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2
2
2
d
d
ds
d s
ds

ˆ
ˆ
ˆ
()
t
()
t
()
t
()
t κ t
()
()
t
P
×
P
=
T
×
T
+
N

2
2
dt
dt
dt
dt
dt

3
ds

ˆ
ˆ
()
()
()
=
κ t
TN
t
×
t
.
(11.103)

dt

ˆ N are both unit vectors, the magnitude of their cross product is
unity. Thus, upon using Equation (11.94) to replace the quantity ds dt , we arrive
at the following expression for
ˆ
()
()
Since
T
t
and
()
κ t .
()
′′
()
PP
P
t
×
t
()
κ t
=
(11.104)
3
()
t
Let us consider the curvature of a circle of radius ρ . Such a circle lying in the
x - y plane can be expressed parametrically as
()
P
t ρ t ρ t
=
cos ,
sin , 0
.
(11.105)
Applying Equation (11.104), we see that the curvature is
ρ t ρ t
sin ,
cos , 0
× −
ρ t ρ t
cos ,
sin , 0
1
()
κ t
=
=
,
(11.106)
3
ρ
ρ t ρ t
sin ,
cos , 0
or simply the reciprocal of the radius of the circle. For a general curve, we call
the quantity
()
()
the radius of curvature . As shown in Figure 11.20, the
radius of curvature at a point
ρ t κ t
=
1
()
P
t
corresponds to the radius of a circle that is
()
tangent to the curve at
P
t
and lies in the plane determined by the directions
ˆ N . This plane is called the osculating plane , and the circle is called
the osculating circle .
ˆ
()
()
T
t
and
()
P gives the acceleration of a
particle following the path at time t . Examining Equation (11.102) more closely,
we observe
The second derivative of the position vector
[
]
()
2
()
2
d
d
vt
ˆ
ˆ
()
()
() ()
()
a
t
=
P
t
=
v t
T
t
+
N
t
,
(11.107)
dt
2
dt
ρ t
()
where
v t
=
s t
is the scalar speed at time t . The coefficients a and a de-
fined by
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