\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Precalculus - Logistic Growth

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The exponential growth model given by the equation \(A(t) = A_oe^{kt}\) has one problem when modeling things like population growth, it is unrealistic in that it has uninhibited growth. The logistic model takes care of that problem by taking into account things like limitations on food, space and other resources.

Logistic Growth/Decay Model Equation

\(\displaystyle{ P(t) = \frac{c}{1+ae^{-bt}} }\)

\(a,b,c\) are constants

\(a \gt 0\) and \(c \gt 0\)

\(b \gt 0\) growth; \(b \lt 0\) decay

The constant \(c\) is called the carrying capacity and \(b\) is the growth/decay rate. An inflection point occurs at \(c/2\), which is where \(P(t)\) changes concavity. (You will understand this better in calculus.)

Let's look at the point when \(t=0\), i.e. the initial time.
\(\displaystyle{ P(0) = \frac{c}{1+ae^0} = \frac{c}{1+a} }\)
So, the initial value is \(c/(1+a)\). Since \(e^{-bt}\) is in the denominator, there is not a good way to specify \(P(0)=P_0\) as a single parameter.

We use the logistic model to model population growth and decomposition of materials.

Logistic Growth/Decay Plot

So what does the graph of this equation look like? Here is a simple plot for \(a = b = c = 1\).

Plot 1 - Logistic Decay

This is a typical shape for the logistic decay equation. Notice to the right there is a horizontal asymptote at \(y = 0\). We are not really concerned with what is going to the left of \(t = 0\) since most models have a horizontal scale of time.

When we change \(b\) to \(b = -1\), we get the plot below.

Plot 2 - Logistic Growth

In this case, notice the horizontal asymptote to the right is now \(y = 1\). This implies that we now have some limiting factors that keep the growth of the function in check. This makes sense when modeling for things like population growth since eventually food supplies, space, predators and other factors will keep a population in check.

By adjusting \(a, b\) and \(c\), you can change the value of \(P(0)\), the upper horizontal asymptote and how quickly the graph heads toward the asymptote.

Before you go on, watch this video clip for a more general explanation of the logistic growth/decay function. Be careful to notice that he uses different names for the constants than we have discussed up until now.

What is a Logistic Growth Function?

Work the practice problems.

Practice

For \( \displaystyle{ N(t) = \frac{1000}{20+480e^{-0.8t}} }\), determine \(N\) for \( t = 0 \), \( t = \infty \), \( t = 1 \) and \( t = 5 \).

Problem Statement

For \( \displaystyle{ N(t) = \frac{1000}{20+480e^{-0.8t}} }\), determine \(N\) for \( t = 0 \), \( t = \infty \), \( t = 1 \) and \( t = 5 \).

Solution

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A conservation group releases 100 animals of an endangered species into a game preserve. The organization believes that the preserve has a carrying capacity of 1000 animals and that the growth of the herd will be modeled by the logistics curve given by the equation \(\displaystyle{ p = \frac{1000}{1+9e^{-kt}} }\) and \( t \geq 0 \) where t is measure in years. If the herd size is 134 after 2 years, find the population after 5 years.

Problem Statement

A conservation group releases 100 animals of an endangered species into a game preserve. The organization believes that the preserve has a carrying capacity of 1000 animals and that the growth of the herd will be modeled by the logistics curve given by the equation \(\displaystyle{ p = \frac{1000}{1+9e^{-kt}} }\) and \( t \geq 0 \) where t is measure in years. If the herd size is 134 after 2 years, find the population after 5 years.

Solution

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A baseball stadium seats a total of 46621 fans. At 6pm 5000 seats are occupied. At 7pm, game start, 40000 seats are occupied. Assume a logistic function accurately models the number of seats occupied where \(t\) is the time in hours from the start of the game. Use the given values to algebraically find the particular equation that models this situation and determine the number of seats occupied 2 hours after the game starts.

Problem Statement

A baseball stadium seats a total of 46621 fans. At 6pm 5000 seats are occupied. At 7pm, game start, 40000 seats are occupied. Assume a logistic function accurately models the number of seats occupied where \(t\) is the time in hours from the start of the game. Use the given values to algebraically find the particular equation that models this situation and determine the number of seats occupied 2 hours after the game starts.

Solution

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For a certain population, we use the logistic growth model equation \(\displaystyle{ P(t) = \frac{55.1}{1+9.6e^{-0.02515t}} }\) where \(t\) is the number of years since the beginning of 1900.
a. Evaluate P(0) and describe what this value means.
b. Use the equation to give an approximate population at the beginning of the year 2015.

Problem Statement

For a certain population, we use the logistic growth model equation \(\displaystyle{ P(t) = \frac{55.1}{1+9.6e^{-0.02515t}} }\) where \(t\) is the number of years since the beginning of 1900.
a. Evaluate P(0) and describe what this value means.
b. Use the equation to give an approximate population at the beginning of the year 2015.

Solution

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Biologists stock a lake with 500 fish and estimate the carrying capacity to be 10,000. The number of fish tripled during the first year. Assuming that the size of the fish population satisfies the logistic equation, find an expression for the size of the population after t years. How long will it take population to reach 4,000?

Problem Statement

Biologists stock a lake with 500 fish and estimate the carrying capacity to be 10,000. The number of fish tripled during the first year. Assuming that the size of the fish population satisfies the logistic equation, find an expression for the size of the population after t years. How long will it take population to reach 4,000?

Solution

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100 dogs were released on an island in 2005. By 2012, there were 324 dogs on the island. The island can sustain a maximum of 5000 dogs. Assuming logistic growth, write a general equation that describes the population P(t) at any time t. How many dogs will there be by 2020? In what year will the population reach 1000? How many dogs were born per year in 2016?

Problem Statement

100 dogs were released on an island in 2005. By 2012, there were 324 dogs on the island. The island can sustain a maximum of 5000 dogs. Assuming logistic growth, write a general equation that describes the population P(t) at any time t. How many dogs will there be by 2020? In what year will the population reach 1000? How many dogs were born per year in 2016?

Solution

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Write a logistic growth equation and find the population after 4 years where the initial population is 1000, the carrying capacity is 10,000 and the population after 1 year is 2500.

Problem Statement

Write a logistic growth equation and find the population after 4 years where the initial population is 1000, the carrying capacity is 10,000 and the population after 1 year is 2500.

Solution

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Biologists stocked a lake with 400 fish and estimated the carrying capacity to be 8700. The number of fish doubled in the first year. Assuming the size of the fish population satisfies the logistic equation, find an equation for the size of the population after t years. How long will it take the population to increase to half the carrying capacity?

Problem Statement

Biologists stocked a lake with 400 fish and estimated the carrying capacity to be 8700. The number of fish doubled in the first year. Assuming the size of the fish population satisfies the logistic equation, find an equation for the size of the population after t years. How long will it take the population to increase to half the carrying capacity?

Solution

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Really UNDERSTAND Precalculus

Topics You Need To Understand For This Page

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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