SECTION
14.5
Motion in ThreeSpace
(ET Section 13.5)
543
We now compute
T
(
0
)
:
T
(
0
)
=
r
0
(
0
)
k
r
0
(
0
)
k
=
h
0
,
1
,
0
i
k h
0
,
1
,
0
i k
= h
0
,
1
,
0
i
Finally we fnd
N
=
B
×
T
:
N
(
0
)
=
1
√
5
h
2
,
0
,
1
i × h
0
,
1
,
0
i =
1
√
5
(
2
i
+
k
)
×
j
=
1
√
5
(
2
i
×
j
+
k
×
j
)
=
1
√
5
(
2
k
−
i
)
=
1
√
5
h−
1
,
0
,
2
i
(b)
Di±±erentiating
r
(
t
)
=

t
2
,
t
−
1
,
t
®
gives
r
0
(
t
)
=

2
t
,
−
t
−
2
,
1
®
r
00
(
t
)
=

2
,
2
t
−
3
,
0
®
⇒
r
0
(
1
)
= h
2
,
−
1
,
1
i
r
00
(
1
)
= h
2
,
2
,
0
i
We compute the cross product:
r
0
(
1
)
×
r
00
(
1
)
=
¯
¯
¯
¯
¯
¯
ijk
2
−
11
220
¯
¯
¯
¯
¯
¯
=
¯
¯
¯
¯
−
20
¯
¯
¯
¯
i
−
¯
¯
¯
¯
21
¯
¯
¯
¯
j
+
¯
¯
¯
¯
2
−
1
22
¯
¯
¯
¯
k
=−
2
i
+
2
j
+
6
k
= h−
2
,
2
,
6
i
k
r
0
(
1
)
×
r
00
(
1
)
k=
q
(
−
2
)
2
+
2
2
+
6
2
=
√
44
=
2
√
11
Substituting in (1) gives:
B
(
1
)
=
h−
2
,
2
,
6
i
2
√
11
=
1
√
11
h−
1
,
1
,
3
i
We now fnd
T
(
1
)
:
T
(
1
)
=
r
0
(
1
)
k
r
0
(
1
)
k
=
h
2
,
−
1
,
1
i
√
4
+
1
+
1
=
1
√
6
h
2
,
−
1
,
1
i
Finally we fnd
N
(
1
)
by computing the ±ollowing cross product:
N
(
1
)
=
B
(
1
)
×
T
(
1
)
=
1
√
11
h−
1
,
1
,
3
i ×
1
√
6
h
2
,
−
1
,
1
i =
1
√
66
¯
¯
¯
¯
¯
¯
ij
k
−
113
2
−
¯
¯
¯
¯
¯
¯
=
1
√
66
½ ¯
¯
¯
¯
13
−
¯
¯
¯
¯
i
−
¯
¯
¯
¯
−
¯
¯
¯
¯
j
+
¯
¯
¯
¯
−
2
−
1
¯
¯
¯
¯
k
¾
=
1
√
66
(
4
i
+
7
j
−
k
)
=
1
√
66
h
4
,
7
,
−
1
i
14.5 Motion in ThreeSpace
(ET Section 13.5)
Preliminary Questions
1.
I± a particle travels with constant speed, must its acceleration vector be zero? Explain.
SOLUTION
I± the speed o± the particle is constant, the tangential component,
a
T
(
t
)
=
v
0
(
t
)
, o± the acceleration is zero.
However, the normal component,
a
N
(
t
)
=
κ
(
t
)v(
t
)
2
is not necessarily zero, since the particle may change its direction.
2.
For a particle in uni±orm circular motion around a circle, which o± the vectors
v
(
t
)
or
a
(
t
)
always points toward the
center o± the circle?
For a particle in uni±orm circular motion around a circle, the acceleration vector
a
(
t
)
points towards the
center o± the circle, whereas
v
(
t
)
is tangent to the circle.
3.
Two objects travel to the right along the parabola
y
=
x
2
with nonzero speed. Which o± the ±ollowing must be true?
(a)
Their velocity vectors point in the same direction.
(b)
Their velocity vectors have the same length.
(c)
Their acceleration vectors point in the same direction.
(a)
The velocity vector points in the direction o± motion, hence the velocities o± the two objects point in the same
direction.
(b)
The length o± the velocity vector is the speed. Since the speeds are not necessarily equal, the velocity vectors may
have di±±erent lengths.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document544
CHAPTER 14
CALCULUS OF VECTORVALUED FUNCTIONS
(ET CHAPTER 13)
(c)
The acceleration is determined by the tangential component
v
0
(
t
)
and the normal component
κ
(
t
)v(
t
)
2
.Since
v
and
v
0
may be different for the two objects, the acceleration vectors may have different directions.
4.
Use the decomposition of acceleration into tangential and normal components to explain the following statement: If
the speed is constant, then the acceleration and velocity vectors are orthogonal.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '09
 Park
 Multivariable Calculus, Acceleration, Velocity, ThreeSpace

Click to edit the document details