Problem 16 - Applied 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 dose is \(10 \mathrm{mg}\). (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) Solve the differential equation given in part (a). (c) Use the half-life to find the constant of proportionality, \(k\). (d) How much of the \(10-\mathrm{mg}\) dose is still in the body after 12 hours?

Step-by-Step Solution

Answer

(a) \( \frac{dQ}{dt} = -kQ \); (b) \( Q = Ce^{-kt} \); (c) \( k \approx 0.1824 \); (d) 0.73 mg remains after 12 hours.

Step 1: Write the Differential Equation

The rate at which the drug decreases is proportional to the amount remaining. This is expressed as \( \frac{dQ}{dt} = -kQ \), where \( Q \) is the quantity of the drug, and \( k \) is the constant of proportionality.

Step 2: Solve the Differential Equation

To solve the differential equation \( \frac{dQ}{dt} = -kQ \), separate the variables to get \( \frac{dQ}{Q} = -kdt \). Integrating both sides, we have \( \ln|Q| = -kt + C \). Exponentiating both sides, we obtain \( Q = Ce^{-kt} \), where \( C \) is a constant that can be determined by initial conditions.

Step 3: Use Half-Life to Find k

The half-life \( t_{1/2} \) of a substance is the time it takes for half of the substance to decay. We have \( \frac{Q_0}{2} = Q_0e^{-kt_{1/2}} \). Solving for \( k \), the equation becomes \( e^{kt_{1/2}} = 2 \). Hence, \( k = \frac{\ln(2)}{3.8} \approx 0.1824 \) hours^{-1}.

Step 4: Calculate Amount after 12 Hours

Using the equation from Step 2, \( Q = Ce^{-kt} \), where \( C = 10 \) mg. We substitute \( t = 12 \) and \( k = 0.1824 \): \( Q = 10 \times e^{-0.1824 \times 12} \approx 0.73 \) mg. So, approximately 0.73 mg of the drug remains after 12 hours.

Key Concepts

Half-life

The concept of half-life is commonly used in the context of radioactive decay, but it also applies to any process that follows exponential decay, such as the pharmacokinetics of drugs in the body. Half-life is the time required for half of the initial amount of a substance to be reduced to half its original value. In our example, hydrocodone bitartrate has a half-life of 3.8 hours. This means that every 3.8 hours, the quantity of the drug in the body decreases by half. Understanding half-life helps to predict how long a substance remains active in the body and is crucial for determining dosing intervals for medications. Moreover, the concept of half-life allows us to find the constant of proportionality, which is essential in modeling exponential decay processes mathematically. This constant is derived from the half-life using the formula: \[ k = \frac{\ln(2)}{t_{1/2}} \]where - \(k\) is the constant of proportionality, and- \(t_{1/2}\) is the half-life. For hydrocodone bitartrate, substituting the values gives \(k \approx 0.1824\) hours^{-1}. This enables the prediction of drug levels at any given time.

Proportional Decay

Proportional decay describes a process where the rate of change of a quantity is directly proportional to the amount present. This implies that as the quantity decreases, the rate of decay slows down. In mathematical terms, it is expressed by the differential equation:\[ \frac{dQ}{dt} = -kQ \]where:- \(\frac{dQ}{dt}\) is the rate of change of the quantity,- \(Q\) is the current quantity, and- \(k\) is the constant of proportionality.This principle is not only applicable to drug elimination but also to other natural processes like credit card debt amortization or depreciation of assets. In the case of hydrocodone bitartrate's decay, the negative sign indicates a decrease. The bigger the amount of drug left, the faster it is eliminated.Recognizing this proportional relationship is valuable because it lets us construct mathematical models to predict how the quantity varies with time. This foundational concept underlies many physical and biological processes where resources deplete over time due to their current state.

Exponential Functions

Exponential functions play a crucial role in modeling proportional decay because they describe situations where the rate of change is proportional to the current amount. For hydrocodone's decay, the solution of the differential equation \( \frac{dQ}{dt} = -kQ \) results in the exponential function:\[ Q = Ce^{-kt} \]where:- \(Q\) is the amount of substance remaining,- \(C\) is the initial amount,- \(k\) is the rate constant, and- \(t\) is the time elapsed.These functions have a characteristic curve that decreases rapidly at first and then levels off, never quite reaching zero.Exponential functions are found in diverse fields such as finance and biology. They are advantageous due to their predictable nature and ease of calculation, allowing for efficient prediction of decay patterns.In practical scenarios, such as pharmacokinetics, exponential functions help in understanding how quickly a drug is metabolized and eliminated from the body over time, ensuring proper therapeutic dosing and timing. The exponential decay model predicts the remaining drug quantity, such as how after 12 hours, approximately 0.73 mg of hydrocodone remains, given an initial dose of 10 mg and a decay constant \(k\). This knowledge helps in planning subsequent doses and avoiding overdose.