500
C H A P T E R
14
CALCULUS OF VECTORVALUED FUNCTIONS
(ET CHAPTER 13)
14.4 Curvature
(ET Section 13.4)
Preliminary Questions
1.
What is the unit tangent vector of a line with direction vector
v
=
2
,
1
,
−
2 ?
SOLUTION
A line with direction vector
v
has the parametrization:
r
(
t
)
=
−−→
O P
0
+
t
v
hence, since
−−→
O P
0
and
v
are constant vectors, we have:
r
(
t
)
=
v
Therefore, since
v
=
3, the unit tangent vector is:
T
(
t
)
=
r
(
t
)
r
(
t
)
=
v
v
=
2
/
3
,
1
/
3
,
−
2
/
3
2.
What is the curvature of a circle of radius 4?
SOLUTION
The curvature of a circle of radius
R
is
1
R
, hence the curvature of a circle of radius 4 is
1
4
.
3.
Which has larger curvature, a circle of radius 2 or a circle of radius 4?
SOLUTION
The curvature of a circle of radius 2 is
1
2
, and it is larger than the curvature of a circle of radius 4, which is
1
4
.
4.
What is the curvature of
r
(
t
)
=
2
+
3
t
,
7
t
,
5
−
t
?
SOLUTION
r
(
t
)
parametrizes the line 2
,
0
,
5
+
t
3
,
7
,
−
1 , and a line has zero curvature.
5.
What is the curvature at a point where
T
(
s
)
=
1
,
2
,
3 in an arc length parametrization
r
(
s
)
?
SOLUTION
The curvature is given by the formula:
κ
(
t
)
=
T
(
t
)
r
(
t
)
In an arc length parametrization,
r
(
t
)
=
1 for all
t
, hence the curvature is
κ
(
t
)
=
T
(
t
)
. Using the given information
we obtain the following curvature:
κ
=
1
,
2
,
3
=
1
2
+
2
2
+
3
2
=
√
14
6.
What is the radius of curvature of a circle of radius 4?
SOLUTION
The definition of the osculating circle implies that the osculating circles at the points of a circle, is the
circle itself. Therefore, the radius of curvature is the radius of the circle, that is, 4.
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 Winter '08
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