Math 497C
Sep 17, 2004
1
Curves and Surfaces
Fall 2004, PSU
Lecture Notes 4
1.9
Curves of Constant Curvature
Here we show that the only curves in the plane with constant curvature are
lines and circles. The case of lines occurs precisely when the curvature is
zero:
Exercise 1.
Show that the only curves with constant zero curvature in
R
n
are straight lines. (
Hint
: We may assume that our curve,
α
:
I
→
R
n
has unit
speed. Then
κ
=
α
. So zero curvature implies that
α
= 0. Integrating
the last expression twice yields the desired result.)
So it remains to consider the case where we have a planar curve whose
curvature is equal to some nonzero constant
c
. We claim that in this case the
curve has to be a circle of radius 1
/c
. To this end we introduce the following
definition. If a curve
α
:
I
→
R
n
has nonzero curvature, the
principal normal
vector field of
α
is defined as
N
(
t
) :=
T
(
t
)
T
(
t
)
,
where
T
(
t
) :=
α
(
t
)
/ α
(
t
)
is the tantrix of
α
as we had defined earlier.
Thus the principal normal is the tantrix of the tantrix.
Exercise 2.
Show that
T
(
t
) and
N
(
t
) are orthogonal. (
Hint
: Differentiate
both sides of the expression
T
(
t
)
, T
(
t
) = 1).
So, if
α
is a planar curve,
{
T
(
t
)
, N
(
t
)
}
form a
moving frame
for
R
2
, i.e.,
any element of
R
2
may be written as a linear combination of
T
(
t
) and
N
(
t
)
for any choice of
t
. In particular, we may express the derivatives of
T
and
1
Last revised: September 20, 2004
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N
in terms of this frame. The definition of
N
already yields that, when
α
is
parametrized by arclength,
T
(
t
) =
κ
(
t
)
N
(
t
)
.
To get the corresponding formula for
N
, first observe that
N
(
t
) =
aT
(
t
) +
bN
(
t
)
.
for some
a
and
b
. To find
a
note that, since
T, N
= 0,
T , N
=
−
T, N
.
Thus
α
=
N
(
t
)
, T
(
t
) =
−
T
(
t
)
, N
(
t
) =
−
κ
(
t
)
.
Exercise 3.
Show that
b
= 0. (
Hint:
Differentiate
N
(
t
)
, N
(
t
) = 1).
So we conclude that
N
(
t
) =
−
κ
(
t
)
T
(
t
)
,
where we still assume that
t
is the arclength parameter. The formulas for
the derivative may be expressed in the matrix notation as
T
(
t
)
N
(
t
)
=
κ
(
t
)
0
0
−
κ
(
t
)
T
(
t
)
N
(
t
)
.
Now recall that our main aim here is to classify curves of constant curva
ture in the plane. To this end define the
center of the osculating circle
of
α
as
p
(
t
) :=
α
(
t
) +
1
κ
(
t
)
N
(
t
)
.
The circle which is centered at
p
(
t
) and has radius of 1
/κ
(
t
) is called the
osculating
circle of
α
at time
t
. This is the circle which best approximates
α
up to the second order:
Exercise 4.
Check that the osculating circle of
α
is tangent to
α
at
α
(
t
)
and has the same curvature as
α
at time
t
.
Now note that if
α
is a circle, then it coincides with its own osculating
circle. In particular
p
(
t
) is a fixed point (the center of the circle) and
α
(
t
)
−
p
(
t
) is constant (the radius of the circle). Conversely:
Exercise 5.
Show that if
α
has constant curvature
c
, then (i)
p
(
t
) is a fixed
point, and (ii)
α
(
t
)
−
p
(
t
) = 1
/c
(
Hint
: For part (i) differentiate
p
(
t
); part
(ii) follows immediately from the definition of
p
(
t
)).
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 Math, Geometry, Osculating circle, constant curvature

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