Biology Reference
In-Depth Information
Assume that we examine a number of individuals A
0
¼
A(0) who have
contracted the disease at about the same time. Let A(t) be the number
of individuals who remain sick after time t. Because the per capita rate of
recovery is
b
, the rate of change of A(t) will be:
dA
dt
¼b
A
ð
t
Þ:
The basic principle that we use is that the mean value of a group of
numbers is computed as
X
t
tp
ð
t
Þ
, where p
ð
t
Þ
is the proportion of scores
that have the value t.
To compute the average length of the disease, we begin by dividing
the interval
½
0
;1Þ
into equal (small) subintervals by selecting
0
¼
t
0
<
t
1
<
t
2
< ...
, with t
n
þ
1
t
n
¼ D
t, for each n
¼
0, 1, 2,
...
. (To
be rigorous, we should consider the interval
½
0
;
L
and then take the limit
of our answer as L
!1
.)
The number of individuals who recover between t
n
and t
n
þ
1
is
A
and the approximate length of their infection is t
n
.
Therefore, the proportion of A
0
that become cured between t
n
and t
n
þ
1
is
A
ð
t
n
Þ
A
ð
t
n
þ
1
Þ
ð
t
n
Þ
A
ð
t
n
þ
1
Þ
and, thus, the mean value of the infection is
A
0
approximately:
1
A
0
1
n
A
ð
t
n
Þ
A
ð
t
n
þ
1
Þ
1
A
ð
t
n
Þ
A
ð
t
n
þ
1
Þ
d
t
n
¼
t
n
D
t
:
A
0
D
t
n
¼
0
¼
0
Notice now that, as
D
t
!
0, the approximation above improves.
approaches
A
ð
t
n
Þ
A
ð
t
n
þ
1
Þ
dA
ð
t
Þ
Further, because the expression
D
t
dt
A
0
1
n
and the sum
1
A
ð
t
n
Þ
A
ð
t
n
þ
1
Þ
D
t
n
t approaches
D
t
¼
0
Z
1
dt
Z
1
1
A
0
dA
ð
t
Þ
1
A
0
t
¼
tdA
ð
t
Þ
, we obtain that the average duration
dt
0
0
of the disease is given by:
Z
1
1
A
0
d
¼
tdA
ð
t
Þ:
0
To compute the integral above, we integrate by parts to get:
Z
1
Z
1
Z
1
1
0
tdA
ð
t
Þ¼
tA
ð
t
Þ
A
ð
t
Þ
dt
¼
A
ð
t
Þ
dt. The last equality holds
0
0
0
1
0
¼
because lim t
t
A
ð
t
Þ¼
0 and 0A(0)
¼
0 imply tA
ð
t
Þ
0
ð
lim
t
tA
ð
t
Þ¼
0
!1
!1
holds because A(t) decays exponentially).
