CHAPTER 1
Section 1.
Differential Equation Models
1.1.
Let
y(t)
be the number of bacteria at time
t
. The rate of change of the number of bacteria is
y (t)
. Since this
rate of change is given to be proportional to
y(t)
, the resulting differential equation is
y (t)
=
ky(t)
. Note
that
k
is a positive constant since
y (t)
must be positive (i.e. the number of bacteria is growing).
1.2.
Let
y(t)
be the number of field mice at time
t
. The rate of change of the number of mice is
y (t)
. Since
this rate of change is given to be inversely proportional to the square root of
y(t)
, the resulting differential
equation is
y (t)
=
k/
√
y(t)
. Note that
k
is a positive constant since
y (t)
must be positive (i.e. the number
of mice is growing).
1.3.
Let
y(t)
be the number of ferrets at time
t
. The rate of change of the number of ferrets is
y (t)
. Since this rate of
change is given to be proportional to the product of
y(t)
and the difference between the maximum population
and
y(t)
(i.e. 100
−
y(t)
), the resulting differential equation is
y (t)
=
ky(t)(
100
−
y(t))
. Note that
k
is a
positive constant since
y (t)
must be positive (i.e. the number of ferrets is growing provided
y(t) <
100).
1.4.
Let
y(t)
be the quantity of radioactive substance at time
t
. The rate of change of the material is
y (t)
. Since this
rate of change (decay) is given to be proportional to
y(t)
, the resulting differential equation is
y (t)
= −
ky(t)
.
Note that
k
is a positive constant since
y (t)
must be negative (i.e. the quantity of radioactive material is
decreasing).
1.5.
Let
y(t)
be the quantity of material at time
t
. The rate of change of the material is
y (t)
. Since this rate of change
(decay) is given to be inversely proportional to
y(t)
, the resulting differential equation is
y (t)
= −
k/y(t)
.
Note that
k
is a positive constant since
y (t)
must be negative (i.e. the quantity of material is decreasing).
1.6.
Let
y(t)
be the temperature of the potato at time
t
. The rate of change of the temperature is
y (t)
. Since this
rate of change is given to be proportional to the difference between the potato’s temperature and that of the
surrounding room (i.e.
y(t)
−
65), the resulting differential equation is
y (t)
= −
k(y(t)
−
65
)
. Note that
k
is
a positive constant since
y (t)
must be negative (i.e. the potato is cooling) and since
y(t)
−
65
>
0 (i.e. the
potato is hotter than the surrounding room).
1.7.
Let
y(t)
be the temperature of the thermometer at time
t
. The rate of change of the temperature is
y (t)
.
Since this rate of change is given to be proportional to the difference between the thermometer’s temperature
and that of the surrounding room (i.e. 77
−
y(t)
), the resulting differential equation is
y (t)
=
k(
77
−
y(t))
.
Note that
k
is a positive constant since
y (t)
must be positive (i.e. the thermometer is warming) and since
77
−
y(t) >
0 (i.e. the thermometer is cooler than the surrounding room).
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 Spring '08
 lee
 Rate Of Change, 2, initial condition

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