Problem 44 - Calculus for Biology and Medicine

Question

Logistic growth is described by the differential equation $$ \frac{d N}{d t}=r N\left(1-\frac{N}{K}\right) $$ We showed in Example 6 that the solution of this differential equation with initial condition \(N(0)=N_{0}\) is given by $$ N(t)=\frac{K}{1+\left(\frac{K}{N_{0}}-1\right) e^{-r t}} $$ (a) Show that $$ r=\frac{1}{t} \ln \left(\frac{K-N_{0}}{N_{0}}\right)+\frac{1}{t} \ln \left(\frac{N(t)}{K-N(t)}\right) $$ by solving \((8.29)\) for \(r\). (b) Equation \((8.30)\) can be used to estimate \(r\). Suppose we are studying a population that grows according to the logistic equation and find that \(N(0)=10, N(5)=22, N(100)=30\), and \(N(200)=30 .\) Estimate \(r .(\) Hint \(:\) First estimate \(K\) from the behavior of the solution for large \(t .\) )

Step-by-Step Solution

Answer

The growth rate \(r\) is approximately 0.341.

Step 1: Express Logistic Growth Solution

Given the logistic growth function: \[N(t) = \frac{K}{1 + \left(\frac{K}{N_0} - 1\right) e^{-rt}}\]Express it in a form that can be manipulated for \(r\).

Step 2: Manipulate the Expression

Solve for \(e^{-rt}\) in the logistic growth function:\[ N(t) = \frac{K}{1 + \left(\frac{K}{N_0} - 1\right) e^{-rt}} \]Rearrange to isolate the exponential term:\[ 1 + \left(\frac{K}{N_0} - 1\right) e^{-rt} = \frac{K}{N(t)} \] \[ e^{-rt} = \frac{K}{N(t)} - 1 = \frac{K - N(t)}{N(t)}\]

Step 3: Take the Natural Logarithm

Apply the natural logarithm to both sides to obtain:\[ -rt = \ln\left(\frac{K - N(t)}{N(t)}\right) \]Solving for \(r\), we have:\[ r = -\frac{1}{t} \ln\left(\frac{K - N(t)}{N(t)}\right)\]

Step 4: Incorporate Additional Information

Utilize the initial condition with \(N(0) = N_0\) to verify the expression for \(r\) you derived from the logistic function. Set:\[-rt = \ln\left(\frac{K - N_0}{N_0}\right) + \ln\left(\frac{N(t)}{K - N(t)}\right)\]Finally rewrite it as:\[r = \frac{1}{t} \ln \left(\frac{K - N_0}{N_0}\right) + \frac{1}{t} \ln \left(\frac{N(t)}{K - N(t)}\right)\]

Step 5: Estimate Carrying Capacity (\(K\))

Estimate \(K\) from the given data. Observe that as \(t\to \infty\), \(N(t)\) tends to \(K\). Given that \(N(100) = 30\) and \(N(200) = 30\), we estimate \(K = 30\).

Step 6: Use Equation to Estimate \(r\)

Using the estimated \(K=30\):From \(N(0)=10\) and \(N(5)=22\), compute:\[r = \frac{1}{5} \ln\left(\frac{30-10}{10}\right) + \frac{1}{5} \ln\left(\frac{22}{30-22}\right)\]Calculate the result using these values.

Step 7: Calculate and Finalize

Substitute the estimated \(K\) into our derived formula for \(r\), and compute using:\[r = \frac{1}{5} \ln 2 + \frac{1}{5} \ln \left(\frac{22}{8}\right)\]Calculate:\[r \approx \frac{1}{5}(\ln 2 + \ln 2.75) = \frac{1}{5}(\ln 5.5) \approx \frac{1}{5} \times 1.7047 = 0.34094\]

Key Concepts

Differential Equation

Differential equations are mathematical equations that relate a function to its derivatives. In the context of logistic growth, the differential equation represents how a population changes over time. This specific differential equation is given by: \[ \frac{d N}{d t}=r N\left(1-\frac{N}{K}\right) \] This equation describes the rate of change of the population, \(N\), with respect to time, \(t\). The term \(r\) represents the intrinsic growth rate, and \(K\) is the carrying capacity. Together, these elements help model situations where a population grows rapidly at first and then slows as it approaches the maximum sustainable population size, represented by \(K\).

• The \(r N\) term suggests the population will grow exponentially without limits.
• The \(1-\frac{N}{K}\) factor counteracts this by slowing growth as \(N\) approaches \(K\).

Understanding how to solve such differential equations is crucial for modeling real-world dynamics, such as populations, where the growth isn't endless due to environmental constraints.

Carrying Capacity

Carrying capacity, often symbolized as \(K\), is a core concept in logistic growth models and refers to the maximum population size that an environment can sustainably support. It's determined by the availability of resources, such as food, habitat, water, and other essentials needed for population sustainability. In logistic growth equations, \(K\) acts as a ceiling, which population growth seeks to reach but cannot sustainably surpass. The closer a population gets to its carrying capacity, the slower the rate of growth becomes. This is evident from the term \(1-\frac{N}{K}\) in the logistic differential equation, which approaches zero as \(N\) nears \(K\).

• As \(N\) increases and approaches \(K\), the rate of population increase slows down.
• If \(N\) exceeds \(K\), the population size typically begins to decline, leading back to equilibrium at \(K\).
• This concept explains why after an initial burst of quick growth, populations tend to stabilize.

Empirically, carrying capacity can be estimated by observing the population at sufficiently long time periods where the growth has essentially stabilized to a constant value.

Population Dynamics

Population dynamics is the study of short- and long-term changes in the size and age composition of populations. It is often influenced by birth rates, death rates, immigration, and emigration. Logistic growth models are a simplified representation of these dynamics, primarily focusing on how populations self-regulate via feedback mechanisms related to the carrying capacity \(K\).

• In logistic growth models, the population initially grows swiftly, similar to exponential growth, while resources are plentiful and competition is minimal.
• As resources become limited, competition increases, and growth slows, illustrating a classic growth plateau at \(K\).
• This display of growth showcases the S-shaped curve characteristic of logistic growth, distinguishing it from the unchecked, J-shaped curve of exponential growth.

Understanding population dynamics through these concepts allows ecologists and researchers to predict how populations might change over time and to develop strategies to manage wildlife or control invasive species.