The equation for finding the arc length of a curve is \(\displaystyle{ L = \int_{a}^{b}{ \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2} ~dt } }\). In shorthand notation, you will often see \(\displaystyle{ L = \int_{a}^{b}{ \| \vec{r}'(t) \| ~dt } }\) where \(\displaystyle{ \| \vec{r}'(t) \| = \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2} }\). Of course, we are assuming this is a smooth, continuous curve.
Before we go on, let's work some practice problems.

As you know from parametrics, there is more than one way to describe a parametric curve. Since vector functions are parametrics written in vector form, we can also describe the same curve with a different vector function. For example, to describe the circle with radius 1 we can write \(\vec{r}(t)=\langle \cos(2t), \sin(2t) \rangle\); \(0\leq t\leq \pi\). Another way of describing the same circle is \(\vec{q}(s)=\langle \cos s, \sin s \rangle\); \( 0 \leq s \leq 2\pi \). Now, let's calculate the arc length of both of these vector functions.

For \(\vec{r}(t)\), the derivative is \(\vec{r}'(t) = \langle -2\sin(2t), 2\cos(2t) \rangle\) and so the arc length integral is
\(\displaystyle{ \int_{0}^{\pi}{ \| \vec{r}'(t) \| ~ dt } = \int_{0}^{\pi}{ \sqrt{ (-2\sin 2t)^2 + (2\cos 2t)^2 } ~ dt } = }\) \(\displaystyle{ \int_{0}^{\pi}{2 ~ dt} = }\) \(\displaystyle{ \left. 2t \right|_{0}^{\pi} = 2(\pi) - 2(0) = 2\pi }\)

Now, let's do the same calculation using the other vector function, \(\vec{q}(s)=\langle \cos s, \sin s \rangle\); \( 0 \leq s \leq 2\pi \).
The derivative is \(\vec{q}'(s)= \langle -\sin s, \cos s \rangle \) and so the arc length integral is
\(\displaystyle{ \int_{0}^{2\pi}{ \| \vec{q}'(s) \| ~ ds } = \int_{0}^{2\pi}{ \sqrt{ (-\sin s)^2 + (\cos s)^2 } ~ ds } = }\) \(\displaystyle{ \int_{0}^{2\pi}{ 1~ds } = }\) \(\displaystyle{ s|_{0}^{2\pi} = 2\pi - 0 = 2\pi }\)

As expected the answer (arc length) is the same in both cases. However, let's look closely as the intermediate steps. Notice in the case of \(\vec{r}(t)\), that \(\| \vec{r}'(t) \| = 2\) and for \(\vec{q}(s)\), \(\|\vec{q}'(s)\|=1\). This is significant for our discussion. The fact that \(\|\vec{q}'(s)\|=1\) allows us to say that the arc length for this vector function is equal to the difference in the endpoints, i.e. \(2\pi-0\). That's what makes the parameterization \(\vec{q}(s)\) special. We call this parameterization, the arc length parameterization, specifically for this reason, i.e. \(\|\vec{q}'(s)\|=1\). (The use of the variable s is not random here. Most textbooks use s to signify that they are talking about the arc length parameter.)

So basically, we say that if we know that some curve is described by it's arc length parameter s, we can directly say that the arc length is the difference between the endpoints, without evaluating the integral.

In many cases, it is possible to convert/translate/change a vector function into one using the arc length parameterization. The reason we may want to do this is that there are some calculations that are easier to do if we have the arc length parameterization than if we have a generic parameterization. (See the section on curvature, for example.)

The general steps are
1. Calculate \(\displaystyle{s(t)=\int_{0}^{t}{ \| \vec{r}'(u) \| ~du } }\);
2. Convert t to s in the vector function and the range
Let's do an example of this by using \(\vec{r}(t)=\langle \cos(2t), \sin(2t) \rangle\); \(0\leq t\leq \pi\) to see how we get \(\vec{q}(s)=\langle \cos s, \sin s \rangle\); \( 0 \leq s \leq 2\pi \).

First, we calculate \(\displaystyle{s(t)=\int_{0}^{t}{ \| \vec{r}'(u) \| ~du } }\). Notice we use zero as the lower limit of integration since the range \(0\leq t\leq \pi\) starts with zero. Notice also that the upper limit is t, so we use u as our variable of integration. Any variable of integration would work except for t since t is the upper limit.
In our example, we now have \(\displaystyle{ s(t)=\int_{0}^{t}{ \sqrt{ (-2\sin 2u)^2 + (2\cos 2u)^2 ~du } } = }\) \(\displaystyle{ \int_{0}^{t}{ 2 ~ du} = 2u|_{0}^{t} = 2t }\). So now we have \(s=2t\). We solve this for t to give us \(t=s/2\).

For the second step, we substitute \(s/2\) for \(t\) in our original vector function and range to get \( \vec{r}(s) = \langle \cos s, \sin s \rangle \); \(0\leq s \leq 2\pi\) which is equal to \(\vec{q}(s)\).

Okay, so you now know how to find the arc length parameterization of a curve, but what does it really mean? Why is it important?
The idea is very simple but it is hidden in all the equations. Let's look at the number line below and the example above.

adapted from wikipedia

Let's think about the unit circle for a moment. We can take the unit circle and 'unroll it' and place it on the number line. Each angle will match up perfectly from zero to \(2\pi\). Now, let's look at the equations.
From the previous example, the two parameterizations of the unit circle are

Let's look at the first parameterization of the unit circle, \(\vec{r}(t)=\langle \cos(2t), \sin(2t) \rangle\); \(0\leq t\leq \pi\). Intuitively, if we were given the time \(0\leq t\leq \pi\) to walk along the unit circle, we would have to do it at a rate of 2 steps for each t.

However, the second parameterization is on the range, \(0\leq s\leq 2\pi\). This arc length parameterization means that if we were to think about walking on the unit circle and we were given \(s=2\pi\) time to walk the entire circle, we could walk at a constant speed of exactly 1 to traverse the entire circle, i.e. for every step we take along the curve, we take exactly the same distance along the number line. So the interval \(0\leq s\leq 2\pi\) matches up perfectly with the arc length.

This is true for any curve that is described by the arc length parameter, i.e. you can just unroll the curve and place it on the number line that is the same length as the interval and they will match up one-to-one.

So how does this help us? Well, if you are given an extremely complicated set of equations for a curve and you are asked to find the arc length on an interval, say \(a\leq t \leq b\) but you know that the parameter describing the curve is the arc length parameter, you can say that the arc length is just \(b-a\) and you do not need to do any other work. Pretty cool, eh?

Okay, time for some practice problems.