Equation VI1. Differential Equation During an IV Infusion.
This is the differential equation during the infusion period and it can be integrated to give:
Equation VI2. Integrated Equation for Cp versus Time
Calculate Cp Given k0, kel and V at time t

Calculate k0 required to give Cp at time t

You may notice that Equation VI2 for Cp is quite similar to Equation V8 we had before for the cumulative amount of drug excreted into urine. As you might expect the plot of Cp would be similar in shape.
Figure VI1. Linear Plot of Cp versus Time During a Continuous Infusion
Plasma Concentration versus Time Plots
This is because the rate of infusion will be constant whereas the rate of elimination will increase as the plasma concentration increases. At steady state the two rates become equal. We can determine the steady state concentration from the differential equation by setting the rate of change of Cp, i.e. dCp/dt = 0.
Then
therefore
Equation VI3. Steady State Concentration after Continuous Infusion
This could also be calculated from the integrated equation by setting e^{ kel * t } = 0 at t = .
We can now calculate the infusion rate necessary to produce some desired steady state plasma level.
For Example:
A desired steady state plasma level of theophylline maybe 15 mg/L. The average halflife of theophylline is about 4 hr and the apparent volume of distribution is about 25 liter. What infusion rate is necessary?
First, kel = 0.693/4 = 0.17 hr^{1}
then k0 = kel * V * Cp = 0.17 * 25 * 15 = 63.8 mg/hr
We would probably use an infusion of 60 mg/hr which would produce a Cp^{ss} value given:
Calculate Cp^{ss} Given k0, kel and V

Calculate k0 required to give Cp^{ss}

Copyright 2002 David W.A. Bourne