Problem 58
Question
In 1934 the biologist G. F. Gause placed 20 paramecia in 5 cubic centimeters of a saline solution with a constant amount of food and measured their growth on a daily basis. He found that the population \(f(t)\) at any time \(t\) (in days) was approximately $$ f(t)=\frac{4490}{1+e^{5.4094-1.0235 t}} \text { for } t \geq 0 $$ Determine the carrying capacity of the medium.
Step-by-Step Solution
Verified Answer
The carrying capacity is 4490 paramecia.
1Step 1: Identify the Logistic Growth Model
Recognize that the given formula for population growth, \( f(t) = \frac{4490}{1+e^{5.4094-1.0235 t}} \), is a logistic growth model. The general form of a logistic growth function is \( f(t) = \frac{L}{1 + e^{k(b-t)}} \), where \( L \) is the carrying capacity.
2Step 2: Determine the Carrying Capacity
In the logistic growth model, the carrying capacity \( L \) represents the maximum population size that the environment can sustain. From the given equation, we identify \( L = 4490 \).
3Step 3: Conclusion
The carrying capacity is derived directly from the coefficient in the numerator of the logistic growth function, which is 4490 in this case. Thus, the carrying capacity of the medium is 4490 paramecia.
Key Concepts
Carrying CapacityPopulation GrowthDifferential Equations
Carrying Capacity
In the context of population biology, the carrying capacity refers to the maximum number of individuals in a species that an environment can support sustainably. It’s like the full capacity of a bus, where no more passengers can board. Carrying capacity is determined by environmental limits, like space, food, and other resources.
For example, a forest might have a carrying capacity for 200 deer due to available food and space. If the population grows beyond this number, resources become scarce, leading to competition, which can lower the population back down. This balancing act helps maintain ecosystem stability.
Within the logistic growth model, carrying capacity appears as a constant "L" in the function. It signifies where the population level stabilizes over time. In the exercise provided, this value is 4490 paramecia, indicating that the medium can support a maximum of 4490 paramecia. As the population approaches this number, the growth rate slows down and eventually levels off.
For example, a forest might have a carrying capacity for 200 deer due to available food and space. If the population grows beyond this number, resources become scarce, leading to competition, which can lower the population back down. This balancing act helps maintain ecosystem stability.
Within the logistic growth model, carrying capacity appears as a constant "L" in the function. It signifies where the population level stabilizes over time. In the exercise provided, this value is 4490 paramecia, indicating that the medium can support a maximum of 4490 paramecia. As the population approaches this number, the growth rate slows down and eventually levels off.
Population Growth
Population growth describes how the number of individuals in a population changes over time. Growth can be linear, exponential, or logistic. The logistic growth model, which is relevant here, is particularly interesting as it accounts for resource limitations, making it more realistic than simple exponential growth.
This model starts with a phase of rapid growth when resources are abundant. However, as the population nears the carrying capacity, resources start to become limited. This limitation slows the growth rate, creating an S-shaped curve typical of logistic growth.
In our scenario with the paramecia, the population was initially small and experienced significant growth. As time progresses and the quantity of available food and space nears depletion, growth slows, preventing the population from surpassing the carrying capacity of 4490 paramecia. This showcases real-world population dynamics, where natural capacity limits control growth.
This model starts with a phase of rapid growth when resources are abundant. However, as the population nears the carrying capacity, resources start to become limited. This limitation slows the growth rate, creating an S-shaped curve typical of logistic growth.
In our scenario with the paramecia, the population was initially small and experienced significant growth. As time progresses and the quantity of available food and space nears depletion, growth slows, preventing the population from surpassing the carrying capacity of 4490 paramecia. This showcases real-world population dynamics, where natural capacity limits control growth.
Differential Equations
Differential equations play a crucial role in expressing population growth mathematically. They provide a way to describe how a population changes over time, considering influences like birth and death rates, as well as carrying capacity.
In the logistic growth model specifically, the differential equation takes the form \[\frac{dP}{dt} = rP\left(1 - \frac{P}{L}\right), \]where:
Understanding differential equations provides invaluable insights into the mechanisms driving population dynamics and can be applied beyond biology, into economics, physics, and other fields where change over time needs to be modeled and predicted.
In the logistic growth model specifically, the differential equation takes the form \[\frac{dP}{dt} = rP\left(1 - \frac{P}{L}\right), \]where:
- \( \frac{dP}{dt} \) is the rate of change of the population over time
- \( r \) is the rate of maximum growth
- \( P \) is the current population size
- \( L \) is the carrying capacity
Understanding differential equations provides invaluable insights into the mechanisms driving population dynamics and can be applied beyond biology, into economics, physics, and other fields where change over time needs to be modeled and predicted.
Other exercises in this chapter
Problem 57
Plot the graph of \(f\). Using the zoom feature of the calculator, approximate to within \(0.1\) all values of \(c\) such that \(f(c)\) is a relative extreme va
View solution Problem 57
Show that the given inequality holds for the given values of \(\mathrm{x}\). $$ \frac{1}{4} x+1 / x>1 \text { for } x>2 $$
View solution Problem 58
A rectangle is inscribed in a semicircle of radius \(r\) with one side lying on the diameter of the semicircle. Find the maximum possible area of the rectangle.
View solution Problem 58
Plot the graph of \(f\). Using the zoom feature of the calculator, approximate to within \(0.1\) all values of \(c\) such that \(f(c)\) is a relative extreme va
View solution