Page 104 - Vector Analysis

100 CHAPTER 4. Vector Calculus

In general, the arc-length parametrization of a rectifiable curve exists, and we have the
following

Theorem 4.14. Let C Ď Rn be a rectifiable simple curve. Then there exists an arc-length
parametrization of C.

Proof. We only prove the case that C has a regular C 1-parametrization γ : [a, b] Ñ Rn.

żt

Let s(t) = }γ 1(t1)}Rn dt1. Note that the s : [a, b] Ñ R is strictly increasing since

a

the fundamental theorem of Calculus implies that s1(t) = }γ 1(t)}Rn ą 0 for all t P [a, b].

The Inverse Function Theorem (Theorem A.10) then guarantees that s has a C 1-inverse

u : [0, ℓ(C)] Ñ [a, b] and we have u1(t) = 1. Define γr = γ ˝ u. Then the chain rule

s 1(u(t))

implies that γ : [0, ℓ(C )] Ñ Rn is a C 1-parametrization of C, and Theorem 4.10 implies that
r

( ) ż s ż s ż s
ℓ γ([0, s])
= }γr 1(t)}Rn dt = }γ 1(u(t))u 1(t)}Rn dt = }γ 1(u(t))}Rn ˇˇu 1(t)ˇˇ dt
r

00 0

= żs s 1(u(t)) ˇˇs 1 dt = żs 1dt = s
1(u(t))ˇˇ
0 0

which implies that γr : [0, ℓ(C)] is an arc-length parametrization of C. ˝

Theorem 4.15. Let C Ď Rn be a C 1-curve with an arc-length parametrization γ : I Ñ Rn.
Then }γ 1(s)}Rn = 1 for all s P I.

Proof. Suppose that I = [a, b]. Since γ : I Ñ Rn is an arc-length parametrization of C, we

must have żs
s ´ a = }γ 1(t)}Rn dt
@s P I.
a

Differentiating both sides of the equality above in t, the fundamental theorem of Calculus

implies that 1 = }γ 1(s)}Rn for all s P I. ˝

4.1.2 The line element and line integrals

Line elements

Definition 4.16. A curve C Ď Rn is said to be piecewise C k (smooth, regular) if there exists
a parametrization γ : [a, b] Ñ Rn and a finite set of points ta = t0 ă t1 ă ¨ ¨ ¨ ă tN = bu
such that γ : [ti, ti+1] Ñ Rn is C k (smooth, regular) for all i P t0, 1, ¨ ¨ ¨ , N ´ 1u.