Problem 33 - Calculus for Biology and Medicine

Question

A two-compartment model of how drugs are absorbed into the body predicts that the amount of drug in the blood will vary with time according to the following function: $$ M(t)=a\left(e^{-k t}-e^{-3 k t}\right), \quad t \geq 0 $$ where \(a>0\) and \(k>0\) are parameters that vary depending on the patient and the type of drug being administered. For parts (a)-(c) of this question you should assume that \(a=1\). (a) Show that \(M(t) \rightarrow 0\) as \(t \rightarrow \infty\) (b) Show that the function \(M(t)\) has a single local maximum, and find the maximum concentration of drug in the patient's blood. (c) Show that the \(M(t)\) has a single inflection point (which you should find). Does the function go from concave up to concave down at this inflection point or vice versa? (d) Would any of your answers to (a)-(c) be changed if \(a\) were not equal to \(1 .\) Which answers?

Step-by-Step Solution

Answer

(a) Limits to 0; (b) Max at \(t=\frac{\ln(3)}{2k}\); (c) Inflection at \(t=\frac{\ln(9)}{2k}\), shifts from concave up to down; (d) Results unchanged by \(a\).

Step 1: Analyze Limit as t Approaches Infinity

To show that \(M(t) \rightarrow 0\) as \(t \rightarrow \infty\): Analyze the limit of each individual term in the function: \(M(t)=a(e^{-kt}-e^{-3kt})\). Since \(e^{-kt}\) and \(e^{-3kt}\) are exponential decay functions, as \(t\) approaches infinity, both \(e^{-kt}\) and \(e^{-3kt}\) approach zero. Therefore, \(M(t) \rightarrow a(0-0) = 0\).

Step 2: Find the First Derivative and Analyze Critical Points

Find the derivative of \(M(t)\):\[\frac{dM}{dt} = a(-ke^{-kt} + 3ke^{-3kt})\]. Set this derivative to zero to find critical points: \(-ke^{-kt} + 3ke^{-3kt} = 0\), which simplifies to \(3e^{-3kt} = e^{-kt}\). This implies \(3 = e^{2kt}\), hence \(2kt = \ln(3)\), giving \(t = \frac{\ln(3)}{2k}\).

Step 3: Determine the Nature of the Critical Point

To confirm that the critical point is a maximum, find the second derivative: \[\frac{d^2M}{dt^2} = a(k^2e^{-kt} - 9k^2e^{-3kt})\]. Evaluate at \(t = \frac{\ln(3)}{2k}\): \[\frac{d^2M}{dt^2} = a(k^2e^{-\frac{\ln(3)}{2}} - 9k^2e^{-\frac{3\ln(3)}{2}})\]. Since the second derivative is less than zero at this point, \(t = \frac{\ln(3)}{2k}\) is a local maximum.

Step 4: Calculate Maximum Concentration

Evaluate \(M(t)\) at \(t=\frac{\ln(3)}{2k}\): \(M(\frac{\ln(3)}{2k}) = a(e^{-\frac{\ln(3)}{2}} - e^{-\frac{3\ln(3)}{2}})\). Simplifying gives \(M(\frac{\ln(3)}{2k}) = a\left(\sqrt{\frac{1}{3}} - \frac{1}{3\sqrt{3}}\right)\).

Step 5: Identify Inflection Point

Set the second derivative \(\frac{d^2M}{dt^2} = 0\) to find the inflection point: \(k^2e^{-kt} = 9k^2e^{-3kt}\), which simplifies to \(9e^{-3kt} = e^{-kt}\), implying \(2kt = \ln(9)\), giving \(t = \frac{\ln(9)}{2k}\).

Step 6: Determine Concavity Around Inflection Point

Check the sign of \(\frac{d^2M}{dt^2}\) to either side of \(t = \frac{\ln(9)}{2k}\): Just before this point \( \frac{d^2M}{dt^2} > 0\), and just after \( \frac{d^2M}{dt^2} < 0\). This indicates a change from concave up to concave down.

Step 7: Consider Effect of Parameter a

If \(a eq 1\), the limits and existence of the critical/inflection points remain unchanged, since \(a\) is a constant scaling factor. The value of the maximum concentration and inflection point will be scaled by \(a\), but the key properties do not change.

Key Concepts

Drug Absorption Kinetics

Drug absorption kinetics is a critical concept in pharmacology, dealing with how a drug is absorbed into the body over time. In many cases, this process is modeled using mathematical functions. These functions help predict the concentration of a drug in the bloodstream at any given time.
In the context of a two-compartment model, the body is divided into two parts or compartments, typically the central compartment (the bloodstream) and the peripheral compartment (other tissues). This model provides insights into how quickly a drug enters and leaves these compartments.
The function in the original exercise, \(M(t) = a(e^{-kt} - e^{-3kt})\), represents the concentration of the drug in the central compartment. Here, \(a\) and \(k\) are constants that depend on the patient and the specific drug. Understanding drug absorption kinetics and the related mathematical models helps in designing dosing regimens and predicting drug efficacy and safety.

Exponential Decay

Exponential decay describes a process where a quantity decreases at a rate proportional to its current value. It is often observed in natural processes like radioactive decay and drug elimination from the bloodstream.
In the function \(M(t) = a(e^{-kt} - e^{-3kt})\), both \(e^{-kt}\) and \(e^{-3kt}\) are exponential decay terms. As time \(t\) increases, these terms approach zero. The rate of decay is faster for \(e^{-3kt}\) due to a larger decay constant, \(-3k\), compared to \(-k\).
This concept helps us understand why \(M(t)\) decreases over time, as eventually both exponential terms diminish, leading to \(M(t)\) approaching zero. Exponential decay is crucial for modeling how drugs are cleared from the body, guiding clinicians in determining appropriate dosing frequencies.

Critical Points

Critical points of a function are where its first derivative equals zero or is undefined. These points can indicate local maxima, minima, or saddle points. Analyzing them helps us understand key behaviors of a function.
For the function \(M(t)\), we found the derivative and set it to zero to identify critical points:

• \(\frac{dM}{dt} = a(-ke^{-kt} + 3ke^{-3kt}) = 0\)
• This implies \(3e^{-3kt} = e^{-kt}\), leading to \(t = \frac{\ln(3)}{2k}\)

This critical point is a local maximum because further analysis with the second derivative confirms that \(\frac{d^2M}{dt^2}\) is negative at this point. Understanding critical points allows us to pinpoint when the drug concentration in the blood reaches its peak, helping in determining optimal dosing intervals.

Inflection Point

An inflection point is where a function changes concavity, from concave up to concave down or vice versa. To find inflection points, we look for where the second derivative equals zero.
For the given model, the second derivative is:

• \(\frac{d^2M}{dt^2} = a(k^2e^{-kt} - 9k^2e^{-3kt})\)

Set this to zero:

• \(k^2e^{-kt} = 9k^2e^{-3kt}\), implying \(t = \frac{\ln(9)}{2k}\)

Analyzing the concavity immediately before and after this point:

• Before: \(\frac{d^2M}{dt^2} > 0\) (concave up)
• After: \(\frac{d^2M}{dt^2} < 0\) (concave down)

This tells us that the function transitions from a concave up to concave down shape, a useful insight for understanding how the rate of change in drug concentration shifts over time. Recognizing inflection points aids in assessing when the drug’s effect may begin to diminish, informing therapeutic decision-making.