566
CHAPTER 14
CALCULUS OF VECTOR-VALUED FUNCTIONS
(ET CHAPTER 13)
Since the speed
v(
t
)
is increasing at a rate of
t
ft
/
s
2
,wehave
v
0
(
t
)
=
t
. The car starts at rest hence the initial speed is
v
0
=
0. We now integrate to Fnd
t
)
:
t
)
=
Z
t
0
v
0
(
u
)
du
=
Z
t
0
udu
=
1
2
t
2
+
v
0
=
1
2
t
2
+
0
=
1
2
t
2
The curvature of the circular path is
κ
(
t
)
=
1
R
=
1
1000
. Substituting
v
0
(
t
)
=
t
,
=
1
1000
,and
t
)
=
1
2
t
2
in (2) gives:
a
T
(
t
)
=
t
,
a
N
(
t
)
=
1
1000
±
1
2
t
2
¶
2
=
1
4000
t
4
Combining with (1) gives the following decomposition:
a
(
t
)
=
t
T
(
t
)
+
1
4000
t
4
N
(
t
)
(3)
We now Fnd the unit tangent
T
(
t
)
and the unit normal
N
(
t
)
.
N
T
Q
(starting
point)
P
O
We have (see Fgure):
T
=
D
cos
³
π
2
+
θ
²
,
sin
³
2
+
²E
= h−
sin
,
cos
i
(4)
N
= h
cos
(
+
) ,
sin
(
+
)
i = h−
cos
,
−
sin
i
(5)
We use the arc length formula to Fnd
:
_
PQ
=
Z
t
0
k
r
0
(
u
)
k
=
Z
t
0
u
)
=
Z
t
0
1
2
u
2
=
t
3
6
In addition,
_
=
R
=
1000
. Hence,
1
,
000
=
t
3
6
⇒
=
t
3
6
,
000
Substituting in (4) and (5) yields:
T
=
*
−
sin
t
3
6
,
000
,
cos
t
3
6
,
000
+
;
N
=
*
−
cos
t
3
6
,
000
,
−
sin
t
3
6
,
000
+
(6)
We now combine (3) and (6) to obtain the following decomposition:
a
(
t
)
=
t
*
−
sin
t
3
6
,
000
,
cos
t
3
6
,
000
+
+
1
4
,
000
t
4
*
−
cos
t
3
6
,
000
,
−
sin
t
3
6
,
000
+
At
t
=
3weget:
a
T
=
3
a
N
=
3
4
4
,
000
≈
0
.
02025
T
=
*
−
sin
3
3
6
,
000
,
cos
3
3
6
,
000
+
≈ h−
0
.
0045
,
0
.
9999
i
N
=
*
−
cos
3
3
6
,
000
,
−
sin
3
3
6
,
000
+
≈ h−
0
.
9999
,
−
0
.
0045
i
46.
In the notation of Example 5, Fnd the acceleration vector for a person seated in a car at (a) the highest point of the
ferris wheel and (b) the two points level with the center of the wheel.