Problem 30 - Calculus

Question

Hydrocodone bitartrate is used as a cough suppressant. After the drug is fully absorbed, the quantity of drug in the body decreases at a rate proportional to the amount left in the body. The half-life of hydrocodone bitartrate in the body is 3.8 hours, and the usual oral dose is 10 mg. (a) Write a differential equation for the quantity, \(Q,\) of hydrocodone bitartrate in the body at time \(t,\) in hours since the drug was fully absorbed. (b) Find the equilibrium solution of the differential equation. Based on the context, do you expect the equilibrium to be stable or unstable? (c) Solve the differential equation given in part (a). (d) Use the half-life to find the constant of proportionality, \(k\) (e) How much of the 10 mg dose is still in the body after 12 hours?

Step-by-Step Solution

Answer

After 12 hours, about 0.83 mg of the drug remains in the body.

Step 1: Understanding the Proportional Decrease

The rate of decrease of the quantity of hydrocodone, \(Q\), is proportional to the amount left in the body, \(Q\). Mathematically, this is expressed by the differential equation \(\frac{dQ}{dt} = -kQ\), where \(k\) is the constant of proportionality.

Step 2: Finding Equilibrium Solution

The equilibrium solution to a differential equation is when \(\frac{dQ}{dt} = 0\). From \(-kQ = 0\), solving for \(Q\) gives \(Q = 0\). In the context of the drug, this means that eventually all the drug is eliminated from the body, which is stable.

Step 3: Solving the Differential Equation

The differential equation \(\frac{dQ}{dt} = -kQ\) is separable. Separating the variables and integrating, we get \(\int \frac{1}{Q} dQ = -k \int dt\), which results in \(\ln |Q| = -kt + C\). Exponentiating both sides, we find \(Q(t) = Ce^{-kt}\), where \(C\) is the initial amount of drug.

Step 4: Finding the Constant of Proportionality

Using the half-life, we know \(Q(3.8) = \frac{1}{2}C\). Substituting into \(Ce^{-k \, 3.8} = \frac{1}{2}C\), canceling \(C\), and solving, we get \(e^{-3.8k} = \frac{1}{2}\), so \(-3.8k = \ln\left( \frac{1}{2} \right)\). Thus, \(k = \frac{\ln(2)}{3.8}\approx 0.1824\).

Step 5: Calculating Drug Quantity After 12 Hours

We use \(Q(t) = 10e^{-0.1824t}\) with \(t = 12\) hours. So, \(Q(12) = 10e^{-0.1824 \cdot 12}\approx 0.8309 \, \text{mg}\). After 12 hours, about 0.83 mg of the drug remains in the body.

Key Concepts

Exponential Decay

Exponential decay is a commonly encountered concept in differential equations and is particularly relevant in processes like the reduction of drug concentrations in the body over time. In this context, we deal with the drug hydrocodone bitartrate which decreases at a rate proportional to its current amount. This characteristic implies that at any given time, the rate of decrease is a constant fraction of the drug left. This phenomenon can be mathematically described using the differential equation \(\frac{dQ}{dt} = -kQ\), where \(Q\) is the quantity of the drug, and \(k\) is a positive constant known as the decay constant.
This equation tells us that the change in \(Q\) is negative, indicating a decrease, and is proportional to \(Q\) itself. As the process continues, less and less drug remains, making the decrease slower over time. This is a signature of exponential decay, where the concentration drops rapidly at first and then slowly approaches zero.

Half-Life

Half-life is a critical concept tied closely to the idea of exponential decay. It is defined as the time required for a quantity to reduce to half its initial amount. In pharmacology, the half-life is used to describe how quickly a drug is metabolized and eliminated from the body. For hydrocodone bitartrate, the half-life is 3.8 hours, meaning every 3.8 hours, the body's drug concentration reduces by half.
The half-life is crucial for determining the constant of proportionality, \(k\), in our exponential decay equation. This value is found by using the formula for half-life in terms of the decay constant: \(t_{1/2} = \frac{\ln(2)}{k}\). Solving for \(k\) with the given half-life, we obtain \(k = \frac{\ln(2)}{3.8} \approx 0.1824\). Knowing \(k\) allows us to predict how the concentration of a drug decreases over any period.

Drug Absorption

Drug absorption is the process by which a drug is taken into the body and reaches the bloodstream. In this exercise, we assume that the drug is already fully absorbed when we begin to analyze its decay. This simplification allows us to focus entirely on how the drug's concentration decreases over time, ignoring other factors that involve initial absorption processes.
Understanding drug absorption and the time frame of decay helps in designing dosage regimens. Accurate knowledge about how quickly a drug is absorbed and its half-life informs how often doses need to be administered to maintain a therapeutic level in the body, without reaching toxic levels or becoming ineffective.

Proportional Rate

The concept of a proportional rate is central to understanding the nature of the differential equation \(\frac{dQ}{dt} = -kQ\). The term "proportional" indicates that the rate of decline in the drug's concentration is directly related to its current amount. Simply put, as the amount of the drug decreases, so does the speed at which it continues to decline.
This relationship forms the basis for solutions to many real-world problems, where dynamics are governed by proportional relationships. In our drug model, the proportional rate ensures that the drug's concentration decreases in a way that follows exponential decay, reflecting a continuous, naturally occurring process, such as radioactive decay or cooling of a hot object. Understanding this helps predict how much of the drug will remain at any point in time, as demonstrated when calculating the remaining drug 12 hours after administration.