Problem 14 - Applied Calculus

Question

A patient is given the drug theophylline intravenously at a rate of \(43.2 \mathrm{mg} /\) hour to relieve acute asthma. The rate at which the drug leaves the patient's body is proportional to the quantity there, with proportionality constant \(0.082\) if time, \(t\), is in hours. The patient's body contains none of the drug initially. (a) Describe in words how you expect the quantity of theophylline in the patient to vary with time. (b) Write a differential equation satisfied by the quantity of theophylline in the body, \(Q(t)\). (c) Solve the differential equation and graph the solution. What happens to the quantity in the long run?

Step-by-Step Solution

Answer

(a) Quantity increases and stabilizes. (b) \( \frac{dQ}{dt} = 43.2 - 0.082Q \). (c) Solution stabilizes at 526.83 mg as \( t \to \infty \).

Step 1: Understanding the Problem

The patient receives theophylline at a constant rate and the drug leaves the body at a rate proportional to its quantity. We need to find how the drug amount changes over time.

Step 2: Formulating the Differential Equation

Let \( Q(t) \) represent the quantity of theophylline in the body at time \( t \). The rate of change of \( Q \) with respect to time \( t \) can be expressed by the differential equation: \( \frac{dQ}{dt} = 43.2 - 0.082Q \). Here, 43.2 is the rate at which the drug is administered and \( 0.082Q \) is the rate at which it is eliminated from the body.

Step 3: Solving the Differential Equation

The differential equation \( \frac{dQ}{dt} = 43.2 - 0.082Q \) is separable. To solve it, rearrange terms and integrate: \( \int \frac{1}{43.2 - 0.082Q} dQ = \int dt \). This leads to \( -\frac{1}{0.082} \ln|43.2 - 0.082Q| = t + C \), where \( C \) is the integration constant. Solving for \( Q(t) \), we get \( Q(t) = \frac{43.2}{0.082} - C e^{-0.082t} \). As initially \( Q(0) = 0 \), solve \( 0 = \frac{43.2}{0.082} - C \) to find \( C \), thus \( Q(t) = \frac{43.2}{0.082} (1 - e^{-0.082t}) \).

Step 4: Analyzing the Long-Term Behavior

As \( t \to \infty \), the term \( e^{-0.082t} \to 0 \), so \( Q(t) \to \frac{43.2}{0.082} \). This means the drug amount stabilizes at\( 526.83 \, \mathrm{mg} \), a steady state due to balance between administration and elimination.

Key Concepts

Pharmacokinetics

Pharmacokinetics refers to the study of how drugs move through the body over time. It involves understanding the mechanisms of drug absorption, distribution, metabolism, and excretion. In the example of theophylline used to treat asthma, pharmacokinetics helps us model how the drug enters the bloodstream, circulates, and eventually is eliminated by the body at a rate proportional to its current concentration.
When administering drugs like theophylline intravenously, it's crucial to consider the steady rate of drug infusion versus the body's rate of elimination. Understanding this balance allows physicians to maintain effective therapeutic levels of the drug without reaching toxicity.
The initial absence of the drug means the body must build up to reach a stable level where input equals output. This dynamic highlights the drug's journey through the body and the necessity of monitoring patient responses to optimize treatment.

Separable Differential Equations

Separable differential equations are a type of differential equation where variables can be separated on either side of the equation for simpler integration and solution derivation. They are particularly useful in pharmacokinetics to model scenarios where the rate of change depends on the current amount of a substance, as demonstrated in our example with theophylline.
The differential equation given: \( \frac{dQ}{dt} = 43.2 - 0.082Q \) can be rearranged to allow separation of variables. This allows us to integrate both sides independently. By performing integration, you manage to isolate the variable \( Q \) and solve explicitly for time \( t \).
In this scenario, the separation leads to:

• The integral on the left-hand side concerning \( Q \)
• The integral on the right-hand side concerning \( t \)

At the heart of this is the property of exponentiation due to negative proportionality constants, characteristic of exponential decay in pharmacokinetics.

Long-term Behavior Analysis

Understanding the long-term behavior of pharmacokinetic systems is essential for predicting how a drug will act inside the body over extended periods. This involves using the solution obtained from the differential equation to ascertain where the system stabilizes or reaches a steady state.
In the context of our differential equation, \( Q(t) = \frac{43.2}{0.082} (1 - e^{-0.082t}) \), we predict what happens as time approaches infinity. Given the nature of the exponential function \( e^{-0.082t} \), it approaches zero, simplifying \( Q(t) \) to a constant value.
This behavior reveals that theophylline reaches a steady-state concentration of approximately 526.83 mg. This state represents a balance where the rate of drug administration equals the rate at which the body eliminates it, important for maintaining therapeutic effects without alteration. Predicting this long-term behavior informs dosing schedules and ensures continual efficacy and safety of drug treatments.