Problem 24 - Applied Calculus

Question

A drug is administered intravenously at a constant rate of \(r\) mg/hour and is excreted at a rate proportional to the quantity present, with constant of proportionality \(\alpha>0\). (a) Solve a differential equation for the quantity, \(Q\), in milligrams, of the drug in the body at time \(t\) hours. Assume there is no drug in the body initially. Your answer will contain \(r\) and \(\alpha .\) Graph \(Q\) against \(t\) What is \(Q_{\infty}\), the limiting long-run value of \(Q\) ? (b) What effect does doubling \(r\) have on \(Q_{\infty} ?\) What effect does doubling \(r\) have on the time to reach half the limiting value, \(\frac{1}{2} Q_{\infty} ?\) (c) What effect does doubling \(\alpha\) have on \(Q_{\infty} ?\) On the time to reach \(\frac{1}{2} Q \infty ?\)

Step-by-Step Solution

Answer

(a) Solve differential equation: \( Q(t) = \frac{r}{\alpha}(1 - e^{-\alpha t}) \), \( Q_{\infty} = \frac{r}{\alpha} \). (b) Doubling \( r \) doubles \( Q_{\infty} \); time to half \( Q_{\infty} \) is unchanged. (c) Doubling \( \alpha \) halves \( Q_{\infty} \); time to half \( Q_{\infty} \) decreases.

Step 1: Formulate the Differential Equation

Start by setting up the differential equation. The rate of change of the drug quantity \( Q \) in the body is the difference between the rate of administration \( r \) and the rate of excretion \( \alpha Q \). This can be expressed as:\[ \frac{dQ}{dt} = r - \alpha Q. \]

Step 2: Solve the Differential Equation

To solve for \( Q \), we can separate the variables and integrate. Re-arrange the equation:\[ \frac{1}{r-\alpha Q} \frac{dQ}{dt} = 1. \]Integrating both sides with respect to \( t \), we get:\[ \int \frac{1}{r-\alpha Q} dQ = \int dt. \]This results in:\[ -\frac{1}{\alpha} \ln |r - \alpha Q| = t + C, \] where \( C \) is the integration constant. Apply the initial condition \( Q(0) = 0 \) to find \( C \):\[ -\frac{1}{\alpha} \ln(r) = C. \]Thus the solution is:\[ Q(t) = \frac{r}{\alpha} (1 - e^{-\alpha t}). \]

Step 3: Determine the Limiting Value \( Q_{\infty} \)

The limiting long-run value \( Q_{\infty} \) occurs as \( t \to \infty \). Therefore, we calculate:\[ Q_{\infty} = \lim_{t \to \infty} \frac{r}{\alpha} (1 - e^{-\alpha t}) = \frac{r}{\alpha}. \]

Step 4: Doubling \( r \) and Its Impact

If \( r \) is doubled in the expression for \( Q_{\infty} \), the new \( Q_{\infty} \) becomes \( \frac{2r}{\alpha} \). As for the time to reach \( \frac{1}{2} Q_{\infty} \):\[\frac{r}{\alpha} (1 - e^{-\alpha t}) = \frac{1}{2} \frac{r}{\alpha} \Rightarrow e^{-\alpha t} = \frac{1}{2}.\] Solving for \( t \), we have \( t = \frac{\ln(2)}{\alpha} \), which is independent of \( r \). Hence, the time to reach half the limiting value is unaffected.

Step 5: Doubling \( \alpha \) and Its Impact

When \( \alpha \) is doubled, \( Q_{\infty} \) is halved to \( \frac{r}{2\alpha} \). For the time to reach \( \frac{1}{2} Q_{\infty} \):\[\frac{r}{\alpha} (1 - e^{-2\alpha t}) = \frac{1}{2} \frac{r}{2\alpha} \Rightarrow e^{-2\alpha t} = \frac{3}{4}.\] Solving for \( t \) gives \( t = \frac{\ln(4/3)}{2\alpha} \), which means the time decreases. Doubling \( \alpha \) decreases both \( Q_{\infty} \) and the time to reach half of that value.

Key Concepts

Differential Equations

Understanding differential equations is essential for solving problems where the rate of change depends on the current value of the variable. In our drug administration problem, we set up a differential equation to describe how the drug quantity, \( Q \), changes over time. It is expressed as \( \frac{dQ}{dt} = r - \alpha Q \). Here, \( r \) stands for the constant rate of drug administration, and \( \alpha Q \) represents the rate at which the drug is excreted based on the amount already in the body.
To tackle this differential equation, we take it step by step: we separate variables, integrate, and utilize initial conditions. The solution uncovers how \( Q \) dynamically changes as time progresses.
Grasping differential equations allows us to model real-world processes that involve rates of change, such as population growth, cooling, and in this case, drug metabolism.

Integration Techniques

Integration is a powerful mathematical tool used to solve differential equations by finding the antiderivative of a function. In this exercise, after rearranging our differential equation to \( \frac{1}{r-\alpha Q} \frac{dQ}{dt} = 1 \), we use integration to both sides to begin solving for \( Q \).
By integrating, we aim to find a function whose derivative gives us the initial setup of the differential equation.

• The left side integrates as \( \int \frac{1}{r-\alpha Q} \, dQ \) resulting in \( -\frac{1}{\alpha} \ln |r - \alpha Q| \).
• The right side integrates as \( \int dt \) simply results in \( t \).

Integration helps unravel the behavior of the drug quantity over time, crafting a pathway from the differential equation to the explicit solution \( Q(t) = \frac{r}{\alpha} (1 - e^{-\alpha t}) \).
Mastering integration techniques, especially in connection with exponential terms, is vital in calculus, providing the key to unlock solutions in varied applications.

Exponential Functions

Exponential functions are central to many processes that involve growth and decay, like this drug absorption problem. The solution to our differential equation, \( Q(t) = \frac{r}{\alpha} (1 - e^{-\alpha t}) \), features an exponential term \( e^{-\alpha t} \). This term dictates how quickly the drug quantity, \( Q \), approaches its limiting value over time.
In this context:

• The term \( e^{-\alpha t} \) reflects the decay process, emphasizing how the drug's excretion affects the remaining quantity.
• As time \( t \) increases, \( e^{-\alpha t} \) diminishes, letting \( Q(t) \) near its maximum \( \frac{r}{\alpha} \), known as \( Q_{\infty} \).

Understanding exponential functions equips one with the ability to predict how systems behave over time, like the eventual stabilizing of drug levels in the body. It's crucial for handling problems involving compounded interest, radioactive decay, and in this case, drug kinetics.