Derivation for Logarithmic Trapezoidal AUC Calculation

Derivation for Logarithmic Trapezoidal AUC Calculation

Calculating area under the curve requires the use of two separate equations: one is follows the “linear trapezoidal rule” and the other follows the “logarithmic trapezoidal rule.” These equations are normally presented in textbooks without derivations so all you have to do is insert the concentrations and times and you can calculate the area under the curve. I want to show you how those equations are developed so that you understand what you are doing when you calculate the area under the curve (AUC).

Let’s start with the linear trapezoidal method. We will work with a specific type of trapezoid like the one shown here:

Trapezoid

Imagine that the bottom of the trapezoid is the x-axis from time point 1 to time point 2, and the top of the trapezoid starts at concentration 1 and falls to concentration 2. This trapezoid is identical to a partial AUC from a concentration-time curve. The area of a trapezoid is document elsewhere, but let’s review it. First, we have the base of “b” which equals time point 2 – time point 1. Then we have two heights. The one on the left is h1 and the one on the right is h2, with h1>h2. We can divide the trapezoid into two shapes, a rectangle and a triangle. The rectangle has a base of b and a height of h2 (the shorter side). The rectangle has an area of b*h2. The triangle has one side of b and the other is h1-h2 (the difference between the two concentrations). The area of the triangle is 1/2*b*(h1-h2). Thus the total area of the trapezoid is:

AUC = b*h2+\frac{1}{2}*b*(h1-h2)

by expanding this, we get the following:

AUC = b*h2+\frac{1}{2}*b*h1-\frac{1}{2}*b*h2

then we can combine terms to get the following:

AUC =\frac{1}{2}*b*h2+\frac{1}{2}*b*h1

and simplifying we get the following:

AUC =\frac{1}{2}*b*(h1+h2)

By inserting more common PK variables, we get the equation for calculating AUC using the linear trapezoidal method:

AUC =\frac{1}{2}*(t_2 - t_1)*(C_1+C_2)

or

AUC_{lin} =\frac{1}{2}*(\Delta t)*(C_1+C_2)

Now, let’s consider the logarithmic AUC calculation. In this calculation, the trapezoid is nearly identical, except the sloping line from C1 to C2 does not follow a linear path. Instead it follows a logarithmic decline expressed by the following equation:

C = a*e^{-b*t}

and we know that at t1, C = C1 and at t2, C = C2. To simplify the math, let’s assume that t1 = 0. That means that the constant a = C1. We get that by substituting t1=0 for t in the above equation and solving for a. That simplification allows us to now write the following equation:

C_2 = C1*e^{-b*t_2}

Now we solve for b:

\frac{C_2}{C_1} = e^{-b*t_2}

ln(\frac{C_2}{C_1}) = -b*t_2

b = -\frac{1}{t_2}*ln(\frac{C_2}{C_1})

Now that we have found the constants, we can rewrite the equation for the curve:

C = C_1*e^{\frac{1}{t_2}*ln(\frac{C_2}{C_1})*t}

To calculate the area under the curve, we integrate this equation from t1 to t2 (we have assumed that t1=0 for this example to make the math easier):

AUC =\int_{t_1}^{t_2} C\,dt

AUC =\int_{t_1}^{t_2}C_1*e^{\frac{1}{t_2}*ln(\frac{C_2}{C_1})*t}\,dt

Note that

\int a*e^{-bx} = -\frac{a}{b}*e^{-bx}

So we get the following:

AUC =\int_{t_1}^{t_2}C_1*e^{-\frac{1}{t_2}*ln(\frac{C_2}{C_1})*t}\,dt

AUC = \frac{C_1*t_2}{ln(\frac{C_2}{C_1})}*e^{\frac{1}{t_2}*ln(\frac{C_2}{C_1})*t}\Big|_{t_1}^{t_2}

Because t1 = 0, we get the following:

AUC = \frac{C_1*t_2}{ln(\frac{C_2}{C_1})}*[e^{-ln(\frac{C_2}{C_1})}-1]

which can be simplified as follows:

AUC = \frac{C_1*t_2}{ln(\frac{C_2}{C_1})}*[-\frac{C_2}{C_1}-1]

AUC = -\frac{C_1*t_2}{ln(\frac{C_2}{C_1})}*[1-\frac{C_2}{C_1}]

AUC = -\frac{C_1*t_2}{ln(\frac{C_2}{C_1})}*[\frac{C_1}{C_1}-\frac{C_2}{C_1}]

AUC = -\frac{C_1*t_2}{ln(\frac{C_2}{C_1})}*[\frac{C_1-C_2}{C_1}]

And finally, removing the negative by flipping the ratio in the natural logarithm to give the final equation:

AUC = \frac{t_2}{ln(\frac{C_1}{C_2})}*({C_1-C_2})

This was done with the simplification that t1 = 0 … we can generalize this for any situation by replacing t2 with the time difference (t2-t1):

AUC_{log} = \frac{\Delta t}{ln(\frac{C_1}{C_2})}*({C_1-C_2})

and here is the equation for the linear AUC:

AUC_{lin} =\frac{1}{2}*(\Delta t)*(C_1+C_2)

As you can see, they are similar. In all cases AUClog is less than AUClin due to the logarithmic decline rather than linear decline.

To learn about how we’ve improved Phoenix to make performing NCA and PK/PD modeling even easier, please watch this webinar I gave on the latest enhancements to Phoenix.

Nathan Teuscher

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Nathan Teuscher

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Dr. Nathan Teuscher is the Vice President of Pharmacometric Solutions at Certara. He is an expert in clinical pharmacology, pharmacometrics, pharmacokinetics and pharmacodynamics and was trained by David Smith at the University of Michigan. Dr. Teuscher has held leadership positions in biotechnology, pharmaceutical and contract research companies. In 2008 he established the Learn PKPD.com website to share his knowledge with the community. Prior to coming to Certara, he was the Founder and President of PK/PD Associates.