Solutions for the second midterm
1.
Let
C
be the helix given by
x
(
t
) = cos
t,
y
(
t
) = sin
t,
z
(
t
) =
t
for
0
≤
t
≤
2
π
.
(a)
Integrate
f
(
x, y, z
) =
x
+ 3
z
2
along
C
with respect to arc length.
We have been given a parametrization; using it, we have
~
r
0
(
t
) = sin
t~
ı
+ cos
t~
+
~
k
and

~
r
0
(
t
)

=
p
sin
2
t
+ cos
2
t
+ 1 =
√
2
.
Writing the function in terms of this parametrization gives
f
(
x
(
t
)
, y
(
t
)
, z
(
t
)) = cos
t
+3
t
2
.
We have
Z
C
f ds
=
Z
2
π
0
(
cos
t
+ 3
t
2
)
√
2
dt
=
√
2
±
sin
t
+
t
3
²
2
π
0
= 8
√
2
π
3
.
(b)
Compute
R
C
~
F
·
d~
r
where
~
F
(
x, y, z
) =
z
2
~
ı
+
z~
+(2
xz
+
y
)
~
k
. (Hint:
~
F
is conservative.)
Computing the integral directly doesn’t look promising, so instead we follow the hint and
ﬁnd the potential function. Integrating with respect to
x
gives
f
=
xz
2
+
g
(
y, z
). Diﬀer
entiating with respect to
y
gives
∂g/∂y
=
z
and thus
g
(
y, z
) =
yz
+
h
(
z
). Diﬀerentiating
with respect to
z
gives 2
xz
+
y
+
h
0
(
z
) = 2
xz
+
y
, so that
h
(
z
) =
K
for some constant
K
.
We conclude that the potential is
f
=
xz
2
+
yz
+
K
for any constant
K
.
Note that the curve
C
begins at (1
,
0
,
0) and ends at (1
,
0
,
2
π
). We have
Z
C
~
F
·
d~
r
=
f
(1
,
0
,
2
π
)

f
(1
,
0
,
0) = 4
π
2

0 = 4
π
2
.
2.
Let
C
be the curve consisting of the following four line segments: from
(1
,
0
,
0)
to
(1
,
1
,
0)
, followed by
(1
,
1
,
0)
to
(0
,
1
,
1)
, followed by
(0
,
1
,
1)
to
(0
,
0
,
1)
, followed by
(0
,
0
,
1)
to
(1
,
0
,
0)
. In particular,
C
is a rectangle which lies in the plane
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '07
 NEEL
 Calculus, Arc Length, Vector Calculus, Cos, 3 k, STOKE

Click to edit the document details