This preview shows pages 1–2. Sign up to view the full content.
723
SECTION 7.3
.
.
Direction Fields and Euler’s Method
587
78.
The differential equation
y
±
=−
ay
ln(
y
/
b
) (for positive con
stants
a
and
b
) arises in the study of the growth of some animal
tumors. Solve the differential equation and sketch several mem
bers of the family of solutions. What adjective (e.g., rapid,
moderate, slow) would you use to characterize this type of
growth?
79.
Use the technique of exercise 77 to show that solutions of
y
±
ln(
y
/
b
) for positive constants
a
and
b
have at most
one inﬂection point, which occurs at
y
=
b
/
a
.
EXPLORATORY EXERCISES
1.
Look up the census ﬁgures for the U.S. population starting
in 1790. (You can ﬁnd this information in any library, in
virtually any almanac or encyclopedia.) Plot the data on a
graph of population versus time. Does this look like the so
lution of a logistic equation? Brieﬂy explain. If you wanted
to model these data with a logistic function, you would need
to estimate values for
k
and
M
.As shown in exercise 77,
M
equals twice the height of the inﬂection point. Explain why
(for the logistic curve) the inﬂection point represents the point
of maximum slope. Estimate this for the population data. To
estimate
r
, note that for small populations, the logistic equation
y
±
=
ry
(1
−
y
/
k
)
≈
. Then
r
equals the rate of exponential
increase. Show that for the ﬁrst 50 years, the U.S. population
growth was approximately exponential and ﬁnd the percentage
increase as an estimate of
r
.With these values of
r
and
M
and
the initial population in 1790, ﬁnd a function describing the
population. Test this model by comparing actual populations
to predicted populations for several years.
2.
An object traveling through the air is acted on by grav
ity (acting vertically), air resistance (acting in the direc
tion opposite velocity) and other forces (such as a motor).
An equation for the horizontal motion of a jet plane is
v
±
=
c
−
f
(
v
)
/
m
, where
c
is the thrust of the motor and
f
(
v
)i
s the air resistance force. For some ranges of ve
locity, the air resistance actually
drops
substantially for
higher velocities as the air around the object becomes tur
bulent. For example, suppose that
v
±
=
32,000
−
f
(
v
), where
f
(
v
)
=
±
0
.
8
v
2
if 0
≤
v
≤
100
0
.
2
v
2
if 100
<v
.To solve the initial value
problem
v
±
=
32,000
−
f
(
v
)
,v
(0)
=
0, start with the initial
value problem
v
±
=
32,000
−
0
.
8
v
2
(0)
=
0. Solve this IVP
²
Hint:
1
40,000
−
v
2
=
1
400
³
1
200
+
v
+
1
200
−
v
´
µ
and de
termine the time
t
such that
v
(
t
)
=
100. From this time forward,
the equation becomes
v
±
=
32,000
−
0
.
2
v
2
. Solve the IVP
v
±
=
32,000
−
0
.
2
v
2
(0)
=
100. Put this solution together
with the previous solution to piece together a solution valid
for all time.
3.
Solve the initial value problems
dy
dt
=
2(1
−
y
)(2
−
y
)(3
−
y
)
with (a)
y
(0)
=
0, (b)
y
(0)
=
1
.
5, (c)
y
(0)
=
2
.
5 and
(d)
y
(0)
=
4. State as completely as possible how the limit
lim
t
→∞
y
(
t
) depends on
y
(0).
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 '10
 atsumi

Click to edit the document details