342 CHAPTER 16. SPACE CURVES

Corollary 16.4.4 Let f : (a,b) → R be continuous and one to one. Suppose f ′ (x1)

exists for some x1 ∈ (a,b) and f ′ (x1) ̸= 0. Then(

f−1)′( f (x1)) exists and is given by the

formula(

f−1)′( f (x1)) =

1f ′(x1)

.

Proof: From the definition of the derivative and continuity of f−1,

limf (x)→ f (x1)

f−1 ( f (x))− f−1 ( f (x1))

f (x)− f (x1)= lim

x→x1

x− x1

f (x)− f (x1)=

1f ′ (x1)

.

16.4.2 Independence of Parametrization

Theorem 16.4.5 Let φ : [a,b] → [c,d] be one to one and suppose φ′ exists and is

continuous on [a,b]. Then if f is a continuous function defined on [c,d]∫ d

cf (s) ds =

∫ b

af (φ (t))

∣∣φ ′ (t)∣∣ dt

Proof: Let F ′ (s) = f (s). (For example, let F (s) =∫ s

a f (r) dr.) Then the first integralequals F (d)−F (c) by the fundamental theorem of calculus. Since φ is one to one, itfollows from Lemma 16.4.1 above that φ is either strictly increasing or strictly decreasing.Suppose φ is strictly decreasing. Then φ (a) = d and φ (b) = c. Therefore, φ

′ ≤ 0 and thesecond integral equals

−∫ b

af (φ (t))φ

′ (t) dt =∫ a

b

ddt

(F (φ (t))) dt = F (φ (a))−F (φ (b)) = F (d)−F (c) .

The case when φ is increasing is similar but easier.

Lemma 16.4.6 Let f : [a,b]→C, g : [c,d]→C be parameterizations of a smooth curvewhich satisfy conditions 1 - 5. Then ϕ(t) ≡ g−1 ◦f (t) is 1− 1 on (a,b), continuous on[a,b], and either strictly increasing or strictly decreasing on [a,b].

Proof: It is obvious φ is 1− 1 on (a,b) from the conditions f and g satisfy. It onlyremains to verify continuity on [a,b] because then the final claim follows from Lemma16.4.1. If φ is not continuous on [a,b], then there exists a sequence, {tn} ⊆ [a,b] suchthat tn → t but φ (tn) fails to converge to φ (t). Therefore, for some ε > 0, there exists asubsequence, still denoted by n such that |φ (tn)−φ (t)| ≥ ε . By sequential compactnessof [c,d], there is a further subsequence, still denoted by n, such that {φ (tn)} converges to apoint s, of [c,d] which is not equal to φ (t). Thus g−1 ◦f (tn)→ s while tn → t. Therefore,the continuity of f and g imply f (tn)→ g (s) and f (tn)→ f (t). Thus, g (s) = f (t), sos = g−1 ◦f (t) = φ (t), a contradiction. Therefore, φ is continuous as claimed.

Theorem 16.4.7 The length of a smooth curve is not dependent on which paramet-rization is used.

Proof: Let C be the curve and suppose f : [a,b] → C and g : [c,d] → C both satisfyconditions 1 - 5. Is it true that

∫ ba

∣∣f ′ (t)∣∣ dt =

∫ dc |g′ (s)| ds?

Let φ (t)≡ g−1◦f (t) for t ∈ [a,b]. I want to show that φ is C1 on an interval of the form[a+δ ,b−δ ]. By the above lemma, φ is either strictly increasing or strictly decreasing on