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Arc length
Let $f$ be a parametric curve
$$f : {\bf R} \rightarrow {\bf R}^n $$
and $b > a$. Define the arc length of the curve from $f(a)$ to $f(b)$.
We "approximate" the curve with segments. Pick $s$ points
$$A_1, A_2, ..., A_s$$
on the curve with
The length of segment $[A_i, A_{i+1}]$ is
$$||A_i - A_{i+1}||.$$
Then the total length is
$$L({A_i}) \leq \displaystyle\sum_{i=1}^s ||A_i - A_{i+1}||.$$
Suppose the "true" length is equal to $l$. Then, by the Triangle Inequality, we have
$$L \leq l.$$
Definition. Define the arc length as the least upper bound of the set
$$L({ A_i }). $$
Consider the length of a curve $f: [a, b] \rightarrow {\bf R}^n$. By theorem, the length always exists. If the length is finite,
$$L( f; [a, b] ) \neq \infty, $$
then $f$ is called rectifiable.
Theorem. Suppose
$$f: [a, b] \rightarrow {\bf R}^n$$
is rectifiable and
$${\gamma}: [c, d] \rightarrow [a, b]$$
is monotone and continuous. Then
$$g = f {\gamma}: [c, d] \rightarrow {\bf R}^n$$
is rectifiable, and
$$L( f; [a, b] ) = L ( g; [c, d] ).$$
This theorem is about an re-parametrization of the curve.
$L( f; [a, b] )$ is approximated by the sum discussed above:
$$L({ A_i }) = \displaystyle\sum_{i=0}^{n-1} || f(a_{i+1}) - f(a_i) ||.$$
The idea is to turn this into an integral
$$\displaystyle\int_a^b h(x) dx.$$
As usual, this is done by recognizing the Riemann sum of some function, $h$:
$$\displaystyle\sum_{i=0}^{n-1} h(a_i) {\Delta}x, {\rm \hspace{3pt} where \hspace{3pt}} {\Delta}x = a_{i+1} - a_i.$$
To get to this point we re-write the sum
$$L({ A_i }) = \displaystyle\sum_{i=0}^{n-1} || f(a_{i+1}) - f(a_i) || \frac{{\Delta}x}{{\Delta}x}$$
to make
$$|| f(a_{i+1}) - f(a_i) || (\frac{1}{\Delta x}) = h(a_i).$$
Further
$$L({ A_i }) = \displaystyle\sum_{i=0}^{n-1} \frac{|| f(a_{i+1}) - f(a_i) ||}{a_{i+1} - a_i} {\Delta}x,$$
so that
$$\frac{|| f(a_{i+1}) - f(a_i) ||}{a_{i+1} - a_i} \rightarrow || f'(x) ||.$$
The limit is the following Riemann integral:
$$L( f; [a, b] ) = \displaystyle\int_a^b || f'(x) || dx.$$
Example. The length of a circle of radius $r$ is equal to $2{\pi}r$. How do we know? Describe the circle as
$$f(t) = r (\cos t, \sin t), t \in [0, 2{\pi}].$$
Now
$$f'(t) = r ( - \sin t, \cos t)$$
and
$$|| f'(t) || = r ( \sin^2 t + \cos^2 t)^{\frac{1}{2}} = r.$$
Then
$$L = \displaystyle\int_0^{2 \pi} r dt = rt |_0^{2 \pi} = r (2{\pi} - 0) = 2{\pi}r.$$