Math 275 – February 9

By ketchers

We will briefly comment on vector function integration 11.2: 3, 11, 19, 25 and move on to arc length and arc length parameterization in 11.3.

Recall that a curve is a continuous $\vec f:I \to \mathbb R^n$ where $I$ is an interval. Even continuous curve can be a bit complicated, for example, the Peano curve, fills space! – hence clearly has no well defined notion of length.The Peano curve is really wierd since it is a “one dimensional” object that has volume! A slightly less wierd curve with no notion of length is the Koch curve – since each little bit is infinitely long.

One way that a curve can have a well defined notion of lenght is that it be rectifiable. To define rectifiability we need to recall some notions learned – I hope – when you learned integration. A partition $P$ of an interval [a,b] is a sequence $a=t_0<t_1<\cdots<t_{n-1}<t_n=b$ , the size of the parition is $n$ , and is denoted $|P|=n$ , the norm of the partition is the maximum of the lengths of the subintervals determined by the partition and is denoted $||P||$ . Given a partition $P$ of [a,b] we get an approximation to the length of the curve by $L_P=\sum_{i=1}^n |\vec f(t_i)-\vec f(t_{i-1})|$ . A partition $P'$ is a refinement of $P$ if $P\subset P'$ . clearly:

if $P$ and $P'$ are two partitions we can find $P''$ refining both
if $P'$ is a refinement of $P$ , then $0\leq L_P\leq L_{P'}$

So we would have a reasonable notion of length if $L_C= \sup\{ L_P : P \text{ is a partition of } [a,b] \}$ is finite. (This is the same as demanding $\lim_{||P||\to 0} \sum_{i=1}^{|P|}|\vec r(t_i)-\vec r(t_{i-1})|$ exists.)

Definition. Let $\vec f:[a,b]\to \mathbb R^3$ be continuous (a curve). Then $\vec f$ is rectifiable iff $L_C$ exists. If $\vec f$ is rectifiable, then $L_C$ is the length of $\vec f$ .

If $\vec f:[a,b]\to \mathbb R^3$ is rectifiable and $t\in[a,b]$ , then set $s(t)$ to be the length of the curve got by restricting $f$ to [a,t]. For $c<d$ in [a,b], $s(d)-s(c)$ is the length of the curve got by restricting $\vec f$ to [c,d]. It is not hard to see that $s:[a,b]\to [0,L_C]$ is continuous and non-decreasing.

There is an exact characterization of rectifiability, but that is for some later course. The following gives a sufficient (not necessary) condition:

Theorem. If $f$ is continuously differentiable on [a,b], then $\vec f$ is rectifiable. Moreover, $s(t)=\int_a^t |\vec f'(u)|\, du$ so $s$ is continuously differentiable.

Proof. For rectififiability we simply need to show that there is a $L>0$ such that for any partition of [a,b], $L_P<L$ . We have

$L_P=\sum_{i=1}^{|P|} \sqrt{(f_1(t_i)-f_1(t_{i-1})^2+(f_2(t_i)-f_2(t_{i-1})^2+(f_3(t_i)-f_3(t_{i-1})^2}$

Since each $f_j$ is differentiable on ${[t_i,t_{i-1}]}$ MVT gives $c^j_i\in(t_i,t_{i-1})$ such that $f_j(t_i)-f_j(t_{i-1})=f'_j(c^j_i)(t_i-t_{i-1})$ . So we get

$L_P=\sum_{i=1}^{|P|} \sqrt{[f_1'(c^1_i)(t_i-t_{i-1})]^2+[f_2'(c^2_i)(t_i-t_{i-1})]^2+[f_3'(c^3_i)(t_i-t_{i-1})]^2}=\sum_{i=1}^{|P|}\sqrt{[f_1'(c^1_i)]^2+[f_2'(c^2_i)]^2+[f_3'(c^3_i)]^2}(t_i=t_{i-1})$

Using the triangle inequality we have $\sqrt{[f_1'(c^1_i)]^2+[f_2'(c^2_i)]^2+[f_3'(c^3_i)]^2}<|f_1'(c^1_i)|+|f_2'(c^2_i)|+|f_3'(c^3_i)|$ so

$L_P\leq \sum_{i=1}^{|P|} (|f_1'(c^1_i)|+|f_2'(c^2_i)|+|f_3'(c^3_i)|)(t_i-t_{i-1})$

Since $f_i'$ is continuous on [a,b] it has a maximum value (EVT) so in particular $f_i'(t) \leq M_i$ for all $t\in [a,b]$ . Thus

$L_C\leq \sum_{i=1}^{|P|} (M_1+M_2+M_3)(t_i-t_{i-1}) = (M_1+M_2+M_3)(b-a)$

So we have the desired upper bound.

For the rest we actually compute $\frac{ds}{dt}(t)$ from the definition of derivative. The key point is that as above, from MVT for the interval ${[t,t+h]}$ , we have $c^j$ from MVT so that

$\frac{|\vec f(t+h) - \vec f(t)|}{h}\leq \frac{s(t+h)-s(t)}{h} \leq \sqrt{[f_1'(c^1)]^2+[f_2'(c^2)]^2+[f_3'(c^3)]^2}$

$\lim_{h\to 0^+}\frac{|\vec f(t+h) - \vec f(t)|}{h}\leq \lim_{h\to 0^+}\frac{s(t+h)-s(t)}{h} \leq \lim_{h\to 0^+}\sqrt{[f_1'(c^1)]^2+[f_2'(c^2)]^2+[f_3'(c^3)]^2}$

$|\vec f'(t)|\leq s'(t) \leq \lim_{h\to 0^+}\sqrt{[f_1'(c^1)]^2+[f_2'(c^2)]^2+[f_3'(c^3)]^2}$

But $t<c^j<t+h$ so $c^j\to t$ as $h\to 0$ and since $f'_j$ is continuous at $t$ we have for each j, $\lim_{h\to 0^+} f'_j(c^j)=f'_j(t)$ , so $\lim_{h\to 0^+}\sqrt{[f_1'(c^1)]^2+[f_2'(c^2)]^2+[f_3'(c^3)]^2}= \sqrt{[f_1'(t)]^2+[f_2'(t)]^2+[f_3'(t)]^2}=|\vec f'(t)|$ . The same argument works for $h\to 0^-$ and so we have:

$|\vec f'(t)|\leq s'(t) \leq |\vec f'(t)|$

hence $s'(t)=|\vec f'(t)|$ . QED

This entry was posted on February 12, 2009 at 6:39 am and is filed under Math 275, Spring 2009. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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Math 275 – February 9

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