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) \(Q(t) = \frac{r}{\alpha} (1 - e^{-\alpha t})\), \(Q_{\infty} = \frac{r}{\alpha}\). (b) Doubling \(r\) doubles \(Q_{\infty}\), no effect on time. (c) Doubling \(\alpha\) halves \(Q_{\infty}\) and decreases time.

Step 1: Formulate the Differential Equation

The rate of change of the quantity, \(Q\), of drug in the body is the rate of administration minus the rate of excretion, which is given by: \[ \frac{dQ}{dt} = r - \alpha Q. \] This is a first-order linear differential equation where \(r\) is the constant rate of administration and \(\alpha Q\) is the rate of excretion.

Step 2: Solve the Differential Equation

To solve \( \frac{dQ}{dt} = r - \alpha Q \), we can separate variables or use an integrating factor:\[ \frac{dQ}{dt} + \alpha Q = r. \]Using an integrating factor, multiply through by \(e^{\alpha t}\):\[ e^{\alpha t} \frac{dQ}{dt} + \alpha e^{\alpha t} Q = r e^{\alpha t}. \]The left-hand side is the derivative of \(e^{\alpha t}Q\):\[ \frac{d}{dt}(e^{\alpha t}Q) = r e^{\alpha t}. \]Integrate both sides with respect to \(t\):\[ e^{\alpha t}Q = \frac{r}{\alpha}e^{\alpha t} + C. \]Solving for \(Q\):\[ Q = \frac{r}{\alpha} + Ce^{-\alpha t}. \]

Step 3: Apply Initial Condition

Given that initially, there is no drug in the body (\(Q(0) = 0\)), substitute \(t = 0\) in the equation:\[ 0 = \frac{r}{\alpha} + C e^0. \]So, \(C = -\frac{r}{\alpha}\).Thus, the solution becomes:\[ Q(t) = \frac{r}{\alpha} (1 - e^{-\alpha t}). \]

Step 4: Determine Limiting Long-Run Value

The long-run value of \(Q\) or \(Q_{\infty}\) as \(t \to \infty\) is:\[ \lim_{t \to \infty} Q(t) = \frac{r}{\alpha} (1 - e^{-\alpha t}) = \frac{r}{\alpha}. \]

Step 5: Analyze the Effect of Doubling r

Doubling \(r\) leads to doubling \(Q_{\infty}\) because \(Q_{\infty} = \frac{r}{\alpha}\), so \(Q_{\infty} \) will change to \(\frac{2r}{\alpha}\). However, the time to reach \(\frac{1}{2}Q_{\infty}\), determined by \(1 - e^{-\alpha t} = \frac{1}{2}\), remains the same because the equation solvable for time is independent of \(r\).

Step 6: Analyze the Effect of Doubling α

Doubling \(\alpha\) leads to decreasing \(Q_{\infty}\) by half since \(Q_{\infty} = \frac{r}{\alpha}\), so \(Q_{\infty}\) becomes \(\frac{r}{2\alpha}\). The time to reach \(\frac{1}{2}Q_{\infty}\) decreases because the eponential decay rate increases.

Key Concepts

Rate of Change

Understanding the rate of change in differential equations is crucial because it describes how a quantity varies over time. In our exercise, the rate of change of the drug quantity, \( Q \), in the body involves both addition and subtraction.
The drug enters the body at a constant rate, \( r \), often measured in mg/hour. On the other hand, the drug is also excreted, and this happens at a rate proportional to the quantity already present in the body, expressed as \( \alpha Q \).
This means that the more drug you have, the faster it leaves the body. Mathematically, this scenario is represented by the differential equation:

• \( \frac{dQ}{dt} = r - \alpha Q \)

This equation tells us how to adjust \( Q \) every instant based on the current amount present and the rate it's excreted. It's a dynamic balance between how fast the drug is administered versus how fast it exits the system.

Exponential Decay

Exponential decay is a pattern where quantities decrease proportional to their current value. This concept plays a significant role in our differential equation, specifically in how the drug is excreted from the body.
When solving the equation, such as \( \frac{dQ}{dt} = r - \alpha Q \), you'll encounter a function of time that decreases exponentially, \( e^{-\alpha t} \).
This term models the natural decay of the drug in the body. The solution after applying initial conditions is:

• \( Q(t) = \frac{r}{\alpha} (1 - e^{-\alpha t}) \)

The exponential decay \( e^{-\alpha t} \) represents how fast the leftover drug amount reduces over time.
The larger the value of \( \alpha \), the quicker the drug decreases, making it significant in predicting how long the drug affects the body. Over long durations, as \( t \to \infty \), \( e^{-\alpha t} \) approaches zero. Therefore, \( Q(t) \) approaches \( \frac{r}{\alpha} \), referred to as \( Q_{\infty} \), the long-term steady-state value.

Integration Techniques

To solve differential equations, particularly those involving the rate of change, integration techniques are essential. Our differential equation, \( \frac{dQ}{dt} = r - \alpha Q \), requires careful maneuvering to find \( Q(t) \).
We used an integrating factor method, a common technique for first-order linear differential equations. By multiplying through by \( e^{\alpha t} \), the equation simplifies into a more solvable form:

• \( e^{\alpha t} \frac{dQ}{dt} + \alpha e^{\alpha t} Q = r e^{\alpha t} \)
• \( \frac{d}{dt}(e^{\alpha t}Q) = r e^{\alpha t} \)

Integrating both sides with respect to time results in:

• \( e^{\alpha t}Q = \frac{r}{\alpha}e^{\alpha t} + C \)

This equation sets the stage to solve for \( Q \), giving us:

• \( Q(t) = \frac{r}{\alpha} + Ce^{-\alpha t} \)

Applying initial conditions helps to solve for \( C \), completing the integration process. Mastery of these techniques allows seamless navigation through differential equations, opening insights into how time influences change in physical systems.