2. arclength
Let Γ be a parametrized C1-curve in R2 with parametrizations x = x(t), y = y(t), for a ≤ t ≤ b.
Let P = {a = t0< t1< · · · < tn = b} be a partition of [a, b]. Denote (x(ti), y(ti)) by Pifor 0 ≤ i ≤ n.
We define the approximate arclength of Γ with respect to the partition P by s(Γ, P ) =
n
X
i=1
PiPi−1=
n
X
i=1
p(x(ti) − x(ti−1))2+ (y(ti) − y(ti−1))2.
Theorem 2.1. Let Γ be a C1 plane curve and I =
Z b a
px0(t)2+ y0(t)2dt.
For each > 0, there exists δ> 0 such that
|s(Γ, P ) − I| < whenever kP k < δ.
Proof. By mean value theorem, there exist ci, di∈ (ti−1, ti) such that
x(ti) − x(ti−1) = x0(ci)(ti− ti−1), y(ti) − y(ti−1) = y0(di)(ti− ti−1) for 1 ≤ i ≤ n. Then s(Γ, P ) can be rewritten as
s(Γ, P ) =
n
X
i=1
px0(ci)2+ y0(di)2(ti− ti−1).
Let us rewrite the sum as s(Γ, P ) =
n
X
i=1
px0(ci)2+ y0(ci)2(ti− ti−1) (2.1)
+
n
X
i=1
(p
x0(ci)2+ y0(di)2−p
x0(ci)2+ y0(ci)2)(ti− ti−1).
(2.2)
Since both x, y ∈ C1[a, b], x0, y0 ∈ C[a, b]. The function f (t) = px0(t)2+ y0(t)2 is continuous on [a, b]. Thus f is Riemann integrable over [a, b]. For > 0 we may choose δ0 such that whenever kP k < δ0, we have
n
X
i=1
f (ci)(ti− ti−1) − I
< 2. Here the Riemann sumPn
i=1f (ci)(ti− ti−1) equals to the right hand side of (2.1).
1
2
For each a, b1, b2∈ R, we can prove the following triangle inequality (2.3)
q
a2+ b21− q
a2+ b22
≤ |b1− b2|.
Using (2.4), we know (2.4)
px0(ci)2+ y0(di)2−p
x0(ci)2+ y0(ci)2
≤ |y0(ci) − y0(di)| for 1 ≤ i ≤ n.
Multiplying (2.4) by ti− ti−1and summing them up, we obtain
n
X
i=1
px0(ci)2+ y0(di)2−p
x0(ci)2+ y0(ci)2
(ti− ti−1) ≤
n
X
i=1
|y0(ci) − y0(di)|(ti− ti−1).
Since y0 is continuous on [a, b], it is uniformly continuous on [a, b]. For the given > 0, there exists δ00 such that |y0(s) − y0(t)| < /2(b − a) whenever |s − t| < δ00. Take δ = min{δ0, δ00}. Whenever kP k < δ, |ci− di| ≤ |ti− ti−1| ≤ kP k < δ. In this case, we have |y0(ci) − y0(di)| < /2(b − a) for 1 ≤ i ≤ n. Hence
n
X
i=1
|y0(ci) − y0(di)| (ti− ti−1) < 2(b − a)
n
X
i=1
(ti− ti−1) = 2. Here we use the factPn
i=1(ti− ti−1) = b − a. Thus
|s(Γ, P ) − I| ≤
n
X
i=1
px0(ci)2+ y0(ci)2(ti− ti−1) − I +
n
X
i=1
px0(ci)2+ y0(di)2−p
x0(ci)2+ y0(ci)2
(ti− ti−1)
< 2 +
2 = .
Definition 2.1. The arclength of a C1-curve Γ is defined to be
l(Γ) = Z b
a
px0(t)2+ y0(t)2dt.
A general definition of arc length of a (plane) curve Γ is given as follows. We define l(Γ) = sup
P
s(Γ, P)
where sup runs over all partition of [a, b]. We say that Γ is rectifiable if l(Γ) < ∞ and nonrectifiable otherwise. In this section, we proved that if Γ is C1, then Γ is rectifiable with
l(Γ) = Z b
a
px0(t)2+ y0(t)2dt.
Example 2.1. The Cr : x2+ y2 = r2 be the circle centered at (0, 0) of radius r. Then Cr can be parametrized by the functions
x(t) = r cos t, y(t) = r sin t, for t ∈ [0, 2π].
Find its perimeter (arclength).
Solution:
We compute x0(t) = −r sin t and y0(t) = r cos t for t ∈ [0, 2π]. Hence px0(t)2+ y0(t)2 = r for t ∈ [0, 2π]. Thus the arclength of Cris
l(Cr) = Z 2π
0
rdt = 2πr.
3
Example 2.2. Suppose Γ is a parametrized plane curve given by x = cos3t, y = sin3t, for t ∈ [0, 2π].
Find its arclength.
Solution:
We compute x0(t) = −3 cos2t sin t and y0(t) = 3 sin2t cos t. Thus px0(t)2+ y0(t)2 = 3| cos t sin t|
for t ∈ [0, 2π]. Hence the arclength of Γ is given by l(Γ) =
Z 2π 0
3| cos t sin t|dt = 3 · 4 Z π2
0
sin t cos tdt = 6 Z π2
0
sin 2tdt = −3 cos 2t|0π2 = 6.
Remark. If Γ is a plane curve defined by the graph of a C1-function y = f (x) for a ≤ x ≤ b, then its arclength is given by
(2.5) l(Γ) =
Z b a
p1 + f0(x)2dx,
and if Γ is defined by the graph of a C1-function x = g(y) for c ≤ y ≤ d, then its arclength is given by
(2.6) l(Γ) =
Z b a
p1 + g0(y)2dy.
Example 2.3. Find the arclength of
y = ln(cos x), for 0 ≤ x ≤ π 3. Solution:
We know d
dxln cos x = − tan x. Using (2.5), its arclength is given by the integral Z π3
0
p1 + tan2xdx = Z π3
0
sec xdx = ln | tan x + sec x||
π 3
0 = ln(2 +√ 3).
Example 2.4. Find the arclength of
x = 2y3/2, for 0 ≤ d ≤ 1.
Solution:
Using (5), its arclength is given by the integral Z 1
0
p1 + 9ydy = 2
3 · 9(1 + 9y)32
1
0
= 2 27(10√
10 − 1).
