The way I displayed the distance traveled in the last activity implied that arc length can be used to define a (scalar-valued) function. (BTW [b][color=#ff0000]scalar-valued [/color][/b]is a way of saying the codomain of a function is [math]\mathbb{R}[/math], as opposed to paths which are [b][color=#ff0000]vector-valued[/color][/b] functions). [br][br]If [math]\vec{c}:\left[a,b\right]\to\mathbb{R}[/math] is a differentiable path we can define a scalar-valued function as follows:[br][br][math]s\left(t\right)=\int_a^t\left|\left|\vec{c}\,'\left(u\right)\right|\right|du[/math][br][br]
By the Fundamental Theorem of Calculus, what is [math]s'\left(t\right)[/math]?
[math]s'\left(t\right)=\left|\left|\vec{c}\,'\left(t\right)\right|\right|[/math]
Since [math]s'\left(t\right)\ge0[/math] always (since speed by definition cannot be negative), we know that [math]s\left(t\right)[/math] is an increasing function. If we further impose the requirement that [math]\vec{c}[/math] is a [i]regular[/i] curve (recall regular means the velocity never vanishes), then we find that the arc length function is injective. This allows us to use the arc length function to reparameterize the image curve to achieve a [b][color=#ff0000]unit-speed[/color][/b] (aka [color=#ff0000][b]arc length[/b][/color]) parameterization - that is a parameterization whose speed is constantly 1.
Consider the example [math]\vec{c}\left(t\right)=\left(1-e^t,3+e^t\right),t\in\left(0,5\right)[/math][br]Compute the speed of this path as a function of [math]t[/math].
[math]\left|\left|\vec{c}\,'\left(t\right)\right|\right|=\sqrt{\left(-e^t\right)^2+\left(e^t\right)^2}=e^t\sqrt{2}[/math]
Now find an expression for [math]s\left(t\right)=\int_0^te^u\sqrt{2}du[/math]
[math]s\left(t\right)=e^t\sqrt{2}-\sqrt{2}[/math]
[math]s\left(t\right)[/math] is invertible as desired. Find an inverse function [math]t=\varphi\left(s\right)[/math] so that [math]s\left(\varphi\left(s\right)\right)=s[/math] and [math]\varphi\left(s\left(t\right)\right)=t[/math].
[math]t=\varphi\left(s\right)=\ln\left(\frac{s+\sqrt{2}}{\sqrt{2}}\right)[/math]
Finally, find an arc length parameterization of the image curve of [math]\vec{c}\left(t\right)[/math]
[math]\vec{c}\left(s\right)=\left(1-\frac{s+\sqrt{2}}{\sqrt{2}},3+\frac{s+\sqrt{2}}{\sqrt{2}}\right),s\in\left[0,e^5\sqrt{2}-\sqrt{2}\right][/math]
In the GeoGebra applet below you can see the two parameterizations of the image curve of [math]\vec{c}\left(t\right)[/math]. How can you tell the second parameterization is an arc length parameterization?