Arc length In $L_\infty$ Metric

Relevant wiki: Arc Length and Surface Area - Problem Solving

Basically, just compute the length of an arc normally by considering the derivative. $\frac{d}{dt} (t, t^2) = (1, 2t)$ , so the length is $\int_{-1}^1 |(1, 2t)| \, dt$ (on any metric!). The only difference is that $| \cdot |$ is defined differently; in this case, it's the $L_\infty$ metric, so $|(1, 2t)| = \max \{1, |2t|\}$ . Now it's standard stuff.

---

The notation $| \cdot |$ is known as the norm, and is defined as $|x| = d(0, x)$ on the appropriate metric. On the standard Euclidean metric, $|(x,y)| = \sqrt{x^2 + y^2}$ . But here we'll work on the $L_\infty$ metric, so $|(x,y)| = \max \{x,y\}$ . It can be proven that $|a-b| = d(a,b)$ . It can also be proven that $\lim_{h \to 0} \frac{|f(x+h) - f(x)|}{|(x+h) - x|} = |f'(x)|$ for "reasonably good" function $f$ no matter what the metric is.

Let $f : [-1, 1] \to \mathbb{R}^2$ be defined as $f(t) = (t, t^2)$ ; we want to compute the arc length of $f$ . Note that $f$ is continuously differentiable (in some sense "smooth"), so that we can define the arc length.

As with computing arc length on the Euclidean metric, this length can be approximated by taking a finite number of regularly-spaced points and taking straight-line distances between them. That is, for a fixed $N$ , define $t_i = -1 + \frac{2i}{N}$ for all $i = 0, 1, 2, \ldots, N$ . Then the approximation is $\displaystyle\sum_{i=1}^N |f(t_{i-1}) - f(t_i)|$ , and the length of the arc is that approximation taken as $N \to \infty$ .

Now, $|f(t_{i-1}) - f(t_i)| = \dfrac{|f(t_{i-1}) - f(t_i)|}{|t_{i-1} - t_i|} \cdot |t_{i-1} - t_i|$ . As $N \to \infty$ , we have $|t_{i-1} - t_i| = \frac{2}{N} \to 0$ , so the left term is simply the definition of derivative of $f(t_{i-1})$ . Thus

$\displaystyle\begin{aligned} \lim_{N \to \infty} \sum_{i=1}^N |f(t_{i-1}) - f(t_i)| &= \lim_{N \to \infty} \sum_{i=1}^N \frac{|f(t_{i-1}) - f(t_i)|}{|t_{i-1} - t_i|} \cdot |t_{i-1} - t_i| \\ &= \int_{t_0}^{t_N} |f'(t)| \, dt \\ &= \int_{-1}^1 |f'(t)| \, dt \end{aligned}$

Now, $f'(t) = (1, 2t)$ , and $|f'(t)| = |(1, 2t)| = \max \{1, |2t|\}$ . Thus we can break the integral:

$\displaystyle\begin{aligned} \int_{-1}^1 |f'(t)| \, dt &= \int_{-1}^1 \max \{1, |2t|\} \, dt \\ &= \int_{-1}^{-0.5} \max \{1, |2t|\} \, dt + \int_{-0.5}^{0.5} \max \{1, |2t|\} \, dt + \int_{0.5}^{1} \max \{1, |2t|\} \, dt \\ &= \int_{-1}^{-0.5} -2t \, dt + \int_{-0.5}^{0.5} 1 \, dt + \int_{0.5}^{1} 2t \, dt \\ &= 0.75 + 1 + 0.75 \\ &= \boxed{2.5} \end{aligned}$

From the metric, $ds = \max \{ |dt|,|d(t^2)| \} = \max \{ dt,2|t| \ dt \}$ . For $t \in [-1,1]$ , $2|t| \ dt \leq dt$ iff $-\dfrac{1}{2} \leq t \leq \dfrac{1}{2}$ . Hence,

$\begin{aligned} \int_{-1}^1 ds &= \int_{-1}^1 \max \{ dt,2|t| \ dt \} \\ &= \int_{-1}^{-1/2} -2t \ dt + \int_{-1/2}^{1/2} \ dt + \int_{1/2}^1 2t \ dt \\ &= \frac{3}{4} + 1 + \frac{3}{4} \\ &= \frac{5}{2}. \end{aligned}$