Biomedical Engineering Reference
In-Depth Information
z, and the time between dosages is
T
. With the first dose administered at
t
¼
0 and zero
initial conditions, we have
e
K
10
t
q
ðÞ¼
z
1
for 0
<
t
<
T
. The maximum quantity in the compartment is at
t
¼
0, and the minimum is at
t
¼
T
.At
t
¼
T
, the second dose is given, and according to superposition, the quantity is
e
K
10
t
þ
e
K
10
ð
t
T
Þ
q
ðÞ¼
z
z
1
e
K
10
2
T
e
K
10
T
þ
for
T
t
<
2
T
, with a maximum of z
z at
t
¼
T
, and a minimum of z
at
e
K
10
T
e
K
10
T
1
e
K
10
T
T
.
þ
z
¼
z
þ
z
t
¼
2
In general, for any interval (
n
1)
T
t
<
nT
, we have
e
K
10
t
þ
e
K
10
ð
t
T
Þ
e
K
10
ð
t
n
ð
2
Þ
T
Þ
e
K
10
ð
t
n
ð
1
Þ
T
Þ
q
ðÞ¼
z
z
þþ
z
þ
z
1
with the maximum in the interval at
t
¼
(
n
1)
T
of
e
K
10
n
1
ð
Þ
T
e
K
10
n
2
ð
Þ
T
e
K
10
T
þ
q
ð
ð
n
1
Þ
T
Þ ¼
z
þ
z
þþ
z
z
ð
7
:
41
Þ
1
max
Equation (7.41) is written in closed form as
e
K
10
nT
1
q
1
max
ð
ð
n
1
Þ
T
Þ ¼
z
ð
7
:
42
Þ
1
e
K
10
T
t
¼
nT
,
To find the minimum quantity at
e
K
10
nT
Þ
T
e
K
10
2
T
e
K
10
T
q
1
min
nT
e
K
10
n
1
ð
ð
Þ¼
z
þ
z
þþ
z
þ
z
ð
7
:
43
Þ
Next, we write a closed form expression for Eq. (7.43), giving
e
K
10
nT
1
nT
e
K
10
T
q
ð
Þ¼
z
ð
7
:
44
Þ
1
min
e
K
10
T
1
As
n
!1
, Eq. (7.42) approaches
z
q
1
max
¼
ð
7
:
45
Þ
1
e
K
10
T
and Eq. (7.43) approaches
e
K
10
T
z
q
¼
ð
7
:
46
Þ
1
min
1
e
K
10
T
With the parameter values shown in Figure 7.12, the maximum approaches
z
q
¼
e
K
10
T
¼
3
:
86
1
max
1
and the minimum approaches
e
K
10
T
z
q
1
min
¼
e
K
10
T
¼
2
:
86
1
To convert the quantities into concentrations, simply divide them by the volume of the com-
partment.
