## Integrated equation

We can calculate the line (in Figure VIII-1) using the integrated form of the
equation. Starting with the differential equation we can substitute Xg = Xg
*e^{-ka * t}. The integration process won't be described here but a
reference, which describes the Laplace transform method of integration
[1], can be presented. This method makes
integration as *easy* as the
logarithmic transform makes multiplication and division easier.

If we use F * DOSE for Xg^{0} where F is the fraction of the dose
absorbed, the integrated equation for Cp versus time is :-

**Equation VIII-4**

Notice that
the right hand side of this equation (Equation VIII-4) is a constant multiplied by
the difference of two exponential terms. **A biexponential equation**.

We can plot Cp as a constant times the difference between two exponential
curves (see Figure II-1). If we plot each exponential separately.

**Figure VIII-3, Linear Plot of e**^{-ka x t} versus Time for Two Exponential Terms

Notice that the difference starts at zero, increases, and finally decreases again.

**Plotting this difference **__by__

gives Cp versus time.

**Figure VIII-4, Linear Plot of Cp versus Time**
We can calculate the plasma
concentration at anytime if we know the values of all the parameters of
Equation VIII-4.

We can also calculate the time of peak concentration using the equation:-

As an example we could calculate the peak plasma concentration given that F =
0.9, DOSE = 600 mg, ka = 1.0 hr^{-1},
kel = 0.15 hr^{-1},
and V = 30 liter.

= 2.23 hour

= 21.18 x [ 0.7157 - 0.1075] = 12.9 mg/L

As another example we could consider what would happen with ka = 0.2 hr^{-1}
instead of 1.0 hr^{-1}

= 5.75 hour

= 72 x (0.4221 - 0.3166) = 7.6 mg/L lower and slower than before

**Error Message** Value is not a numeric literal probably means that one of the parameter fields is empty or a value is inappropriate.

This page was last modified: 17 March 2005
Copyright 2001 David W.A. Bourne