# Geometry/Differential Geometry/Basic Curves

The differential geometry of curves is usual starting point of students in field of differential geometry which is the field concerned with studying curves, surfaces, etc. with the use of the concept of derivatives in calculus. Thus, implicit in the discussion, we assume that the defining functions are sufficiently differentiable, i.e., they have no corners or cusps, etc. Curves are usually studied as subsets of an ambient space with a notion of equivalence. For example, one may study curves in the plane, the usual three dimensional space, curves on a sphere, etc. The most common notion of equivalence is that of rigid or Euclidean motion where two curves may be brought into alignment by a rotation and a translation. There are other interesting notions, however. In particular, in affine differential geometry of curves, two curves are equivalent if the may be brought into alignment through a rotation and a linear transformation. Special affine differential geometry considers two curves equivalent if they may be brought into alignment with a translation and linear transformation of determinant one. All ellipses in the plane are equivalent in affine geometry and are equivalent in special affine geometry if their interior has the same area. We will concentrate on the equivalence under Euclidean motions. In all these notions of equivalency, the ambient space is equipped with some additional structure. In the case of Euclidean motion equivalency, the additional structure is the inner or dot product of vectors.

Plane curves: Curves may defined parametrically, say ${\displaystyle (x(t),y(t))=(\cos(t),\sin(t))}$ or as the level set of a function ${\displaystyle f(x,y)=c}$, e.g., ${\displaystyle \{(x,y)|x^{2}+y^{2}=1\}}$. These, of course, both define the circle of radius one. The third method of defining curves is that of a graph, ${\displaystyle (x,y=f(x))}$. We will find the parametrically formulation usually easier to work with. Any graph type curve has the parametrization ${\displaystyle (x,y)=(t,f(t))}$. For the most part, we will not concern ourselves with the "speed" of curve, i.e., the actual parametrization of the curve. For example, the map ${\displaystyle t\to 2t}$ defines a curve that traverses the same path only twice as quickly.

Given a plane curve ${\displaystyle \mathbf {x} (t)=(x(t),y(t))}$, one may consider its velocity which is simply the component-wise derivative, ${\displaystyle \mathbf {x} '(t)=(x'(t),y'(t))}$. If one considers a rather simple reparametrization ${\displaystyle t\to e^{tan(t)}}$, one can quickly obtain derivatives which become quite ugly and unwieldy even starting with "easy" curves such as the circle. Thus recognizing that one is dealing with a circle might not be obvious if one looks at these derivatives.

Given a curve with a specific parametrization, there is a special reparametrization (almost unique) that eliminates this freedom that causes our headaches. Again, let ${\displaystyle \mathbf {x} (t)=(x(t),y(t))}$ be our curve. We want ${\displaystyle s}$ to be the reparametrization ${\displaystyle t=t(s)}$ so that the new curve has speed one, i.e., the magnitude of the vector velocity ${\displaystyle ={\sqrt {x'(s)^{2}+y'(s)^{2}}}=1}$. One may determine the function ${\displaystyle t(s)}$ with the chain rule. In order for the curve ${\displaystyle \mathbf {x} (s)=(x(t(s)),y(t(s)))}$ to have unit speed, by the chain rule, we need

${\displaystyle ({\frac {dx}{dt}}{\frac {dt}{ds}})^{2}+({\frac {dy}{dt}}{\frac {dt}{ds}})^{2}=1}$ or

${\displaystyle {\frac {dt}{ds}}={\frac {1}{\sqrt {({\frac {dx}{dt}})^{2}+({\frac {dy}{dt}})^{2}}}}}$.

The later is a differential equation, which, for all intents and purposes, cannot be solved explicitly except in very special circumstances. It will, by the standard theory of differential equations, have a solution away from points where the velocity vanishes. We will develop the theory assuming that this equation has been solved but then show how to work in other, non-unit-speed, parameterizations.

Thus, we assume that ${\displaystyle \mathbf {x} =\mathbf {x} (s)=(x(s),y(s))}$. With this parametrization, we have ${\displaystyle \mathbf {x} '(s)=(x'(s),y'(s))}$ is a unit vector, the tangent unit vector. Let us call this vector ${\displaystyle \mathbf {e} _{1}}$, i.e., ${\displaystyle \mathbf {e} _{1}(s)=(x'(s),y'(s))}$. We define unit normal vector ${\displaystyle \mathbf {e} _{2}}$ be the unit vector obtained by rotating ${\displaystyle \mathbf {e} _{1}}$ by 90° counter clockwise:

${\displaystyle \mathbf {e} _{2}(s)=(-y'(s),x'(s))}$.

Because we are in the plane, all vectors that are perpendicular to the vector ${\displaystyle \mathbf {e} _{1}}$ are necessarily some scalar multiple of ${\displaystyle \mathbf {e} _{2}}$. We use this observation as follows: The scalar function of ${\displaystyle s}$ given by ${\displaystyle \mathbf {e} _{1}(s)\cdot \mathbf {e} _{1}(s)}$ is constant (equal to one). Thus, its derivative vanishes:

${\displaystyle {\frac {d}{ds}}(\mathbf {e} _{1}(s)\cdot \mathbf {e} _{1}(s))=2\,\mathbf {e} _{1}(s)\cdot {\frac {d}{ds}}\mathbf {e} _{1}(s)=0}$.

Thus, the vector valued function ${\displaystyle {\frac {d}{ds}}\mathbf {e} _{1}(s)=(x''(s),y''(s))}$ is perpendicular to ${\displaystyle \mathbf {e} _{1}(s)}$, and thus by the above observation, we may write:

${\displaystyle {\frac {d}{ds}}\mathbf {e} _{1}(s)=\kappa (s)\,\mathbf {e} _{2}(s)}$

for some function ${\displaystyle \kappa (s)}$. The function ${\displaystyle \kappa (s)}$ is intrinsic and may be understood as the amount that the unit vector swings into the unit normal direction.

An example: We consider a circle of radius ${\displaystyle r}$ centered around the origin. This may be parametrized by ${\displaystyle (r\,\cos(t),r\,\sin(t))}$. One can solve the differential equation defining ${\displaystyle s}$ in this case and the unit speed parametrization is given by ${\displaystyle (r\,\cos(s/r),r\,\sin(s/r))}$. (This may have guessed as well.) The unit tangent and normal is given by ${\displaystyle \mathbf {e} _{1}=\cos(s/r),\sin(s/r),\mathbf {e} _{2}=(-\sin(s/r),\cos(s/r))}$. One then has:

${\displaystyle {\frac {d}{ds}}\mathbf {e} _{1}(s)=(1/r)(-\sin(s/r),\cos(s/r))=\kappa (s)\,\mathbf {e} _{2}(s)}$

with ${\displaystyle \kappa (s)}$ being the constant function ${\displaystyle 1/r}$. This gives another interpretation of the function ${\displaystyle \kappa (s)}$ as the reciprocal of the radius of the best fitting or osculating circle.

The function ${\displaystyle \kappa (s)}$ characterizes curves up Euclidean motion with the following result:

Theorem: If two curves ${\displaystyle \mathbf {x} _{1}(s)=(x_{1}(s),y_{1}(s))}$ and ${\displaystyle \mathbf {x} _{2}(s)=(x_{1}(s),y_{2}(s))}$ have the same curvature function ${\displaystyle \kappa (s)=\kappa _{1}(s)=\kappa _{2}(s)}$ then necessarily there exists a rigid motion involving a rotation and translation taking the curve ${\displaystyle \mathbf {x} _{1}}$ to ${\displaystyle \mathbf {x} _{2}}$.