Arc Length

Using the length formula $L = \int |\gamma'(t)|\,dt$ for $C^1$ curves, show completely that $\gamma(t) = (R\cos t, R\sin t)$ on $t \in [\alpha, \beta]$ has length $L = R(\beta - \alpha)$ . Then specialize to $R = 1$ on $[0, \pi/2]$ and to the full circle.

\gamma'(t) = (-R\sin t, R\cos t), \qquad |\gamma'(t)| = R

L = \int_\alpha^\beta |\gamma'(t)|\,dt = R(\beta - \alpha)

The quarter circle at

n = 16

: the polygonal length

L_{16} \approx 1.570166

approaches the exact value

\pi/2 \approx 1.570796

from below — the formula below gives that exact value.

Show proof

Differentiate componentwise: from $\gamma(t) = (R\cos t, R\sin t)$ we get $\gamma'(t) = (-R\sin t, R\cos t)$ . Since $\gamma$ is a $C^1$ curve, the length formula $L = \int_\alpha^\beta |\gamma'(t)|\,dt$ applies.
The speed is constant: $|\gamma'(t)| = \sqrt{(-R\sin t)^2 + (R\cos t)^2} = \sqrt{R^2\sin^2 t + R^2\cos^2 t} = R\sqrt{\sin^2 t + \cos^2 t} = R$ (for $R > 0$ ). The Pythagorean identity $\sin^2 t + \cos^2 t = 1$ cancels the angle dependence exactly.
Integrating, $L = \int_\alpha^\beta |\gamma'(t)|\,dt = \int_\alpha^\beta R\,dt = R(\beta - \alpha)$ . So the length of a circular arc is the radius times the subtended angle.
Specialize: with $R = 1$ and $[\alpha, \beta] = [0, \pi/2]$ , $L = 1 \cdot (\pi/2 - 0) = \pi/2$ — exactly the quarter circle of the scene. For the full circle, $[\alpha, \beta] = [0, 2\pi]$ , $L = R \cdot 2\pi = 2\pi R$ , the familiar circumference.

Prove completely that refinement never shortens a polygonal length: if $P'$ refines $P$ then $L(P) \le L(P')$ . Deduce that $L(P) \le L := \sup_P L(P)$ for every $P$ , and that the values along dyadic refinements increase while staying below $L$ .

|\gamma(t_i) - \gamma(t_{i-1})| \le |\gamma(t_i) - \gamma(s)| + |\gamma(s) - \gamma(t_{i-1})|

P \subseteq P' \implies L(P) \le L(P') \le L = \sup_P L(P)

The parabola at

n = 4

: the model values

L_4 \approx 1.474280

< L_8 \approx 1.477778

< L \approx 1.478943

— refinement lengthening the path, visible numerically.

Show proof

First insert a single point. Putting $s$ with $t_{i-1} < s < t_i$ into the subinterval $[t_{i-1}, t_i]$ replaces that subinterval's contribution $|\gamma(t_i) - \gamma(t_{i-1})|$ by $|\gamma(t_i) - \gamma(s)| + |\gamma(s) - \gamma(t_{i-1})|$ . Every other subinterval contributes exactly as before.
Apply the triangle inequality in $\mathbb{R}^2$ , $|u + v| \le |u| + |v|$ , with $u = \gamma(t_i) - \gamma(s)$ and $v = \gamma(s) - \gamma(t_{i-1})$ . Then $u + v = \gamma(t_i) - \gamma(t_{i-1})$ , so $|\gamma(t_i) - \gamma(t_{i-1})| \le |\gamma(t_i) - \gamma(s)| + |\gamma(s) - \gamma(t_{i-1})|$ . Hence the new contribution is at least the old one: inserting one point never shortens the polygonal length. A single straight piece cannot beat the two bent pieces replacing it.
Induction on finitely many inserted points gives the general case. If $P'$ refines $P$ , insert the points of $P' \setminus P$ one at a time to pass from $P$ to $P'$ ; the length never decreases at any step, so $L(P) \le L(P')$ .
Consequently every polygonal value lies below the supremum: $L(P) \le L := \sup_P L(P)$ for all $P$ , by the very definition of a supremum. In particular, along dyadic refinements of the uniform partition — each step inserting exactly the midpoints — the values increase: $L_n \le L_{2n} \le L_{4n} \le \cdots \le L$ . The scene's $L_4 \approx 1.474280 < L_8 \approx 1.477778$ is one slice of this monotonicity.

Let $f(x) = x\sin(1/x)$ on $(0, 1]$ and $f(0) = 0$ . Prove completely that although $f$ is continuous on $[0, 1]$ , the length of its graph is infinite — the graph is not rectifiable. Use the points $t_k = \frac{2}{\pi(2k+1)}$ for $k = 0, 1, \dots, K$ .

t_k = \frac{2}{\pi(2k+1)}, \qquad f(t_k) = (-1)^k t_k

L \ge \sum_{k=0}^{K-1} (t_k + t_{k+1}) \ge \frac{1}{\pi}\sum_{k=0}^{K-1} \frac{1}{k+1} \to \infty

The scene's smooth curves (the circular arc and the parabola) are all rectifiable — contrast them with the graph of

x\sin(1/x)

in this problem: continuous, yet of infinite length.

Show proof

Evaluate at the points. With $t_k = \frac{2}{\pi(2k+1)}$ we have $1/t_k = \pi(2k+1)/2 = \pi k + \pi/2$ , so $\sin(1/t_k) = \sin(\pi k + \pi/2) = (-1)^k$ , and hence $f(t_k) = t_k \sin(1/t_k) = (-1)^k t_k$ . Mind the order: $t_k$ decreases as $k$ grows, so $t_K < t_{K-1} < \cdots < t_0 = 2/\pi < 1$ . In the partition, list these points in increasing order, $t_K < \cdots < t_0$ , adding $0$ and $1$ if needed (adding points never shortens a polygonal length, by the previous problem).
Consecutive values alternate in sign: $f(t_k) = (-1)^k t_k$ and $f(t_{k+1}) = (-1)^{k+1} t_{k+1}$ , so $|f(t_k) - f(t_{k+1})| = t_k + t_{k+1} \ge t_k$ . The polygonal segment over $[t_{k+1}, t_k]$ has length $\sqrt{(t_k - t_{k+1})^2 + (f(t_k) - f(t_{k+1}))^2} \ge |f(t_k) - f(t_{k+1})|$ — the vertical drop alone is a lower bound for the segment.
Summing over $k = 0, \dots, K-1$ gives $L \ge \sum_{k=0}^{K-1} (t_k + t_{k+1}) \ge \sum_{k=0}^{K-1} t_k = \sum_{k=0}^{K-1} \frac{2}{\pi(2k+1)}$ . Now $\frac{2}{2k+1} \ge \frac{1}{k+1}$ , because cross-multiplying gives $2k + 2 \ge 2k + 1$ . Hence $L \ge \frac{1}{\pi}\sum_{k=0}^{K-1} \frac{1}{k+1} = \frac{1}{\pi} H_K$ .
The harmonic sum $H_K = \sum_{k=0}^{K-1} \frac{1}{k+1}$ diverges as $K \to \infty$ (divergence of the harmonic series). Since $K$ can be taken arbitrarily large, for every finite $M$ there is a partition with $L \ge \frac{1}{\pi} H_K > M$ . Thus $L = \sup_P L(P) = +\infty$ : the graph is not rectifiable. Yet $f$ is continuous on $[0, 1]$ — a product of continuous functions on $(0, 1]$ , and at $0$ , $|f(x)| \le x \to 0 = f(0)$ . Continuity alone does not guarantee finite length.

Examples & Exercises