{"id":370,"date":"2016-10-11T20:43:54","date_gmt":"2016-10-11T20:43:54","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/math4libarts\/?post_type=chapter&#038;p=370"},"modified":"2019-05-30T16:43:37","modified_gmt":"2019-05-30T16:43:37","slug":"logistic-growth","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/chapter\/logistic-growth\/","title":{"raw":"Logistic Growth","rendered":"Logistic Growth"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Evaluate and rewrite logarithms using the properties of logarithms<\/li>\r\n \t<li>Use the properties of logarithms to solve exponential models\u00a0for time<\/li>\r\n \t<li>Identify the carrying capacity in a logistic growth model<\/li>\r\n \t<li>Use a logistic growth model to predict growth<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Limits on Exponential Growth<\/h2>\r\nIn our basic exponential growth scenario, we had a recursive equation of the form\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n <\/sub><\/em>= <em>P<sub>\u00adn-1 <\/sub><\/em>+ <em>r P<sub>\u00adn-1<\/sub><\/em><\/p>\r\nIn a confined environment, however, the growth rate may not remain constant. In a lake, for example, there is some <em>maximum sustainable population<\/em> of fish, also called a <strong>carrying capacity<\/strong>.\r\n<div class=\"textbox\">\r\n<h3>Carrying Capacity<\/h3>\r\nThe <strong>carrying capacity<\/strong>, or <strong>maximum sustainable population<\/strong>, is the largest population that an environment can support.\r\n\r\n<\/div>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/20205537\/fishes-1711002_1280.jpg\"><img class=\"aligncenter size-large wp-image-892\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/20205537\/fishes-1711002_1280-1024x682.jpg\" alt=\"Dense koi fish population in water\" width=\"1024\" height=\"682\" \/><\/a>\r\n\r\nFor our fish, the carrying capacity is the largest population that the resources in the lake can sustain. If the population in the lake is far below the carrying capacity, then we would expect the population to grow essentially exponentially. However, as the population approaches the carrying capacity, there will be a scarcity of food and space available, and the growth rate will decrease. If the population exceeds the carrying capacity, there won\u2019t be enough resources to sustain all the fish and there will be a negative growth rate, causing the population to decrease back to the carrying capacity.\r\n\r\nIf the carrying capacity was 5000, the growth rate might vary something like that in the graph shown.<img class=\"aligncenter wp-image-371\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11203520\/populationgrowthrate.png\" alt=\"Graph. Vertical measures Growth Rate, in increments of .1, from -0.1 to 0.1. Horizontal measures Population, in increments of 5000, from 0 to 10000. There's a diagonal line sloping down from 0.1 at 0 to -0.1 at 10000.\" width=\"350\" height=\"246\" \/>\r\n\r\nNote that this is a linear equation with intercept at 0.1 and slope [latex]-\\frac{0.1}{5000}[\/latex], so we could write an equation for this adjusted growth rate as:\r\n<p style=\"text-align: center;\"><em>r<sub>adjusted <\/sub><\/em>= [latex]0.1-\\frac{0.1}{5000}P=0.1\\left(1-\\frac{P}{5000}\\right)[\/latex]<\/p>\r\nSubstituting this in to our original exponential growth model for <em>r<\/em> gives\r\n<p style=\"text-align: center;\">[latex]{{P}_{n}}={{P}_{n-1}}+0.1\\left(1-\\frac{{{P}_{n-1}}}{5000}\\right){{P}_{n-1}}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">View the following for a detailed explanation of the concept.<\/p>\r\nhttps:\/\/youtu.be\/-6VLXCTkP_c\r\n<div class=\"textbox\">\r\n<h3>Logistic Growth<\/h3>\r\nIf a population is growing in a constrained environment with carrying capacity <em>K<\/em>, and absent constraint would grow exponentially with growth rate <em>r<\/em>, then the population behavior can be described by the logistic growth model:\r\n\r\n[latex]{{P}_{n}}={{P}_{n-1}}+r\\left(1-\\frac{{{P}_{n-1}}}{K}\\right){{P}_{n-1}}[\/latex]\r\n\r\n<\/div>\r\nUnlike linear and exponential growth, logistic growth behaves differently if the populations grow steadily throughout the year or if they have one breeding time per year. The recursive formula provided above models generational growth, where there is one breeding time per year (or, at least a finite number); there is no explicit formula for this type of logistic growth.\r\n<div class=\"textbox exercises\">\r\n<h3>Examples<\/h3>\r\nA forest is currently home to a population of 200 rabbits. The forest is estimated to be able to sustain a population of 2000 rabbits. Absent any restrictions, the rabbits would grow by 50% per year. Predict the future population using the logistic growth model.\r\n[reveal-answer q=\"543594\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"543594\"]\r\n\r\nModeling this with a logistic growth model, <em>r <\/em> = 0.50, <em>K<\/em> = 2000, and <em>P<sub>\u00ad0 <\/sub><\/em>= 200. Calculating the next year:\r\n<p style=\"text-align: center;\">[latex]{{P}_{1}}={{P}_{0}}+0.50\\left(1-\\frac{{{P}_{0}}}{2000}\\right){{P}_{0}}=200+0.50\\left(1-\\frac{200}{2000}\\right)200=290[\/latex]<\/p>\r\nWe can use this to calculate the following year:\r\n<p style=\"text-align: center;\">[latex]{{P}_{2}}={{P}_{1}}+0.50\\left(1-\\frac{{{P}_{1}}}{2000}\\right){{P}_{1}}=290+0.50\\left(1-\\frac{290}{2000}\\right)290\\approx414[\/latex]<\/p>\r\nA calculator was used to compute several more values:\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>n<\/em><\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<td>4<\/td>\r\n<td>5<\/td>\r\n<td>6<\/td>\r\n<td>7<\/td>\r\n<td>8<\/td>\r\n<td>9<\/td>\r\n<td>10<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>P<sub>n<\/sub><\/em><\/td>\r\n<td>200<\/td>\r\n<td>290<\/td>\r\n<td>414<\/td>\r\n<td>578<\/td>\r\n<td>784<\/td>\r\n<td>1022<\/td>\r\n<td>1272<\/td>\r\n<td>1503<\/td>\r\n<td>1690<\/td>\r\n<td>1821<\/td>\r\n<td>1902<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nPlotting these values, we can see that the population starts to increase faster and the graph curves upwards during the first few years, like exponential growth, but then the growth slows down as the population approaches the carrying capacity.\r\n\r\n<img class=\"aligncenter wp-image-372\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11203921\/populationyears.png\" alt=\"Graph. Vertical measures Population, in increments of 500, from 0 to 2000. Horizontal measures Years, in increments of 1, from 0 to 10. The line increases quickly and then tapers, similar to the first half of a bell curve.\" width=\"350\" height=\"283\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\nView more about this example below.\r\n\r\nhttps:\/\/youtu.be\/dPOlEgJ2QX0\r\n\r\n<hr \/>\r\n\r\nOn an island that can support a population of 1000 lizards, there is currently a population of 600. These lizards have a lot of offspring and not a lot of natural predators, so have very high growth rate, around 150%. Calculating out the next couple generations:\r\n<p style=\"text-align: center;\">[latex]{{P}_{1}}={{P}_{0}}+1.50\\left(1-\\frac{{{P}_{0}}}{1000}\\right){{P}_{0}}=600+1.50\\left(1-\\frac{600}{1000}\\right)600=960[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{{P}_{2}}={{P}_{1}}+1.50\\left(1-\\frac{{{P}_{1}}}{1000}\\right){{P}_{1}}=960+1.50\\left(1-\\frac{960}{1000}\\right)960=1018[\/latex]<\/p>\r\nInterestingly, even though the factor that limits the growth rate slowed the growth a lot, the population still overshot the carrying capacity. We would expect the population to decline the next year.\r\n<p style=\"text-align: center;\">[latex]{{P}_{3}}={{P}_{2}}+1.50\\left(1-\\frac{{{P}_{3}}}{1000}\\right){{P}_{3}}=1018+1.50\\left(1-\\frac{1018}{1000}\\right)1018=991[\/latex]<\/p>\r\nCalculating out a few more years and plotting the results, we see the population wavers above and below the carrying capacity, but eventually settles down, leaving a steady population near the carrying capacity.\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21175945\/Screen-Shot-2016-12-21-at-12.59.24-PM.png\"><img class=\"aligncenter wp-image-910\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21175945\/Screen-Shot-2016-12-21-at-12.59.24-PM.png\" alt=\"Graph. Vertical measures Population, in increments of 200 from 0 to 1200. Horizontal measures Years, in increments of 1 from 0 to 10. Year 0 shows population of 600, jumping to ~1000 in year 1, a little over 1000 in year 2, and staying close to 1000 in every subsequent year. \" width=\"350\" height=\"277\" \/><\/a>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nA field currently contains 20 mint plants. Absent constraints, the number of plants would increase by 70% each year, but the field can only support a maximum population of 300 plants. Use the logistic model to predict the population in the next three years.\r\n[reveal-answer q=\"726473\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"726473\"][latex]P_1=P_0+0.70(1-\\frac{P_0}{300})P_0=20+0.70(1-\\frac{20}{300})20=33[\/latex]\r\n\r\n[latex]P_2=54[\/latex]\r\n\r\n[latex]P_3=85[\/latex]\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=54627&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"100%\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nOn a neighboring island to the one from the previous example, there is another population of lizards, but the growth rate is even higher \u2013 about 205%.\r\n\r\nCalculating out several generations and plotting the results, we get a surprise: the population seems to be oscillating between two values, a pattern called a 2-cycle.<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21184924\/Screen-Shot-2016-12-21-at-1.47.24-PM.png\"><img class=\"aligncenter wp-image-911\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21184924\/Screen-Shot-2016-12-21-at-1.47.24-PM.png\" alt=\"Graph. Vertical measures Population, in increments of 200 from 0 to 1200. Horizontal measures Years, in increments of 1 from 0 to 10. Year 0 shows population of 600, jumping to ~1100 in year 1, down to ~900 in year 2, and vacillating from 1100 to 800 in alternating years through the rest of the graph.\" width=\"350\" height=\"276\" \/><\/a>\r\n\r\nWhile it would be tempting to treat this only as a strange side effect of mathematics, this has actually been observed in nature. Researchers from the University of California observed a stable 2-cycle in a lizard population in California.[footnote]<a href=\"http:\/\/users.rcn.com\/jkimball.ma.ultranet\/BiologyPages\/P\/Populations2.html\" target=\"_blank\" rel=\"noopener\">http:\/\/users.rcn.com\/jkimball.ma.ultranet\/BiologyPages\/P\/Populations2.html<\/a>[\/footnote]\r\n\r\nTaking this even further, we get more and more extreme behaviors as the growth rate increases higher. It is possible to get stable 4-cycles, 8-cycles, and higher. Quickly, though, the behavior approaches chaos (remember the movie <em>Jurassic Park<\/em>?).\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21185438\/Screen-Shot-2016-12-21-at-1.50.01-PM.png\"><img class=\"aligncenter wp-image-913 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21185438\/Screen-Shot-2016-12-21-at-1.50.01-PM.png\" alt=\"Two graphs. Left graph, labeled \u201cr=2.46 A 4-cycle.\u201d Vertical measures Population, in increments of 200 from 0 to 1400. Horizontal measures Years, in increments of 1 from 0 to 10. The plotted line moves in sharp up and down cycles, from ~700 in year 0, 1200 in year 1, 600 in year 2, 1200 in year 3, ~600 in year 4, and so forth. Right graph, labeled \u201cr=2.9 Chaos!\u201d Vertical measures Population, in increments of 200 from 0 to 1400. Horizontal measures Years, in increments of 5 from 0 to 30. The plotted line moves in sharp up and down cycles of varying lengths, forming an erratic back and forth between highs and lows.\" width=\"613\" height=\"279\" \/><\/a>\r\n\r\nAll of the lizard island examples are discussed in this video.\r\n\r\nhttps:\/\/youtu.be\/fuJF_uZGoFc\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6589&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"100%\"><\/iframe>\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Evaluate and rewrite logarithms using the properties of logarithms<\/li>\n<li>Use the properties of logarithms to solve exponential models\u00a0for time<\/li>\n<li>Identify the carrying capacity in a logistic growth model<\/li>\n<li>Use a logistic growth model to predict growth<\/li>\n<\/ul>\n<\/div>\n<h2>Limits on Exponential Growth<\/h2>\n<p>In our basic exponential growth scenario, we had a recursive equation of the form<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n <\/sub><\/em>= <em>P<sub>\u00adn-1 <\/sub><\/em>+ <em>r P<sub>\u00adn-1<\/sub><\/em><\/p>\n<p>In a confined environment, however, the growth rate may not remain constant. In a lake, for example, there is some <em>maximum sustainable population<\/em> of fish, also called a <strong>carrying capacity<\/strong>.<\/p>\n<div class=\"textbox\">\n<h3>Carrying Capacity<\/h3>\n<p>The <strong>carrying capacity<\/strong>, or <strong>maximum sustainable population<\/strong>, is the largest population that an environment can support.<\/p>\n<\/div>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/20205537\/fishes-1711002_1280.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-large wp-image-892\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/20205537\/fishes-1711002_1280-1024x682.jpg\" alt=\"Dense koi fish population in water\" width=\"1024\" height=\"682\" \/><\/a><\/p>\n<p>For our fish, the carrying capacity is the largest population that the resources in the lake can sustain. If the population in the lake is far below the carrying capacity, then we would expect the population to grow essentially exponentially. However, as the population approaches the carrying capacity, there will be a scarcity of food and space available, and the growth rate will decrease. If the population exceeds the carrying capacity, there won\u2019t be enough resources to sustain all the fish and there will be a negative growth rate, causing the population to decrease back to the carrying capacity.<\/p>\n<p>If the carrying capacity was 5000, the growth rate might vary something like that in the graph shown.<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-371\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11203520\/populationgrowthrate.png\" alt=\"Graph. Vertical measures Growth Rate, in increments of .1, from -0.1 to 0.1. Horizontal measures Population, in increments of 5000, from 0 to 10000. There's a diagonal line sloping down from 0.1 at 0 to -0.1 at 10000.\" width=\"350\" height=\"246\" \/><\/p>\n<p>Note that this is a linear equation with intercept at 0.1 and slope [latex]-\\frac{0.1}{5000}[\/latex], so we could write an equation for this adjusted growth rate as:<\/p>\n<p style=\"text-align: center;\"><em>r<sub>adjusted <\/sub><\/em>= [latex]0.1-\\frac{0.1}{5000}P=0.1\\left(1-\\frac{P}{5000}\\right)[\/latex]<\/p>\n<p>Substituting this in to our original exponential growth model for <em>r<\/em> gives<\/p>\n<p style=\"text-align: center;\">[latex]{{P}_{n}}={{P}_{n-1}}+0.1\\left(1-\\frac{{{P}_{n-1}}}{5000}\\right){{P}_{n-1}}[\/latex]<\/p>\n<p style=\"text-align: left;\">View the following for a detailed explanation of the concept.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Logistic model\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/-6VLXCTkP_c?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox\">\n<h3>Logistic Growth<\/h3>\n<p>If a population is growing in a constrained environment with carrying capacity <em>K<\/em>, and absent constraint would grow exponentially with growth rate <em>r<\/em>, then the population behavior can be described by the logistic growth model:<\/p>\n<p>[latex]{{P}_{n}}={{P}_{n-1}}+r\\left(1-\\frac{{{P}_{n-1}}}{K}\\right){{P}_{n-1}}[\/latex]<\/p>\n<\/div>\n<p>Unlike linear and exponential growth, logistic growth behaves differently if the populations grow steadily throughout the year or if they have one breeding time per year. The recursive formula provided above models generational growth, where there is one breeding time per year (or, at least a finite number); there is no explicit formula for this type of logistic growth.<\/p>\n<div class=\"textbox exercises\">\n<h3>Examples<\/h3>\n<p>A forest is currently home to a population of 200 rabbits. The forest is estimated to be able to sustain a population of 2000 rabbits. Absent any restrictions, the rabbits would grow by 50% per year. Predict the future population using the logistic growth model.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q543594\">Show Solution<\/span><\/p>\n<div id=\"q543594\" class=\"hidden-answer\" style=\"display: none\">\n<p>Modeling this with a logistic growth model, <em>r <\/em> = 0.50, <em>K<\/em> = 2000, and <em>P<sub>\u00ad0 <\/sub><\/em>= 200. Calculating the next year:<\/p>\n<p style=\"text-align: center;\">[latex]{{P}_{1}}={{P}_{0}}+0.50\\left(1-\\frac{{{P}_{0}}}{2000}\\right){{P}_{0}}=200+0.50\\left(1-\\frac{200}{2000}\\right)200=290[\/latex]<\/p>\n<p>We can use this to calculate the following year:<\/p>\n<p style=\"text-align: center;\">[latex]{{P}_{2}}={{P}_{1}}+0.50\\left(1-\\frac{{{P}_{1}}}{2000}\\right){{P}_{1}}=290+0.50\\left(1-\\frac{290}{2000}\\right)290\\approx414[\/latex]<\/p>\n<p>A calculator was used to compute several more values:<\/p>\n<table>\n<tbody>\n<tr>\n<td><em>n<\/em><\/td>\n<td>0<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<td>5<\/td>\n<td>6<\/td>\n<td>7<\/td>\n<td>8<\/td>\n<td>9<\/td>\n<td>10<\/td>\n<\/tr>\n<tr>\n<td><em>P<sub>n<\/sub><\/em><\/td>\n<td>200<\/td>\n<td>290<\/td>\n<td>414<\/td>\n<td>578<\/td>\n<td>784<\/td>\n<td>1022<\/td>\n<td>1272<\/td>\n<td>1503<\/td>\n<td>1690<\/td>\n<td>1821<\/td>\n<td>1902<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Plotting these values, we can see that the population starts to increase faster and the graph curves upwards during the first few years, like exponential growth, but then the growth slows down as the population approaches the carrying capacity.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-372\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11203921\/populationyears.png\" alt=\"Graph. Vertical measures Population, in increments of 500, from 0 to 2000. Horizontal measures Years, in increments of 1, from 0 to 10. The line increases quickly and then tapers, similar to the first half of a bell curve.\" width=\"350\" height=\"283\" \/><\/p>\n<\/div>\n<\/div>\n<p>View more about this example below.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Logistic growth of rabbits\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/dPOlEgJ2QX0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<hr \/>\n<p>On an island that can support a population of 1000 lizards, there is currently a population of 600. These lizards have a lot of offspring and not a lot of natural predators, so have very high growth rate, around 150%. Calculating out the next couple generations:<\/p>\n<p style=\"text-align: center;\">[latex]{{P}_{1}}={{P}_{0}}+1.50\\left(1-\\frac{{{P}_{0}}}{1000}\\right){{P}_{0}}=600+1.50\\left(1-\\frac{600}{1000}\\right)600=960[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{{P}_{2}}={{P}_{1}}+1.50\\left(1-\\frac{{{P}_{1}}}{1000}\\right){{P}_{1}}=960+1.50\\left(1-\\frac{960}{1000}\\right)960=1018[\/latex]<\/p>\n<p>Interestingly, even though the factor that limits the growth rate slowed the growth a lot, the population still overshot the carrying capacity. We would expect the population to decline the next year.<\/p>\n<p style=\"text-align: center;\">[latex]{{P}_{3}}={{P}_{2}}+1.50\\left(1-\\frac{{{P}_{3}}}{1000}\\right){{P}_{3}}=1018+1.50\\left(1-\\frac{1018}{1000}\\right)1018=991[\/latex]<\/p>\n<p>Calculating out a few more years and plotting the results, we see the population wavers above and below the carrying capacity, but eventually settles down, leaving a steady population near the carrying capacity.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21175945\/Screen-Shot-2016-12-21-at-12.59.24-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-910\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21175945\/Screen-Shot-2016-12-21-at-12.59.24-PM.png\" alt=\"Graph. Vertical measures Population, in increments of 200 from 0 to 1200. Horizontal measures Years, in increments of 1 from 0 to 10. Year 0 shows population of 600, jumping to ~1000 in year 1, a little over 1000 in year 2, and staying close to 1000 in every subsequent year.\" width=\"350\" height=\"277\" \/><\/a><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>A field currently contains 20 mint plants. Absent constraints, the number of plants would increase by 70% each year, but the field can only support a maximum population of 300 plants. Use the logistic model to predict the population in the next three years.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q726473\">Show Solution<\/span><\/p>\n<div id=\"q726473\" class=\"hidden-answer\" style=\"display: none\">[latex]P_1=P_0+0.70(1-\\frac{P_0}{300})P_0=20+0.70(1-\\frac{20}{300})20=33[\/latex]<\/p>\n<p>[latex]P_2=54[\/latex]<\/p>\n<p>[latex]P_3=85[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=54627&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"100%\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>On a neighboring island to the one from the previous example, there is another population of lizards, but the growth rate is even higher \u2013 about 205%.<\/p>\n<p>Calculating out several generations and plotting the results, we get a surprise: the population seems to be oscillating between two values, a pattern called a 2-cycle.<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21184924\/Screen-Shot-2016-12-21-at-1.47.24-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-911\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21184924\/Screen-Shot-2016-12-21-at-1.47.24-PM.png\" alt=\"Graph. Vertical measures Population, in increments of 200 from 0 to 1200. Horizontal measures Years, in increments of 1 from 0 to 10. Year 0 shows population of 600, jumping to ~1100 in year 1, down to ~900 in year 2, and vacillating from 1100 to 800 in alternating years through the rest of the graph.\" width=\"350\" height=\"276\" \/><\/a><\/p>\n<p>While it would be tempting to treat this only as a strange side effect of mathematics, this has actually been observed in nature. Researchers from the University of California observed a stable 2-cycle in a lizard population in California.<a class=\"footnote\" title=\"http:\/\/users.rcn.com\/jkimball.ma.ultranet\/BiologyPages\/P\/Populations2.html\" id=\"return-footnote-370-1\" href=\"#footnote-370-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a><\/p>\n<p>Taking this even further, we get more and more extreme behaviors as the growth rate increases higher. It is possible to get stable 4-cycles, 8-cycles, and higher. Quickly, though, the behavior approaches chaos (remember the movie <em>Jurassic Park<\/em>?).<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21185438\/Screen-Shot-2016-12-21-at-1.50.01-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-913 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21185438\/Screen-Shot-2016-12-21-at-1.50.01-PM.png\" alt=\"Two graphs. Left graph, labeled \u201cr=2.46 A 4-cycle.\u201d Vertical measures Population, in increments of 200 from 0 to 1400. Horizontal measures Years, in increments of 1 from 0 to 10. The plotted line moves in sharp up and down cycles, from ~700 in year 0, 1200 in year 1, 600 in year 2, 1200 in year 3, ~600 in year 4, and so forth. Right graph, labeled \u201cr=2.9 Chaos!\u201d Vertical measures Population, in increments of 200 from 0 to 1400. Horizontal measures Years, in increments of 5 from 0 to 30. The plotted line moves in sharp up and down cycles of varying lengths, forming an erratic back and forth between highs and lows.\" width=\"613\" height=\"279\" \/><\/a><\/p>\n<p>All of the lizard island examples are discussed in this video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Logistic growth of lizards\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/fuJF_uZGoFc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6589&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"100%\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-370\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Logistic Growth. <strong>Authored by<\/strong>: David Lippman. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\">http:\/\/www.opentextbookstore.com\/mathinsociety\/<\/a>. <strong>Project<\/strong>: Math in Society. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><li>fishes-colourful-beautiful-koi. <strong>Authored by<\/strong>: sharonang. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/pixabay.com\/en\/fishes-colourful-beautiful-koi-1711002\/\">https:\/\/pixabay.com\/en\/fishes-colourful-beautiful-koi-1711002\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/cc0\">CC0: No Rights Reserved<\/a><\/em><\/li><li>Logistic model. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/-6VLXCTkP_c\">https:\/\/youtu.be\/-6VLXCTkP_c<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Logistic growth of rabbits. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/dPOlEgJ2QX0\">https:\/\/youtu.be\/dPOlEgJ2QX0<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Logistic growth of lizards. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/fuJF_uZGoFc\">https:\/\/youtu.be\/fuJF_uZGoFc<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 54627. <strong>Authored by<\/strong>: Hartley,Josiah. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License<\/li><li>Question ID 6589. <strong>Authored by<\/strong>: Lippman,David. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section><hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-370-1\"><a href=\"http:\/\/users.rcn.com\/jkimball.ma.ultranet\/BiologyPages\/P\/Populations2.html\" target=\"_blank\" rel=\"noopener\">http:\/\/users.rcn.com\/jkimball.ma.ultranet\/BiologyPages\/P\/Populations2.html<\/a> <a href=\"#return-footnote-370-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":20,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Logistic Growth\",\"author\":\"David 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