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In this section we will consider the case of a smooth curve in a two-dimensional space. This could be, for example, the trajectory of a particle, the motion of a robot, or the contour of an object in an image. We can represent a curve, C, in the following way:

$ C=\left\{ p(t):\big( x(t),y(t)\big) |t\in I\right\}, $

where I is some open interval of the real number set $ \mathbb{R} $. This is called a `parametric representation' of the curve, where t is the parameter.

Imagine we are drawing the curve in the xy-plane with a pen, and define $ p(t) $ as the position of the pen at time t. Fig. 1 is an example of a curve parameterization. However, you may be able to recall from your multivariable calculus course that the same curve C can have infinitely many parameterizations that produce the same graph. The easiest way to parametrize a graph described by y = f(x) is to choose a parameter and set it equal to x, then express y in terms of that parameter. For example, the straight line y=2x in Fig. 2 may be written as:

(a) $ \left\{ \begin{align} x(t)&=& t \\ y(t)&=&2t\end{align} \right. $, or (b) $ \left\{\begin{align} x(t)&=&t^3 \\y(t)&=&2t^3\end{align}\right. $.

In other words, we are choosing any allowable parametrization we want for x (i.e. it has the same range), and then replacing that parameterization for x inside f(x) to obtain a parameterization for y.

Question: Why is $ x(t) = t^2 $ NOT a valid parameterization?

Answer: Negative values are not in the range of $ t^2 $.

Note that as we vary the parameter t, each of these parameterizations will give different $ (x,y) $ values at the same value t, but they will eventually trace out the same graph. In other words, the way we traverse the graph is independent of its shape. As we mentioned before there are many possible parameterizations, and it stands to reason that not all parameterizations are created equal. A particularly useful parameter, `arc length', will lead us to what has been labeled the `canonical form' of a curve and is introduced below.

Fig 1:Animated example of curve parameterization

Fig 2: Straight line through the origin with slope 2

First, we need to define the tangent vector of the curve. If the parametric equations x(t) and y(t) are differentiable, then we may define the tangent vector to the curve C at the point (x(t),y(t)) as:

$ p^{\prime}(t)=\frac{d}{dt}p(t)=\left( \frac{d}{dt}x(t),\frac{d}{dt}y(t)\right). $

The vector $ p^{\prime}(t) $ is tangent to the graph of the curve C at the point (x(t),y(t)). The magnitude of the tangent vector,

$ \| p^{\prime}(t)\| =\sqrt{\left( \frac{d}{dt}x(t)\right)^2+\left( \frac{d}{dt}y(t)\right)^2} $

corresponds to the speed at which we are tracing the curve. For example, using the parameterizations for the straight line y=2x in Fig.\eqref{fig:notation_4}, we obtain the following speeds: (a) $ \| p^{\prime} (t)\| =\| (1,2)\|=\sqrt{5} $ and (b) $ \| p^{\prime} (t)\| =\| (3,24)\| =\sqrt{585} $. From these examples we can see that the speed is not equal to the slope of the straight line, it is dependent upon the parameterization of the curve; this is a key point.

The `arc length' s, is a parameterization such that the speed is equal to one, i.e. $ \| \frac{d}{ds}p(s)\| =1 $ for all s. Hence, for any given parameter t, we can compute the derivative of p(t) with respect to the arc length s as follows:

$ \label{eq:derivative1} \frac{d}{ds}p(t)=\frac{p^{\prime}(t)}{\| p^{\prime}(t)\|}=\frac{\big( x^{\prime}(t),y^{\prime}(t)\big)}{\sqrt{\big( x^{\prime}(t)\big) ^2+\big( y^{\prime}(t)\big) ^2}}. $

On the other hand, we can also compute the same derivative with the chain rule:

$ \label{eq:derivative2} \frac{d}{ds}p(t)=\frac{d}{dt}p(t)\cdot\frac{dt}{ds}= p^{\prime}(t)\cdot\frac{dt}{ds}. $

Comparing Eq.\eqref{eq:derivative1} and Eq.\eqref{eq:derivative2}, we can derive that: $ \frac{dt}{ds}=\frac{1}{\sqrt{\big( x^{\prime}(t)\big) ^2+\big( y^{\prime}(t)\big) ^2}}\Rightarrow ds=\sqrt{\big( x^{\prime}(t)\big) ^2+\big( y^{\prime}(t)\big) ^2}dt=\|p^{\prime}(t)\| dt, $

since $ \| \frac{d}{ds}p(t)\| =1 $. Then, we can compute the length of the curve p(t) from t=a to t=b:

$ \int^{s(b)}_{s(a)}\| p^{\prime}(s)\| ds=\int^{s(b)}_{s(a)}1\cdot ds=\int^{b}_{a}\| p^{\prime}(t)\| dt=s(b)-s(a), $

where $s(t)$ is the parameter substitution function from t to s. For example, if p(s) is parameterized by arc length, the length of the curve from point p(2) to p(17.5) is 17.5-2=15.5.

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OLD STUFF

Notation

For a two-dimensional (2D) space $ \mathbb{R}^2 $, for example the X-Y plane, we can represent a smooth curve, C, in it as following:

$ C=\left\{ p(t):\big( x(t),y(t)\big) |t\in I\right\} $

, where I is some open interval of real number set $ \mathbb{R} $. The smooth curve indicates that the curve is differentiable everywhere (at least 3 times). Imaging we are drawing the this curve C in that plane with a pen, and then p(t) is the position of the pen at time t. Fig.1 and Figs.2 are two examples. However, the same curve C can have many different expressions. For example, the straight line that passes through the origin with slope 2, i.e. the straight in Fig.3, can be written as (a) $ \left\{\begin{array}{l}x(t)=t \\y(t)=2t\end{array}\right. $, (b) $ \left\{\begin{array}{l}x(t)=t^2 \\y(t)=2t^2\end{array}\right. $, or (c) $ \left\{\begin{array}{l}x(t)=\cos (t) \\y(t)=2\cos (t)\end{array}\right. $. Therefore, the spacing in time has nothing to do with the spacing in space.

Fig 1: Example of curve parameterization

Fig 2.a: The points in $X$-$Y$ plane

Fig 2.b: X(t)

Fig 2.c: Y(t)

Fig 3: Straight line go through origin with slope 2

In order to make the formulas for the same curve unanimous, we are going to introduce the concept of "arclength". First, we need to define the speed vector of the curve. For the curve C defined in above equation, its speed vector is defined as:

$ p^{\prime}(t)=\frac{d}{dt}p(t)=\left( \frac{d}{dt}x(t),\frac{d}{dt}y(t)\right). $

The speed, i.e. the magnitude of this speed vector, is:

$ \| p^{\prime}(t)\| =\sqrt{\left( \frac{d}{dt}x(t)\right)^2+\left( \frac{d}{dt}y(t)\right)^2}. $

For the same expressions of the straight line y = 2x above, we know the corresponding speeds are: (a) $ \|p^{\prime} (t)\| =\| (1,2)\|=\sqrt{5} $, (b) $ \|p^{\prime} (t)\| =\| (2t,4t)\| =\sqrt{20} |t| $, and (c) $ \| p^{\prime} (t)\| =\| \big( -\sin(t),-2\text{sin} (t) \big)\| =\sqrt{5} |\sin(t)| $. From these examples, we can see that the speed is not equal to the slope of the straight line.

Now, we will uniquely parameterize the curve with respect to the "arclength" s, such that $ \| \frac{d}{ds}p(s)\| =1 $, $ \forall s $, i.e. the unit speed parameterization. Hence, with any given parameter t, we can compute the derivative of p(t) with respect to the arclength s as following: $ \frac{d}{ds}p(t)=\frac{p^{\prime}(t)}{\| p^{\prime}(t)\|}=\frac{\big( x^{\prime}(t),y^{\prime}(t)\big)}{\sqrt{\big( x^{\prime}(t)\big) ^2+\big( y^{\prime}(t)\big) ^2}}. $

In other hand, we can also compute the same derivative with the chain rule:

$ \frac{d}{ds}p(t)=\frac{d}{dt}p(t)\cdot\frac{dt}{ds}=p^{\prime}(t)\cdot\frac{dt}{ds}. $

Comparing above two equations, we can derive that:

$ \frac{dt}{ds}=\frac{1}{\sqrt{\big( x^{\prime}(t)\big) ^2+\big( y^{\prime}(t)\big) ^2}}\Rightarrow ds=\sqrt{\big( x^{\prime}(t)\big) ^2+\big( y^{\prime}(t)\big) ^2}dt=\|p^{\prime}(t)\| dt, $

since $ \| \frac{d}{ds}p(t)\| =1 $. Then, we can easily compute the length of the curve p(t) from t = a to t = b: $ \int^{s(b)}_{s(a)}\|p^{\prime}(s)\| ds=\int^{s(b)}_{s(a)}1\cdot ds=\int^{b}_{a}\|p^{\prime}(t)\| dt=s(b)-s(a), $

where s(t) is the parameter substitution function from t to s. For example, if p(s) is parameterized by arclength, the length of the curve from point p(2) to p(17.5) is 17.5 − 2 = 15.5.

Back to "Boutin lectures on Intro Differential geometry by Wu and Black"