Problem 9 - Calculus for Biology and Medicine

Question

Compartment models are used to model the flow of traffic between different roads, by treating each road as a compartment. As an example, consider how the number of cars on a freeway on-ramp, \(N(t)\), changes with time. For a simplified model let's assume that cars join the on-ramp at a constant rate \(q\) (that is, \(q\) cars join the on-ramp in one unit of time). Cars then leave the on-ramp by entering the freeway itself. Assume that a fraction \(f\) of the cars on the on-ramp enter the freeway in one unit of time. (a) Derive a differential equation for \(N(t) .\) Your differential equation will include the unknown constants \(f\) and \(q\). (b) Analyze your model from part (a) to find the equilibrium number of cars on the on-ramp, and determine whether this equilibrium is stable or unstable. (c) Suppose that the maximum capacity of the on-ramp is 90 cars, and the rate at which cars flow onto the on-ramp is \(q=60\) cars per min. Find the value of \(f\) that is needed to keep \(N\) below the on-ramp's capacity.

Step-by-Step Solution

Answer

(a) \( \frac{dN}{dt} = q - fN(t) \); (b) \( N_{eq} = \frac{q}{f} \), stable; (c) \( f > \frac{2}{3} \).

Step 1: Derive the Differential Equation for N(t)

The number of cars on the ramp, \( N(t) \), changes over time as cars flow in and out. Cars enter at a constant rate \( q \) and leave at a rate proportional to their number, \( fN(t) \). Thus, the rate of change of \( N(t) \) is expressed by the differential equation:\[ \frac{dN}{dt} = q - fN(t) \]

Step 2: Find the Equilibrium State

The equilibrium state occurs when the rate of change of \( N(t) \) is zero; i.e., \( \frac{dN}{dt} = 0 \). Substitute this into the equation from Step 1:\[ 0 = q - fN(t) \]Solving for \( N(t) \), we find the equilibrium number of cars as:\[ N_{eq} = \frac{q}{f} \]

Step 3: Determine Stability of the Equilibrium

To determine stability, consider small perturbations from \( N_{eq} \). If \( \frac{dN}{dt} < 0 \) when \( N > N_{eq} \) and \( \frac{dN}{dt} > 0 \) when \( N < N_{eq} \), the equilibrium is stable. Since \( \frac{dN}{dt} = q - fN(t) \), increasing \( N(t) \) above \( N_{eq} \) makes \( \frac{dN}{dt} \) negative, and decreasing \( N(t) \) makes \( \frac{dN}{dt} \) positive. Thus, the equilibrium is stable.

Step 4: Apply the Model to Capacity Constraints

To ensure \( N(t) < 90 \), set the equilibrium \( N_{eq} = \frac{q}{f} \) to be less than 90. Given \( q = 60 \), solve:\[ \frac{60}{f} < 90 \]Rearrange to find \( f \):\[ f > \frac{60}{90} = \frac{2}{3} \]

Step 5: Conclusion

Thus, the value of \( f \) that ensures the number of cars never exceeds the maximum capacity is \( f > \frac{2}{3} \).

Key Concepts

Differential Equations

Differential equations are crucial for modeling how things change over time. In this particular compartment model, we are examining the change in the number of cars on a freeway on-ramp, denoted as \(N(t)\), with respect to time. The key to understanding such a system lies in grasping how factors contributing to the change are mathematically represented.

The differential equation used in this scenario is \( \frac{dN}{dt} = q - fN(t) \). This equation succinctly captures two primary processes:

• Cars entering the on-ramp at a constant rate \(q\) per unit time.
• Cars leaving the on-ramp at a rate proportional to the number of cars currently on the ramp, which is \(fN(t)\).

In essence, this equation provides a balance of flow, accounting for inflow and outflow. Solving this differential equation allows us to predict the behavior of \(N(t)\) over time, helping us understand how traffic builds up or diminishes on the on-ramp.

It is essential to set up the equation correctly to account for all factors. By manipulating variables like \(q\) and \(f\), different traffic scenarios can be modeled effectively, which is a powerful tool in traffic engineering.

Equilibrium Analysis

Equilibrium analysis focuses on understanding the state of the system when no further changes occur over time, meaning the system is at rest. For the on-ramp traffic model, this occurs when \(\frac{dN}{dt} = 0\). This condition simplifies our differential equation to \(q - fN(t) = 0\).

Solving for \(N(t)\), we find the equilibrium number of cars, denoted as \(N_{eq}\), given by:
\[ N_{eq} = \frac{q}{f} \]
This relation directly ties the equilibrium number of cars to the inflow rate \(q\) and the fraction of cars leaving \(f\). The equilibrium point is where the number of cars entering equals the number of cars leaving. Hence, there is no net change in the number of cars.

Knowing the equilibrium value is critical, especially in planning and designing traffic systems, to ensure that on-ramp capacities are not exceeded during peak times.

Stability Analysis

Determining the stability of an equilibrium point allows us to predict how the system will respond to small disturbances. In traffic modeling, it pertains to understanding if a few extra or fewer cars on the ramp will lead back to the equilibrium state or if it will further deviate.

The stability of the equilibrium was analyzed by looking at the behavior of \(\frac{dN}{dt}\) around \(N_{eq}\).

• If \(N(t) > N_{eq}\), then \(\frac{dN}{dt}\) is negative, meaning the number of cars will decrease toward \(N_{eq}\).
• If \(N(t) < N_{eq}\), then \(\frac{dN}{dt}\) is positive, indicating that the number of cars will increase toward \(N_{eq}\).

This gives us that the equilibrium is stable because regardless of whether \(N(t)\) is momentarily increased or decreased, it naturally returns to equilibrium. Stable traffic systems are vital as they ensure that slight increases in traffic can be absorbed without leading to congestion or delays. This understanding aids in devising systems that can handle real-world variability efficiently.