14. a. Let
a
(
t
) be the amount of pollutant, and the concentration
c
(
t
) is the concentration
of pollutant (in ppb).
The change in amount = the amount entering  the amount leaving.
The change in amount,
a
0
(
t
), has units (mass/day). The amount entering is
f
1
Q
1
+
f
2
Q
2
=
4000
·
18+2500
·
4 = 82
,
000 ppb
·
m
3
/day (mass/day), while the amount leaving is (
f
1
+
f
2
)
c
(
t
) =
6500
c
(
t
) ppb
·
m
3
/day (mass/day). Thus, the differential equation for the change in amount is
da
(
t
)
dt
= 82
,
000

6500
c
(
t
)
.
The relation between the amount and concentration is
c
(
t
) =
a
(
t
)
V
=
a
(
t
)
3000000
and
c
0
(
t
) =
a
(
t
)
3000000
,
so the concentration differential equation:
dc
(
t
)
dt
=
82
,
000

6500
c
(
t
)
3
,
000
,
000

13
6000
c

164
13
≈ 
0
.
0021667(
c

12
.
615)
.
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With the initial condition
c
(0) = 0, we make the substitution
z
(
t
) =
c
(
t
)

12
.
615, so
z
(0) =

12
.
615 and the differential equation is
dz
dt
=

0
.
0021667
z,
z
(0) =

12
.
615
,
which gives
z
(
t
) =

12
.
615
e

0
.
0021667
t
=
c
(
t
)

12
.
615. Thus,
c
(
t
) = 12
.
615
1

e

0
.
0021667
t
.
b. We solve
c
(
t
) = 12
.
615
(
1

e

0
.
0021667
t
)
= 4, so
e
0
.
0021667
t
=
12
.
615
8
.
615
= 1
.
4643.
Thus,
t
=
ln(1
.
4643)
0
.
0021667
≈
176
.
01 days.
Hence, the concentration reaches 4 ppb at
t
≈
176
.
01 days.
The
limiting concentration is
lim
t
→∞
c
(
t
) = 12
.
615 ppb
.
This easily follows because the exponential tends to zero for large
t
.
15. a. The DE
dP
dt
= (
b

at
)
P
is both a linear and a separable equation.
From our linear
techniques, we write
dP
dt
+ (
at

b
)
P
= 0, which has an integrating factor
μ
(
t
) =
exp
Z
(
at

b
)
dt
=
e
at
2
2

bt
.
Thus,
d
dt
e
at
2
2

bt
P
(
t
)
= 0
or
P
(
t
) =
Ce
bt

at
2
2
.
With the initial condition,
P
(0) = 6
.
94 =
C
, so
P
(
t
) = 6
.
94
e
bt

1
2
at
2
.
From the data,
P
(20) = 7
.
47 and
P
(40) = 7
.
72.
7
.
47 = 6
.
94
e
20
b

200
a
and
7
.
72 = 6
.
94
e
40
b

800
a
.
Taking logarithms,
ln
7
.
47
6
.
94
= 20
b

200
a
and
ln
7
.
72
6
.
94
= 40
b

800
a.
Taking the 2
nd
equation times 0.5 and adding to the first gives
200
a
=
ln
7
.
47
6
.
94

0
.
5 ln
7
.
72
6
.
94
a
=
0
.
005 ln
7
.
47
√
7
.
72
×
6
.
94
≈
0
.
000101685
.
It follows that
b
= 0
.
0046965.
b. The year 2000 is
t
= 50.
P
(50) = 6
.
94
e
0
.
0046965(50)

0
.
000050843(50)
2
≈
7
.
7293 million.
The
actual census data is 8.13 million, so the percent error is 100
(7
.
73

8
.
13)
8
.
13
% =

4
.
93% from the
actual census data.
c. The maximum population occurs when
dP
dt
= (
b

at
)
P
= 0, so
t
=
b
a
≈
46
.
19 yr. This gives
P
max
= 6
.
94
e
0
.
004697(46
.
19)

0
.
00005084(46
.
19)
2
≈
7
.
735
.
Therefore, according to the model, the maximum population occurs in about 1996, and is about
7.735 million. This is clearly not correct based on the census data in 2000.
17. The electric circuit with
R
= 10Ω,
C
= 0
.
1F, and
E
(
t
) = 40V is a linear differential
equation:
10
dQ
dt
+
1
0
.
1
Q
= 40
or
dQ
dt
+
Q
= 4
,
which has an integrating factor
μ
(
t
) =
e
t
. Thus,
d
dt
e
t
Q
= 4
e
t
,
so
e
t
Q
(
t
) = 4
e
t
+
C.
With the initial condition,
Q
(0) = 0, the solution becomes
Q
(
t
) = 4

4
e

t
.
18. Programs are provided to solve this problem along with solutions expected in the written
work.
a. The least squares best fit to the data is found with MatLab. The sum of square errors is
computed with:
1
function
J = sum
vonB(p , tdata , ldata )
2
% von
Bertalanffy
eqn
3
model = p (1)
*
(1

exp
(

p (2)
*
tdata ) ) ;
% Model
eqn
with
parameters
4
error
= model

ldata ;
% Error
between
model
and
data
5
J =
error
*
error
’ ;
% Sum
of
square
e r r o r s
6
end
The least sum of square errors uses the command line:
[p1,J,flag] = fminsearch(
@
sum
vonB,[1.9,0.2],[],tdfish,ldfish)
where
tdfish
and
ldfish
are the age and length data for the fish.
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 staff