{"id":1679,"date":"2015-11-12T18:30:46","date_gmt":"2015-11-12T18:30:46","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&#038;p=1679"},"modified":"2017-04-03T17:40:14","modified_gmt":"2017-04-03T17:40:14","slug":"use-logistic-growth-models","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/use-logistic-growth-models\/","title":{"raw":"Use logistic-growth models","rendered":"Use logistic-growth models"},"content":{"raw":"<p id=\"fs-id1165137737682\">Exponential growth cannot continue forever. Exponential models, while they may be useful in the short term, tend to fall apart the longer they continue. Consider an aspiring writer who writes a single line on day one and plans to double the number of lines she writes each day for a month. By the end of the month, she must write over 17 billion lines, or one-half-billion pages. It is impractical, if not impossible, for anyone to write that much in such a short period of time. Eventually, an exponential model must begin to approach some limiting value, and then the growth is forced to slow. For this reason, it is often better to use a model with an upper bound instead of an <strong>exponential growth<\/strong> model, though the exponential growth model is still useful over a short term, before approaching the limiting value.<\/p>\r\n<p id=\"fs-id1165135194441\">The <strong>logistic growth model<\/strong> is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model\u2019s upper bound, called the <strong>carrying capacity<\/strong>. For constants <em>a<\/em>, <em>b<\/em>, and <em>c<\/em>, the logistic growth of a population over time <em>x<\/em>\u00a0is represented by the model<\/p>\r\n\r\n<div id=\"eip-391\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/div>\r\n<p id=\"fs-id1165135439859\">Figure 6\u00a0shows how the growth rate changes over time. The graph increases from left to right, but the growth rate only increases until it reaches its point of maximum growth rate, at which point the rate of increase decreases.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202016\/CNX_Precalc_Figure_04_07_0062.jpg\" alt=\"Graph of f(x)=c\/(1+ae^(-tx)). The carrying capacity is the asymptote at y=c. The initial value of population is (0, c\/(1+a)). The point of maximum growth is (ln(a)\/b, c\/2).\" width=\"487\" height=\"367\" data-media-type=\"image\/jpg\"\/><b>Figure 6<\/b>[\/caption]\r\n\r\n<div id=\"fs-id1165135207602\" class=\"note textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"A General Note\">\r\n<h3 class=\"title\" data-type=\"title\">A General Note: Logistic Growth<\/h3>\r\n<p id=\"fs-id1165135511485\">The logistic growth model is<\/p>\r\n\r\n<div id=\"eip-id1165134239632\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/div>\r\n<p id=\"eip-id1165135264518\">where<\/p>\r\n\r\n<div id=\"fs-id1165137526317\" data-type=\"list\" data-list-type=\"bulleted\">\r\n<ul><li data-type=\"item\">[latex]\\frac{c}{1+a}[\/latex] is the initial value<\/li>\r\n\t<li data-type=\"item\"><em>c<\/em>\u00a0is the <em data-effect=\"italics\">carrying capacity<\/em>, or <em data-effect=\"italics\">limiting value<\/em><\/li>\r\n\t<li data-type=\"item\"><em>b<\/em>\u00a0is a constant determined by the rate of growth.<\/li>\r\n<\/ul><\/div>\r\n<\/div>\r\n<div id=\"Example_04_07_06\" class=\"example\" data-type=\"example\">\r\n<div id=\"fs-id1165135572080\" class=\"exercise\" data-type=\"exercise\">\r\n<div id=\"fs-id1165135572082\" class=\"problem textbox shaded\" data-type=\"problem\">\r\n<h3 data-type=\"title\">Example 5: Using the Logistic-Growth Model<\/h3>\r\n<p id=\"fs-id1165135572087\">An influenza epidemic spreads through a population rapidly, at a rate that depends on two factors: The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model a good one to study the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.<\/p>\r\n<p id=\"fs-id1165134194949\">For example, at time <em>t\u00a0<\/em>= 0 there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is <em>b\u00a0<\/em>= 0.6030. Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135471237\" class=\"solution textbox shaded\" data-type=\"solution\">\r\n<h3 id=\"fs-id1165135471239\">Solution<\/h3>\r\nWe substitute the given data into the logistic growth model\r\n<div id=\"eip-id1165133200610\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/div>\r\n<p id=\"fs-id1165137698371\">Because at most 1,000 people, the entire population of the community, can get the flu, we know the limiting value is <em>c\u00a0<\/em>= 1000. To find <em>a<\/em>, we use the formula that the number of cases at time <em>t\u00a0<\/em>= 0 is [latex]\\frac{c}{1+a}=1[\/latex], from which it follows that <em>a\u00a0<\/em>= 999. This model predicts that, after ten days, the number of people who have had the flu is [latex]f\\left(x\\right)=\\frac{1000}{1+999{e}^{-0.6030x}}\\approx 293.8[\/latex]. Because the actual number must be a whole number (a person has either had the flu or not) we round to 294. In the long term, the number of people who will contract the flu is the limiting value, <em>c\u00a0<\/em>= 1000.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135445930\" class=\"commentary\" style=\"text-align: center;\" data-type=\"commentary\">\r\n<h3 data-type=\"title\">Analysis of the Solution<\/h3>\r\n<p id=\"fs-id1165135445936\" style=\"text-align: left;\">Remember that, because we are dealing with a virus, we cannot predict with certainty the number of people infected. The model only approximates the number of people infected and will not give us exact or actual values.<\/p>\r\n<p id=\"fs-id1165135344064\" style=\"text-align: left;\">Figure 7\u00a0gives a good picture of how this model fits the data.<\/p>\r\n\r\n<figure class=\"medium\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202017\/CNX_Precalc_Figure_04_07_0072.jpg\" alt=\"Graph of f(x)=1000\/(1+999e^(-0.5030x)) with the y-axis labeled as\" data-media-type=\"image\/jpg\"\/><\/figure><figure id=\"CNX_Precalc_Figure_04_07_007\" class=\"medium\"><figcaption><strong>Figure 7.\u00a0<\/strong>The graph of [latex]f\\left(x\\right)=\\frac{1000}{1+999{e}^{-0.6030x}}[\/latex]<\/figcaption><\/figure><\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It 5<\/h3>\r\n<p id=\"fs-id1165135689538\">Using the model in Example 5, estimate the number of cases of flu on day 15.<\/p>\r\n<a href=\"https:\/\/courses.candelalearning.com\/osprecalc\/chapter\/solutions-23\/\" target=\"_blank\">Solution<\/a>\r\n\r\n<\/div>","rendered":"<p id=\"fs-id1165137737682\">Exponential growth cannot continue forever. Exponential models, while they may be useful in the short term, tend to fall apart the longer they continue. Consider an aspiring writer who writes a single line on day one and plans to double the number of lines she writes each day for a month. By the end of the month, she must write over 17 billion lines, or one-half-billion pages. It is impractical, if not impossible, for anyone to write that much in such a short period of time. Eventually, an exponential model must begin to approach some limiting value, and then the growth is forced to slow. For this reason, it is often better to use a model with an upper bound instead of an <strong>exponential growth<\/strong> model, though the exponential growth model is still useful over a short term, before approaching the limiting value.<\/p>\n<p id=\"fs-id1165135194441\">The <strong>logistic growth model<\/strong> is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model\u2019s upper bound, called the <strong>carrying capacity<\/strong>. For constants <em>a<\/em>, <em>b<\/em>, and <em>c<\/em>, the logistic growth of a population over time <em>x<\/em>\u00a0is represented by the model<\/p>\n<div id=\"eip-391\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/div>\n<p id=\"fs-id1165135439859\">Figure 6\u00a0shows how the growth rate changes over time. The graph increases from left to right, but the growth rate only increases until it reaches its point of maximum growth rate, at which point the rate of increase decreases.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202016\/CNX_Precalc_Figure_04_07_0062.jpg\" alt=\"Graph of f(x)=c\/(1+ae^(-tx)). The carrying capacity is the asymptote at y=c. The initial value of population is (0, c\/(1+a)). The point of maximum growth is (ln(a)\/b, c\/2).\" width=\"487\" height=\"367\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/b><\/p>\n<\/div>\n<div id=\"fs-id1165135207602\" class=\"note textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"A General Note\">\n<h3 class=\"title\" data-type=\"title\">A General Note: Logistic Growth<\/h3>\n<p id=\"fs-id1165135511485\">The logistic growth model is<\/p>\n<div id=\"eip-id1165134239632\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/div>\n<p id=\"eip-id1165135264518\">where<\/p>\n<div id=\"fs-id1165137526317\" data-type=\"list\" data-list-type=\"bulleted\">\n<ul>\n<li data-type=\"item\">[latex]\\frac{c}{1+a}[\/latex] is the initial value<\/li>\n<li data-type=\"item\"><em>c<\/em>\u00a0is the <em data-effect=\"italics\">carrying capacity<\/em>, or <em data-effect=\"italics\">limiting value<\/em><\/li>\n<li data-type=\"item\"><em>b<\/em>\u00a0is a constant determined by the rate of growth.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"Example_04_07_06\" class=\"example\" data-type=\"example\">\n<div id=\"fs-id1165135572080\" class=\"exercise\" data-type=\"exercise\">\n<div id=\"fs-id1165135572082\" class=\"problem textbox shaded\" data-type=\"problem\">\n<h3 data-type=\"title\">Example 5: Using the Logistic-Growth Model<\/h3>\n<p id=\"fs-id1165135572087\">An influenza epidemic spreads through a population rapidly, at a rate that depends on two factors: The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model a good one to study the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.<\/p>\n<p id=\"fs-id1165134194949\">For example, at time <em>t\u00a0<\/em>= 0 there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is <em>b\u00a0<\/em>= 0.6030. Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed.<\/p>\n<\/div>\n<div id=\"fs-id1165135471237\" class=\"solution textbox shaded\" data-type=\"solution\">\n<h3 id=\"fs-id1165135471239\">Solution<\/h3>\n<p>We substitute the given data into the logistic growth model<\/p>\n<div id=\"eip-id1165133200610\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/div>\n<p id=\"fs-id1165137698371\">Because at most 1,000 people, the entire population of the community, can get the flu, we know the limiting value is <em>c\u00a0<\/em>= 1000. To find <em>a<\/em>, we use the formula that the number of cases at time <em>t\u00a0<\/em>= 0 is [latex]\\frac{c}{1+a}=1[\/latex], from which it follows that <em>a\u00a0<\/em>= 999. This model predicts that, after ten days, the number of people who have had the flu is [latex]f\\left(x\\right)=\\frac{1000}{1+999{e}^{-0.6030x}}\\approx 293.8[\/latex]. Because the actual number must be a whole number (a person has either had the flu or not) we round to 294. In the long term, the number of people who will contract the flu is the limiting value, <em>c\u00a0<\/em>= 1000.<\/p>\n<\/div>\n<div id=\"fs-id1165135445930\" class=\"commentary\" style=\"text-align: center;\" data-type=\"commentary\">\n<h3 data-type=\"title\">Analysis of the Solution<\/h3>\n<p id=\"fs-id1165135445936\" style=\"text-align: left;\">Remember that, because we are dealing with a virus, we cannot predict with certainty the number of people infected. The model only approximates the number of people infected and will not give us exact or actual values.<\/p>\n<p id=\"fs-id1165135344064\" style=\"text-align: left;\">Figure 7\u00a0gives a good picture of how this model fits the data.<\/p>\n<figure class=\"medium\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202017\/CNX_Precalc_Figure_04_07_0072.jpg\" alt=\"Graph of f(x)=1000\/(1+999e^(-0.5030x)) with the y-axis labeled as\" data-media-type=\"image\/jpg\" \/><\/figure>\n<figure id=\"CNX_Precalc_Figure_04_07_007\" class=\"medium\"><figcaption><strong>Figure 7.\u00a0<\/strong>The graph of [latex]f\\left(x\\right)=\\frac{1000}{1+999{e}^{-0.6030x}}[\/latex]<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 5<\/h3>\n<p id=\"fs-id1165135689538\">Using the model in Example 5, estimate the number of cases of flu on day 15.<\/p>\n<p><a href=\"https:\/\/courses.candelalearning.com\/osprecalc\/chapter\/solutions-23\/\" target=\"_blank\">Solution<\/a><\/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-1679\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: Jay Abramson, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175<\/a>. <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>: Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":276,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"Jay Abramson, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1679","chapter","type-chapter","status-publish","hentry"],"part":1668,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1679","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/users\/276"}],"version-history":[{"count":2,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1679\/revisions"}],"predecessor-version":[{"id":3086,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1679\/revisions\/3086"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/1668"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1679\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/media?parent=1679"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1679"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/contributor?post=1679"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/license?post=1679"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}