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
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