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**Figure 7.5.1 A Two Compartment Pharmacokinetic Model**

Following an IV bolus dose the drug is placed in component 1. Drug concentration in this component achieves equilibrium quickly. More slowly the drug either distributes into and out of various tissues and body fluids designated component 2 or is eliminated from the body as described by the first order rate constant, kel. The distribution processes are usually described using first order rate constants. Alternately, the elimination and distribution could be describe by clearance terms. For the purposes of this derivation we will confine the mathematical model to use rate constants. The differential equations for both components are shown below:

**Equations 7.5.1 Differential Equations for the Two Compartment Model after an IV Bolus Dose**

Using the rules from Chapter 2 it is possible to write the Laplace transform of each of the equations.

**Equations 7.5.2 Laplace Transforms for the Equations 7.5.1**

Since the amount of drug in component 2 is zero at time 0 the term, X^{0}_{2} can be set to 0. Equation 7.5.2.b becomes:

**Equation 7.5.3 Laplace Transform of the Amount in Component 2**

With X^{0}_{1} the Dose we can substitute the value in Equation 7.5.3 for the Laplace of the amount in component 2 into Equation 7.5.2.b to continue the derivation of the Laplace of component 1.

**Equation 7.5.4 Deriving the Laplace of the Amount in Component 1**

And now the fun stuff. We can make a substitution to simplify Equation 7.5.4. Looking at:

**Equation 7.5.5 Introducing the terms α and β**

Looking at Equation 7.5.5 and the last term in Equation 7.5.4 we can make the substitutions:

α + β = k12 + kel + k21

and α x β = kel x k21

to derive the final equation for the Laplace of the amount in component 1.

**Equation 7.5.6 Laplace of the Amount in Component 1**

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