4.4: Natural Growth and Logistic Growth
In this chapter, we have been looking at linear and exponential growth. Another very useful tool for modeling population growth is the natural growth model. This model uses base e , an irrational number, as the base of the exponent instead of \((1+r)\). You may remember learning about \(e\) in a previous class, as an exponential function and the base of the natural logarithm.
The Natural Growth Model is
\[P(t)=P_{0} e^{k t} \nonumber \]
where \(P_{0}\) is the initial population, \(k\) is the growth rate per unit of time, and \(t\) is the number of time periods.
Given \(P_{0} > 0\), if k > 0 , this is an exponential growth model, if k < 0 , this is an exponential decay model.
When a certain drug is administered to a patient, the number of milligrams remaining in the bloodstream after t hours is given by the model
\[P(t) = 40e^{-.25t} \nonumber \]
How many milligrams are in the blood after two hours?
Solution
To solve this problem, we use the given equation with t = 2
\[\begin{align*} P(2) &= 40e^{-.25(2)} \\ P(2) &= 24.26 \end{align*} \nonumber \]
There are approximately 24.6 milligrams of the drug in the patient’s bloodstream after two hours.
In the next example, we can see that the exponential growth model does not reflect an accurate picture of population growth for natural populations.
Bob has an ant problem. On the first day of May, Bob discovers he has a small red ant hill in his back yard, with a population of about 100 ants. If conditions are just right red ant colonies have a growth rate of 240% per year during the first four years. If Bob does nothing, how many ants will he have next May? How many in five years?
Solution
We solve this problem using the natural growth model.
\[P(t) = 100e^{2.4t} \nonumber \]
In one year, t = 1, we have
\[P(1) = 100e^{2.4(1)} = 1102 \text{ ants} \nonumber \]
In one year, t = 5, we have
\[P(5) = 100e^{2.4(5)} = 16,275,479 \text{ ants} \nonumber \]
That is a lot of ants! Bob will not let this happen in his back yard!
Note: The population of ants in Bob’s back yard follows an exponential (or natural) growth model.
The problem with exponential growth is that the population grows without bound and, at some point, the model will no longer predict what is actually happening since the amount of resources available is limited. Populations cannot continue to grow on a purely physical level, eventually death occurs and a limiting population is reached.
Another growth model for living organisms in the logistic growth model. The logistic growth model has a maximum population called the carrying capacity. As the population grows, the number of individuals in the population grows to the carrying capacity and stays there. This is the maximum population the environment can sustain.
\[P(t) = \dfrac{M}{1+ke^{-ct}} \nonumber \]
where M , c , and k are positive constants and t is the number of time periods.
Figure \(\PageIndex{1}\): Comparison of Exponential Growth and Logistic Growth
The horizontal line K on this graph illustrates the carrying capacity. However, this book uses M to represent the carrying capacity rather than K.
(Logistic Growth Image 1, n.d.)
(Logistic Growth Image 2, n.d.)
The graph for logistic growth starts with a small population. When the population is small, the growth is fast because there is more elbow room in the environment. As the population approaches the carrying capacity, the growth slows.
The population of an endangered bird species on an island grows according to the logistic growth model.
\[P(t) = \dfrac{3640}{1+25e^{-0.04t}} \nonumber \]
Identify the initial population. What will be the bird population in five years? What will be the population in 150 years? What will be the population in 500 years?
Solution
We know the initial population,\(P_{0}\), occurs when \(t = 0\).
\[P_{0} = P(0) = \dfrac{3640}{1+25e^{-0.04(0)}} = 140 \nonumber \]
Calculate the population in five years, when \(t = 5\).
\[P(5) = \dfrac{3640}{1+25e^{-0.04(5)}} = 169.6 \nonumber \]
The island will be home to approximately 170 birds in five years
Calculate the population in 150 years, when \(t = 150\).
\[P(150) = \dfrac{3640}{1+25e^{-0.04(150)}} = 3427.6 \nonumber \]
The island will be home to approximately 3428 birds in 150 years.
Calculate the population in 500 years, when \(t = 500\).
\[P(500) = \dfrac{3640}{1+25e^{-0.04(500)}} = 3640.0 \nonumber \]
The island will be home to approximately 3640 birds in 500 years.
This example shows that the population grows quickly between five years and 150 years, with an overall increase of over 3000 birds; but, slows dramatically between 150 years and 500 years (a longer span of time) with an increase of just over 200 birds.
The student population at NAU can be modeled by the logistic growth model below, with initial population taken from the early 1960’s. We will use 1960 as the initial population date.
\[P(t) = \dfrac{30,000}{1+5e^{-0.06t}} \nonumber \]
Determine the initial population and find the population of NAU in 2014. What will be NAU’s population in 2050? From this model, what do you think is the carrying capacity of NAU?
Solution
We solve this problem by substituting in different values of time.
When \(t = 0\), we get the initial population \(P_{0}\).
\[P_{0} = P(0) = \dfrac{30,000}{1+5e^{-0.06(0)}} = \dfrac{30,000}{6} = 5000 \nonumber \]
The initial population of NAU in 1960 was 5000 students.
In the year 2014, 54 years have elapsed so, \(t = 54\).
\[P(54) = \dfrac{30,000}{1+5e^{-0.06(54)}} = \dfrac{30,000}{1+5e^{-3.24}} = \dfrac{30,000}{1.19582} = 25,087 \nonumber \]
There are 25,087 NAU students in 2014.
In 2050, 90 years have elapsed so, \(t = 90\).
\[P(90) = \dfrac{30,000}{1+5e^{-0.06(90)}} = \dfrac{30,000}{1+5e^{-5.4}} = 29,337 \nonumber \]
There are 29,337 NAU students in 2050.
Finally, to predict the carrying capacity, look at the population 200 years from 1960, when \(t = 200\).
\[P(200) = \dfrac{30,000}{1+5e^{-0.06(200)}} = \dfrac{30,000}{1+5e^{-12}} = \dfrac{30,000}{1.00003} = 29,999 \nonumber \]
Thus, the carrying capacity of NAU is 30,000 students.
It appears that the numerator of the logistic growth model, M, is the carrying capacity.
Given the logistic growth model \(P(t) = \dfrac{M}{1+ke^{-ct}}\), the carrying capacity of the population is \(M\). \(M\), the carrying capacity, is the maximum population possible within a certain habitat.
Suppose that in a certain fish hatchery, the fish population is modeled by the logistic growth model where \(t\) is measured in years.
\[P(t) = \dfrac{12,000}{1+11e^{-0.2t}} \nonumber \]
What is the carrying capacity of the fish hatchery? How long will it take for the population to reach 6000 fish?
Solution
The carrying capacity of the fish hatchery is \(M = 12,000\) fish.
Now, we need to find the number of years it takes for the hatchery to reach a population of 6000 fish. We must solve for \(t\) when \(P(t) = 6000\).
\[6000 =\dfrac{12,000}{1+11e^{-0.2t}} \nonumber \]
\[\begin{align*} (1+11e^{-0.2t}) \cdot 6000 &= \dfrac{12,000}{1+11e^{-0.2t}} \cdot (1+11e^{-0.2t}) \\ (1+11e^{-0.2t}) \cdot 6000 &= 12,000 \\ \dfrac{(1+11e^{-0.2t}) \cdot \cancel{6000}}{\cancel{6000}} &= \dfrac{12,000}{6000} \\ 1+11e^{-0.2t} &= 2 \\ 11e^{-0.2t} &= 1 \\ e^{-0.2t} &= \dfrac{1}{11} = 0.090909 \end{align*} \nonumber \]
Take the natural logarithm (ln on the calculator) of both sides of the equation.
\[\begin{align*} \text{ln} e^{-0.2t} &= \text{ln} 0.090909 \\ \text{ln}e^{-0.2t} &= -0.2t \text{ by the rules of logarithms.} \\ -0.2t &= \text{ln}0.090909 \\ t &= \dfrac{\text{ln}0.090909}{-0.2} \\ t&= 11.999\end{align*} \nonumber \]
It will take approximately 12 years for the hatchery to reach 6000 fish.