)\) in the orientation of \(\mathbb{E}^2\). notice that we are assuming that \(k \neq 0\). Show that

(a) The indicatrix of tangents is a regular parametrized curve.

(b) \(dt/ds = (d\theta/ds)n\), that is, \(k = d\theta/ds\).

(a) Easy to see since \(|t'(s)|=|k(s)|\neq 0\).

(b) Let

\[ t(s)=[\cos \theta(s), \sin \theta(s)], \]

then

\[ t'(s)=[-\sin \theta(s), \cos\theta(s) ]\theta'(s)=\theta'(s)n, \]

since we have \(n = j t\) (rotation by \(\pi/2\)).

\(\square\)

Tortion

Already we have \(t'=kn\). Now we do not check \(n'\), but check \(b(s):=t(s)\times n(s)\), which is called binormal vector at \(s\), also a unit vector representing the osculating plane. \(|b'(s)|\) meausres the rate of change of the neighboring osculating plane. We claim that \(b'(s)\) is parallel with \(n(s)\). Indeed,

\[ b' = t' \times n + t\times n'=t\times n'. \]

The focus is not \(n'\), but \(t\perp b'\). With \(b'\perp b\), we have the result.

Extract the number out and define \(b'(s)=\tau(s)n(s)\) to be the following concept.

Definition of tortion

Let \(\alpha:I\rightarrow \mathbb{E}^3\) be a curve parametrized be arc length \(s\in I\) such that \(\alpha''(s)\neq 0\). The number \(\tau(s):=|b'(s)|\) is called the tortion of \(\alpha\) at \(s\).

Also, \(\alpha\) is a plane curve, if and only if \(|b'(s)|\equiv 0\). Necessarily speaking, it is easy. On the other hand, we shall show that \(b(s):=b_0\), take a inner product \(\alpha(s)\cdot b_0\) and check it is also a constant, which is exactly the parametrized form of plane.

Tortion could be either positive or negative.

Frenet trihedron

We have associated three orthonomal unit vector \(t(s), n(s), b(s)\), which is referred to as the Frenet trihedron at \(s\). And we already known that \(t'=kn\), and \(b'=\tau n\), for \(n=b\times t\) we

\[ n'=b'\times t + b\times t'=-\tau b -k t. \]

We call the above three equations the Frenet formulas.

Now we give the core theorem of this chapter.

Fundamental theorem of the local theory of curves

Given differentiable functions \(k(s)>0\) and \(\tau(s)\), \(s\in I\), there exsits a regular parametrized curve \(\alpha: I\rightarrow \mathbb{E}^3\) such that \(s\) is the arc length, \(k(s)\) is the curvature, and \(\tau(s)\) is the tortion of \(\alpha\).

Moreover, any other curve \(\overline{\alpha}\), which satisfies the same condition, differs from \(\alpha\) by a rigid motion. That is, there exists a orthonormal map \(\rho\) with positive determinant, and a translation vector \(c\), such that \(\overline{\alpha}=\rho \circ \alpha +c\).

This is by ODEs theory.

For plane curve, we could have a simpler version of the above theorem.

Example. Given a function \(k(s)\), show that the parametrized plane curve have \(k\) as curvature is given by

\[ \alpha(s)=\left( \int \cos \theta(s)ds +a, \int \sin \theta(s)ds +b\right) \]

where \(\theta(s)=\int k(s)ds + \varphi\). The curve is determined up to a translation of the vector \((a,b)\) and a rotation of the angle \(\varphi\).

Calculations

For general regular parametrized curve \(\alpha(t)\), we have the following formula for calculating the geometric variables.

Calculations of curvature

(i) generally speaking, we have

\[ \kappa(t) = \frac{|\alpha'(t)\times \alpha''(t)|}{|\alpha'(t)|^3}. \]

(ii) for plane curve \(\alpha'(t)=[x(t), y(t)]\), we have signed curvature

\[ \kappa(t) = \frac{x'y''-y'x''}{|x'^2+y'^2|^{3/2}}. \]

Calculation of tortion

Generally speaking, we have

\[ \tau(t)=-\frac{(\alpha'(t), \alpha''(t), \alpha'''(t))}{|\alpha'(t)\times \alpha''(t)|^2}. \]

The local canonical form

Let \(\alpha:I\rightarrow\mathbb{E}^3\) be a curve parametrized by arc length without singular points of order \(1\). We now consider the equations in a neighborhood of \(s_0\) using the trihedron \(t(s_0)\), \(n(s_0)\), \(b(s_0)\) as a basis for \(\mathbb{E}^3\). We may assume without loss of generality, that \(s_0=0\), and consider Taylor expansion

\[ \alpha(s)-\alpha(0)=\alpha'(0)s + \frac{1}{2}\alpha''(0)s^2 + \frac{1}{6}\alpha'''(0)s^3 + R(s), \]

where \(R(s)/s^3\rightarrow \pmb{0}\) as \(s\rightarrow 0\). Using \(\alpha'(0)=t\), \(\alpha''(0)=kn\), and \(\alpha'''(0)=(kn)'=k'n-k^2t-k\tau b\), we rewrite the above equation sorted by \(t,n,b\)

\[ \alpha(s)-\alpha(0)=\left(s-\frac{1}{6}k^2s^3\right)t + \left(\frac{1}{2}ks^2-\frac{1}{6}k's^3\right)n -\frac{1}{6}k\tau s^3 b+ R(s), \]

with

\[ \begin{cases} x(s)&=s-\frac{1}{6}k^2s^3 + R_x\\ y(s)&=\frac{1}{2}ks^2-\frac{1}{6}k's^3 + R_y\\ z(s)&=-\frac{1}{6}k\tau s^3 + R_z \end{cases} \]

which is called the local canonical form of \(\alpha\).

Classical form of curves

Example. Given the parametrized curve

\[ \alpha(s)=\left(a\cos \frac{s}{c}, a\sin \frac{s}{c}, b\frac{s}{c}\right) \]

where \(c^2=a^2+b^2\). Show that

(a) \(s\) is the arc length. (i.e. \(|\alpha'(s)|=1\)).

(b)

\[ \kappa(s)=\frac{|a|}{c^2},\quad \tau(s)=\frac{b}{c^2}. \]

Example. A curve \(\alpha\) is called a helix if the tangent line of \(\alpha\) make a constant angle with a fixed direction. Assume \(\tau(s)\neq 0\), \(s\in I\), show the following statements are equivalent:

(i) \(\alpha\) is a helix,

(ii) \(\kappa/\tau=const\).

(iii) the lines containing \(n(s)\) and passing \(\alpha(s)\) are parallel to a fixed plane.

(iv) the lines containing \(b(s)\) and passing \(\alpha(s)\) make a constant angle with a fixed direction.