Problem 39 - Calculus for Biology and Medicine

Question

Suppose that \(N(t)\) denotes the size of a population at time \(t\) and that $$\frac{d N}{d t}=1.5 N\left(1-\frac{N}{50}\right)$$ (a) Solve this differential equation when \(N(0)=10\). (b) Solve this differential equation when \(N(0)=90\). (c) Graph your solutions in (a) and (b) in the same coordinate system. (d) Find \(\lim _{t \rightarrow \infty} N(t)\) for your solutions in (a) and (b).

Step-by-Step Solution

Answer

Both population solutions approach a value of 50 as time goes to infinity.

Step 1: Rewrite the Differential Equation

Start by identifying the given differential equation: \(\frac{dN}{dt} = 1.5N\left(1 - \frac{N}{50}\right)\). This is a separable differential equation.

Step 2: Separate Variables

Rearrange the equation to separate the variables \(N\) and \(t\): \[\frac{dN}{N(1 - \frac{N}{50})} = 1.5 \, dt.\]

Step 3: Integrate Both Sides

Perform partial fraction decomposition on the left side and integrate both sides. The partial fraction decomposition gives:\[\frac{1}{N(1 - \frac{N}{50})} = \frac{1}{N} + \frac{1}{50-N}.\]Thus, integrate:\[\int \left(\frac{1}{N} + \frac{1}{50-N}\right) dN = 1.5 \int \, dt.\]This results in:\[\ln |N| - \ln |50-N| = 1.5t + C.\]

Step 4: Simplify the Integration Result

Combine the logarithms:\[\ln\left|\frac{N}{50-N}\right| = 1.5t + C.\]Exponentiate both sides to eliminate the logarithm:\[\frac{N}{50-N} = Ce^{1.5t}.\]

Step 5: Solve for N(t)

Solve for \(N(t)\) in terms of \(t\):\[N(t) = \frac{50Ce^{1.5t}}{1 + Ce^{1.5t}}.\]

Step 6: Determine Constant C for Part (a)

Using initial condition \(N(0) = 10\), substitute into the equation:\[10 = \frac{50C}{1+C}.\]Solve for \(C\):\[C = \frac{1}{4}.\]

Step 7: Determine Solution for Part (a)

With the value of \(C\), the solution becomes:\[N(t) = \frac{50 \cdot \frac{1}{4} \, e^{1.5t}}{1 + \frac{1}{4} \, e^{1.5t}},\]or\[N(t) = \frac{12.5e^{1.5t}}{1 + 0.25e^{1.5t}}.\]

Step 8: Determine Constant C for Part (b)

Using initial condition \(N(0) = 90\), substitute into the equation:\[90 = \frac{50C}{1+C}.\]Solve for \(C\):\[C = 9.\]

Step 9: Determine Solution for Part (b)

With the value of \(C\), the solution becomes:\[N(t) = \frac{450e^{1.5t}}{1 + 9e^{1.5t}}.\]

Step 10: Graph the Solutions

Plot both solutions on the same set of axes: \(N(t) = \frac{12.5e^{1.5t}}{1 + 0.25e^{1.5t}}\) from part (a) and \(N(t) = \frac{450e^{1.5t}}{1 + 9e^{1.5t}}\) from part (b), showing their behavior over time.

Step 11: Find Long-Term Behavior

Evaluate \(\lim_{t \to \infty} N(t)\) for both solutions:For part (a):\[\lim_{t \to \infty} \frac{12.5e^{1.5t}}{1 + 0.25e^{1.5t}} = 50.\]For part (b):\[\lim_{t \to \infty} \frac{450e^{1.5t}}{1 + 9e^{1.5t}} = 50.\]

Key Concepts

Population Growth

Population growth is a fascinating topic in the study of differential equations, often modeling how populations expand or contract over time depending on various factors. In this exercise, we examine a specific case where the growth rate depends not just on the current population size, but also on how close the population is to a limit. The model used here is the logistic model.

In general, population growth can be described by differential equations which are mathematical expressions involving functions and their derivatives. These equations can model how different populations grow or shrink, taking into account factors like reproduction rates and resources.

In our example, the differential equation provided is \[ \frac{dN}{dt} = 1.5N\left(1 - \frac{N}{50}\right), \] which portrays a more nuanced growth than simple exponential growth, showing how the growth rate decelerates as the population nears a certain limit, known as the carrying capacity. This deceleration is due to limited resources, space, or other environmental constraints.

Logistic Model

The logistic model is an improvement on the simple exponential growth model as it includes a limiting factor which accounts for the carrying capacity of the environment. This ensures that as population increases, growth slows down, preventing the population from growing indefinitely.

The logistic model is represented by the equation:\[ \frac{dN}{dt} = rN\left(1 - \frac{N}{K}\right), \] where:

• \( r \) is the intrinsic growth rate of the population.
• \( K \) is the carrying capacity of the environment, which is the maximum population size the environment can sustain.

In our given example with \( K = 50 \)and \( r = 1.5 \),the equation emphasizes how the growth rate diminishes as \( N \)approaches \( K \). This pattern reflects realistic scenarios of growth, often encountered in threshold-limited environments, and underlies the importance of logistic functions in understanding real-world phenomena like population dynamics.

Limiting Behavior

The concept of limiting behavior is crucial when analyzing long-term trends in population models like the logistic model. It refers to the behavior of a function as time progresses towards infinity, indicating the eventual outcome of the system.

In a logistic model, the limiting behavior is highlighted by the carrying capacity \( K \).This is because as time \( t \)progresses and \( N(t) \)reaches a steady state, the population size stabilizes around this value. Performing the limit operation on our solution functions for part (a) and (b): \( \lim_{t \to \infty} N(t) = K \) returns the equilibrium state.

In our exercise, we computed:

• For initial condition \( N(0) = 10 \), the long-term outcome is \( 50 \).
• For initial condition \( N(0) = 90 \), the population also settles at \( 50 \).

Understanding this limiting behavior assists in predicting the stable size of the population, emphasizing why looking at \( \lim_{t \to \infty} N(t) \) is a powerful tool in population dynamics.