file for printingMATHEMATICS IN
PHARMACOKINETICS
by Jeff Stark
What and Why (A second attempt to make it
clearer)
file for printing
We have used equations for concentration (C) as a function of
time (t). We will continue to use these equations since the
plasma concentrations of drugs will be important in determining
amount of dose, frequency of dose, etc. From these
concentration/time equations we can determine the elimination
rate constant (ke), the half-life of the drug (t1/2),
and the area under the curve (AUC), and predict concentrations at
given time points.
The rate of decrease in concentration (C) with time can be
described by the equation
where n is the "order" of the rate process. We will
consider two cases: zero-order (n=0) and first-order (n=1).
If n = 0, the rate expression is (from above)
or
We can then say that the rate of decrease in
concentration is independent of concentration and depends
only on the rate constant k. So, in a zero-order process, the
same amount of drug will disappear in a given amount of time
regardless of how much drug is present.
e.g. If k = 2 mg/l/hr, my concentration will decrease by 2mg/l
every hour whether the starting concentration is 10 mg/l or 100
mg/l.
This process of constant change will show a linear plot when
graphing C vs t.
Since equations for straight lines have the same form (y = mx
+ b), we can easily write down an equation for C in terms of t
from the information in this graph. We can also obtain an
equation for C(t) by solving the zero-order rate equation given
earlier (i.e. "solve the differential equation").
Recall the equation
Rearranging dC = -kdt
We now need to integrate (to remove the differential and obtain an equation for C). The limits of integration are typically C: C0C
and t: 0 t
This will give us an equation where the concentration is C0
at t=0 and C at time t.
Integrating
= 
=
= 
C-C0 = -k(t-0)
= -kt
solving for C gives
From this we see that the y-intercept is C0 (the
initial concentration) and the slope is -k (the negative of the
rate constant). This is a rather straight-forward way of
obtaining k (m = -k).
Note: The "rate of change" can be found by taking
the derivative (d/dt).
,
which is what we started with.
What about the half-life t1/2? The half-life gives
us an idea of how long the drug will stay in the body. Would we
expect the t1/2 to be dependent or independent
of the drug concentration?
Recall that t1/2 of a drug is the time required for
half of the drug to go away. Since the rate of decrease (-dC/dt)
for a zero-order process is independent of concentration, we see
that the more drug we start off with, the more time is required
for half to be removed.
e.g. Say that the rate of decrease is 2 mg/l/hr as before. If
our initial concentration is C0=100 mg/l, it will take
a long time for half of this to go away (and have a concentration
C = 50 mg/l). However, if C0 = 10 mg/l, it will take a
much shorter time to reach C = 5 mg/l. Thus, t1/2 is concentration-dependent
for a zero-order process.
We can prove this by solving our equation C(t) for t1/2
.
At t1/2 , C = C0/2 (by definition of
half-life)
The general equation C = C0-kt becomes
at t1/2
Solving for t1/2 ,
= 
= -kt1/2
= -kt1/2
= t1/2
= -kCn
If n = 1, we have
= -kC
Thus, the rate of change depends on both the rate constant and
concentration. So, the amount of drug that goes away in a given
time period depends on how much drug we start with.
A typical 1st order plot is
This curve can be transformed to a linear plot by using ln C
instead of C (i.e. taking the natural logarithm of our
concentrations and graphing this value vs t).
Here again we see a straight line which should have an equation of the form
y = mx + b.
We need to solve the differential equation to obtain an
equation for c in terms of t.
Starting with the rate expression,
= -kC
dC = -kCdt
We need to divide through by C (to get the C's and t's on
opposite sides of the equation),
dC/C = -kdt
Integrating (again with C:C0 t and t: 0 t
= 
= 
lnC-lnC0 = -kt
lnC = lnC0 - kt (this is the equation for the
straight line seen when plotting ln C vs t)
We can transform this equation to obtain C rather than lnC by:
ClnC = elnC0-kt
C = 
C = C0e-kt
So, a first-order process shows an exponential decay.
Note: We could have plotted log C vs t and still had a linear
plot. We can convert from lnX to log X by
the equation is
log C = 
AUC can be estimated as before: 
Half-life: Is t1/2 dependent or independent of
concentration? Consider the plot of C vs t:
It seems that no matter where we start in the concentration
curve, it takes the same amount of time for half the drug to
disappear. Let's prove this.
Recall at t1/2 , C = C0/2
Putting this into our equation gives
C0/2 = C0e-kt1/2
Dividing through by C0 and solving for t1/2
1/2 = e-kt1/2
ln(1/2) = -kt1/2
t1/2 = 
t1/2 = 0.693/k
Thus, t1/2 is independent of concentration for a
first--order process.
What about area under the curve (AUC)?
This is an important parameter since it combines information
on concentrations achieved and the length of time the drug
stays around.
To determine AUC, we integrate our equation for C over some
time interval. Often t: 0. We can use two methods to determine
AUC: the trapezoidal rule and integration. AUCt (that
is, the area under the curve from some time t to infinity) is
estimated by integration.
AUCt = Ct/k
If t = 0, AUC0 = C0/k
| Summary | Zero-order | First-order |
| rate expression: | dC/dt = -k | dC/dt =  |
| solve the differential equation | ||
| equation for C | C = C0-kt | C = C0e-kt |
| to find k, plot: | C vs t | lnC vs t |
| slope: m = -k | slope: m = -k | |
| t1/2: | t1/2 = C0/2k | t1/2 = 0.693/k |
