Ex2-9

Solution to Chapter 2, Exercise 9
Question: Let Cbe a curve with parametric equations x=\phi(t)and y=\psi(t), a \leq t \leq b.
(a) If \phiand \psi are of bounded variation and continuous, show that L=\lim_{|\Gamma|\to 0}l(\Gamma).
(b) If \phiand \psi are continuously differentiable, show that L=\int_a^b([\phi'(t)]^2+[\psi'(t)]^2)^{1/2}dt.
Proof:
(a) If \phiand \psiare of bounded variation, we know that Cis rectifiable. We want to prove that for any M<L, there exists \delta>0such that if |\Gamma|<\delta, then l(\Gamma)>M. Choose \mu>0so that M+\mu<L. Then, choose a partition \bar{\Gamma}of [a,b]such that L(\bar{\Gamma})>M+\mu and let \bar{\Gamma}=\{\bar{t_0},\bar{t_1},...,\bar{t_k}\}.
\phiand \psi are uniformly continuous since they are continuous on a compact set. Then there exists \delta'>0such that if |s-t|<\delta', then
(i) |\phi(s)-\phi(t)|<\frac{\mu}{\sqrt{8}(k+1)}
(ii) |\psi(s)-\psi(t)|<\frac{\mu}{\sqrt{8}(k+1)}
for all s,t\in[a,b].
Let \Gamma=\{t_0,t_1,...,t_m\}be any partition that satisfies
(iii) |\Gamma|<\delta'
(iv) |\Gamma|<\min\{\overline{t_i}-\overline{t_{i-1}}\vert1\leq i \leq m\}
We have
\begin{align}
l(\Gamma)&=\sum_{i=1}^m ([\phi(t_i)-\phi (t_{i-1})]^2+[\psi(t_i)-\psi (t_{i-1})]^2)^{1/2}\\
&=\sum' + \sum ~~~~~~~~~\text{(1)}
\end{align}
where \sum' is extended over all isuch that (t_{i-1},t_i)does not contain some \bar{t_j}while \sum ''is extended over all isuch that (t_{i-1},t_i)contains some \bar{t_j}. Notice that any (t_{i-1},t_i) can contain at most one \bar{t_j} by (iv), and the number of terms of \sum ''is at most k+1.
Now we consider the partition \Gamma \cup \bar{\Gamma}. l(\Gamma \cup \bar{\Gamma})can be written as
l(\Gamma \cup \bar{\Gamma})=\sum' + \sum '''
where \sum 'is the same as before while \sum '''is obtained from \sum ''by replacing each ([\phi(t_i)-\phi (t_{i-1})]^2+[\psi(t_i)-\psi (t_{i-1})]^2)^{1/2}term by ([\phi(t_i)-\phi (\bar{t_{j}})]^2+[\psi(t_i)-\psi (\bar{t_{j}})]^2)^{1/2} + ([\phi(\bar{t_j})-\phi (t_{i-1})]^2+[\psi(\bar{t_j})-\psi (t_{i-1})]^2)^{1/2}.
Then each term in \sum ''will be smaller than or equal to those in \sum '''and hence \sum ' \geq \sum . Thus, l(\Gamma \cup \bar{\Gamma}) \geq l(\Gamma). Similarly, we can show that l(\Gamma \cup \bar{\Gamma}) \geq l(\bar{\Gamma}).
\sum '''has at most 2(k+1) terms and by (i),(ii),(iii), we have
\begin{align}
\sum &< \left( \frac{\mu^2}{8(k+1)^2}+\frac{\mu^2}{8(k+1)^2} \right)^{1/2}2(k+1)\\
&=\left( \frac{\mu^2}{4(k+1)^2} \right)^{1/2}2(k+1)\\
&=\mu.
\end{align}
Since l(\Gamma \cup \bar{\Gamma}) = \sum ' + \sum ''', we have
\begin{align}
\sum ' &= l(\Gamma \cup \bar{\Gamma}) - \sum \\
&> l(\Gamma \cup \bar{\Gamma}) - \mu.
\end{align}
Recall that \bar{\Gamma} has been chosen so that l(\bar{\Gamma})>M+\mu. Then
\begin{align}
l(\Gamma) &= \sum ' + \sum \\
&\geq \sum '\\
&> l(\Gamma \cup \bar{\Gamma}) - \mu\\
&\geq l(\bar{\Gamma}) - \mu\\
&> M+\mu-\mu\\
&=M
\end{align}
Take \delta=\min\{ \delta',\min\{\overline{t_i}-\overline{t_{i-1}}\}\}, then for any \Gamma with |\Gamma|<\delta, we will have l(\Gamma)>M.
(b) Let \Gamma=\{t_0,t_1,...,t_m\}. By Mean Value Theorem,
\begin{align}
\phi(t_i)-\phi(t_{i-1})&=\phi'(c_i)(t_i-t_{i-1})\\
\psi(t_i)-\psi(t_{i-1})&=\psi'(d_i)(t_i-t_{i-1})
\end{align}
for some c_iand d_i in (t_{i-1},t_i). Then
\begin{align}
l(\Gamma)&=\sum_{i=1}^m ([\phi(t_i)-\phi (t_{i-1})]^2+[\psi(t_i)-\psi (t_{i-1})]^2)^{1/2}\\
&=\sum_{i=1}^m ([\phi'(c_i)(t_i-t_{i-1})]^2+[\psi'(d_i)(t_i-t_{i-1})]^2)^{1/2}\\
&=\sum_{i=1}^m ([\phi'(c_i)]^2+[\psi'(d_i)]^2)^{1/2}(t_i-t_{i-1})
\end{align}
If we take the limit as |\Gamma| \to 0on both sides, the left hand side converges to L by part (a) while the limit of the right hand side is the integral
\int_a^b ([\phi'(t)]^2+[\psi'(t)]^2)^{1/2}~dt