A fly buzzing around the room, a person riding a roller coaster, and a satellite orbiting
the earth all have something in common. They are moving over some sort of curve in
three dimensions.
Denote by R
(t)
the position vector of the point on the curve which occurs at time t.
Assume that R^{′},R^{′′} exist and are continuous. Thus R^{′} = v, the velocity and R^{′′} = a is
defined as the acceleration.
PICT
Lemma 16.4.1Define T
(t)
≡ R^{′}
(t)
∕
|R ′(t)|
. Then
|T (t)|
= 1 and ifT^{′}
(t)
≠0, then there exists a unit vector N
(t)
perpendicular to T
(t)
and a scalarvalued function κ
(t)
, with T^{′}
(t)
= κ
(t)
|v |
N
(t)
.
Proof:It follows from the definition that
|T|
= 1. Therefore, T ⋅ T = 1 and so, upon
differentiating both sides,
T′ ⋅T + T ⋅T ′ = 2T′ ⋅T = 0.
Therefore, T^{′} is perpendicular to T. Let N
(t)
|T′|
≡ T^{′}. Note that if
|T ′|
= 0, you could
let N
(t)
be any unit vector. Then letting κ
(t)
be defined such that
|T′|
≡ κ
(t)
|v (t)|
, it
follows
T ′(t) = |T′(t)|N (t) = κ(t)|v (t)|N (t). ■
Definition 16.4.2The vector T
(t)
is called theunit tangent vectorand the vector N
(t)
is called theprincipal normal. The function κ
(t)
in the abovelemma is called the curvature.The radius of curvature is defined as ρ = 1∕κ.The plane determined by the two vectors T and N in the case where T^{′}≠0 is calledtheosculating^{1} plane.It identifies a particular plane which is in a sense tangent to this space curve.
The important thing about this is that it is possible to write the acceleration as the
sum of two vectors, one perpendicular to the direction of motion and the other in the
direction of motion.
Theorem 16.4.3For R
(t)
the position vector of a space curve, the accelerationis given by the formula
d |v| 2
a =-dt-T + κ|v| N ≡ aTT + aN N. (16.11)
(16.11)
Furthermore, a_{T}^{2} + a_{N}^{2} =
|a|
^{2}.
Proof:
dv d d d|v| d|v|
a = ---= -- (R ′) =--(|v|T) = ---T + |v|T′ = ---T + |v|2κN.
dt dt dt dt dt
This proves the first part.
For the second part,
|a|2 = (aTT + aN N) ⋅(aTT + aN N)
= a2T ⋅T + 2a a T ⋅N + a2N ⋅N = a2 + a2
T N T N T N
because T ⋅ N = 0. ■
From 16.11 and the geometric properties of the cross product,
a× v = κ|v|2 N × v
Hence, using the geometric description of the cross product again using that the angle
between N and T is 90^{∘},
Finally, it is good to point out that the curvature is a property of the curve itself, and
does not depend on the parametrization of the curve. If the curve is given by two
different vector valued functions R
From this, it is possible to give an important formula from physics. Suppose an object
orbits a point at constant speed v and it travels over a circle of radius r. In the above
notation,
|v|
= v. What is the centripetal acceleration of this object? You may know
from a physics class that the answer is v^{2}∕r where r is the radius. This follows from the
above quite easily. First, what is the curvature of a circle of radius r? A parameterization
of such a curve is