Problem 43 - Calculus: Early Transcendentals

Question

A glucose solution is administered intravenously into the bloodstream at a constant rate \( r. \) As the glucose is added, it is converted into other substances and removed from the bloodstream at a rate that is proportional to the concentration at that time. Thus a model for the concentration \( C = C(t) \) of the glucose solution in the bloodstream is \( \frac {dC}{dt} = r - kC \) where \( k \) is a positive constant. (a) Suppose that the concentration at time \( t = 0 \) is \( C_o. \) Determine the concentration at any time \( t \) by solving the differential equation. (b) Assuming that \( C_o < r/k, \) find lim \( _{t \to \infty} C(t) \) and interpret your answer.

Step-by-Step Solution

Answer

The concentration reaches \( \frac{r}{k} \) as \( t \to \infty \), achieving steady-state equilibrium.

Step 1: Understanding the Differential Equation

The given differential equation is \( \frac{dC}{dt} = r - kC \). Here, \( C(t) \) is the concentration of glucose in the bloodstream, \( r \) is the rate at which glucose is added, and \( k \) is a proportionality constant for its removal. We need to solve this equation with the initial condition \( C(0) = C_0 \).

Step 2: Separating Variables and Integrating

Rearrange the terms to separate the variables: \( \frac{1}{r - kC} \frac{dC}{dt} = 1 \). Integrate both sides with respect to their respective variables. The left side integrates with respect to \( C \) and the right side with respect to \( t \).

Step 3: Solving the Integral

Integrating \( \int \frac{1}{r - kC} dC \) gives \( -\frac{1}{k} \ln|r - kC| \), and integrating \( \int dt \) gives \( t + C_1 \), where \( C_1 \) is an integration constant.

Step 4: Applying the Initial Condition

Insert the initial condition \( C(0) = C_0 \) into the integrated equation: \[-\frac{1}{k} \ln|r - kC_0| = 0 + C_1 \].This helps us find the value of the integration constant \( C_1 \).

Step 5: Solving for C(t)

Use the result from the previous steps to express \( C(t) \): \[ C(t) = \frac{r}{k} + (C_0 - \frac{r}{k})e^{-kt} \]. This is the general solution to the differential equation.

Step 6: Finding the Limit as t Approaches Infinity

To find \( \lim_{t \to \infty} C(t) \), observe that as \( t \to \infty \), the exponential term \( e^{-kt} \to 0 \). Thus, \[ \lim_{t \to \infty} C(t) = \frac{r}{k} \].

Step 7: Interpretation of the Limit

The limit \( \frac{r}{k} \) represents a steady state or equilibrium concentration in the bloodstream. It means that regardless of the initial concentration, the system will stabilize to this concentration over time.

Key Concepts

Glucose Solution

In the medical field, a glucose solution is often given intravenously to ensure that patients get the necessary glucose, especially when they cannot intake food or have additional energy requirements. When you introduce glucose into the bloodstream at a constant rate, it is absorbed and used by the body. In this problem, the glucose solution is modeled using differential equations to represent its concentration in the bloodstream over time, accounting for both its continuous addition and natural removal from the body.

Concentration Modeling

The process of concentration modeling involves using mathematical equations to describe how the concentration of a substance changes over time. Here, the concentration of glucose in the bloodstream is modeled by the differential equation \( \frac{dC}{dt} = r - kC \). This equation represents a balance between two processes: the constant rate \( r \) at which glucose is added to the blood, and the rate \( kC \) at which it is metabolized and removed based on its current concentration.

Steady State Analysis

A steady state in a system is when the system stabilizes so that its variables no longer change as time progresses. For this glucose solution model, the steady state is reached when the concentration of glucose in the bloodstream no longer changes, meaning \( \frac{dC}{dt} = 0 \). Solving for \( C \) when \( \frac{dC}{dt} = 0 \) gives \( C = \frac{r}{k} \). This indicates a steady concentration. At this point, the rate at which glucose is added and removed from the system is perfectly balanced.

Exponential Decay

Exponential decay describes a process where a quantity decreases at a rate proportional to its current value. In this glucose concentration model, the term \( (C_0 - \frac{r}{k})e^{-kt} \) represents exponential decay over time. As \( t \to \infty \), \( e^{-kt} \to 0 \), meaning the impact of the starting concentration \( C_0 \) diminishes, leading the system toward the steady state \( \frac{r}{k} \). This demonstrates that even if the initial concentration differs, eventually, the system reaches the equilibrium state due to exponential decay.