1.3.3 Creating an integral
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.
Intuitively now, we should define the length of the curve as . 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 secant vector: To simplify the notation, let's name the end-points of the sub-division of by the following scheme: etc... Thus we can rewrite the secant vector as: Find an expression for in terms of the component functions of .
At this point to continue we'll need to apply an additional restriction to our path : must be a differentiable path with continuous component functions. We now apply the MVT to invoke the existence of a value with the property: Use this to write in terms of the velocity vector to .
So to wrap this discussion up, we now can say that if is a differentiable path from into , then between the two numbers and there is a number with the property:
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 . We just have to scale by that "small change" to reflect that the change in time is less than one unit.
In the GeoGebra applet below you can adjust the slider to shrink the value of . You'll notice that I've drawn originating from . It would probably be more correct to have this vector originate from , but since the interval is collapsing these points will converge anyway.