{"id":2316,"date":"2024-02-22T18:11:06","date_gmt":"2024-02-22T18:11:06","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/?post_type=chapter&#038;p=2316"},"modified":"2024-02-22T19:46:32","modified_gmt":"2024-02-22T19:46:32","slug":"logistic-models","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/logistic-models\/","title":{"raw":"Bounded Exponential Growth","rendered":"Bounded Exponential Growth"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use Newton's Law of Cooling.<\/li>\r\n \t<li>Use a logistic growth model.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Bounded Exponential Growth and Decay<\/h2>\r\n<h3>Using Newton\u2019s Law of Cooling<\/h3>\r\nExponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object\u2019s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a <strong>vertical shift<\/strong> of the generic <strong>exponential decay function<\/strong>. This translation leads to <strong>Newton\u2019s Law of Cooling<\/strong>, the scientific formula for temperature as a function of time as an object\u2019s temperature is equalized with the ambient temperature.\r\n\r\nThe formula is derived as follows:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}T\\left(t\\right)=A{b}^{ct}+{T}_{s}\\hfill &amp; \\hfill \\\\ T\\left(t\\right)=A{e}^{\\mathrm{ln}\\left({b}^{ct}\\right)}+{T}_{s}\\hfill &amp; \\text{Properties of logarithms}.\\hfill \\\\ T\\left(t\\right)=A{e}^{ct\\mathrm{ln}b}+{T}_{s}\\hfill &amp; \\text{Properties of logarithms}.\\hfill \\\\ T\\left(t\\right)=A{e}^{kt}+{T}_{s}\\hfill &amp; \\text{Rename the constant }c \\mathrm{ln} b,\\text{ calling it }k.\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: Newton\u2019s Law of Cooling<\/h3>\r\nThe temperature of an object, <em>T<\/em>, in surrounding air with temperature [latex]{T}_{s}[\/latex] will behave according to the formula\r\n\r\n[latex]T\\left(t\\right)=A{e}^{kt}+{T}_{s}[\/latex]\r\nwhere\r\n<ul>\r\n \t<li><em>t<\/em>\u00a0is time<\/li>\r\n \t<li><em>A<\/em>\u00a0is the difference between the initial temperature of the object and the surroundings<\/li>\r\n \t<li><em>k<\/em>\u00a0is a constant, the continuous rate of cooling of the object<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a set of conditions, apply Newton\u2019s Law of Cooling<\/h3>\r\n<ol>\r\n \t<li>Set [latex]{T}_{s}[\/latex] equal to the <em>y<\/em>-coordinate of the horizontal asymptote (usually the ambient temperature).<\/li>\r\n \t<li>Substitute the given values into the continuous growth formula [latex]T\\left(t\\right)=A{e}^{k}{}^{t}+{T}_{s}[\/latex] to find the parameters <em>A<\/em>\u00a0and <em>k<\/em>.<\/li>\r\n \t<li>Substitute in the desired time to find the temperature or the desired temperature to find the time.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using Newton\u2019s Law of Cooling<\/h3>\r\nA cheesecake is taken out of the oven with an ideal internal temperature of [latex]165^\\circ\\text{F}[\/latex] and is placed into a [latex]35^\\circ\\text{F}[\/latex] refrigerator. After 10 minutes, the cheesecake has cooled to [latex]150^\\circ\\text{F}[\/latex]. If we must wait until the cheesecake has cooled to [latex]70^\\circ\\text{F}[\/latex] before we eat it, how long will we have to wait?\r\n\r\n[reveal-answer q=\"705928\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"705928\"]\r\n\r\nBecause the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake\u2019s temperature will decay exponentially toward 35, following the equation\r\n<p style=\"text-align: center;\">[latex]T\\left(t\\right)=A{e}^{kt}+35[\/latex]<\/p>\r\nWe know the initial temperature was 165, so [latex]T\\left(0\\right)=165[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}165=A{e}^{k0}+35\\hfill &amp; \\text{Substitute }\\left(0,165\\right).\\hfill \\\\ A=130\\hfill &amp; \\text{Solve for }A.\\hfill \\end{array}[\/latex]<\/p>\r\nWe were given another data point, [latex]T\\left(10\\right)=150[\/latex], which we can use to solve for <em>k<\/em>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }150=130{e}^{k10}+35\\hfill &amp; \\text{Substitute (10, 150)}.\\hfill \\\\ \\text{ }115=130{e}^{k10}\\hfill &amp; \\text{Subtract 35 from both sides}.\\hfill \\\\ \\text{ }\\frac{115}{130}={e}^{10k}\\hfill &amp; \\text{Divide both sides by 130}.\\hfill \\\\ \\text{ }\\mathrm{ln}\\left(\\frac{115}{130}\\right)=10k\\hfill &amp; \\text{Take the natural log of both sides}.\\hfill \\\\ \\text{ }k=\\frac{\\mathrm{ln}\\left(\\frac{115}{130}\\right)}{10}=-0.0123\\hfill &amp; \\text{Divide both sides by the coefficient of }k.\\hfill \\end{array}[\/latex]<\/p>\r\nThis gives us the equation for the cooling of the cheesecake: [latex]T\\left(t\\right)=130{e}^{-0.0123t}+35[\/latex].\r\n\r\nNow we can solve for the time it will take for the temperature to cool to 70 degrees.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}70=130{e}^{-0.0123t}+35\\hfill &amp; \\text{Substitute in 70 for }T\\left(t\\right).\\hfill \\\\ 35=130{e}^{-0.0123t}\\hfill &amp; \\text{Subtract 35 from both sides}.\\hfill \\\\ \\frac{35}{130}={e}^{-0.0123t}\\hfill &amp; \\text{Divide both sides by 130}.\\hfill \\\\ \\mathrm{ln}\\left(\\frac{35}{130}\\right)=-0.0123t\\hfill &amp; \\text{Take the natural log of both sides}.\\hfill \\\\ t=\\frac{\\mathrm{ln}\\left(\\frac{35}{130}\\right)}{-0.0123}\\approx 106.68\\hfill &amp; \\text{Divide both sides by the coefficient of }t.\\hfill \\end{array}[\/latex]<\/p>\r\nIt will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to [latex]70^\\circ\\text{F}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nA pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees?\r\n\r\n[reveal-answer q=\"846066\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"846066\"]6.026 hours[\/hidden-answer]\r\n<iframe id=\"mom150\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=114392&amp;theme=oea&amp;iframe_resize_id=mom150\" width=\"100%\" height=\"300\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Logistic Models<\/h2>\r\nExponential 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 although the exponential growth model is still useful over a short term before approaching the limiting value.\r\n\r\nThe <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\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/p>\r\nThe graph below shows 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 the rate of increase decreases.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181346\/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\" \/>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Logistic Growth<\/h3>\r\nThe logistic growth model is\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/p>\r\nwhere\r\n<ul>\r\n \t<li>[latex]\\frac{c}{1+a}[\/latex] is the initial value<\/li>\r\n \t<li><em>c<\/em>\u00a0is the <em>carrying capacity\u00a0<\/em>or <em>limiting value<\/em><\/li>\r\n \t<li><em>b<\/em>\u00a0is a constant determined by the rate of growth.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Logistic-Growth Model<\/h3>\r\nAn 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 good for studying the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.\r\n\r\nFor 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.\r\n\r\n[reveal-answer q=\"346660\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"346660\"]\r\n\r\nWe substitute the given data into the logistic growth model\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/p>\r\nBecause 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.\r\n<h4>Analysis of the Solution<\/h4>\r\nRemember 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.\r\n\r\nThe graph below gives a good picture of how this model fits the data.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181349\/CNX_Precalc_Figure_04_07_0072.jpg\" alt=\"Graph of f(x)=1000\/(1+999e^(-0.5030x)) with the y-axis labeled as\" width=\"731\" height=\"492\" \/> The graph of [latex]f\\left(x\\right)=\\frac{1000}{1+999{e}^{-0.6030x}}[\/latex].[\/caption][\/hidden-answer]<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUsing the model in the previous example, estimate the number of cases of flu on day 15.\r\n\r\n[reveal-answer q=\"421768\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"421768\"]\r\n\r\n895 cases on day 15[\/hidden-answer]\r\n<iframe id=\"mom150\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5801&amp;theme=oea&amp;iframe_resize_id=mom150\" width=\"100%\" height=\"300\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Key Equations<\/h2>\r\n<table style=\"height: 138px; width: 736px;\" summary=\"..\">\r\n<tbody>\r\n<tr style=\"height: 30px;\">\r\n<td style=\"height: 30px; width: 90.0469px;\">Newton's Law of Cooling<\/td>\r\n<td style=\"height: 30px; width: 619.953px;\">[latex]T(t)={A}_{0}{e}^{kt}+T_s[\/latex], where [latex]T_s[\/latex] is the ambient temperature,\u00a0[latex]A=T(0)-T_s[\/latex], and\u00a0[latex]k[\/latex] is the continuous rate of cooling.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 28px;\">\r\n<td style=\"width: 90.0469px; height: 28px;\">Logistic Growth Model<\/td>\r\n<td style=\"width: 619.953px; height: 28px;\">[latex]f(x)=\\frac{c}{1+ae^{-bx}}[\/latex], where [latex]\\frac{c}{1+a}[\/latex] is the initial value, <em style=\"font-size: inherit;\">c<\/em><span style=\"font-size: inherit;\">\u00a0is the carrying capacity, or limiting value, and <\/span><em style=\"font-size: inherit;\">b<\/em><span style=\"font-size: inherit;\">\u00a0is a constant determined by the rate of growth<\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">Key Concepts<\/span>\r\n<ul>\r\n \t<li>We can use Newton\u2019s Law of Cooling to find how long it will take for a cooling object to reach a desired temperature or to find what temperature an object will be after a given time.<\/li>\r\n \t<li>We can use logistic growth functions to model real-world situations where the rate of growth changes over time, such as population growth, spread of disease, and spread of rumors.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt><strong>carrying capacity<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137838640\">in a logistic model, the limiting value of the output<\/dd>\r\n<\/dl>\r\n<dl class=\"definition\">\r\n \t<dt>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\"><\/dl>\r\n<\/dt>\r\n<\/dl>\r\n<dl class=\"definition\">\r\n \t<dt>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt><strong>logistic growth model<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137838640\">a function of the form [latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex] where [latex]\\frac{c}{1+a}[\/latex] is the initial value, <em>c<\/em>\u00a0is the carrying capacity, or limiting value, and <em>b<\/em>\u00a0is a constant determined by the rate of growth<\/dd>\r\n<\/dl>\r\n<\/dt>\r\n \t<dt><strong>Newton\u2019s Law of Cooling<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137838640\">the scientific formula for temperature as a function of time as an object\u2019s temperature is equalized with the ambient temperature<\/dd>\r\n<\/dl>\r\n<\/dt>\r\n \t<dd id=\"fs-id1165137838640\"><\/dd>\r\n<\/dl>\r\n<\/dt>\r\n<\/dl>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use Newton&#8217;s Law of Cooling.<\/li>\n<li>Use a logistic growth model.<\/li>\n<\/ul>\n<\/div>\n<h2>Bounded Exponential Growth and Decay<\/h2>\n<h3>Using Newton\u2019s Law of Cooling<\/h3>\n<p>Exponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object\u2019s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a <strong>vertical shift<\/strong> of the generic <strong>exponential decay function<\/strong>. This translation leads to <strong>Newton\u2019s Law of Cooling<\/strong>, the scientific formula for temperature as a function of time as an object\u2019s temperature is equalized with the ambient temperature.<\/p>\n<p>The formula is derived as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}T\\left(t\\right)=A{b}^{ct}+{T}_{s}\\hfill & \\hfill \\\\ T\\left(t\\right)=A{e}^{\\mathrm{ln}\\left({b}^{ct}\\right)}+{T}_{s}\\hfill & \\text{Properties of logarithms}.\\hfill \\\\ T\\left(t\\right)=A{e}^{ct\\mathrm{ln}b}+{T}_{s}\\hfill & \\text{Properties of logarithms}.\\hfill \\\\ T\\left(t\\right)=A{e}^{kt}+{T}_{s}\\hfill & \\text{Rename the constant }c \\mathrm{ln} b,\\text{ calling it }k.\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Newton\u2019s Law of Cooling<\/h3>\n<p>The temperature of an object, <em>T<\/em>, in surrounding air with temperature [latex]{T}_{s}[\/latex] will behave according to the formula<\/p>\n<p>[latex]T\\left(t\\right)=A{e}^{kt}+{T}_{s}[\/latex]<br \/>\nwhere<\/p>\n<ul>\n<li><em>t<\/em>\u00a0is time<\/li>\n<li><em>A<\/em>\u00a0is the difference between the initial temperature of the object and the surroundings<\/li>\n<li><em>k<\/em>\u00a0is a constant, the continuous rate of cooling of the object<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a set of conditions, apply Newton\u2019s Law of Cooling<\/h3>\n<ol>\n<li>Set [latex]{T}_{s}[\/latex] equal to the <em>y<\/em>-coordinate of the horizontal asymptote (usually the ambient temperature).<\/li>\n<li>Substitute the given values into the continuous growth formula [latex]T\\left(t\\right)=A{e}^{k}{}^{t}+{T}_{s}[\/latex] to find the parameters <em>A<\/em>\u00a0and <em>k<\/em>.<\/li>\n<li>Substitute in the desired time to find the temperature or the desired temperature to find the time.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using Newton\u2019s Law of Cooling<\/h3>\n<p>A cheesecake is taken out of the oven with an ideal internal temperature of [latex]165^\\circ\\text{F}[\/latex] and is placed into a [latex]35^\\circ\\text{F}[\/latex] refrigerator. After 10 minutes, the cheesecake has cooled to [latex]150^\\circ\\text{F}[\/latex]. If we must wait until the cheesecake has cooled to [latex]70^\\circ\\text{F}[\/latex] before we eat it, how long will we have to wait?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q705928\">Show Solution<\/span><\/p>\n<div id=\"q705928\" class=\"hidden-answer\" style=\"display: none\">\n<p>Because the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake\u2019s temperature will decay exponentially toward 35, following the equation<\/p>\n<p style=\"text-align: center;\">[latex]T\\left(t\\right)=A{e}^{kt}+35[\/latex]<\/p>\n<p>We know the initial temperature was 165, so [latex]T\\left(0\\right)=165[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}165=A{e}^{k0}+35\\hfill & \\text{Substitute }\\left(0,165\\right).\\hfill \\\\ A=130\\hfill & \\text{Solve for }A.\\hfill \\end{array}[\/latex]<\/p>\n<p>We were given another data point, [latex]T\\left(10\\right)=150[\/latex], which we can use to solve for <em>k<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }150=130{e}^{k10}+35\\hfill & \\text{Substitute (10, 150)}.\\hfill \\\\ \\text{ }115=130{e}^{k10}\\hfill & \\text{Subtract 35 from both sides}.\\hfill \\\\ \\text{ }\\frac{115}{130}={e}^{10k}\\hfill & \\text{Divide both sides by 130}.\\hfill \\\\ \\text{ }\\mathrm{ln}\\left(\\frac{115}{130}\\right)=10k\\hfill & \\text{Take the natural log of both sides}.\\hfill \\\\ \\text{ }k=\\frac{\\mathrm{ln}\\left(\\frac{115}{130}\\right)}{10}=-0.0123\\hfill & \\text{Divide both sides by the coefficient of }k.\\hfill \\end{array}[\/latex]<\/p>\n<p>This gives us the equation for the cooling of the cheesecake: [latex]T\\left(t\\right)=130{e}^{-0.0123t}+35[\/latex].<\/p>\n<p>Now we can solve for the time it will take for the temperature to cool to 70 degrees.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}70=130{e}^{-0.0123t}+35\\hfill & \\text{Substitute in 70 for }T\\left(t\\right).\\hfill \\\\ 35=130{e}^{-0.0123t}\\hfill & \\text{Subtract 35 from both sides}.\\hfill \\\\ \\frac{35}{130}={e}^{-0.0123t}\\hfill & \\text{Divide both sides by 130}.\\hfill \\\\ \\mathrm{ln}\\left(\\frac{35}{130}\\right)=-0.0123t\\hfill & \\text{Take the natural log of both sides}.\\hfill \\\\ t=\\frac{\\mathrm{ln}\\left(\\frac{35}{130}\\right)}{-0.0123}\\approx 106.68\\hfill & \\text{Divide both sides by the coefficient of }t.\\hfill \\end{array}[\/latex]<\/p>\n<p>It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to [latex]70^\\circ\\text{F}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q846066\">Show Solution<\/span><\/p>\n<div id=\"q846066\" class=\"hidden-answer\" style=\"display: none\">6.026 hours<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom150\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=114392&amp;theme=oea&amp;iframe_resize_id=mom150\" width=\"100%\" height=\"300\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<h2>Logistic Models<\/h2>\n<p>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 although the exponential growth model is still useful over a short term before approaching the limiting value.<\/p>\n<p>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<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/p>\n<p>The graph below shows 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 the rate of increase decreases.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181346\/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\" \/><\/p>\n<div class=\"textbox\">\n<h3>A General Note: Logistic Growth<\/h3>\n<p>The logistic growth model is<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/p>\n<p>where<\/p>\n<ul>\n<li>[latex]\\frac{c}{1+a}[\/latex] is the initial value<\/li>\n<li><em>c<\/em>\u00a0is the <em>carrying capacity\u00a0<\/em>or <em>limiting value<\/em><\/li>\n<li><em>b<\/em>\u00a0is a constant determined by the rate of growth.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Logistic-Growth Model<\/h3>\n<p>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 good for studying the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.<\/p>\n<p>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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q346660\">Show Solution<\/span><\/p>\n<div id=\"q346660\" class=\"hidden-answer\" style=\"display: none\">\n<p>We substitute the given data into the logistic growth model<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/p>\n<p>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<h4>Analysis of the Solution<\/h4>\n<p>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>The graph below gives a good picture of how this model fits the data.<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181349\/CNX_Precalc_Figure_04_07_0072.jpg\" alt=\"Graph of f(x)=1000\/(1+999e^(-0.5030x)) with the y-axis labeled as\" width=\"731\" height=\"492\" \/><\/p>\n<p class=\"wp-caption-text\">The graph of [latex]f\\left(x\\right)=\\frac{1000}{1+999{e}^{-0.6030x}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Using the model in the previous example, estimate the number of cases of flu on day 15.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q421768\">Show Solution<\/span><\/p>\n<div id=\"q421768\" class=\"hidden-answer\" style=\"display: none\">\n<p>895 cases on day 15<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom150\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5801&amp;theme=oea&amp;iframe_resize_id=mom150\" width=\"100%\" height=\"300\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<h2>Key Equations<\/h2>\n<table style=\"height: 138px; width: 736px;\" summary=\"..\">\n<tbody>\n<tr style=\"height: 30px;\">\n<td style=\"height: 30px; width: 90.0469px;\">Newton&#8217;s Law of Cooling<\/td>\n<td style=\"height: 30px; width: 619.953px;\">[latex]T(t)={A}_{0}{e}^{kt}+T_s[\/latex], where [latex]T_s[\/latex] is the ambient temperature,\u00a0[latex]A=T(0)-T_s[\/latex], and\u00a0[latex]k[\/latex] is the continuous rate of cooling.<\/td>\n<\/tr>\n<tr style=\"height: 28px;\">\n<td style=\"width: 90.0469px; height: 28px;\">Logistic Growth Model<\/td>\n<td style=\"width: 619.953px; height: 28px;\">[latex]f(x)=\\frac{c}{1+ae^{-bx}}[\/latex], where [latex]\\frac{c}{1+a}[\/latex] is the initial value, <em style=\"font-size: inherit;\">c<\/em><span style=\"font-size: inherit;\">\u00a0is the carrying capacity, or limiting value, and <\/span><em style=\"font-size: inherit;\">b<\/em><span style=\"font-size: inherit;\">\u00a0is a constant determined by the rate of growth<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><span style=\"color: #077fab; font-size: 1.15em; font-weight: 600;\">Key Concepts<\/span><\/p>\n<ul>\n<li>We can use Newton\u2019s Law of Cooling to find how long it will take for a cooling object to reach a desired temperature or to find what temperature an object will be after a given time.<\/li>\n<li>We can use logistic growth functions to model real-world situations where the rate of growth changes over time, such as population growth, spread of disease, and spread of rumors.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165137838635\" class=\"definition\">\n<dt><strong>carrying capacity<\/strong><\/dt>\n<dd id=\"fs-id1165137838640\">in a logistic model, the limiting value of the output<\/dd>\n<\/dl>\n<dl class=\"definition\">\n<dt>\n<\/dt>\n<\/dl>\n<dl class=\"definition\">\n<dt>\n<\/dt>\n<dt>\n<\/dt>\n<dt>\n<\/dt>\n<dt><strong>logistic growth model<\/strong><\/dt>\n<dd id=\"fs-id1165137838640\">a function of the form [latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex] where [latex]\\frac{c}{1+a}[\/latex] is the initial value, <em>c<\/em>\u00a0is the carrying capacity, or limiting value, and <em>b<\/em>\u00a0is a constant determined by the rate of growth<\/dd>\n<\/dl>\n<p> \t<strong>Newton\u2019s Law of Cooling<\/strong><br \/>\n \tthe scientific formula for temperature as a function of time as an object\u2019s temperature is equalized with the ambient temperature<\/p>\n","protected":false},"author":756952,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2316","chapter","type-chapter","status-publish","hentry"],"part":1875,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapters\/2316","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/wp\/v2\/users\/756952"}],"version-history":[{"count":8,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapters\/2316\/revisions"}],"predecessor-version":[{"id":2337,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapters\/2316\/revisions\/2337"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/parts\/1875"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapters\/2316\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/wp\/v2\/media?parent=2316"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=2316"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/wp\/v2\/contributor?post=2316"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/wp\/v2\/license?post=2316"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}