If you are interested in more on space curves, you should read this section. Otherwise,
proceed to the exercises. Denote by R
(s)
the function which takes s to a point on this
curve where s is arc length. Thus R
(s)
equals the point on the curve which occurs when
you have traveled a distance of s along the curve from one end. This is known as the
parametrization of the curve in terms of arc length. Note also that it incorporates an
orientation on the curve because there are exactly two ends you could begin measuring
length from. In this section, assume anything about smoothness and continuity to make
the following manipulations valid. In particular, assume that R^{′} exists and is
continuous.
Lemma 16.5.1Define T
(s)
≡ R^{′}
(s)
. Then
|T (s)|
= 1 and if T^{′}
(s)
≠0, thenthere exists a unit vector N
(s)
perpendicular to T
(s)
and a scalar valued functionκ
(s)
with T^{′}
(s)
= κ
(s)
N
(s)
.
Proof:First, s = ∫_{0}^{s}
|R ′(r)|
dr because of the definition of arc length. Therefore,
from the fundamental theorem of calculus, 1 =
|R′(s)|
=
|T (s)|
. Therefore, T ⋅ T = 1
and so upon differentiating this on both sides, yields T^{′}⋅T + T ⋅ T^{′} = 0 which shows
T ⋅ T^{′} = 0. Therefore, the vector T^{′} is perpendicular to the vector T. In case T^{′}
(s)
≠0,
let N
(s)
=
-T′(s)
|T′(s)|
and so T^{′}
(s)
=
′
|T (s)|
N
(s)
, showing the scalar valued function is
κ
(s)
=
′
|T (s)|
. ■
The radius of curvature is defined as ρ =
1
κ
. Thus at points where there is
a lot of curvature, the radius of curvature is small and at points where the
curvature is small, the radius of curvature is large. The plane determined by the two
vectors T and N is called the osculating plane. It identifies a particular plane
which is in a sense tangent to this space curve. In the case where
′
|T (s)|
= 0
near the point of interest, T
(s)
equals a constant and so the space curve is a
straight line which it would be supposed has no curvature. Also, the principal
normal is undefined in this case. This makes sense because if there is no curving
going on, there is no special direction normal to the curve at such points which
could be distinguished from any other direction normal to the curve. In the case
where
|T′(s)|
= 0, κ
(s)
= 0 and the radius of curvature would be considered
infinite.
Definition 16.5.2The vector T
(s)
is called theunit tangent vector andthe vector N
(s)
is called theprincipal normal. The function κ
(s)
in the abovelemma is called the curvature.When T^{′}
(s)
≠0 so the principal normal is defined,the vector B
(s)
≡ T
(s)
× N
(s)
is called the binormal.
The binormal is normal to the osculating plane and B^{′} tells how fast this vector
changes. Thus it measures the rate at which the curve twists.
Lemma 16.5.3Let R
(s)
be a parametrization of a space curve with respect toarc length and let the vectorsT,N,and B be as defined above. Then B^{′} = T × N^{′}and there exists a scalar function τ
(s)
such that B^{′} = τN.
Proof:From the definition of B = T × N, and you can differentiate both sides and
get B^{′} = T^{′}×N + T × N^{′}. Now recall that T^{′} is a multiple called curvature multiplied
by N so the vectors T^{′} and N have the same direction, so B^{′} = T × N^{′}. Therefore, B^{′} is
either zero or is perpendicular to T. But also, from the definition of B,B is a unit vector
and so B
(s)
⋅B
(s)
= 1. Differentiating this, B^{′}
(s)
⋅B
(s)
+ B
(s)
⋅B^{′}
(s)
= 0 showing that
B^{′} is perpendicular to B also. Therefore, B^{′} is a vector which is perpendicular to
both vectors T and B and since this is in three dimensions, B^{′} must be some
scalar multiple of N, and this multiple is called τ. Thus B^{′} = τN as claimed.
■
Lets go over this last claim a little more. The following situation is obtained. There
are two vectors T and B which are perpendicular to each other and both B^{′} and
N are perpendicular to these two vectors, hence perpendicular to the plane
determined by them. Therefore, B^{′} must be a multiple of N. Take a piece of paper,
draw two unit vectors on it which are perpendicular. Then you can see that any
two vectors which are perpendicular to this plane must be multiples of each
other.
The scalar function τ is called the torsion. In case T^{′} = 0, none of this is defined
because in this case there is not a well defined osculating plane. The conclusion of the
following theorem is called the Serret Frenet formulas.
Theorem 16.5.4(Serret Frenet) Let R
(s)
be the parametrization with respectto arc length of a space curve and T
(s)
= R^{′}
(s)
is the unit tangent vector. Suppose
|T ′(s)|
≠0 so the principal normal N
(s)
=
T′′(s)-
|T (s)|
is defined. The binormal is the vectorB ≡ T × NsoT,N,Bforms a right handed system of unit vectors each of which isperpendicular to every other. Then the following system of differential equations holds inℝ^{9}.
B ′ = τN,T ′ = κN, N′ = − κT − τB
where κ is the curvature and is nonnegative and τ is thetorsion.
Proof: κ ≥ 0 because κ =
|T′(s)|
. The first two equations are already
established. To get the third, note that B × T = N which follows because T,N,B
is given to form a right handed system of unit vectors each perpendicular to
the others. (Use your right hand.) Now take the derivative of this expression.
thus
N′ = B′ × T + B × T′ = τN × T+ κB × N.
Now recall again that T,N,B is a right hand system. Thus
N × T = − B, B × N = − T.
This establishes the Frenet Serret formulas. ■
This is an important example of a system of differential equations in ℝ^{9}. It is a
remarkable result because it says that from knowledge of the two scalar functions τ and
κ, and initial values for B,T, and N when s = 0 you can obtain the binormal, unit
tangent, and principal normal vectors. It is just the solution of an initial value problem
although this is for a vector valued rather than scalar valued function. Having done this,
you can reconstruct the entire space curve starting at some point R_{0} because
R^{′}
(s)
= T
(s)
and so R
(s)
= R_{0} + ∫_{0}^{s}T
(r)
dr. There are ways to solve such a system of
equations numerically and even draw the graph of the resulting curve but this is not a
topic for this book.