10. a. The change in amount of phosphate,
P
(
t
), is found by adding the amount entering and
subtracting the amount leaving.
dP
dt
= 200
·
10

200
·
c
(
t
)
,
where
c
(
t
) is the concentration in the lake with
c
(
t
) =
P
(
t
)
/
10
,
000. By dividing the equation by
the volume, the concentration equation is given by
dc
dt
= 0
.
2

0
.
02
c
=

0
.
02(
c

10)
,
c
(0) = 0
.
With the substitution
z
(
t
) =
c
(
t
)

10, the equation above reduces to the problem
dz
dt
=

0
.
02
z,
z
(0) =

10
,
which has the solution
z
(
t
) =

10
e

0
.
02
t
. Thus, the concentration is given by
c
(
t
) = 10

10
e

0
.
02
t
.
b. The differential equation describing the growth of the algae is given by
dA
dt
= 0
.
5(1

e

0
.
02
t
)
A
2
/
3
.
By separating variables, we see
Z
A

2
/
3
dA
=
0
.
5
Z
(1

e

0
.
02
t
)
dt
3
A
1
/
3
(
t
)
=
0
.
5(
t
+ 50
e

0
.
02
t
) +
C
A
(
t
)
=
0
.
5(
t
+ 50
e

0
.
02
t
) +
C
3
!
3
From the initial condition
A
(0) = 1000, we have 1000 =
25+
C
3
3
. It follows that
C
= 5, so
A
(
t
) =
t
+ 50
e

0
.
02
t
+ 10
6
!
3
.
11. a. Write the differential equation
dw
dt
=

0
.
2(
w

80), then
z
(
t
) =
w
(
t
)

80. It follows that
dz
dt
=

0
.
2
z,
z
(0) =

80
,
with the solution
z
(
t
) =

80
e

0
.
2
t
=
w
(
t
)

80. Thus,
w
(
t
) = 80
1

e

0
.
2
t
.
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For a 40 kg alligator,
w
(
t
) = 40 = 80
(
1

e

0
.
2
t
)
or 40 = 80
e

0
.
2
t
, so
e
0
.
2
t
= 2 or 0
.
2
t
= ln(2).
Thus,
t
= 5 ln(2)
≈
3
.
47 years.
b. The pesticide accumulation is given by
dP
dt
= 600
80
1

e

0
.
2
t
,
P
(0) = 0
.
The solution is given by
P
(
t
) = 48
,
000
Z
1

e

0
.
2
t
dt
= 48
,
000
t
+ 5
e

0
.
2
t
+
C.
The initial condition gives
P
(0) = 0 = 240
,
000 +
C
, so
C
=

240
,
000. Hence,
P
(
t
) = 48
,
000
t
+ 5
e

0
.
2
t

240
,
000
.
The amount of pesticide in the alligator at age 5 is
P
(5) = 48
,
000
(
5 + 5
e

1
)

240
,
000 =
240
,
000
e

1
≈
88291
μ
g.
c. The pesticide concentration for a 5 year old alligator is
c
(5) =
P
(5)
1000
w
(5)
=
88
,
291
80
,
000 (1

e

1
)
≈
1
.
75 ppm
.
12. a. The differential equation can be written:
dc
dt
=

0
.
004(
c

15)
,
so we make the substitution
z
(
t
) =
c
(
t
)

15.
Since
c
(0) = 0, it follows that
z
(0) =

15.
The
solution of the substituted equation is given by:
z
(
t
)
=

15
e

0
.
004
t
=
c
(
t
)

15
c
(
t
)
=
15

15
e

0
.
004
t
.
The limiting concentration satisfies:
lim
t
→∞
c
(
t
) = 15 mg
/
m
3
.
b. We begin by separating variables, which gives:
Z
dc
c

15
=

0
.
001
Z
(4

cos(0
.
0172
t
))
dt
ln(
c
(
t
)

15)
=

0
.
001
4
t

sin(0
.
0172
t
)
0
.
0172
+
C
c
(
t
)
=
15 +
Ae

0
.
001
(
4
t

sin(0
.
0172
t
)
0
.
0172
)
It is easy to see that the initial condition
c
(0) = 0 implies that
A
=

15. Thus, the solution to this
problem is given by:
c
(
t
) = 15

15
e

0
.
001(4
t

58
.
14 sin(0
.
0172
t
))
13. a. We separate variables, so
Z
M

3
/
4
dM
=

k
Z
dt
or
4
M
1
/
4
=

kt
+ 4
C
M
(
t
) =
C

k
4
t
4
From the initial condition,
M
(0) = 16 =
C
4
, it follows that
C
= 2.
From the information that
M
(10) = 1 = (2

10
k/
4)
4
, we have
k
= 0
.
4, so
M
(
t
) = (2

0
.
1
t
)
4
.
The fruit vanishes in 20 days.
b. We separate variables again to find:
Z
M

3
/
4
dM
=

0
.
8
Z
e

0
.
02
t
dt
or
4
M
1
/
4
=
0
.
8
0
.
02
e

0
.
02
t
+ 4
C
M
(
t
) =
10
e

0
.
02
t
+
C
4
.
From the initial condition,
M
(0) = 16 = (10 +
C
)
4
, it follows that
C
=

8, so
M
(
t
) =
10
e

0
.
02
t

8
4
.
Solving 10
e

0
.
02
t
= 8, which is when the fruit vanishes, we find
t
= 50 ln(5
/
4).
Thus, the fruit
vanishes in 11.157 days.
14. a. The general solution to the Malthusian growth problem with the initial condition
P
(0) = 60
is
P
(
t
) = 60
e
rt
.
We are given that 2 weeks later
P
(2) = 80 = 60
e
2
r
, so it follows that
r
=
1
2
ln
4
3
= 0
.
14384. This
gives the solution:
P
(
t
) = 60
e
0
.
14384
t
.
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 Fall '08
 staff