Problem 26
Question
In Exercises \(23-26,\) the logistic equation describes the growth of a population \(P,\) where \(t\) is measured in years. In each case, find (a) the carrying capacity of the population, (b) the size of the population when it is growing the fastest, and (c) the rate at which the population is growing when it is growing the fastest. $$\frac{d P}{d t}=10^{-5} P(5000-P)$$
Step-by-Step Solution
Verified Answer
The carrying capacity of the population is 5000. The size of the population when it is growing the fastest is 2500. The rate at which the population is growing the fastest is 3.125 populations per year.
1Step 1: Determine the Carrying Capacity
In the logistic differential equation, the carrying capacity is the value of \( P \) that makes \( \frac{dP}{dt} = 0 \). So it is the value of \( P \) when \( P(5000-P) = 0 \). This happens when \( P = 0 \) or when \( P = 5000 \). Since we are considering a growing population and the carrying capacity must be a positive value, the carrying capacity of this population is \( P = 5000 \). So the carrying capacity is 5000.
2Step 2: Determine the Size of the Population when it is Growing the Fastest
The population grows fastest at half the carrying capacity. Therefore, the size of the population is \( P = \frac{M}{2} = \frac{5000}{2} = 2500 \). So, the population size grows fastest when it reaches 2500.
3Step 3: Determine the Rate at which the Population is Growing the Fastest
To find the rate at which the population is growing its fastest, substitute \( P = 2500 \) into the population equation: \( \frac{dP}{dt} = 10^{-5} P(5000-P) = 10^{-5} * 2500 * (5000 - 2500) \). Simplifying this results in \( \frac{dP}{dt} = 3.125 \) per year. So the fastest rate of growth is 3.125 populations per year.
Key Concepts
Carrying CapacityPopulation Growth RateDifferential Equation
Carrying Capacity
In the context of population growth, the carrying capacity refers to the maximum population size that an environment can sustain indefinitely. When a population reaches this size, it stabilizes due to limitations such as resources, space, and other environmental factors.
In logistic growth models, the carrying capacity is represented by the variable that nullifies the growth differential equation, which makes the rate of change of the population zero.
This is calculated in scenarios where the environment can no longer support additional population growth.
For the equation \( \frac{dP}{dt}=10^{-5}P(5000-P) \), the carrying capacity is 5000.
This means that when the population reaches 5000, it will stop growing as the environment can no longer support more individuals without deterioration of the habitat.
In logistic growth models, the carrying capacity is represented by the variable that nullifies the growth differential equation, which makes the rate of change of the population zero.
This is calculated in scenarios where the environment can no longer support additional population growth.
For the equation \( \frac{dP}{dt}=10^{-5}P(5000-P) \), the carrying capacity is 5000.
This means that when the population reaches 5000, it will stop growing as the environment can no longer support more individuals without deterioration of the habitat.
- The carrying capacity is a crucial part of understanding population dynamics as it represents the limits of growth.
- It is determined by factors such as food availability, habitat space, and competition among the population.
- Logistic growth models account for these real-world constraints, unlike exponential models which assume unlimited growth potential.
Population Growth Rate
The population growth rate is the speed at which the number of individuals in a population increases over a specific period. It's often expressed as a percentage and is a vital measure in population studies to understand how quickly a population is expanding.
In the logistic growth model, the population growth rate is influenced by both the current population size and the carrying capacity.
Initially, the rate of growth increases rapidly, but it slows down as the population approaches the carrying capacity due to limiting factors.
At half the carrying capacity, the population experiences its maximum growth rate.
For example, in our exercise, when the population is 2500 (half of 5000), it is growing at its fastest rate.
In the logistic growth model, the population growth rate is influenced by both the current population size and the carrying capacity.
Initially, the rate of growth increases rapidly, but it slows down as the population approaches the carrying capacity due to limiting factors.
At half the carrying capacity, the population experiences its maximum growth rate.
For example, in our exercise, when the population is 2500 (half of 5000), it is growing at its fastest rate.
- Half the carrying capacity is a critical point because the environmental resistance is moderate enough to allow for accelerated growth.
- As populations grow, the resources become limited, slowing down the rate at which the population can grow further.
- Understanding this rate helps ecologists and planners in managing resources and ensuring sustainable growth.
Differential Equation
A differential equation is a mathematical equation that relates a function with its derivatives, expressing how the rate of change of a quantity depends on other quantities.
In the context of population growth, a differential equation is used to model how populations change over time considering various factors like birth and death rates and resource availability.
In the exercise provided, the logistic differential equation \( \frac{dP}{dt}=10^{-5}P(5000-P) \) models the population growth rate.
The equation incorporates both the population, \( P \), and its carrying capacity (5000 in this case), representing how growth slows as \( P \) approaches the carrying capacity.
In the context of population growth, a differential equation is used to model how populations change over time considering various factors like birth and death rates and resource availability.
In the exercise provided, the logistic differential equation \( \frac{dP}{dt}=10^{-5}P(5000-P) \) models the population growth rate.
The equation incorporates both the population, \( P \), and its carrying capacity (5000 in this case), representing how growth slows as \( P \) approaches the carrying capacity.
- Differential equations are powerful tools in predicting how populations evolve over time under given constraints.
- Logistic equations include the concept of carrying capacity, providing more realistic growth predictions than simple exponential models.
- Solving these equations allows us to understand the dynamics of population changes in various ecological and environmental settings.
Other exercises in this chapter
Problem 26
In Exercises \(25-46,\) use substitution to evaluate the integral. $$\int \sec ^{2}(x+2) d x$$
View solution Problem 26
In Exercises \(25-28\) , evaluate the integral analytically. Support your answer using NINT. $$\int_{0}^{\pi / 2} x^{3} \cos 2 x d x$$
View solution Problem 26
Polonium-21O The number of radioactive atoms remaining after \(t\) days in a sample of polonium- 210 that starts with \(y_{0}\) radioactive atoms is \(y=y_{0} e
View solution Problem 27
In Exercises \(25-46,\) use substitution to evaluate the integral. $$\int \sqrt{\tan x} \sec ^{2} x d x$$
View solution