1130
CHAPTER 14
.
.
Vector Calculus
1416
easier vector ﬁeld,
±
x
,
2
x
−
y
²
. First, note that if
x
³
(
t
)
=
x
and
y
³
(
t
)
=
2
x
−
y
, then
dy
dx
=
2
x
−
y
x
=
2
−
y
x
.
The ﬂow lines will be the graphs of functions
y
(
x
) such
that
y
³
(
x
)
=
2
−
y
x
,or
y
³
+
1
x
y
=
2. The lefthand side of the
equation should look a little like a product rule. Our main
goal is to multiply by a term called an
integrating factor,
to make the lefthand side exactly a product rule derivative.
It turns out that for the equation
y
³
+
f
(
x
)
y
=
g
(
x
), an in
tegrating factor is
e
±
f
(
x
)
.In the present case, for
x
>
0, we
have
e
±
1
/
xdx
=
e
ln
x
=
x
.(We have chosen the integration con
stant to be 0 to keep the integrating factor simple.) Multiply
both sides of the equation by
x
and show that
xy
³
+
y
=
2
x
.
Show that
³
+
y
=
(
)
³
. From (
)
³
=
2
x
, integrate to get
=
x
2
+
c
y
=
x
+
c
x
.To ﬁnd a ﬂow line passing through
the point (1, 2), show that
c
=
1 and thus,
y
=
x
+
1
x
.To ﬁnd a
ﬂow line passing through the point (1, 1), show that
c
=
0 and
thus,
y
=
x
.Sketch the vector ﬁeld and highlight the curves
y
=
x
+
1
x
and
y
=
x
.
14.2
LINE INTEGRALS
In section 5.6, we used integration to ﬁnd the mass of a thin rod with variable mass density.
There, we had observed that if the rod extends from
x
=
a
to
x
=
b
and has mass density
function
ρ
(
x
), then the mass of the rod is given by
±
b
a
ρ
(
x
)
. This deﬁnition is ﬁne for
objects that are essentially onedimensional, but what if we wanted to ﬁnd the mass of a
helical spring (see Figure 14.11)? Calculus is remarkable in that the same technique can
solve a wide variety of problems. As you should expect by now, we will derive a solution
by ﬁrst approximating the curve with line segments and then taking a limit.
FIGURE 14.11
A helical spring
(
x
n
,
y
n
,
z
n
)
(
x
0
,
y
0
,
z
0
)
(
x
2
,
y
2
,
z
2
)
(
x
1
,
y
1
,
z
1
)
FIGURE 14.12
Partitioned curve
In this threedimensional setting, the density function has the form
ρ
(
x
,
y
,
z
) (where
ρ
is measured in units of mass per unit length). We assume that the object is in the
shape of a curve
C
in three dimensions with endpoints (
a
,
b
,
c
) and (
d
,
e
,
f
). Further,
we assume that the curve is
oriented,
which means that there is a direction to the curve.
Fo
rexample, the curve
C
might start at (
a
,
b
,
c
) and end at (
d
,
e
,
f
). We ﬁrst partition
the curve into
n
pieces with endpoints (
a
,
b
,
c
)
=
(
x
0
,
y
0
,
z
0
)
,
(
x
1
,
y
1
,
z
1
)
,
(
x
2
,
y
2
,
z
2
)
,...,
(
x
n
,
y
n
,
z
n
)
=
(
d
,
e
,
f
), as indicated in Figure 14.12. We will use the shorthand
P
i
to denote
the point (
x
i
,
y
i
,
z
i
), and
C
i
for the section of the curve
C
extending from
P
i
−
1
to
P
i
, for
each
i
=
1
,
2
n
. Our initial objective is to approximate the mass of the portion of the
object along
C
i
. Note that if the segment
C
i
is small enough, we can consider the density
to be constant on
C
i
.In this case, the mass of this segment would simply be the product
of the density and the length of
C
i
.For some point (
x
∗
i
,
y
∗
i
,
z
∗
i
)on
C
i
,we approximate the
density on
C
i
by
ρ
(
x
∗
i
,
y
∗
i
,
z
∗
i
). The mass of the section
C
i
is then approximately
ρ
(
x
∗
i
,
y
∗
i
,
z
∗
i
)
±
s
i
,
where
±
s
i
represents the arc length of
C
i
. The mass
m
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 Spring '10
 CARTER
 CN

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