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 ∫ b ′
f (s)ds = f (ϕ(t))|ϕ (t)|dt
c a
Proof:Let F^{′}
(s)
= f
(s)
. (For example, let F
(s)
= ∫_{a}^{s}f
(r)
dr.) Then the first integral
equals F
(d)
− F
(c)
by the fundamental theorem of calculus. Since ϕ is one to one, it
follows from Lemma 16.2.1 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 d-
− a f (ϕ(t))ϕ (t)dt = b dt (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 16.2.6Letf :
[a,b]
→ C,g :
[c,d]
→ C be parameterizations of asmooth curve which satisfy conditions 1- 5. Thenϕ
(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 only
remains to verify continuity on
[a,b]
because then the final claim follows from Lemma
16.2.1. 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]
, 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 16.2.7The length of a smooth curve is not dependent on whichparametrization 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
hi(gi(s)) = s.
This implies that for s ∈ I,
h ′i(gi(s)) = -′1-. (16.9)
gi(s)
(16.9)
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′(fi(t))f′(t) = h′(gi(s))f′(t) =-fi (t) (16.10)
i i i i g′i(ϕ (t))
(16.10)
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∘
( −1 )
g ∘ f
(t)
= g
(ϕ(t))
, and it was just shown that ϕ^{′} is a continuous
function on
[a − δ,b+ δ]
. It follows from the chain rule applied to the components that
f^{′}