Problem 27
Question
Consider the logistic growth function $$ Q(t)=\frac{A}{1+B e^{-k r}} $$ Suppose the population is \(Q_{1}\) when \(t=t_{1}\) and \(Q_{2}\) when \(t=t_{2}\). Show that the value of \(k\) is $$ k=\frac{1}{t_{2}-t_{1}} \ln \left[\frac{Q_{2}\left(A-Q_{1}\right)}{Q_{1}\left(A-Q_{2}\right)}\right] $$
Step-by-Step Solution
Verified Answer
To find the value of \(k\) in the logistic growth function \(Q(t) = \frac{A}{1+B e^{-kr}}\), given that the population is \(Q_1\) when \(t = t_1\) and \(Q_2\) when \(t = t_2\), follow these steps:
1. Plug the points \((t_1, Q_1)\) and \((t_2, Q_2)\) into the logistic growth function.
2. Eliminate the variable \(B\) by solving one equation for \(B\) and substituting this expression into the other equation.
3. Solve for \(k\) and simplify, resulting in the formula:
\[ k = \frac{1}{t_2 - t_1} \ln\left[\frac{Q_2 (A - Q_1)}{Q_1 (A - Q_2)}\right] \]
1Step 1: Plug the points into the logistic growth function
We are given that the population is \(Q_1\) when \(t = t_1\) and \(Q_2\) when \(t = t_2\). Plug these points into the logistic growth function:
For \(t = t_1\), we have:
\[ Q_1 = \frac{A}{1 + B e^{-k t_1}}\]
For \(t = t_2\), we have:
\[ Q_2 = \frac{A}{1 + B e^{-k t_2}}\]
2Step 2: Eliminate the variable B
We now have two equations with two unknowns, \(A\) and \(B\). We will eliminate the variable \(B\) by solving the first equation for \(B\) and then substituting this expression in the second equation.
From the first equation:
\[ B = \frac{A - Q_1}{Q_1} \cdot e^{kt_1} \]
Now substitute this into the second equation:
\[ Q_2 = \frac{A}{1 + \left(\frac{A - Q_1}{Q_1} \cdot e^{kt_1}\right) e^{-k t_2}} \]
3Step 3: Solve for k
Now we will solve for \(k\). First, simplify the denominator in the expression for \(Q_2\):
\[ Q_2 = \frac{A}{1 + \frac{A - Q_1}{Q_1} \cdot e^{k(t_1 - t_2)}} \]
Now, we will clear the fraction by multiplying both sides by the denominator:
\[ Q_2 ( 1 + \frac{A - Q_1}{Q_1} \cdot e^{k(t_1 - t_2)}) = A \]
Subtract \(Q_2\) from both sides:
\[ Q_2 \frac{A - Q_1}{Q_1} \cdot e^{k(t_1 - t_2)} = A - Q_2 \]
Rearrange to isolate the exponential term:
\[ e^{k(t_1 - t_2)} = \frac{Q_1(A - Q_2)}{Q_2(A - Q_1)} \]
Now, we can take the natural logarithm of both sides to solve for \(k\):
\[ k(t_1 - t_2) = \ln\left[\frac{Q_1(A - Q_2)}{Q_2(A - Q_1)}\right] \]
Divide by \((t_1 - t_2)\) to find the value of \(k\):
\[ k = \frac{1}{t_2 - t_1} \ln\left[\frac{Q_2 (A - Q_1)}{Q_1 (A - Q_2)}\right] \]
Thus, we have found the value of \(k\) as desired.
Key Concepts
Understanding Differential EquationsThe Role of Population DynamicsThe Nature of Exponential Functions
Understanding Differential Equations
Differential equations are mathematical expressions that relate a function with its derivatives. They are powerful tools used to model various real-world phenomena. A key aspect of these equations is that they describe the rate at which something changes. For instance, in logistics growth models, differential equations can precisely depict how a population evolves over time.
These equations can be classified into different types, such as ordinary differential equations (ODEs) or partial differential equations (PDEs). In the context of logistic growth, we often deal with ODEs because they involve ordinary derivatives.
The logistic growth function itself is derived from a differential equation. It shows how growth is not just a simple linear increase but keeps adjusting based on current conditions. The classic logistic differential equation is:
These equations can be classified into different types, such as ordinary differential equations (ODEs) or partial differential equations (PDEs). In the context of logistic growth, we often deal with ODEs because they involve ordinary derivatives.
The logistic growth function itself is derived from a differential equation. It shows how growth is not just a simple linear increase but keeps adjusting based on current conditions. The classic logistic differential equation is:
- \( \frac{dQ}{dt} = rQ \left(1 - \frac{Q}{A}\right) \)
- Here, \( Q \) is the population at time \( t \), \( r \) is the growth rate, and \( A \) is the carrying capacity.
The Role of Population Dynamics
Population dynamics is the study of how populations change over time. It involves understanding the complex factors that affect population growth, including birth rates, death rates, and carrying capacities.
One of the simplest models used to describe population dynamics is the logistic growth model. This model highlights that resources are limited, which curtails the indefinite exponential growth of populations. Initially, populations may grow exponentially, but this growth slows down as they approach a maximum limit, known as the carrying capacity.
The logistic growth equation accounts for this by using a proportion of the population:
One of the simplest models used to describe population dynamics is the logistic growth model. This model highlights that resources are limited, which curtails the indefinite exponential growth of populations. Initially, populations may grow exponentially, but this growth slows down as they approach a maximum limit, known as the carrying capacity.
The logistic growth equation accounts for this by using a proportion of the population:
- The term \( \left(1 - \frac{Q}{A}\right) \) represents free resources as a fraction of carrying capacity.
- As \( Q \) approaches \( A \), resources become scarce, and growth slows.
The Nature of Exponential Functions
Exponential functions are a critical component of the logistic growth model. These functions are of the form \( f(t) = Ce^{kt} \), where \( e \) is the base of the natural logarithm, \( C \) is a constant, and \( k \) is the growth rate constant. Understanding exponential functions helps us grasp rapid changes, such as population growth initially being exponential before constraints are applied.
Logistic growth equations incorporate exponential components to model how populations increase before tapering off. In our problem, the term \( e^{-kr} \) in the logistic growth function alters how quickly the population approaches the carrying capacity.
Logistic growth equations incorporate exponential components to model how populations increase before tapering off. In our problem, the term \( e^{-kr} \) in the logistic growth function alters how quickly the population approaches the carrying capacity.
- Rapid growth phases are represented by \( e^{kt} \) when resources are plentiful.
- The exponential decay component \( e^{-kr} \) contextualizes the reduction in growth rate as resources dwindle.
Other exercises in this chapter
Problem 26
Use the laws of logarithms to expand and simplify the expression. $$\ln x(x+1)(x+2)$$
View solution Problem 26
Sketch the graphs of the given functions on the same axes. \(y=4^{0.5 x}\) and \(y=4^{-0.5 x}\)
View solution Problem 27
Use the laws of logarithms to expand and simplify the expression. $$\ln \frac{x^{1 / 2}}{x^{2} \sqrt{1+x^{2}}}$$
View solution Problem 27
Sketch the graphs of the given functions on the same axes. \(y=4^{0.5 x}, y=4^{x}\), and \(y=4^{2 x}\)
View solution