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

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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 half-life 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: zero-order (n=) and firs-order (n=1). Zero-order 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 zero-order 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\PKMATH-sark.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 zero-order 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\PKMATH-sark.DO 2

3 = - k y b + m x From his we see ha he y-inercep is (he iniial concenraion) and he slope is -k (he negaive of he rae consan). This is a raher sraigh-forward 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 half-life 1/2? The half-life 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 zero-order 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 concenraion-dependen for a zero-order process. We can prove his by solving our equaion () for 1/2. A 1/2, = /2 (by definiion of half-life) The general equaion = -k becomes k = / a 1/2 D:\PHA412\INTRODU\PKMATH-sark.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\PKMATH-sark.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] ln-ln 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 ln-k D:\PHA412\INTRODU\PKMATH-sark.DO 5

6 = e e ln k = e -k So, a firs-order 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 = Half-life: 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\PKMATH-sark.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 firs--order 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\PKMATH-sark.DO 7

8 If =, AU = /k Summary. Zero-order Firs-order 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\PKMATH-sark.DO 8

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