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Logistic Growth and Decay

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Alexandra Shlosman

on 16 March 2015

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Transcript of Logistic Growth and Decay

Logistic Decay Examples
Wood Products
P.F. Verhulst was a Belgian mathematician who studied growth and decay in the 19th century.
How this relates to class
In class we learned about exponential growth and decay which has the same purpose as logistic growth and decay, except logistic growth and decay has more exact calculations. This topic uses properties of logarithms, exponential growth, exponential decay, and the natural base e.
Logistic Growth Examples
Rabbit Population
Logistic Growth and Decay
The EFISCEN wood product model classifies wood products according to their life span. There are four classifications: short (1 year), medium short (4 years), medium long (16 years), long (50 years). Based on data obtained from the European Forest Institute the percentage of remaining wood products after "t" years for wood products with long life-spans (such as those used in the building industry) is given by:
P(t) = 100.3952/(1+0.0316e^0.0581t )
2 years ago, a population of rabbits in Japan on Rabbit Island was 500. Rabbit Island is a predator- free zone, making it inevitable for growth to occur. The population has grown to 5,000.

What will the population of rabbits be 1 year from now?
How long will it take for the population to grow from 500 to 10,000?
How to Solve
Logistic Growth and Decay
Given information is P_0(initial population) is 500 rabbits. First step is to solve for "k" the growth constant. Second step is answering the first question, which asks what the population will be in 1 year, meaning when t=1. Third step is answering the second question, which asks for the time it takes for the population to grow from 500 to 10,000 which is telling to solve for t.
1) Solve for k

k=0.5 ln⁡〖(10)

1. Domain: {all real numbers}
Range: {0,c} where c is the carrying
2. No x-intercepts; y-intercept is P(0)
3. 2 Horizontal Asymptotes at y=0 and y=c
4. P(t) is an increasing function if b>0 and a decreasing function if b<0
5. Inflection point where P(t)= 0.5 of c
growth: curve up to curve down
decay: curve down to curve up
6. Smooth continuous graphs
*Typically used for calculation of population growth
Differential Equation
If dy/dt=ky, then y=Ae^kt for some constant A.
The rate of change is y=ky
"k" is the constant variable
Differential Eq. of Growth
Differential Eq. of Decay
Differential Equation of Growth:
"k" is the growth constant

Exponential Growth Equation:
P(t)= P_0e^kt
P(t) is the population at time "t"
P_0 is the initial population

The Differential Equation of Growth:
"-k" is the decay constant

Exponential Decay Equation:
P(t)= P_0e^-kt
In logistic growth, there are limitations that are taken into account. They are represented by either the variable of "c" or "k"
Growth Model
Logistic functions provide a more realistic model of population growth.
P(t) = c/1 + ae^bt
t= time (in years, months, days, etc.)
P(t)= population after "t" years
a= the number that shifts the graph
b= positive growth rate
c= carrying capacity

*Typically used with radioactive elements
Education and teaching
Secondary education at Athenaeum of Brussels
Studied at the University of Ghent and received a degree in exact sciences
Taught calculus at the Belgium Military Academy
Became professor of mathematics at Université Libre of Brussels
Elected president of the Belgium Academy of Science
Published a historical essay on an eighteenth century patriot
Contributions to logistic
growth and decay

This graph compares exponential growth and logistic growth and shows how logistic growth is more realistic because of the limitations or "carrying capacity"
"Logistic Growth - Boundless Open Textbook." Boundless. N.p., n.d. Web. 08 Mar. 2015.
Alexandra Shlosman
Mrs. Slavoski
Algebra 2B Per.4
15 March 2015

When the "carrying capacity" "c" is between 0 and 1, the graph represents logistic decay.

When the "carrying capacity" " is greater than 1, the graph represents logistic growth
"Logistic Functions." N.p., n.d. Web. 08 Mar. 2015 <http://wmueller.com/precalculus/families/1_81.html>.
"Logistic Population Growth." Encyclopedia Britannica Online. Encyclopedia Britannica, n.d. Web. 09 Mar. 2015. <http://www.britannica.com/EBchecked/topic/470416/population-ecology/70579/Logistic-population-growth>.
This is a graph of logistic growth which follows the basic equation of f(x)=1/(1+e^-x)
Levenbach, Hans. "Forecasting Trending Time Series with Relative Growth Rate Models." Technometrics 18.3 (1976): 261-72. Web. 9 Mar. 2015. <http://www.pps.k12.or.us/schools/benson/files/htrinh/5144_Demana_Ch03pp275-348.pdf>.

First published Logistic Growth Equation in 1838
Used data of several countries in comparison to Belgium to estimate the unknown parameters
Continued population studies in 1845
When "carrying capacity" "c" is between
0 and 1, decay is represented.
324. "Ch.4 Exponential and Logarithmic Functions." 4.8 (n.d.): n. pag.
Web. 10 Mar. 2015. <http://faculty.uncfsu.edu/fnani/FicamsFrontpage/ch4.8.pdf>.
b>0 for growth and b<0 for decay where
c is greater than 0 for both functions.
1) What is the decay rate?
2)What is the percentage of remaining wood
products after 10 years?
3) How long does it take for the percentage
of remaining wood products to reach 50%?
In this problem, we will be using the logistic decay model where P(t)=c/1+ae^-bt
The decay rate is equal to the absolute value of "b".
The value of "b" in the given equation is 0.0581.
The question asks to solve when "t" is 10 years.

P(10)=100.3952/(1+0.0316e^(0.0581(10)) )
P(10) is approximately 95.0, so 95% of the wood products remain after 10 years
To solve, make P(t)=50 and solve for "t".
P(10)=100.3952/(1+0.0316e^0.0581(10) )=~

50=100.3952/(1+0.0316e^0.0581t )
100.3952=50(1+0.0316e^0.0581t )
t=59.6 years
It will take approximately 59 years and 7 months for the percentage of wood products remaining to reach 50%.
324. "Ch.4 Exponential and Logarithmic Functions." 4.8 (n.d.): n. pag.
Web. 10 Mar. 2015. <http://faculty.uncfsu.edu/fnani/FicamsFrontpage/ch4.8.pdf>.
2) Population in 1 year

Since the data was taken 2 years ago and is asking for the population 1 year later, the time will actually be 3.

P(3)=500e^(0.5 ln⁡(10)×3)

The population of rabbits is approximately 15,811.
3) Time for population to grow from 500 to 10,000

1000=500e^(0.5 ln⁡(10)t)
20=e^(0.5 ln⁡(10)t)
0.5 ln⁡(10)t=ln⁡(20)
t=(ln⁡(20))/(0.5 ln⁡(10) )
t=2 years,7 months,6 days

The time for the population to grow from 500 to 10,000 is approximately 2 years, 7 months, and 6 days.

1. Banner, Adrian. The Calculus Lifesaver. Princeton: Princeton UP, 2007. Print.
2. "Calculus 131, Supplemental Sections 11.1-11.2 Logistic Growth" . (n.d.): n. pag. University of Maryland. Web. 8 Mar. 2015.
3. Exponential Functions: Population Growth, Radioactive Decay, and More (n.d.): n. pag. University of North Texas. Web. 8 Mar. 2015.
4. La Rosa, Myrna. 6.8, Section, Exponential Growth And, Decay Models, and Newton’s Law. Models OBJECTIVE 1 (n.d.): n. pag. Web. 22 Feb. 2015.
5. Levenbach, Hans. "Forecasting Trending Time Series with Relative Growth Rate Models." Technometrics 18.3 (1976): 261-72. Web. 8 Mar. 2015.
6. "Logistic Growth - Boundless Open Textbook." Boundless. N.p., n.d. Web. 08 Mar. 2015.
7. "Logistic Growth, Part 1." Logistic Growth, Part 1. Duke University, n.d. Web. 22 Feb. 2015.
8. "Logistic Population Growth." Encyclopedia Britannica Online. Encyclopedia Britannica, n.d. Web. 8 Mar. 2015.
9. Mueller, William. "Logistic Functions." Logistic Functions. N.p., n.d. Web. 8 Mar. 2015.
10. "Pierre François Verhulst." Verhulst Biography. University of St. Andrews, Scotland, Jan. 2014. Web. 8 Mar. 2015.

Works Cited
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