Comments on: Pharmacology calculation! need immedite help please!Can anyone help solving it? http://www.abortoterapeutico.org/therapeutic-agents/pharmacology-calculation-need-immedite-help-pleasecan-anyone-help-solving-it Wed, 12 May 2010 06:52:59 -0400 http://wordpress.org/?v=2.8.4 hourly 1 By: grimmyTea http://www.abortoterapeutico.org/therapeutic-agents/pharmacology-calculation-need-immedite-help-pleasecan-anyone-help-solving-it/comment-page-1#comment-1322 grimmyTea Wed, 24 Feb 2010 19:54:59 +0000 http://www.abortoterapeutico.org/therapeutic-agents/pharmacology-calculation-need-immedite-help-pleasecan-anyone-help-solving-it#comment-1322 This question is really more appropriate for the engineering or math sections. Consider the human body as a box. Drug is going in and drug is coming out. Your ultimate goal for this question is to figure out what the concentration of the drug is inside the box at any given time (and you will want it to be in the range of 0.06-1.3 mg/L). The only info you have are the rates at which things are going in and coming out. These won't tell you what the concentration is, but they will tell you how it is changing over time--and you can figure out the concentration based on this since you know the starting condition. First, what is going into the box at any given time? (1) D * ba * exp(-ka * t) / V where D is the dosage (either 200 or 400 mg), ba is the bioavailability (20%), ka is the absorption constant (either 0.167/hr or 0.083/hr), t is time, and V is volume of distribution (56 L). Second, what is coming out of the box at any given time? (2) C * ke where C is the concentration, and ke is the elimination constant (0.28/hr). The change in concentration with time is equal to [stuff going in] - [stuff coming out]. (3) dC/dt = [D * ba * exp(-ka * t) / V] - [C * ke] rearrange to: dC + ke*C*dt = D * ba * exp(-ka * t) / V * dt If you have been assigned this problem, then you must know how to solve differential equations. Analytically, you can solve this first-order linear differential equation using an integrating factor of exp(ke*t). (4) dC*exp(ke*t) + ke*C*dt*exp(ke*t) = D*ba*exp(-ka*t)/V*dt*exp(ke*t) The left side of the equation (4) is the differential of C*exp(ke*t) (that's the whole trick behind the use of integrating factors). So when you integrate both sides of the equation and you get: (5) C*exp(ke*t) = A + Int{ D*ba*exp(-ka*t)/V *dt*exp(ke*t)} where A is a constant of integration and Int{ denotes the integral operator. Solve this to get (6) C = {A + D*ba*exp[(ke-ka)*t] / V / (ke-ka)} / exp(ke*t) You could also solve equation (3) computationally, but why bother? You can solve for A using either the initial condition of C(0) = 0 or C(0) = 0.06. Just make sure to specify which one you used. Now you have the function C(t). You can plot it using the two possible values for D and the two possible values of ka over a time period of 24 hours. You should find that D=200 mg and ka=0.167/hr are the only values that keep the plasma concentration within the therapeutic range.<br><b>References : </b><br> This question is really more appropriate for the engineering or math sections. Consider the human body as a box. Drug is going in and drug is coming out. Your ultimate goal for this question is to figure out what the concentration of the drug is inside the box at any given time (and you will want it to be in the range of 0.06-1.3 mg/L). The only info you have are the rates at which things are going in and coming out. These won’t tell you what the concentration is, but they will tell you how it is changing over time–and you can figure out the concentration based on this since you know the starting condition.

First, what is going into the box at any given time?
(1) D * ba * exp(-ka * t) / V
where D is the dosage (either 200 or 400 mg), ba is the bioavailability (20%), ka is the absorption constant (either 0.167/hr or 0.083/hr), t is time, and V is volume of distribution (56 L).

Second, what is coming out of the box at any given time?
(2) C * ke
where C is the concentration, and ke is the elimination constant (0.28/hr).

The change in concentration with time is equal to [stuff going in] – [stuff coming out].
(3) dC/dt = [D * ba * exp(-ka * t) / V] – [C * ke]

rearrange to:
dC + ke*C*dt = D * ba * exp(-ka * t) / V * dt

If you have been assigned this problem, then you must know how to solve differential equations. Analytically, you can solve this first-order linear differential equation using an integrating factor of exp(ke*t).

(4) dC*exp(ke*t) + ke*C*dt*exp(ke*t) = D*ba*exp(-ka*t)/V*dt*exp(ke*t)

The left side of the equation (4) is the differential of C*exp(ke*t) (that’s the whole trick behind the use of integrating factors). So when you integrate both sides of the equation and you get:

(5) C*exp(ke*t) = A + Int{ D*ba*exp(-ka*t)/V *dt*exp(ke*t)}
where A is a constant of integration and Int{ denotes the integral operator.

Solve this to get
(6) C = {A + D*ba*exp[(ke-ka)*t] / V / (ke-ka)} / exp(ke*t)
You could also solve equation (3) computationally, but why bother?

You can solve for A using either the initial condition of C(0) = 0 or C(0) = 0.06. Just make sure to specify which one you used.

Now you have the function C(t). You can plot it using the two possible values for D and the two possible values of ka over a time period of 24 hours. You should find that D=200 mg and ka=0.167/hr are the only values that keep the plasma concentration within the therapeutic range.
References :

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