Mathematics in Pharmacokinetics What and Why (A second attempt to make it clearer)


 Kristopher Howard
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1 Mahemaics in Pharmacokineics Wha and Why (A second aemp o make i clearer) We have used equaions for concenraion () as a funcion of ime (). We will coninue o use hese equaions since he plasma concenraions of drugs will be imporan in deermining amoun of dose, frequency of dose, ec. From hese concenraion/ime equaions we can deermine he eliminaion rae consan (ke), he halflife of he drug ( 1/2 ), and he area under he curve (AU), and predic concenraions a given ime poins. The rae of decrease in concenraion () wih ime can be described by he equaion d = k n, where n is he order of he rae process. We will consider wo cases: zeroorder (n=) and firsorder (n=1). Zeroorder If n =, he rae expression is (from above) or d d = k = k 1 = k We can hen say ha he rae of decrease in concenraion is independen of concenraion and depends only on he rae consan k. So, in a zeroorder process, he same amoun of drug will disappear in a given amoun of ime regardless of how much drug is presen. e.g. If k = 2 mg/l/hr, my concenraion will decrease by 2mg/l every hour wheher he saring concenraion is 1 mg/l or 1 mg/l. D:\PHA412\INTRODU\PKMATHsark.DO 1
2 This process of consan change will show a linear plo when graphing vs. Iniial conc. (o) slope: m = k Since equaions for sraigh lines have he same form (y = mx + b), we can easily wrie down an equaion for in erms of from he informaion in his graph. We can also obain an equaion for () by solving he zeroorder rae equaion given earlier (i.e. solve he differenial equaion ). Recall he equaion d = k Rearranging = kd We now need o inegrae (o remove he differenial and obain an equaion for ). The limis of inegraion are ypically : and : This will give us an equaion where he concenraion is a = and a ime. Inegraing = kd = k d ] = ] k  = k() = k solving for gives D:\PHA412\INTRODU\PKMATHsark.DO 2
3 =  k y b + m x From his we see ha he yinercep is (he iniial concenraion) and he slope is k (he negaive of he rae consan). This is a raher sraighforward way of obaining k (m = k). Noe: The rae of change can be found by aking he derivaive (d/d). d d = ( k) = k, which is wha we sared wih. d Wha abou he halflife 1/2? The halflife gives us an idea of how long he drug will say in he body. Would we expec he 1/2 o be dependen or independen of he drug concenraion? Recall ha 1/2 of a drug is he ime required for half of he drug o go away. Since he rae of decrease (/d) for a zeroorder process is independen of concenraion, we see ha he more drug we sar off wih, he more ime is required for half o be removed. e.g. Say ha he rae of decrease is 2 mg/l/hr as before. If our iniial concenraion is =1 mg/l, i will ake a long ime for half of his o go away (and have a concenraion = 5 mg/l). However, if = 1 mg/l, i will ake a much shorer ime o reach = 5 mg/l. Thus, 1/2 is concenraiondependen for a zeroorder process. We can prove his by solving our equaion () for 1/2. A 1/2, = /2 (by definiion of halflife) The general equaion = k becomes k = / a 1/2 D:\PHA412\INTRODU\PKMATHsark.DO 3
4 Solving for 1/2, Firs Order 2 = k 1/ 2 2 = k 1/2 2 2k = k 1/2 = 1/2 d = k n If n = 1, we have d = k Thus, he rae of change depends on boh he rae consan and concenraion. So, he amoun of drug ha goes away in a given ime period depends on how much drug we sar wih. A ypical 1s order plo is iniial conc. ( ) k is expressed somehow in he curve. How can we find k? This curve can be ransformed o a linear plo by using ln insead of (i.e. aking he naural logarihm of our concenraions and graphing his value vs ). D:\PHA412\INTRODU\PKMATHsark.DO 4
5 ln ln slope: m = k (proved laer) Here again we see a sraigh line which should have an equaion of he form y = mx + b. We need o solve he differenial equaion o obain an equaion for c in erms of. Saring wih he rae expression, d = k = kd We need o divide hrough by (o ge he s and s on opposie sides of he equaion), / = kd Inegraing (again wih : and : = kd ln] lnln ln = k] = k = ln  k (his is he equaion for he sraigh line seen when ploing ln vs ) We can ransform his equaion o obain raher han ln by: ln = e lnk D:\PHA412\INTRODU\PKMATHsark.DO 5
6 = e e ln k = e k So, a firsorder process shows an exponenial decay. Noe: We could have ploed log vs and sill had a linear plo. We can conver from lnx o log X by log log slope: m =  k ) 2.33 he equaion is log = log k AU can be esimaed as before: AU = Halflife: Is 1/2 dependen or independen of concenraion? onsider he plo of vs : 1 mg/l rapid change a large conc less a smaller conc. 1/2 1/2 I seems ha no maer where we sar in he concenraion curve, i akes he same amoun of ime for half he drug o disappear. Le s prove his. Recall a 1/2, = /2 D:\PHA412\INTRODU\PKMATHsark.DO 6
7 Puing his ino our equaion gives /2 = e k 1/2 Dividing hrough by and solving for 1/2 1/2 = e k 1/2 ln(1/2) = k 1/2 1/2 = ln 1 / 2 k 1/2 =.693/k Thus, 1/2 is independen of concenraion for a firsorder process. AU Wha abou area under he curve (AU)? AU This is an imporan parameer since i combines informaion on concenraions achieved and he lengh of ime he drug says around. To deermine AU, we inegrae our equaion for over some ime inerval. Ofen :. We can use wo mehods o deermine AU: he rapezoidal rule and inegraion. AU (ha is, he area under he curve from some ime o infiniy) is esimaed by inegraion. AU = /k D:\PHA412\INTRODU\PKMATHsark.DO 7
8 If =, AU = /k Summary. Zeroorder Firsorder rae expression: /d = k /d = k solve he differenial equaion equaion for = k = e k o find k, plo: vs ln vs slope: m = k slope: m = k 1/2 : 1/2 = /2k 1/2 =.693/k D:\PHA412\INTRODU\PKMATHsark.DO 8
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