Gauthier, Joseph – Homework 9 – Due: Oct 26 2007, 3:00 am – Inst: JEGilbert
1
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The due time is Central
time.
001
(part 1 oF 1) 10 points
Determine the vector
I
=
Z
1
0
r
(
t
)
dt
when
r
(
t
) =
D
4
1 +
t
2
,
2
t
1 +
t
2
,
6
(1 +
t
)
2
E
.
1. I
=
h
π,
ln 2
,
3
i
correct
2. I
=
h
4
,
2
π,
6 ln 2
i
3. I
=
h
4 ln 2
,
2
,
6
π
i
4. I
=
h
ln 2
, π,
3
i
5. I
=
h
4
,
ln 2
,
6
π
i
6. I
=
h
π,
2
,
3 ln 2
i
Explanation:
±or a vector Function
r
(
t
) =
h
f
(
t
)
, g
(
t
)
, h
(
t
)
i
,
the components oF the vector
I
=
Z
1
0
r
(
t
)
dt
are given by
Z
1
0
f
(
t
)
dt,
Z
1
0
g
(
t
)
dt,
Z
1
0
h
(
t
)
dt,
respectively. But when
r
(
t
) =
D
4
1 +
t
2
,
2
t
1 +
t
2
,
6
(1 +
t
)
2
E
,
we see that
Z
1
0
f
(
t
)
dt
=
Z
1
0
4
1 +
t
2
dt
=
h
4 tan

1
t
i
1
0
=
π ,
while
Z
1
0
g
(
t
)
dt
=
Z
1
0
2
t
1 +
t
2
dt
=
h
ln(1 +
t
2
)
i
1
0
= ln 2
,
and
Z
1
0
h
(
t
)
dt
=
Z
1
0
6
(1 +
t
)
2
dt
=
h

6
1 +
t
i
1
0
= 3
.
Consequently,
I
=
h
π,
ln 2
,
3
i
.
keywords: vector Function, defnite integral,
inverse trig integral, log Function, substitution
002
(part 1 oF 1) 10 points
±ind the arc length oF the curve
r
(
t
) =
h
3 sin 2
t,
8
t,
3 cos 2
t
i
between
r
(0) and
r
(3).
1.
arc length = 3
√
82
2.
arc length = 24
3.
arc length = 3
√
73
4.
arc length = 3
√
91
5.
arc length = 30
correct
Explanation:
The length oF the curve
r
(
t
) between
r
(
t
0
)
and
r
(
t
1
) is given by the integral
L
=
Z
t
1
t
0

r
0
(
t
)

dt.