2Curves and Line
IA Vector Calculus
2.1 Parametrised curves, lengths and arc length
There are many ways we can described a curve. We can, say, describe it by
a equation that the points on the curve satisfy. For example, a circle can be
described by
x
2
+
y
2
= 1. However, this is not a good way to do so, as it is
rather difficult to work with. It is also often difficult to find a closed form like
this for a curve.
Instead, we can imagine the curve to be specified by a particle moving along
the path. So it is represented by a function
x
:
R → R
n
, and the curve itself is
the image of the function. This is known as a parametrisation of a curve. In
addition to simplified notation, this also has the benefit of giving the curve an
orientation.
Definition
(Parametrisation of curve)
.
Given a curve
C
in
R
n
, a parametrisation
of it is a continuous and invertible function
r
:
D → R
n
for some
D ⊆ R
whose
image is C.
r
0
(
u
) is a vector tangent to the curve at each point. A parametrization is
regular if r
0
(u) 6= 0 for all u.
Clearly, a curve can have many different parametrizations.
Example. The curve
1
4
x
2
+ y
2
= 1, y ≥ 0, z = 3.
can be parametrised by 2 cos u
ˆ
i + sin u
ˆ
j + 3
ˆ
k
If we change
u
(and hence
r
) by a small amount, then the distance
|δr|
is
roughly equal to the change in arclength
δs
. So
δs
=
|δr|
+
o
(
δr
). Then we have
Proposition. Let s denote the arclength of a curve r(u). Then
ds
du
= ±
dr
du
= ±|r
0
(u)|
with the sign depending on whether it is in the direction of increasing or decreasing
arclength.
Example. Consider a helix described by r(u) = (3 cos u, 3 sin u, 4u). Then
r
0
(u) = (−3 sin u, 3 cos u, 4)
ds
du
= |r
0
(u)| =
p
3
2
+ 4
2
= 5
So s = 5u. i.e. the arclength from r(0) and r(u) is s = 5u.
We can change parametrisation of
r
by taking an invertible smooth function
u 7→ ˜u
, and have a new parametrization
r
(
˜u
) =
r
(
˜u
(
u
)). Then by the chain rule,
dr
du
=
dr
d˜u
×
d˜u
du
dr
d˜u
=
dr
du
/
d˜u
du
It is often convenient to use the arclength
s
as the parameter. Then the tangent
vector will always have unit length since the proposition above yields
|r
0
(s)| =
ds
ds
= 1.
We call d
s
the scalar line element, which will be used when we consider integrals.
Definition (Scalar line element). The scalar line element of C is ds.
Proposition. ds = ±|r
0
(u)|du