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**Figure 19.2.1 Two Compartment Pharmacokinetic Model**

Clearance Model

Equilibrium Model

**Equation 19.2.1 Differential Equation for the Central Compartment**

The kel • X1 term describes elimination of the drug from the central compartment, while the k12 • X1 and k21 • X2 terms describe the distribution of drug between the central and peripheral compartments. Writing differential equations can be reviewed in Chapter 2.

**Equation 19.2.2 Integrated Equation for Plasma Concentration versus Time**

**
Equation 19.2.3 Integrated Equation for Cp versus Time including k21 and V _{1}
**

with α > β and

**Equation 19.2.4 Calculating values for A and B**

The A, B, α, and β terms were derived from the micro-constants during the integration process. They are functions of the micro-constant k12, k21, kel and V1

Using the substitutions for the sum and product of α and β.

α + β = kel + k12 + k21

α • β = kel • k21

If we know the values of kel, k12 and k21 we can calculate α + β as well as α • β. Substituting these values into Equation 19.2.3 gives us values for α and β.

**Equation 19.2.5 Converting from kel, k12 & k21 to α & β**

Note, in this equation, α is calculated when '+' is used in the numerator and β is calculated when '-' is used in place of the '±'. Thus α is greater than β.

Once we have values for α and β we can calculate values for A and B using Equation 19.2.4.

Since α + β = kel + k12 + k21 = 0.192 + 1.86 + 1.68 = 3.732

and

α x β = kel x k21 = 0.192 x 1.68 = 0.32256

Now:

α = [(a+b) + sqrt((a+b)^{2} - 4xaxb)]/2 = [3.732 + sqrt(3.732^{2} - 4x0.32256)]/2 = [3.732 + 3.5549]/2 = 3.643 hr^{-1}

β = [3.732 - 3.5549]/2 = 0.08853 hr^{-1}

A = Dose x (α - k21)/[V1 x (α - b)] = 500 x (3.643 - 1.68)/[90.5 x 0.39 x (3.643 - 0.08853)] = 500 x 1.963/[35.295 x 3.55447] = 7.824 mg/L

B = Dose x (k21 - b)/[V1 x (α - b)] = 500 x (1.68 - 0.08853)/[35.295 x 3.55447] = 6.343 mg/L

The last step is

Cp = α x e^{(-α x t)} + β x e^{(-β x t)} = 7.824 x e^{(-3.643 x 1.5)} + 6.343 x e^{(-0.08853 x 1.5)} = 7.824 x 0.004234 + 6.343 x 0.8756 = 0.0331 + 5.5542 = 5.59 mg/L

Later in this chapter we will use equations for the reverse process of converting α, β, A and B into values for k12, k21, kel and V_{1}.

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