Based on the previous exploration we will add the additional condition that our parameterization must be injective. This will prevent our estimate from incorrectly double counting sections of the curve.[br][br]Intuitively now, we should define the length of the curve as [math]L\left(C\right)=\lim_{n\to\infty}\sum\left|\left|\vec{s}_i\right|\right|[/math]. Hopefully you recognize this as resembling an integral, but it's not quite in the right form yet. An integral is the limit of a sum of products - the product of a scalar valued function and a small change in the input (rectangles, remember? length times width). We're missing the small change in input.
Consider the [math]i^{\text{th}}[/math] secant vector:[br][br][math]\vec{s}_i=\vec{c}\left(a+i\Delta t\right)-\vec{c}\left(a+\left(i-1\right)\Delta t\right)[/math][br][br]To simplify the notation, let's name the end-points of the sub-division of [math]\left[a,b\right][/math] by the following scheme:[br][math]t_0=a[/math][br][math]t_1=a+\Delta t[/math][br][math]t_2=a+2\Delta t[/math][br][math]\vdots[/math][br]etc...[br][br]Thus we can rewrite the [math]i^{\text{th}}[/math] secant vector as:[br][math]\vec{s}_i=\vec{c}\left(t_i\right)-\vec{c}\left(t_{i-1}\right)[/math][br][br]Find an expression for [math]\vec{s}_i[/math] in terms of the component functions of [math]\vec{c}[/math].
[math]\vec{s}_i=\left(x\left(t_i\right)-x\left(t_{i-1}\right),y\left(t_i\right)-y\left(t_{i-1}\right)\right)[/math][br][br]From here I am going to ask you to apply the mean value theorem, but to do so you'll need to remember what that theorem says and also what conditions need to be in place to apply it.
At this point to continue we'll need to apply an additional restriction to our path [math]\vec{c}[/math]: [math]\vec{c}[/math] must be a differentiable path with continuous component functions. [br][br]We now apply the MVT to invoke the existence of a value [math]t^{\cdot}_i[/math] with the property:[br][br][math]x'\left(t_i^{\cdot}\right)=\frac{x\left(t_i\right)-x\left(t_{i-1}\right)}{t_i-t_{i-1}}[/math][br][math]y'\left(t_i^{\cdot}\right)=\frac{y\left(t_i\right)-y\left(t_{i-1}\right)}{t_i-t_{i-1}}[/math][br][br]Use this to write [math]\vec{s}_i[/math] in terms of the velocity vector to [math]\vec{c}[/math].
[math]\vec{s}_i=\Delta t\vec{c}'\left(t_i^{\cdot}\right)[/math][br][br]Okay actually this argument has a pretty glaring flaw. Do you see it?
So to wrap this discussion up, we now can say that if [math]\vec{c}[/math] is a differentiable path from [math]\left[a,b\right][/math] into [math]\mathbb{R}^2[/math], then between the two numbers [math]t_{i-1}[/math] and [math]t_i[/math] there is a number [math]t_i^{\cdot}[/math] with the property:[br][math]\vec{s}_i=\Delta t\vec{c}'\left(t_i^{\cdot}\right)[/math][br][br]If you think about the meaning of a derivative, this should feel intuitive. The velocity vector represents the movement of a particle along the path for one unit of time if we suddenly eliminated any turning or velocity changes. What we've deduced is that this vector does a good job of "pushing" a point along the curve for a small change in [math]t[/math]. We just have to scale by that "small change" to reflect that the change in time is less than one unit.[br][br]In the GeoGebra applet below you can adjust the slider to shrink the value of [math]\Delta t[/math]. You'll notice that I've drawn [math]\vec{c}'\left(t_{_i}^{\cdot}\right)[/math] originating from [math]\vec{c}\left(t_{i-1}\right)[/math]. It would probably be more correct to have this vector originate from [math]\vec{c}\left(t_i^{\cdot}\right)[/math], but since the interval [math]\left[t_{i-1},t_i\right][/math] is collapsing these points will converge anyway.