## Derivation

### From here, the overall rate of elimination of drug from the body may be written as: dDB/dt = -kelDB (2)     DB = DBOexp(-kelt)   lnDB = lnDBO - kelt   (take antilog) {note: ln(AeB) = ln(A) + B} The value of exp(-kelt) varies from 1 to 0 as time varies from 0 to infinity.  When time is between 0 and infinity, the fraction of the dose remaining in the body is equal to exp(-kelt). When deriving kinetic equations, exponential rate expressions can be substituted into differential equations that can then be integrated.  An example of this would be from equation (1) describing the rate of appearance of unchanged drug in urine.  This can be written as dDU/dt = + keDB  by substitution of (2) into this equation gives: dDU/dt = +ke[DBO exp(-kelt)] dDU = +ke[DBO exp(-kelt)]dt through integration we get: DU(t)= (-ke/kel)DBOexp(-kelt) + constant at t=0, DU=0, and exp(-kelt)=1; therefore, the constant equals (ke/kel)DBO, and (3)     DU(t)= (ke/kel)DBO[1-exp(-kelt)] when elimination is complete at t = infinity, the total amount of drug excreted unchanged in urine(DU(infinity)) is: DU(infinity) = (ke/kel)DBO(1-0) (4)     DU(infinity)/DBO = ke/kel  = fe Equation (4) shows that the fraction of the dose appearing as unchanged drug in urine (fe) is equal to the fraction of the kel attributable to ke.  DM(infinity) and km can be calculated in the same manner. By substitution of (3) and (4), we get DU(infinity) - DU(t) = DU(infinity) exp(-kelt) And taking the log of both sides we get (5)

Note: all rate constants (ke, km) can be calculated solely on urinary data without knowledge of plasma level.

B.7a-2