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 10.2.1Define T
(s)
≡ R^{′}
(s)
. Then
|T (s)|
= 1 and if T^{′}
(s)
≠0, then there exists a unitvector N
(s)
perpendicular to T
(s)
and a scalar valued function κ
(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|T-(′(s)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 10.2.2The vector T
(s)
is called theunit tangent vector and the vector N
(s)
is calledtheprincipal normal. The function κ
(s)
in the above lemma is called the curvature.WhenT^{′}
(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 10.2.3Let R
(s)
be a parametrization of a space curve with respect to arc length and letthe vectorsT,N,and B be as defined above. Then B^{′} = T × N^{′}and there exists a scalar functionτ
(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 10.2.4(Serret Frenet) Let R
(s)
be the parametrization with respect to arc length of a spacecurve and T
(s)
= R^{′}
(s)
is the unit tangent vector. Suppose
|T ′(s)|
≠0 so the principal normalN
(s)
=
|TT′′(s(s))|
is defined. The binormal is the vectorB ≡ T × NsoT,N,Bforms a right handed system ofunit vectors each of which is perpendicular to every other. Then the following system of differentialequations holds in ℝ^{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.
The vectors B,T, and N are vectors which are functions of position on the space curve. Often,
especially in applications, you deal with a space curve which is parameterized by a function of t where t is
time. Thus a value of t would correspond to a point on this curve and you could let B
(t)
,T
(t)
, and N
(t)
be the binormal, unit tangent, and principal normal at this point of the curve. The following example is
typical.
Example 10.2.5Given thecircular helix, R
(t)
=
(a cost)
i+
(asin t)
j+
(bt)
k,find the arc lengths
(t)
, the unit tangent vector T
(t)
, the principal normal N
(t)
, the binormal B
(t)
, the curvatureκ
(t)
, and the torsion, τ
(t)
. Here t ∈
[0,T]
.
The arc length is s
(t)
= ∫_{0}^{t}
(√ ------)
a2 + b2
dr =
(√ ------)
a2 + b2
t. Now the tangent vector is obtained using
the chain rule as
dR dR dt 1 1
T = -ds = dt-ds = √-2---2-R′(t) = √--2---2 ((− asint)i+ (a cos t)j+ bk)
a + b a + b
. Note the curvature is constant in
this example. The final task is to find the torsion. Recall that B^{′} = τN where the derivative
on B is taken with respect to arc length. Therefore, remembering that t is a function of s,
An important application of the usefulness of these ideas involves the decomposition of the acceleration
in terms of these vectors of an object moving over a space curve.
Corollary 10.2.6Let R
(t)
be a space curve and denote by v
(t)
the velocity, v
(t)
= R^{′}
(t)
,let v
(t)
≡
|v(t)|
denote the speed, and let a
(t)
denote the acceleration. Then v = vT anda =
dv-
dt
T + κv^{2}N.
Proof: T =
dR--
ds
=
dR--
dt
dt
ds
= v
dt
ds
. Also, s = ∫_{0}^{t}v
(r)
dr and so
ds
dt
= v which implies
dt
ds
=
1
v
. Therefore,
T = v∕v which implies v = vT as claimed.
Now the acceleration is just the derivative of the velocity and so by the Serrat Frenet formulas,
dv dT dv dT dv
a = dt T+ v dt-= dtT + v-ds v = dtT + v2κN
Note how this decomposes the acceleration into a component tangent to the curve and one which is
normal to it. Also note that from the above, v