Biomedical Engineering Reference
In-Depth Information
To determine the time when the maximum drug is in the compartment, Eq. (7.34) is
differentiated with respect to
, set equal to zero, and solved as follows:
t
0
1
q
1
¼
d
q
2
ð
0
Þ
@
A
g
e
g
t
e
K
10
t
dt
K
10
ð
7
:
35
Þ
Þ
K
10
q
ð
0
2
e
g
t
þ
K
e
K
10
t
¼
g
g
10
Setting Eq. (7.35) equal to zero and
t
¼
t
max
gives
¼
Þ
K
10
q
ð
0
2
e
g
t
max
þ
K
10
e
K
10
t
max
g
g
0
or
e
g
t
max
e
K
10
t
max
g
¼
K
ð
7
:
36
Þ
10
e
K
10
t
max
and dividing by g yields
Multiplying both sides of Eq. (7.36) by
¼
K
10
g
e
g
t
max
¼
e
K
10
g
ð
Þ
t
max
e
K
10
t
max
ð
7
:
37
Þ
Taking the logarithm of both sides of Eq. (7.37), we have
K
10
g
ð
K
g
Þ
t
¼
ln
10
max
Solving for
t
max
yields
g
ln
K
10
t
max
¼
ð
7
:
38
Þ
K
g
10
The maximum quantity of drug in the compartment is
Þ¼
q
2
ð
0
Þ
g
e
g
t
max
e
K
10
t
max
0
@
1
A
q
ð
t
1
max
K
K
10
g
10
ln
t
max
¼
K
10
g
0
@
1
A
0
@
1
A
0
@
1
A
g
K
10
ln
K
10
g
K
10
K
10
ln
K
10
g
Þ
K
10
ð
0
q
g
g
2
¼
e
e
g