The length of a path

Parametrized curves

Recall how we parametrized a line. We let x = tm + a. As t varied over the real numbers, the point x traced out a line. We can view the line as being parametrized by the vector-valued function c(t) = tm + a. We have also shown how vector-valued functions can parametrize more complicated curves than lines. You can read about the physical interpretation of a parametrization and its derivative to get a better feel for parametrized curves. The important thing to remember is that a function c(t) traces out a curve as you vary t. We refer to the curve that c(t) traces out as a path.

You could think of a path c : R$\to$R³ as being a wire. For example, c(t) = (cos t, sin t, t), for 0$\le$t$\le$6$\pi$, is the parametrization of a helix. You can view it as a slinky or a spring.

You can drag the blue point to change t. The cyan point indicates the position x = (cos t, sin t, t).

Length of a path (also called arclength)

Now, imagine we wanted to estimate the length of the slinky. Unfortunately, it's difficult to calculate the length of a curved piece of wire. It's much easier to calculate the length of straight pieces of wire. Probably the easiest way to calculate the length of the slinky would be to stretch it out into one straight line. But, if you ever tried to do that with a slinky (or a strong spring), you'd discover that stretching it into a straight line is virtually impossible.

So, if you can't stretch the slinky into one straight line, what could you do to estimate its length? One thing you could do is pretend that the slinky, rather than being a curved wire, was really composed of a bunch of short straight wires. In other words, you could approximate the curved slinky (shown in red) with the line segments shown in green.

The length of the line segments is easy to measure. If you add up the lengths of all the line segments, you'll get an estimate of the length of the slinky. This estimate of the length is shown by the green vertical slider and labeled by L(dt). Clearly, the length of the line segments will underestimate the length of the slinky, which is shown by the red mark on the green slider.

However, if you increase the number of line segments (decreasing the length of each line segment), the total length of the line segments becomes a better estimate of the slinky length. The length of the line segments is controlled by the slider labeled dt (it should be labeled $\Delta$t, but I can't get the character $\Delta$ to appear on the figure). As you drag $\Delta$t toward zero, the length of each line segment shrinks toward zero, the number of line segments increases, and the line segments become closer and closer to the slinky. Consequently, the total length of the line segments approaches the slinky length, as shown by L(dt) approaching the red mark on the green slider.

What's the length of each line segment? If there are n line segments, we could define t₀, t₁,..., t_n so that the first line segment goes from the point c(t₀) to the point c(t₁), the second line segment goes from the point c(t₁) to the point c(t₂), etc. The vector from c(t₀) to c(t₁) is simply c(t₁) - c(t₀), so the length of the line segment must be ||c(t₁) - c(t₀)||. The length of the second line segment is ||c(t₂) - c(t₁)||, etc.

To find the total length of the line segments, we just add up those lengths from all n line segments:

$$\sum_{i}^{n}$$ (1)

Now we do some tricks to put this into a different form. First, if $\Delta$t_i = t_i - t_i-1, then we can rewrite t_i as t_i-1 + $\Delta$t_i. (OK, this seems pointless so far, but hang on, because we are going to do what looks even dumber moves before anything will make sense.) Next, we can divide each term of equation (1) by $\Delta$t_i and multiply it by $\Delta$t_i so that our expression for the length becomes

$$\sum_{i}^{n} \sum_{i}^{n} \Delta$$

$$\sum_{i}^{n} \left\Vert\vphantom{ \frac{\mathbf{c}(t_{i-1} + {\textstyle \Delta}t_i) - \mathbf{c}(t_{i-1})}{{\textstyle \Delta}t_i}}\right. {\frac{{\mathbf{c}(t_{i-1} + {\textstyle \Delta}t_i) - \mathbf{c}(t_{i-1})}}{{{\textstyle \Delta}t_i}}} \left.\vphantom{ \frac{\mathbf{c}(t_{i-1} + {\textstyle \Delta}t_i) - \mathbf{c}(t_{i-1})}{{\textstyle \Delta}t_i}}\right\Vert \Delta$$ (2)

Well, maybe equation (2) doesn't look like much of an improvement on equation (1). But if you were a real math nerd, you might have noticed that the quotient involving c(t_i-1) is exactly the expression used in the limit definition of the derivative c'(t) of a path (if we replace h with $\Delta$t_i). In fact, equation (2) is a Riemann sum for an integral just like you saw with double integrals. If we let the number of line segments increase (as we take the limit $\Delta$t_i$\to$ 0) the quotient becomes c'(t), and equation (2) approaches the integral

$$\int_{a}^{b}$$ (3)

which is the true length of the slinky. The numbers a and b are the values of t at the ends of the slinky (i.e., the numbers so that the slinky is defined by c(t) for a$\le$t$\le$b). In our example, the slinky was defined by c(t) for 0$\le$t$\le$6$\pi$, so we would use a = 0 and b = 6$\pi$.

You may remember that the magnitude of the derivative ||c'(t)|| is the speed of a particle that is at position c(t) at time t. Equation (3) simply says that the total length of the path traced by the particle is the integral of its speed. (This length must, of course, be independent of the particle's speed.)

You can see some examples here.