Let x be a variable that ranges from a to b. How does the length of the path on [a,x] vary with x? The difference quotient gives us the length of the path from x to x+h, over h, as h goes to 0.
Let s be the instantaneous speed at x. For any ε, restrict h so that s does not vary by more than ε through h. As described in the previous paragraph, all segments are limited in their lengths. Every approximation is bounded above by h×(s+ε), and when we divide by h, the approximations are within ε of s. This holds for all ε, so the change in arc length cannot exceed s. For the lower bound, we need to look at the individual components of p. Choose h so that the component derivatives don't change by more than ε. The simplest net has only the endpoints x and x+h. This causes the components to advance by their derivatives, give or take ε. The length of this single segment is at least s - some manageable function of ε, times h. Finer nets produce larger approximations, so we have a lower bound on arc length. Divide by h and the change in arc length is at least s. The derivative of arc length is speed, and by the fundamental theorem of calculus, the length of the path is the integral of speed from a to b. Of course speed may be piecewise continuous, since we can add arc lengths together.
For a simple example, let x = cos(t) and y = sin(t). This path runs around the unit circle forever. The velocity is -sin(t),cos(t), and since sine squared + cosine squared = 1, the speed is always 1. The arc length is merely the time traveled. Thus the circumference of the circle is 2π.
When we defined the trig functions, we found that they traced the circle at a constant speed, but it is possible to define them this way, and derive everything else from there. The reasoning would proceed along these lines. We know that sine and cosine always run around the circle, so the direction of travel is always tangent to the circle. If the speed is always 1 (by definition), the velocity vector is the position vector rotated by 90°. This establishes the derivatives of sine and cosine. From there we can compute the derivative of arcsine, and we're back to the integral definition of sine. But I digress.
If the entire path is scaled by k, the speed is also scaled by k at every point. This multiplies the arc length by k. Rectifyable paths are well behaved. Make everything twice as big, and the path is twice as long.
Sometimes the curve is the graph of a function in the xy plane. In this case x(t) = t and y(t) = f(t). Arc length now has an integrand of sqrt(1+f′(t)2). Consider the parabola y = x2, from 0 to 1. If we lay that curve flat along the x axis, how long would it be? Since f′ is 2x, we have the following integral.
∫ sqrt(1+4x2)
Using some trig substitution, the integral becomes:
2x×sqrt(4x2+1) + log(2x+sqrt(4x2+1)) over 4
Such a complicated formula for a simple question of length. Evaluate at 0 and 1 to get an approximate length of 1.47893. This is slightly larger than sqrt(2), as expected.