# Differential Geometry/Arc Length

The length of a vector function **f** on an interval [a,b] is defined as sup{x|t_{n}∈[a,b], t_{n+1}>t_{n}, x= |**f**(t_{k})-**f**(t_{k-1})|}. If this number is finite, then this function is *rectifiable*.

For continuously differentiable vector functions, the arc length of that vector function on the interval [a,b] would be equal to .

Proof: Consider a partition a=t_{0}<t_{1}<t_{2}<...<t_{n}=b, and call it P_{n}. Let P_{n+1} be the partition P_{n} with an additional point, and let the sequence max{t_{n}-t_{n-1}} go into 0 as n goes to infinity, and let l_{n} be the arc length of the segments by joining the f(x) of the vector function. By the mean value theorem, there exists in the nth partition a number t_{n}' such that

.

Hence,

,

which is equal to

.

The amount

shall be denoted d_{j}. Because of the triangle inequality,

.

Each component is at least once continuously differentiable. There exists thus for any ε>0, there is a δ>0 such that

when

|a-b|<δ.

Therefore, if max{t_{n}-t_{n-1}}<δ, then d_{j}<ε, so that

ε(b-a) which approaches 0 when n approaches infinity.

Thus, the amount

approaches the integral since the right term approaches 0.

If there is another parametric representation from [a',b'], and one obtains another arc length, then

,

indicating that it is the same for any parametric representation.

The function s(t)= where t_{0} is a constant is called the arc length parameter of the curve. Its derivative turns out to be |f'(x)|.