1Euclidean geometry
IB Geometry
1.2 Curves in R
n
Next we will briefly look at curves in
R
n
. This will be very brief, since curves in
R
n
aren’t too interesting. The most interesting fact might be that the shortest
curve between two points is a straight line, but we aren’t even proving that,
because it is left for the example sheet.
Definition (Curve). A curve Γ in R
n
is a continuous map Γ : [a, b] → R
n
.
Here we can think of the curve as the trajectory of a particle moving through
time. Our main objective of this section is to define the length of a curve. We
might want to define the length as
Z
b
a
kΓ
0
(t)k dt,
as is familiar from, say, IA Vector Calculus. However, we can’t do this, since our
definition of a curve does not require Γ to be continuously differentiable. It is
merely required to be continuous. Hence we have to define the length in a more
roundabout way.
Similar to the definition of the Riemann integral, we consider a dissection
D = a = t
0
< t
1
< ··· < t
N
= b of [a, b], and set P
i
= Γ(t
i
). We define
S
D
=
X
i
k
−−−−→
P
i
P
i+1
k.
P
0
P
1
P
2
P
N−1
P
N
Notice that if we add more points to the dissection, then
S
D
will necessarily
increase, by the triangle inequality. So it makes sense to define the length as the
following supremum:
Definition (Length of curve). The length of a curve Γ : [a, b] → R
n
is
` = sup
D
S
D
,
if the supremum exists.
Alternatively, if we let
mesh(D) = max
i
(t
i
− t
i−1
),
then if ` exists, then we have
` = lim
mesh(D)→0
s
D
.
Note also that by definition, we can write
` = inf{
˜
` :
˜
` ≥ S
D
for all D}.
The definition by itself isn’t too helpful, since there is no nice and easy way to
check if the supremum exists. However, differentiability allows us to compute
this easily in the expected way.
Proposition. If Γ is continuously differentiable (i.e.
C
1
), then the length of Γ
is given by
length(Γ) =
Z
b
a
kΓ
0
(t)k dt.
The proof is just a careful check that the definition of the integral coincides
with the definition of length.
Proof.
To simplify notation, we assume
n
= 3. However, the proof works for all
possible dimensions. We write
Γ(t) = (f
1
(t), f
2
(t), f
3
(t)).
For every s 6= t ∈ [a, b], the mean value theorem tells us
f
i
(t) − f
i
(s)
t − s
= f
0
i
(ξ
i
)
for some ξ
i
∈ (s, t), for all i = 1, 2, 3.
Now note that
f
0
i
are continuous on a closed, bounded interval, and hence
uniformly continuous. For all
ε ∈
0, there is some
δ >
0 such that
|t − s| < δ
implies
|f
0
i
(ξ
i
) − f
0
(ξ)| <
ε
3
for all ξ ∈ (s, t). Thus, for any ξ ∈ (s, t), we have
Γ(t) − Γ(s)
t − s
− Γ
0
(ξ)
=
f
0
1
(ξ
1
)
f
0
2
(ξ
2
)
f
0
3
(ξ
3
)
−
f
0
1
(ξ)
f
0
2
(ξ)
f
0
3
(ξ)
≤
ε
3
+
ε
3
+
ε
3
= ε.
In other words,
kΓ(t) − Γ(s) − (t − s)Γ
0
(ξ)k ≤ ε(t − s).
We relabel t = t
i
, s = t
i−1
and ξ =
s+t
2
.
Using the triangle inequality, we have
(t
i
− t
i−1
)
Γ
0
t
i
+ t
i−1
2
− ε(t
i
− t
i−1
) < kΓ(t
i
) − Γ(t
i−1
)k
< (t
i
− t
i−1
)
Γ
0
t
i
+ t
i−1
2
+ ε(t
i
− t
i−1
).
Summing over all i, we obtain
X
i
(t
i
− t
i−1
)
Γ
0
t
i
+ t
i−1
2
− ε(b − a) < S
D
<
X
i
(t
i
− t
i−1
)
Γ
0
t
i
+ t
i−1
2
+ ε(b − a),
which is valid whenever mesh(D) < δ.
Since Γ
0
is continuous, and hence integrable, we know
X
i
(t
i
− t
i−1
)
Γ
0
t
i
+ t
i−1
2
→
Z
b
a
kΓ
0
(t)k dt
as mesh(D) → 0, and
length(Γ) = lim
mesh(D)→0
S
D
=
Z
b
a
kΓ
0
(t)k dt.
This is all we are going to say about Euclidean space.