Problem 42 - Calculus for Biology and Medicine

Question

Logistic growth is described by the differential equation $$\frac{d N}{d t}=r N\left(1-\frac{N}{K}\right)$$ 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 \((8.47)\) for \(r\). (b) Equation \((8.48)\) can be used to estimate \(r\). Suppose we follow 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 \(r\).

Step-by-Step Solution

Answer

Estimated value of \(r\) is approximately 0.3409.

Step 1: Simplify the logistic growth equation

Let's start by simplifying the equation (8.47) which is the solution for the logistic growth:\[N(t)=\frac{K}{1+\left(\frac{K}{N_{0}}-1\right) e^{-r t}}\]Our task is to express \(r\) in terms of known variables.

Step 2: Rearrange for the exponential term

Solve for the exponential term by setting up the equation for \(N(t)\):\[N(t) = \frac{K}{1+\left(\frac{K}{N_{0}}-1\right) e^{-r t}}\]Rearrange to solve for \( e^{-r t} \):\[1+\left(\frac{K}{N_{0}}-1\right) e^{-r t} = \frac{K}{N(t)}\]\[\left(\frac{K}{N_{0}}-1\right) e^{-r t} = \frac{K}{N(t)} - 1\]\[e^{-r t} = \frac{1}{\frac{K}{N_{0}}-1} \left(\frac{K - N(t)}{N(t)}\right)\]

Step 3: Solve for r

Take the natural logarithm of both sides of the equation from **Step 2**:\[-rt = \ln \left(\frac{K-N(t)}{N(t)}\right) - \ln \left(\frac{K-N_{0}}{N_{0}}\right)\]Now, divide by \(-t\) to solve for \(r\):\[r = \frac{1}{t} \left( \ln \left(\frac{K-N_{0}}{N_{0}}\right) + \ln \left(\frac{N(t)}{K-N(t)}\right) \right)\]This proves the equation to find \(r\).

Step 4: Substitute given values to estimate r

We are given:- \(N(0)=10\), \(N(5)=22\), \(N(100)=30\), \(N(200)=30\).We know that as \(t\) becomes large, \(N(t)\rightarrow K\) hence \(K=30\). Substitute these into the equation for \(r\):Using \(t = 5\) and \(N(t) = 22\), \[ r = \frac{1}{5} \left(\ln \left(\frac{30-10}{10}\right) + \ln \left(\frac{22}{30-22}\right) \right)\] Calculate these values:\[ r = \frac{1}{5} \left(\ln(2) + \ln(2.75)\right)\]\[ r = \frac{1}{5} \left(\ln(5.5)\right)\]\[ r = \frac{1}{5} \cdot 1.70474809 \]\[ r \approx 0.3409 \]

Step 5: Final estimation of r

The estimated value of \(r\) based on the population values given is approximately \(0.3409\).

Key Concepts

Differential Equation

A differential equation is a mathematical equation that relates a function with its derivatives. In the context of logistic growth, we are dealing with the differential equation \( \frac{d N}{d t}=r N\left(1-\frac{N}{K}\right) \). This equation models how a population \( N \) changes over time \( t \).
Here, \( r \) is the intrinsic growth rate, which determines how quickly the population grows, and \( K \) is the carrying capacity of the environment, representing the maximum population size that the environment can sustain.
When solving a differential equation, the goal is to find the function \( N(t) \) that satisfies the equation. In this logistic growth scenario, the solution \( N(t)=\frac{K}{1+\left(\frac{K}{N_{0}}-1\right) e^{-r t}} \) gives us a way to predict the population size \( N \) at any given time \( t \), based on initial conditions such as the initial population size \( N_0 \).

• The term \( \left(1 - \frac{N}{K}\right) \) introduces a limiting factor that slows population growth as it approaches \( K \).
• This equation thus ensures that \( N(t) \) will never exceed \( K \), modeling a more realistic growth scenario over time.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm to the base \( e \), where \( e \) is approximately 2.71828. In the context of the logistic growth equation, the natural logarithm is used to solve for the growth rate \( r \).
By taking the logarithm of both sides of an equation, we can transform multiplicative relationships into additive ones, making them easier to handle algebraically. In our specific logistic growth model, we used it to manage the exponential term.
The equation \( r = \frac{1}{t} \left( \ln \left(\frac{K-N_{0}}{N_{0}}\right) + \ln \left(\frac{N(t)}{K-N(t)}\right) \right) \) shows how the natural logarithm is instrumental in isolating \( r \).

• Each part of the expression \( \ln \left(\frac{K-N_{0}}{N_{0}}\right) \) and \( \ln \left(\frac{N(t)}{K-N(t)}\right) \) deals with initial conditions and modifications over time, respectively.
• It helps us equate changes in the population's growth factors, such as initial population and timespan, to resolve for \( r \).

The beauty of natural logarithms lies in their ability to make complex exponential calculations straightforward, allowing us to derive important rates of change in scenarios like population growth.

Population Estimation

Population estimation using differential equations provides a mathematical way to predict future population sizes based on initial conditions and environmental parameters. The logistic growth model is especially useful as it accounts for limiting factors like resources and space. In the exercise at hand, the logistic growth equation helps estimate the population growth rate \( r \) given initial and certain future population sizes.
Here's how we approach the problem:

• We're given specific population sizes at different times: \( N(0)=10 \), \( N(5)=22 \), \( N(100)=30 \), and \( N(200)=30 \).
• Given these values, we recognize that the population reaches its carrying capacity \( K \) at \( N(100)=30 \), so \( K = 30 \).

With this setup, we substitute known values into our equation for \( r \) to estimate the intrinsic growth rate. This allows us to understand how rapidly the population transitions from its initial size to one near the environment's capacity, over a defined time span.
By computing \( r \), we not only learn about past growth but also predict how the population may behave in the future, provided conditions remain the same.