### Learning Outcomes

- Write a logistic growth model
- Evaluate a logistic growth model
- Solve a logistic growth model

We have seen that we often use exponential relationships to describe population growth. But populations do not grow without some limitation on their numbers provided by real constraints in the real world. Predators, disease, and limitations on resources all serve to limit the numbers of any living population.

You used the example of fish in a lake to explored the recursive and explicit forms of the logistic equation in the previous module and derived the following logistic growth model with carrying capacity [latex]K[/latex] and growth rate [latex]r[/latex]:

[latex]{{P}_{n}}={{P}_{n-1}}+r\left(1-\frac{{{P}_{n-1}}}{K}\right){{P}_{n-1}}[/latex].

While this formula will allow you to predict future population recursively for populations that grow discretely with one breeding time per year, it doesn’t present a closed form that you can use to make predictions about populations like fish or people that breed year round. For those situations, we can use a continuous logistic model in the form

[latex]P_{t}=\dfrac{c}{1+\left(\dfrac{c}{P_{0}}-1\right)e^{-rt}}[/latex]

where [latex]t[/latex] stands for time in years, [latex]c[/latex] is the carrying capacity (the maximal population), [latex]P_0[/latex] represents the starting quantity, and [latex]r[/latex] is the rate of growth.

### Recall: operations on Fractions

When simplifying algebraic expressions, we may sometimes need to add, subtract, simplify, multiply, or divide fractions. It is important to be able to do these operations on the fractions without having to convert them to decimals, though in the example below, once it becomes clear that a quantity will reduce to a terminating decimal, it is done so for simplicity’s sake.

- [latex]\dfrac{a}{b}\cdot\dfrac{c}{d} = \dfrac {ac}{bd}[/latex]
- [latex]\dfrac{a}{b}\div\dfrac{c}{d}=\dfrac{a}{b}\cdot\dfrac{d}{c}=\dfrac{ad}{bc}[/latex]
- [latex]\dfrac{a}{b}=\dfrac{a\cdot c}{b \cdot c}[/latex]
- [latex]\dfrac{a}{b}\pm \dfrac{c}{d} = \dfrac{ad \pm bc}{bd}[/latex]
- If [latex]\dfrac{a}{b}=\dfrac{c}{d}[/latex], then [latex]ad = bc[/latex] and [latex]a=\dfrac{bc}{d}[/latex] and [latex]\dfrac{ad}{b}=c[/latex].

### Write a logistic model for a population

We can evaluate the logistic model for an input [latex]t[/latex] given the remaining information or the means to solve for it. Here’s an example.

Biologists stock a lake with 500 trout. They estimate the carrying capacity of the lake to be 6900 trout. That is, due to constraints in the environment, the population will grow no further than a maximum 6900 trout in the lake. The biologists noted that the number of fish grew to be 740 in the first year. We’d like to know how many fish there would be 3 years after stocking the lake. We’d also like to how long it will take for the population to increase to 3450 fish.

We start by writing the form of the logistic model, filling in all the information we can.

[latex]P_{t}=\dfrac{c}{1+\left(\dfrac{c}{P_{0}}-1\right)e^{-rt}}[/latex]

[latex]P_{t}=\dfrac{6900}{1+\left(\dfrac{6900}{500}-1\right)e^{-rt}}[/latex]

We have the carrying capacity, [latex]c[/latex], and the initial amount, but we still need the growth rate, [latex]r[/latex]. We can use the information about the first year growth to obtain it. We know that when [latex]t=1 \text{, } P_{t}=740[/latex].

[latex] \begin{array}{rcl} P_{t} & = & \dfrac{6900}{1+\left(\dfrac{6900}{500}-1\right)e^{-rt}} & \\ \\ 740 & = & \dfrac{6900}{1+\left(\dfrac{6900}{500}-1\right)e^{-r \cdot 1}} & \mbox{substitute first year growth} \\ \\ 740 & = & \dfrac{6900}{1+\left(\dfrac{64}{5}\right)e^{-r}} & \mbox{reduce} \\ \\ 1+\dfrac{64}{5}e^{-r} & = & \dfrac{6900}{740} & \mbox{ rewrite the proportion} \\ \\ e^{-r} & = & \dfrac{357 \cdot 5}{37 \cdot 64} & \mbox{ solve for the exponential term} \\ \\ e^{-r} & = & \dfrac{385}{592} & \\ \\ ln\left(e^{-r}\right)&=& ln\left(\dfrac{385}{592}\right) & \mbox{apply natural log to both sides} \\ \\ -r&=&0.4302633 & \mbox{simplify} \\ \\ r& \approx & 0.43 \\ \\ \end{array}[/latex]

Now that we have the growth rate we can write the formula

[latex]P_{t}=\dfrac{6900}{1+12.8e^{-0.43t}}[/latex].

### Recall evaluating and solving an equation

- To evaluate a function, substitute a number for the input variable and calculate the result. This will yield the output that corresponds to the desired input.
- To solve a function, substitute a number for the output variable then use the properties of equality to isolate the input variable on one side of the equation. This will yield the input that corresponds with the desired output.

Ex. For the function [latex]f(x)=3x[/latex] evaluate [latex]f(7)[/latex] and solve [latex]f(x) = 7[/latex].

- Evaluate [latex]f(7)[/latex].
- [latex]f(7)=3(7) = 21[/latex]
- The point [latex]\left(7, 21\right)[/latex] is on the graph of this function.

- Solve [latex]f(x) = 7[/latex]
- [latex]7 = 3x \Longrightarrow x=\dfrac{7}{3}[/latex]
- The point [latex]\left(\dfrac{7}{3}, 7\right)[/latex] is on the graph of this function.

### Evaluate a logistic growth model

Having found the growth rate and written the model, we can now answer the two questions. The first question, *how many fish will be in the lake 3 years after stocking it*, asks us to evaluate the model for [latex]t=3[/latex].

[latex]P_{3}=\dfrac{6900}{1+12.8e^{-0.43\ast 3}}[/latex]

[latex]P_{3}=439.49 \approx 439 \text{ fish.}[/latex]

### Solve a logistic growth model

The final question, *how long will it take for the population to increase to 3450* *fish*, asks us to solve the model for [latex]P_{t}=3450[/latex], that is, half the carrying capacity.

[latex] \begin{array}{rcl} 3450 & = & \dfrac{6900}{1+12.8e^{-0.43t}} & \\ 1 & = & \dfrac{2}{1+\left(12.8\right)e^{-0.43t}} & \\ 1+\left(12.8\right)e^{-0.43t} & = & 2 & \\ 1+12.8e^{-0.43t} & = & 2 \\ ln\left(e^{-0.43t}\right) & = & ln\left(\dfrac{1}{12.8}\right) \\ -0.43t & = & -2.549445171 \\ t & \approx & 5.928942 \\ t & \approx & \text{ 5.9 years later} \\ \end{array}[/latex]

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