Biology Reference
In-Depth Information
elimination. In Chapter 1, we discussed the equation dC
Þ
describing the rate of change in the hormone concentration caused by
the process of ongoing elimination. If the hormone is being secreted at
the rate S(t), the differential equation
=
dt
¼a
C
ð
t
dC
dt
¼a
C
ð
t
Þþ
S
ð
t
Þ
(10-1)
describes the change in the concentration. Here, as before, C
is the
hormone concentration in the corresponding pool; t is the time; S
ð
t
Þ
is
the secretion rate; and the elimination is supposed to be proportional
(with some rate elimination constant
ð
t
Þ
0) to the available
concentration. Recall that the clearance constant
a >
a >
0 and the half-life
t
of the hormone in the blood are related through
t ¼
ln 2
=a
.
This model extends the model we considered in Chapter 1, Section IX.
That model assumed instantaneous entry of drug into the bloodstream
with every dose, and that doses were administered at equally spaced
time intervals. Equation (10-1) extends this construct by allowing
variable amount and continuous delivery with regard to time.
Also in Chapter 9, we discussed the following convolution integral as
an alternative way to describe the processes of simultaneous
secretion S and elimination E:
Z
t
C
ð
t
Þ¼
S
ðtÞ
E
ð
t
tÞ
d
t ¼ð
S
E
Þð
t
Þ:
(10-2)
1
e
a
t
[hence, E(t
If E(t)
¼
t
)=E(t)E(
t
)] and we know the concentration
at t
¼
0, C
0
¼
C
ð
0
Þ
, Eq. (10-2) can also be given as:
Z
t
C
ð
t
Þ¼
S
ðtÞ
E
ð
t
tÞ
d
t þ
C
ð
0
Þ
E
ð
t
Þ;
0
for positive values of t.
E
XERCISE
10-1
e
a
t
, the
Show that in the special case when S
ð
t
Þ¼
S
¼
const and E
ð
t
Þ¼
expression of the concentration function C
ð
t
Þ
becomes
e
a
t
C
ð
t
Þ¼ð
C
0
S
=aÞ
þ
S
=a:
(10-3)
In Chapter 9, we derived the convolution representation (10-2) of C
ð
t
Þ
as
the limit of a sequence of Riemann sums derived from discrete
approximations. As the next exercise shows, when E
e
a
t
, the
convolution representation in Eq. (10-2) implies Eq. (10-1).
ð
t
Þ¼
