189
Chapter 5
Exercises 5.1
1.
a.
and since
, we have
.
P
(
t
)
=
Ce
.02
t
3
=
P
(0)
=
Ce
(.02)(0)
=
C
P
(
t
)
=
3
e
.02
t
b.
P
(0) = 3, so population was 3 million.
c.
.02
d.
≈
3.52 million
P
(8)
=
3
e
(.02)(8)
e.
≈
.065 million per year or 65,000 per year
′
P
(4)
=
.02
P
=
.02(3
e
(.02)(4)
)
f.
Solve
for
P
(
t
) when
, .02
P
(
t
) = .07
P
(
t
)
≈
3.5 million when it is growing at the
rate of 70,000 per year.
′
P
(
t
)
=
P
(
t
)
′
P
(
t
)
=
.07
2.
a.
and since
, we have
.
P
(
t
)
=
Ce
.55
t
10,000
=
P
=
Ce
(.55)(0)
=
C
P
(
t
)
=
10,000
e
.55
t
b.
P
(0) = 10,000
c.
≈
156,426 bacteria
P
(5)
=
10,000
e
(.55)(5)
d.
.55
e.
= .55
P
(
t
) = (.55)(100,000)
= 55,000 bacteria per hour
′
P
(
t
)
f.
Solve
= .55
P
(
t
) for
P
(
t
) when
= 34,000.
.55
P
(
t
) = 34,000
P
(
t
)
≈
61,818 bacteria
′
P
(
t
)
′
P
(
t
)
3.
a.
, so 5000 cells
were present initially.
P
=
5000
e
(.2)(0)
=
5000
b.
k
= .2, so a differential equation is
= .2
P
(
t
).
′
P
(
t
)
c.
10,000
=
5000
e
.2
t
2
=
e
.2
t
ln 2 = .2
t
t
=
ln2
.2
≈
3.5 hours
d.
20,000
=
5000
e
.2
t
4
=
e
.2
t
ln 4 = .2
t
t
=
ln 4
≈
6.9 hours
4.
a.
; so 300 cells were
present initially.
P
=
300
e
(.01)(0)
=
300
b.
k
= .01, so a differential equation is
′
P
(
t
) = .01
P
(
t
).
c.
600
=
300
e
.01
t
2
=
e
.01
t
ln 2 = .01
t
t
=
.01
≈
69.3 days
d.
1200
=
300
e
.01
t
4
=
e
.01
t
ln 4 = .01
t
t
=
ln 4
≈
138.6 days
5.
Let
P
(
t
) be the population after
t
days,
P
(
t
)
=
P
0
e
kt
.
It is given that
P
(40) = 2
P
(0), i.e.,
P
0
e
40
k
=
2
P
0
e
0(
k
)
=
2
P
0
;
e
40
k
=
2;
ln
e
40
k
=
ln2;
k
=
40
≈
.017
6.
Let
P
(
t
) be the population after
t
years,
P
(
t
)
=
P
0
e
kt
.
It is given that
P
(10) = 3
P
(0), i.e.,
P
0
e
10
k
=
3
P
0
e
k
=
3
P
0;
e
10
k
=
3;
ln
e
10
k
=
ln3;
k
=
ln 3
10
≈
.11.
7.
Let
P
(
t
) be the population after
t
years.
P
(
t
)
=
P
0
e
.05
t
3
P
0
=
P
0
e
.05
t
3
=
e
.05
t
ln 3 = .05
t
t
=
ln3
.05
≈
22 years