Models for Population Growth: Calculus

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In general, population ecologists and experts use population change models to describe population and predict how it will change.

In AP Calculus, you will primarily work with two population change modes: exponential and logistic.

Models for Population Growth in Calculus

In AP Calculus, you will primarily work with two population change models: exponential and logistic.

The major differences between the two models include:

Exponential growth is J-shaped while logistic growth is sigmoid (S-shaped)
Exponential growth depends exclusively on the size of the population, while logistic growth depends on the size of the population, competition, and the number of resources
Exponential growth is applicable to a population that does not have any limitations for growth, while logistic growth is more applicable in the sense that it applies to any population with a carrying capacity

In the following sections, you'll learn more about the two models in depth.

The Exponential Model for Population Growth

Exponential growth describes a particular pattern of data that increases more and more over time. The graph of the data mirrors an exponential function and creates a J-shape.

Models for Population Growth example of graph exponential growth StudySmarter The exponential growth of time vs. size of the population is a J-shaped function - StudySmarter Originals

With regards to population change, exponential growth occurs when an infinite amount of resources are available to the population. Consider our garden example. The population of pests will grow exponentially if there are no limits to how much food the pests can eat from your infinitely huge garden.

The human population currently grows at an exponential rate. However, Earth does not have an infinite amount of resources. Scientists hypothesize that we will eventually reach a "carrying capacity," which we will discuss more in the next section.

The formula for exponential growth is

$P = C e^{k t}$

where $C$ is a constant determined by the initial population, $k$ is the constant of growth and is greater than 0, and $t$ is time.

For more details on exponential growth, see our article on Exponential Growth and Decay

The Logistic Model for Population Growth

Logistic growth describes a pattern of data whose growth rate gets smaller and smaller as the population approaches a certain maximum - often referred to as the carrying capacity. The graph of logistic growth is a sigmoid curve.

Models for Population Growth example of graph of logistic growth StudySmarter Logistic growth of time vs. size of the population is a sigmoid (S-shaped) function - StudySmarter Originals

With regards to population change, logistic growth occurs when there are limited resources available or when there is competition among animals. The population of pests will grow until we introduce pesticides. The carrying capacity allows our garden to thrive by ensuring that the pest population doesn't grow too large while limiting our use of toxic pesticides.

The formula for logistic growth is

$N = \frac{M C}{C + e^{- k t}}$

where $M$ is the carrying capacity, $C$ is a constant determined by the initial population, $k$ is the constant of growth, and $t$ is time. All values are positive.

For details on Logistic population growth, see our article on The Logistic Differential Equation

Models for Population Growth Formulas

Exponential growth

The rate of change of an exponential growth function $P$ can be modeled by the differential equation

$\frac{d P}{d t} = k P$

Logistic growth and decay

The rate of change of a logistic growth function $N$ can be modeled by the differential equation

$\frac{d N}{d t} = k N (1 - \frac{N}{M})$

Examples of population growth model

Example 1

The rate of change of a culture of bacteria is proportional to the population itself. When $t = 0$ , there are 100 bacteria. Two minutes later, at $t = 2$ , there are 300 bacteria. How many bacteria are there at 4 minutes?

Step 1: Decide which model the population growth follows

We are not told of any possible carrying capacity limits in this problem, and the growth rate is proportional to the population of bacteria, so it is safe to assume that these bacteria will follow an exponential growth model.

Step 2: Use the initial condition to find C

Since the population models an exponential growth rate, we know that the population $P$ can be modeled by

$P = C e^{k t}$

To find $C$ , we can plug in our initial condition (0, 100).

$100 = C e^{(0) k} 100 = C (1) 100 = C$

Step 3: Use the second condition to find the growth rate k

To find $k$ , we can plug in the second condition (2, 300).

$300 = 100 e^{2 k} 3 = e^{2 k} \ln (3) = 2 k \frac{\ln (3)}{2} = k$

Step 4: Plug in t = 4 to find the population

$P (4) = 100 e^{\frac{\ln (3)}{2} (4)} P (4) = 900$

Therefore, at 4 minutes, the bacteria population is 900.

Example 2

A population of rabbits has a rate of change of

$\frac{d N}{d t} = 0.05 N (1 - \frac{N}{500})$

where t is a measure in years.

What is the size of the population of rabbits at four years?
How many rabbits will there be at 10 years?
When will the rabbit population reach 400?

Step 1: Extract variables from differential form

From the definition of the differential form of the logistic growth model, we know that $k = 0.05$ , $M = 500$ , and at $t = 0$ , $N = 100$ .

Step 2: Plug in variables to the logistic growth equation

$N = \frac{500 C}{C + e^{- 0.05 t}}$

Step 3: Solve for C with the initial value

$100 = \frac{500 C}{C + e^{- 0.05 (0)}} 100 = \frac{500 C}{C + 1}$

We can use cross multiplication to solve for $C$ .

$100 (C + 1) = 500 C C + 1 = 5 C C = \frac{1}{4}$

Step 4: Plug in t = 4 and t = 10

Plugging in $C$ and to $N$

$N = \frac{500 (\frac{1}{4})}{\frac{1}{4} + e^{- 0.05 t}}$

Now we can plug in $t = 4$ and $t = 10$ .

$N = \frac{500 (\frac{1}{4})}{\frac{1}{4} + e^{- 0.05 (4)}} N = 116.961$

$N = \frac{500 (\frac{1}{4})}{\frac{1}{4} + e^{- 0.05 (10)}} N = 145.948$

After four years, the rabbit population will be about 117. After 10 years, the rabbit population will be about 146.

Step 5: Let N = 400 and solve for t

$400 = \frac{500 (\frac{1}{4})}{\frac{1}{4} + e^{- 0.05 t}} 400 (\frac{1}{4} + e^{- 0.05 t}) = 125 100 + 400 e^{- 0.05 t} = 125 e^{- 0.05 t} = \frac{1}{16} - 0.05 t = \ln (\frac{1}{16}) t = 55.452$

So, it would take the rabbit population about 55.5 years to reach a population of 400.

Models for Population Growth - Key takeaways

Population growth can take on two models: exponential or logistic
- Exponential population growth occurs when there are unlimited resources - the rate of change of the population is strictly based on the size of the population
- Logistic population growth occurs when there are limited resources available and competition to access the resources - the rate of change of the population is based on the size of the population, competition, and the number of resources

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Frequently Asked Questions about Models for Population Growth

What are the three models of population growth?

Population growth can be modeled by either a exponential growth equation or a logistic growth equation.

What is the best model for population growth?

A population's growth model depends on the environment that the population grows in.

How do you make a population growth model?

A population growth model is made by deciding if the population has an exponential growth rate or a logistic growth rate based on the nature of the environment the population grows in. From there, the model is made by plugging in known values to solve for unknowns.

What are the two major types of population models?

The two major types of population models are exponential and logistic.

What are examples of population growth model?

An example of a population growth model is bacteria growing in a petri dish.

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Models for Population Growth

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Models for Population Growth in Calculus

The Exponential Model for Population Growth

The Logistic Model for Population Growth