Problem 58 - Calculus for Biology and Medicine

Question

Tylenol in the Body A patient is taking Tylenol (a painkiller that contains acetaminophen) to treat a fever. The data in this question is taken from Rawlins, Henderson, and Hijab (1977). At \(t=0\) the patient takes their first pill. One hour later the drug has been completely absorbed and the blood concentration, measured in \(\mu \mathrm{g} / \mathrm{ml}\), is 15 . Acetaminophen has first order elimination kinetics; in one hour, \(23 \%\) of the acetaminophen present in the blood is eliminated. (a) Write a recursion relation for the concentration \(c_{t}\) of drug in the patient's blood. For \(t \geq 1\) you may assume for now that no other pills are taken after the first one. (b) Find an explicit formula for \(c_{t}\) as a function of \(t\). (c) Suppose that the patient follows the directions on the pill box and takes another Tylenol pill 4 hours after the first (at time \(t=4\) ). What is the concentration at the time at which the second pill is taken? In others words, what is \(c_{4}\) ? (d) Over the next hour \(15 \mathrm{\mug} / \mathrm{ml}\) of drug enter the patient's bloodstream. So, \(c_{5}\) can be calculated from \(c_{4}\) using the word equation: $$ c_{5}=c_{4}+ $$ nt added \(\quad\) amount eliminated blo Given that the amount added is \(15 \mu \mathrm{g} / \mathrm{ml}\), and the amount eliminated is \(0.23 \cdot c_{4}\), calculate \(c_{5} .\) (e) For \(t=5,6,7,8\) the drug continues to be eliminated at a rate of \(23 \%\) per hour. No pills are taken and no extra drug enters the patient's blood. Compute \(c_{8}\). (f) At time \(t=8\), the patient takes another pill. Calculate \(c_{9} .\) Do not forget to include elimination of drug between \(t=8\) and \(t=9\). (g) We want to calculate the maximum concentration of drug in the patient's blood. We know that concentrations are highest in the hour after a pill is taken, namely at time \(t=1, t=5, t=\) \(9, \ldots\) Define a sequence \(C_{n}\) representing the concentration of the drug one hour after the \(n\) th pill is taken. (h) What terms of the original sequence \(\left\\{c_{r}: t=1,2, \ldots\right\\}\) are \(C_{1}\), \(C_{2}\), and \(C_{3} ?\) (i) Explain why $$ C_{n+1}=(0.77)^{4} \cdot C_{n}+15 $$ and \(c_{1}=15\) (j) From the recursion relation, assuming that the patient continues to take Tylenol pills at 4 -hour intervals, calculate \(C_{1}, C_{2}\), \(C_{3}, C_{4}, C_{5}\), and \(C_{6}\) (k) Does \(C_{n}\) increase indefinitely, or do you think that it converges? (1) By looking for fixing point of the recursion relation in (h), find the limit of \(C_{n}\) as \(n \rightarrow \infty\).

Step-by-Step Solution

Answer

The concentration reaches a steady state around \( 23.9 \, \mu \mathrm{g/ml} \) in the long-term.

Step 1: Define the Problem

We are tasked with analyzing the concentration of acetaminophen (Tylenol) in the bloodstream over time, following first-order elimination kinetics, with parameters provided by Rawlins et al. The patient's blood concentration needs to be calculated at various stages after taking the medication at specific intervals.

Step 2: Define Recursion Relation for Concentration

First, we need to establish a recursion relation for the concentration. At time \( t = 0 \), the concentration \( c_0 = 15 \). After each hour, \( 23\% \) of the drug is eliminated. Thus, the concentration at time \( t \) can be defined recursively as \( c_t = 0.77 \cdot c_{t-1} \) for \( t \geq 1 \).

Step 3: Find Explicit Formula for Concentration

Using the recursion relation, \( c_t = 0.77^t \cdot c_0 = 15 \cdot 0.77^t \). This formula offers a way to compute the blood concentration at any time \( t \), assuming no additional pills are taken.

Step 4: Calculate Concentration at t=4

To find \( c_4 \), we need to plug \( t = 4 \) into the explicit formula: \( c_4 = 15 \cdot 0.77^4 \). Calculating this gives \( c_4 \approx 5.416 \).

Step 5: Calculate Concentration at t=5 after Second Pill

At \( t=5 \), after taking the second pill, the concentration increases by 15 \( \mu g/ml \). So, \( c_5 = c_4 + 15 - 0.23 \cdot c_4 = c_4 \cdot 0.77 + 15 \). Substituting \( c_4 \approx 5.416 \), \( c_5 \approx 5.416 \cdot 0.77 + 15 = 19.16972 \).

Step 6: Calculate Subsequent Concentrations (t=6,7,8)

Applying the first order elimination to \( c_5 \), calculate successive concentrations: \( c_6 = 0.77 \cdot c_5 \), \( c_7 = 0.77 \cdot c_6 \), and \( c_8 = 0.77 \cdot c_7 \). This yields \( c_6 \approx 14.761 \), \( c_7 \approx 11.367 \), and \( c_8 \approx 8.755. \)

Step 7: Calculate Concentration for t=9 after Third Pill

At \( t=9 \), following the third pill intake, the concentration increases by 15 \( \mu g/ml \). So, \( c_9 = 0.77 \cdot c_8 + 15 \approx 22.734 \).

Step 8: Sequence Definitions and Recursion For Maximum Concentrations

The sequence \( C_n + 1 \) is defined for maximum drug concentration an hour after each new pill dose. The recursion \( C_{n+1} = (0.77)^4 \cdot C_n + 15 \) involves calculating \( c_1 = 15 \), \( C_n = 0.77^4 \cdot C_n + 15 \).

Step 9: Calculate C Sequence for Several Intervals

Using the recursion, compute \( C_1, C_2, C_3, C_4, C_5, \) and \( C_6 \). With \( C_1 = 15 \), use recursion to find each subsequent \( C \). We find \( C_2 \approx 18.24 \), \( C_3 \approx 20.377 \), \( C_4 \approx 21.87 \), \( C_5 \approx 23.022 \), and \( C_6 \approx 23.89 \).

Step 10: Determine the Long-term Trend of C

The sequence \( C_n \) converges rather than increasing indefinitely. This sequence stabilizes because of the compounding reduction and a fixed supplement from repeated doses.

Step 11: Find the Limit of the Converging Sequence

To find the limit of \( C_n \) as \( n \rightarrow \infty \), solve this convergence: \( (0.77)^4 \cdot L + 15 = L \). Solving gives the limit \( L = \frac{15}{1-(0.77)^4} \approx 23.9 \).

Key Concepts

Acetaminophen Concentration

Acetaminophen is a medication widely used for relieving pain and reducing fever. When taken, it gets absorbed into the bloodstream, where it reaches a certain concentration level, measured in micrograms per milliliter (1/1). This concentration depends on various factors, including dosage strength, individual metabolism, and the drug's elimination rate from the body. For Tylenol, which contains acetaminophen, understanding how the concentration changes over time is crucial to managing dosage effectively.

To illustrate, when a patient takes a pill, it first raises the acetaminophen concentration in their bloodstream. Then, over time, the body's elimination processes start to decrease this concentration following a rate observed as the first-order elimination kinetics. This concept highlights that a fixed percentage of the drug is removed from the bloodstream per time unit rather than a fixed amount, making comprehension of its concentration dynamics vital for effective medical guidance.

Recursion Relation

A recursion relation is a mathematical expression used to predict the behavior of sequences over time. In the context of acetaminophen concentration, it helps determine how the concentration changes from one hour to the next.

After each pill intake, the concentration starts at an initial value (e.g., 15 1/1) and decreases by a specific percentage as the hours pass. For acetaminophen, about 23% of the drug is eliminated every hour, leading to a recursion relation expressed as:

• \(c_{t} = 0.77 \cdot c_{t-1}\)

where \(c_{t}\) represents the concentration at time \(t\) and \(c_{t-1}\) the concentration at the previous hour.

This relation simplifies the complexity of tracking how drug levels decrease between doses, offering clear insight into effective concentration management.

Exponential Decay

Exponential decay describes how the concentration of a substance, like acetaminophen, decreases over time. This model is relevant when discussing first-order elimination kinetics.

In an exponential decay process, instead of losing a fixed amount of drug per hour, you lose a constant proportion, such as 23% in this case. To model this mathematically, we use the exponential decay formula:

• \(c_{t} = c_{0} \times 0.77^{t}\)

where \(c_{0}\) is the initial concentration and \(0.77\) is the proportion of the drug remaining after each hour. This formula allows you to predict the drug concentration at any time \(t\), aiding in understanding how quickly the medication wears off and the timing of subsequent doses.

Using this formula, healthcare providers can optimize dosage timing to ensure therapeutic levels without causing toxicity or unwanted side effects.

Long-term Equilibrium in Drug Dosage

The concept of long-term equilibrium is crucial when patients are on a regular dosing schedule, like taking Tylenol every four hours. Over time, repeated dosing reaches a balance point where the concentration peaks and troughs become stable.

This balance occurs because each new dose not only adds drug concentration but also coincides with ongoing drug elimination. Formally, this can be described with a recursion formula for maximum concentrations after each dose:

• \(C_{n+1} = (0.77)^4 \cdot C_n + 15\)

where \(C_n\) is the concentration one hour after the \(n\)-th pill.

As the patient continues this regimen, the concentration tends toward a stable limit, known in the given problem to be approximately 23.9 1/1. This equilibrium ensures consistent medicinal effects without excessive buildup, offering insight into properly scheduling dosages to balance efficacy and safety.