Revision as of 02:10, 6 June 2013

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.