by the fundamental theorem of calculus. Since ϕ is one to one, it follows from Lemma 9.6.1
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 the second integral equals
∫ b ∫ a
− f (ϕ(t)) ϕ′(t)dt = d-(F (ϕ (t))) dt = F (ϕ (a))− F (ϕ(b)) = F (d) − F (c).
a b dt
The case when ϕ is increasing is similar but easier. ■
Lemma 9.6.6Letf :
[a,b]
→ C,g :
[c,d]
→ C be parameterizations of a smooth curve which satisfyconditions (1) - (5). Thenϕ
(t)
≡ g^{−1}∘ f
(t)
is 1 − 1 on
(a,b)
, continuous on
[a,b]
, and eitherstrictly 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 only remains to verify
continuity on
[a,b]
because then the final claim follows from Lemma 9.6.1. If ϕ is not continuous on
[a,b]
,
then there exists a sequence,
{tn}
⊆
[a,b]
such that t_{n}→ t but ϕ
(tn)
fails to converge to ϕ
(t)
. Therefore,
for some ε > 0, there exists a subsequence, still denoted by n such that
|ϕ(tn)− ϕ(t)|
≥ ε. By sequential
compactness of
[c,d]
, (See Theorem 8.10.12 on Page 458.) there is a further subsequence, still denoted by
n, such that
{ϕ (tn)}
converges to a point s, of
[c,d]
which is not equal to ϕ
(t)
. Thus g^{−1}∘f
(tn)
→ s while
t_{n}→ t. Therefore, the continuity of f and g imply f
(tn)
→ g
(s)
and f
(tn)
→ f
(t)
. Thus,
g
(s)
= f
(t)
, so s = g^{−1}∘ f
(t)
= ϕ
(t)
, a contradiction. Therefore, ϕ is continuous as claimed.
■
Theorem 9.6.7The length of a smooth curve is not dependent on which parametrization is used.
Proof:Let C be the curve and suppose f :
[a,b]
→ C and g :
[c,d]
→ C both satisfy conditions (1) -
(5). Is it true that ∫_{a}^{b}
′
|f (t)|
dt = ∫_{c}^{d}
′
|g (s)|
ds?
Let ϕ
(t)
≡ g^{−1}∘ f
(t)
for t ∈
[a,b]
. I want to show that ϕ is C^{1} on an interval of the form
[a +δ,b− δ]
. By the above lemma, ϕ is either strictly increasing or strictly decreasing on
[a,b]
.
Suppose for the sake of simplicity that it is strictly increasing. The decreasing case is handled
similarly.
Let s_{0}∈ ϕ
([a + δ,b − δ])
⊂
(c,d)
. Then by assumption 4 for smooth curves, g_{i}^{′}
(s0)
≠0 for some i. By
continuity of g_{i}^{′}, it follows g_{i}^{′}
(s)
≠0 for all s ∈ I where I is an open interval contained in
[c,d]
which
contains s_{0}. It follows from the mean value theorem that on this interval g_{i} is either strictly increasing or
strictly decreasing. Therefore, J ≡ g_{i}
(I)
is also an open interval and you can define a differentiable
function h_{i} : J → I by
h (g (s)) = s.
i i
This implies that for s ∈ I,
′ 1
hi(gi(s)) = g′(s). (9.10)
i
(9.10)
Now letting s = ϕ
(t)
for s ∈ I, it follows t ∈ J_{1}, an open interval. Also, for s and t related this way,
f
(t)
= g
(s)
and so in particular, for s ∈ I, g_{i}
(s)
= f_{i}
(t)
. Consequently,
s = hi(gi(s)) = hi(fi(t)) = ϕ(t)
and so, for t ∈ J_{1},
ϕ′(t) = h′(f (t))f′(t) = h′(g (s))f′(t) =-f′i (t)- (9.11)
i i i i i i g′i(ϕ (t))
(9.11)
which shows that ϕ^{′} exists and is continuous on J_{1}, an open interval containing ϕ^{−1}
(s0)
. Since s_{0} is
arbitrary, this shows ϕ^{′} exists on
[a+ δ,b− δ]
and is continuous there.
Now f
(t)
= g∘
( )
g−1 ∘ f
(t)
= g
(ϕ(t))
, and it was just shown that ϕ^{′} is a continuous function
on