{"id":489,"date":"2025-02-13T19:45:25","date_gmt":"2025-02-13T19:45:25","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/exponential-growth-and-decay-learn-it-1\/"},"modified":"2025-02-13T19:45:25","modified_gmt":"2025-02-13T19:45:25","slug":"exponential-growth-and-decay-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/exponential-growth-and-decay-learn-it-1\/","title":{"raw":"Exponential Growth and Decay: Learn It 1","rendered":"Exponential Growth and Decay: Learn It 1"},"content":{"raw":"\n<section class=\"textbox learningGoals\">\n<ul>\n\t<li>Apply the exponential growth formula to real-world cases like increasing populations or investments<\/li>\n\t<li>Describe how long it takes for quantities to double or reduce by half<\/li>\n\t<li>Implement the exponential decay formula for scenarios like radioactive substances decaying or objects cooling down<\/li>\n<\/ul>\n<\/section>\n<h2>Exponential Growth Model<\/h2>\n<p id=\"fs-id1167794066685\">Many systems exhibit <strong>exponential growth<\/strong>. These systems follow a model of the form [latex]y={y}_{0}{e}^{kt},[\/latex] where [latex]{y}_{0}[\/latex] represents the initial state of the system and [latex]k[\/latex] is a positive constant, called the <strong>growth constant<\/strong>.<\/p>\n<p>Notice that in an exponential growth model, we have<\/p>\n<div id=\"fs-id1167793543548\" class=\"equation\" style=\"text-align: center;\">[latex]{y}^{\\prime }=k{y}_{0}{e}^{kt}=ky[\/latex]<\/div>\n<p id=\"fs-id1167793940940\">That is, the rate of growth is proportional to the current function value. This is a key feature of exponential growth.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>exponential growth model<\/h3>\n<p id=\"fs-id1167793936451\">Systems that exhibit exponential growth increase according to the mathematical model<\/p>\n<div id=\"fs-id1167793951818\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y={y}_{0}{e}^{kt},[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793244434\">where [latex]{y}_{0}[\/latex] represents the initial state of the system and [latex]k&gt;0[\/latex] is a constant, called the growth constant.<\/p>\n<\/section>\n<h3>Population Growth<\/h3>\n<p id=\"fs-id1167793392254\"><span class=\"no-emphasis\">Population growth<\/span> is a common example of exponential growth.<\/p>\n<section class=\"textbox example\">\n<p>Consider a population of bacteria, for instance. It seems plausible that the rate of population growth would be proportional to the size of the population. After all, the more bacteria there are to reproduce, the faster the population grows.<\/p>\n<p>Figure 1 and the table below represent the growth of a population of bacteria with an initial population of [latex]200[\/latex] bacteria and a growth constant of [latex]0.02[\/latex]. Notice that after only [latex]2[\/latex] hours [latex](120[\/latex] minutes), the population is [latex]10[\/latex] times its original size!<\/p>\n\n[caption id=\"\" align=\"alignright\" width=\"293\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213335\/CNX_Calc_Figure_06_08_001.jpg\" alt=\"This figure is a graph. It is the exponential curve for y=200e^0.02t. It is in the first quadrant and an increasing function. It begins on the y-axis.\" width=\"293\" height=\"353\"> Figure 1. An example of exponential growth for bacteria.[\/caption]\n\n<table style=\"width: 50%;\">\n<caption>Exponential Growth of a Bacterial Population<\/caption>\n<thead>\n<tr valign=\"top\">\n<th style=\"text-align: center;\"><strong>Time (min)<\/strong><\/th>\n<th style=\"text-align: center;\"><strong>Population Size (no. of bacteria)<\/strong><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">10<\/td>\n<td style=\"text-align: center;\">244<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">20<\/td>\n<td style=\"text-align: center;\">298<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">30<\/td>\n<td style=\"text-align: center;\">364<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">40<\/td>\n<td style=\"text-align: center;\">445<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">50<\/td>\n<td style=\"text-align: center;\">544<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">60<\/td>\n<td style=\"text-align: center;\">664<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">70<\/td>\n<td style=\"text-align: center;\">811<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">80<\/td>\n<td style=\"text-align: center;\">991<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">90<\/td>\n<td style=\"text-align: center;\">1210<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">100<\/td>\n<td style=\"text-align: center;\">1478<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">110<\/td>\n<td style=\"text-align: center;\">1805<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">120<\/td>\n<td style=\"text-align: center;\">2205<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/section>\n<section class=\"textbox proTip\">\n<p>Note that we are using a continuous function to model what is inherently discrete behavior. At any given time, the real-world population contains a whole number of bacteria, although the model takes on noninteger values. When using exponential growth models, we must always be careful to interpret the function values in the context of the phenomenon we are modeling.<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Consider the population of bacteria described earlier. This population grows according to the function [latex]f(t)=200{e}^{0.02t},[\/latex] where [latex]t[\/latex] is measured in minutes. How many bacteria are present in the population after 5 hours [latex](300[\/latex] minutes)? When does the population reach 100,000 bacteria?<\/p>\n\n[reveal-answer q=\"fs-id1167794170510\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167794170510\"]\n\n<p id=\"fs-id1167794170510\">We have [latex]f(t)=200{e}^{0.02t}.[\/latex] Then<\/p>\n<div id=\"fs-id1167793375750\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(300)=200{e}^{0.02(300)}\\approx 80,686.[\/latex]<\/div>\n<p id=\"fs-id1167793814662\">There are 80,686 bacteria in the population after 5 hours.<\/p>\n<p id=\"fs-id1167793607551\">To find when the population reaches 100,000 bacteria, we solve the equation<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill 100,000&amp; =\\hfill &amp; 200{e}^{0.02t}\\hfill \\\\ \\hfill 500&amp; =\\hfill &amp; {e}^{0.02t}\\hfill \\\\ \\hfill \\text{ln}500&amp; =\\hfill &amp; 0.02t\\hfill \\\\ \\hfill t&amp; =\\hfill &amp; \\frac{\\text{ln}500}{0.02}\\approx 310.73.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167794186931\">The population reaches 100,000 bacteria after 310.73 minutes.<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<p>Let\u2019s now turn our attention to a financial application: compound interest.<\/p>\n<h3>Compound Interest<\/h3>\n<p>Interest that is not compounded is called <span class=\"no-emphasis\" style=\"font-size: 16px; font-weight: 400;\"><em>simple interest<\/em><\/span><span style=\"font-size: 16px; font-weight: 400;\">. Simple interest is paid once, at the end of the specified time period (usually 1 year). <\/span><\/p>\n<section class=\"textbox example\">\n<p><span style=\"font-size: 16px; font-weight: 400;\">If we put [latex]$1000[\/latex] in a savings account earning [latex]2\\%[\/latex] simple interest per year, then at the end of the year we have,<\/span><\/p>\n<div id=\"fs-id1167793294235\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]1000(1+0.02)=$1020.[\/latex]<\/div>\n<\/section>\n<p id=\"fs-id1167793414250\">Compound interest is paid multiple times per year, depending on the compounding period.<\/p>\n<section class=\"textbox example\">\n<p>If a bank compounds the interest every [latex]6[\/latex] months, it credits half of the year\u2019s interest to the account after [latex]6[\/latex] months.<\/p>\n<p>During the second half of the year, the account earns interest not only on the initial [latex]$1000,[\/latex] but also on the interest earned during the first half of the year. Mathematically speaking, at the end of the year, we have<\/p>\n<div id=\"fs-id1167794188340\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]1000{(1+\\frac{0.02}{2})}^{2}=$1020.10.[\/latex]<\/div>\n<p id=\"fs-id1167794136411\">Similarly, if the interest is compounded every [latex]4[\/latex] months, we have<\/p>\n<div id=\"fs-id1167794337261\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]1000{(1+\\frac{0.02}{3})}^{3}=$1020.13,[\/latex]<\/div>\n<p id=\"fs-id1167794002738\">and if the interest is compounded daily [latex](365[\/latex] times per year), we have [latex]$1020.20.[\/latex]<\/p>\n<p>If we extend this concept, so that the interest is compounded continuously, after [latex]t[\/latex] years we have<\/p>\n<div id=\"fs-id1167794059054\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]1000\\underset{n\\to \\infty }{\\text{lim}}{(1+\\frac{0.02}{n})}^{nt}.[\/latex]<\/div>\n<div>\n<p id=\"fs-id1167794172746\">Now let\u2019s manipulate this expression so that we have an exponential growth function. Recall that the number [latex]e[\/latex] can be expressed as a limit:<\/p>\n<div id=\"fs-id1167794075520\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e=\\underset{m\\to \\infty }{\\text{lim}}{(1+\\frac{1}{m})}^{m}.[\/latex]<\/div>\n<p id=\"fs-id1167793281296\">Based on this, we want the expression inside the parentheses to have the form [latex](1+1\\text{\/}m).[\/latex] Let [latex]n=0.02m.[\/latex] Note that as [latex]n\\to \\infty ,[\/latex] [latex]m\\to \\infty [\/latex] as well. Then we get<\/p>\n<div id=\"fs-id1167794119172\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]1000\\underset{n\\to \\infty }{\\text{lim}}{(1+\\frac{0.02}{n})}^{nt}=1000\\underset{m\\to \\infty }{\\text{lim}}{(1+\\frac{0.02}{0.02m})}^{0.02mt}=1000{\\left[\\underset{m\\to \\infty }{\\text{lim}}{(1+\\frac{1}{m})}^{m}\\right]}^{0.02t}.[\/latex]<\/div>\n<p id=\"fs-id1167793444559\">We recognize the limit inside the brackets as the number [latex]e.[\/latex] So, the balance in our bank account after [latex]t[\/latex] years is given by [latex]1000{e}^{0.02t}.[\/latex]<\/p>\n<div id=\"fs-id1167793257856\" class=\"equation unnumbered\" style=\"text-align: center;\">&nbsp;<\/div>\n<\/div>\n<\/section>\n<p>Generalizing this concept, we see that if a bank account with an initial balance of [latex]$P[\/latex] earns interest at a rate of [latex]r\\text{%},[\/latex] compounded continuously, then the balance of the account after [latex]t[\/latex] years is<\/p>\n<div id=\"fs-id1167793257856\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Balance}=P{e}^{rt}.[\/latex]<\/div>\n<div>\n<section class=\"textbox example\">\n<p>A 25-year-old student is offered an opportunity to invest some money in a retirement account that pays [latex]5\\%[\/latex] annual interest compounded continuously. How much does the student need to invest today to have [latex]$1[\/latex] million when she retires at age [latex]65?[\/latex]<\/p>\n<p>What if she could earn [latex]6\\%[\/latex] annual interest compounded continuously instead?<\/p>\n\n[reveal-answer q=\"fs-id1167793245842\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793245842\"]\n\n<p id=\"fs-id1167793245842\">We have<\/p>\n<div id=\"fs-id1167793245845\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill 1,000,000&amp; =\\hfill &amp; P{e}^{0.05(40)}\\hfill \\\\ \\hfill P&amp; =\\hfill &amp; 135,335.28.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793832121\">She must invest [latex]$135,335.28[\/latex] at [latex]5\\%[\/latex] interest.<\/p>\n<p id=\"fs-id1167793883723\">If, instead, she is able to earn [latex]6\\text{%},[\/latex] then the equation becomes<\/p>\n<div id=\"fs-id1167793952541\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill 1,000,000&amp; =\\hfill &amp; P{e}^{0.06(40)}\\hfill \\\\ \\hfill P&amp; =\\hfill &amp; 90,717.95.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793956677\">In this case, she needs to invest only [latex]$90,717.95.[\/latex] This is roughly two-thirds the amount she needs to invest at [latex]5\\text{%}.[\/latex] The fact that the interest is compounded continuously greatly magnifies the effect of the 1% increase in interest rate.<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox tryIt\">\n<p>[ohm_question hide_question_numbers=1]64715[\/ohm_question]<\/p>\n<\/section>\n<h3>Doubling Time<\/h3>\n<p id=\"fs-id1167793977352\">If a quantity grows exponentially, the time it takes for the quantity to double remains constant. In other words, it takes the same amount of time for a population of bacteria to grow from [latex]100[\/latex] to [latex] 200[\/latex] bacteria as it does to grow from [latex]10,000[\/latex] to [latex]20,000[\/latex] bacteria. This time is called the <strong>doubling time<\/strong>.<\/p>\n<p>To calculate the doubling time, we want to know when the quantity reaches twice its original size. So we have<\/p>\n<div id=\"fs-id1167793719089\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill 2{y}_{0}&amp; =\\hfill &amp; {y}_{0}{e}^{kt}\\hfill \\\\ \\hfill 2&amp; =\\hfill &amp; {e}^{kt}\\hfill \\\\ \\hfill \\text{ln}2&amp; =\\hfill &amp; kt\\hfill \\\\ \\hfill t&amp; =\\hfill &amp; \\frac{\\text{ln}2}{k}.\\hfill \\end{array}[\/latex]<\/div>\n<section class=\"textbox keyTakeaway\">\n<h3>doubling time<\/h3>\n<p id=\"fs-id1167794171070\">If a quantity grows exponentially, the doubling time is the amount of time it takes the quantity to double. It is given by<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Doubling time}=\\frac{\\text{ln}2}{k}.[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Assume a population of fish grows exponentially. A pond is stocked initially with [latex]500[\/latex] fish. After [latex]6[\/latex] months, there are [latex]1000[\/latex] fish in the pond. The owner will allow his friends and neighbors to fish on his pond after the fish population reaches [latex]10,000[\/latex]. When will the owner\u2019s friends be allowed to fish?<\/p>\n\n[reveal-answer q=\"fs-id1167793367799\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793367799\"]\n\n<p id=\"fs-id1167793367799\">We know it takes the population of fish [latex]6[\/latex] months to double in size. So, if [latex]t[\/latex] represents time in months, by the doubling-time formula, we have [latex]6=(\\text{ln}2)\\text{\/}k.[\/latex] Then, [latex]k=(\\text{ln}2)\\text{\/}6.[\/latex] Thus, the population is given by [latex]y=500{e}^{((\\text{ln}2)\\text{\/}6)t}.[\/latex] To figure out when the population reaches [latex]10,000[\/latex] fish, we must solve the following equation:<\/p>\n<div id=\"fs-id1167793937195\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill 10,000&amp; =\\hfill &amp; 500{e}^{(\\text{ln}2\\text{\/}6)t}\\hfill \\\\ \\hfill 20&amp; =\\hfill &amp; {e}^{(\\text{ln}2\\text{\/}6)t}\\hfill \\\\ \\hfill \\text{ln}20&amp; =\\hfill &amp; (\\frac{\\text{ln}2}{6})t\\hfill \\\\ \\hfill t&amp; =\\hfill &amp; \\frac{6(\\text{ln}20)}{\\text{ln}2}\\approx 25.93.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793495113\">The owner\u2019s friends have to wait [latex]25.93[\/latex] months (a little more than [latex]2[\/latex] years) to fish in the pond.<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox tryIt\">\n<p>[ohm_question hide_question_numbers=1]100768[\/ohm_question]<\/p>\n<\/section>\n<\/div>\n","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Apply the exponential growth formula to real-world cases like increasing populations or investments<\/li>\n<li>Describe how long it takes for quantities to double or reduce by half<\/li>\n<li>Implement the exponential decay formula for scenarios like radioactive substances decaying or objects cooling down<\/li>\n<\/ul>\n<\/section>\n<h2>Exponential Growth Model<\/h2>\n<p id=\"fs-id1167794066685\">Many systems exhibit <strong>exponential growth<\/strong>. These systems follow a model of the form [latex]y={y}_{0}{e}^{kt},[\/latex] where [latex]{y}_{0}[\/latex] represents the initial state of the system and [latex]k[\/latex] is a positive constant, called the <strong>growth constant<\/strong>.<\/p>\n<p>Notice that in an exponential growth model, we have<\/p>\n<div id=\"fs-id1167793543548\" class=\"equation\" style=\"text-align: center;\">[latex]{y}^{\\prime }=k{y}_{0}{e}^{kt}=ky[\/latex]<\/div>\n<p id=\"fs-id1167793940940\">That is, the rate of growth is proportional to the current function value. This is a key feature of exponential growth.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>exponential growth model<\/h3>\n<p id=\"fs-id1167793936451\">Systems that exhibit exponential growth increase according to the mathematical model<\/p>\n<div id=\"fs-id1167793951818\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y={y}_{0}{e}^{kt},[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793244434\">where [latex]{y}_{0}[\/latex] represents the initial state of the system and [latex]k>0[\/latex] is a constant, called the growth constant.<\/p>\n<\/section>\n<h3>Population Growth<\/h3>\n<p id=\"fs-id1167793392254\"><span class=\"no-emphasis\">Population growth<\/span> is a common example of exponential growth.<\/p>\n<section class=\"textbox example\">\n<p>Consider a population of bacteria, for instance. It seems plausible that the rate of population growth would be proportional to the size of the population. After all, the more bacteria there are to reproduce, the faster the population grows.<\/p>\n<p>Figure 1 and the table below represent the growth of a population of bacteria with an initial population of [latex]200[\/latex] bacteria and a growth constant of [latex]0.02[\/latex]. Notice that after only [latex]2[\/latex] hours [latex](120[\/latex] minutes), the population is [latex]10[\/latex] times its original size!<\/p>\n<figure style=\"width: 293px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213335\/CNX_Calc_Figure_06_08_001.jpg\" alt=\"This figure is a graph. It is the exponential curve for y=200e^0.02t. It is in the first quadrant and an increasing function. It begins on the y-axis.\" width=\"293\" height=\"353\" \/><figcaption class=\"wp-caption-text\">Figure 1. An example of exponential growth for bacteria.<\/figcaption><\/figure>\n<table style=\"width: 50%;\">\n<caption>Exponential Growth of a Bacterial Population<\/caption>\n<thead>\n<tr valign=\"top\">\n<th style=\"text-align: center;\"><strong>Time (min)<\/strong><\/th>\n<th style=\"text-align: center;\"><strong>Population Size (no. of bacteria)<\/strong><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">10<\/td>\n<td style=\"text-align: center;\">244<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">20<\/td>\n<td style=\"text-align: center;\">298<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">30<\/td>\n<td style=\"text-align: center;\">364<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">40<\/td>\n<td style=\"text-align: center;\">445<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">50<\/td>\n<td style=\"text-align: center;\">544<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">60<\/td>\n<td style=\"text-align: center;\">664<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">70<\/td>\n<td style=\"text-align: center;\">811<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">80<\/td>\n<td style=\"text-align: center;\">991<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">90<\/td>\n<td style=\"text-align: center;\">1210<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">100<\/td>\n<td style=\"text-align: center;\">1478<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">110<\/td>\n<td style=\"text-align: center;\">1805<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center;\">120<\/td>\n<td style=\"text-align: center;\">2205<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/section>\n<section class=\"textbox proTip\">\n<p>Note that we are using a continuous function to model what is inherently discrete behavior. At any given time, the real-world population contains a whole number of bacteria, although the model takes on noninteger values. When using exponential growth models, we must always be careful to interpret the function values in the context of the phenomenon we are modeling.<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Consider the population of bacteria described earlier. This population grows according to the function [latex]f(t)=200{e}^{0.02t},[\/latex] where [latex]t[\/latex] is measured in minutes. How many bacteria are present in the population after 5 hours [latex](300[\/latex] minutes)? When does the population reach 100,000 bacteria?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794170510\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167794170510\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794170510\">We have [latex]f(t)=200{e}^{0.02t}.[\/latex] Then<\/p>\n<div id=\"fs-id1167793375750\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(300)=200{e}^{0.02(300)}\\approx 80,686.[\/latex]<\/div>\n<p id=\"fs-id1167793814662\">There are 80,686 bacteria in the population after 5 hours.<\/p>\n<p id=\"fs-id1167793607551\">To find when the population reaches 100,000 bacteria, we solve the equation<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill 100,000& =\\hfill & 200{e}^{0.02t}\\hfill \\\\ \\hfill 500& =\\hfill & {e}^{0.02t}\\hfill \\\\ \\hfill \\text{ln}500& =\\hfill & 0.02t\\hfill \\\\ \\hfill t& =\\hfill & \\frac{\\text{ln}500}{0.02}\\approx 310.73.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167794186931\">The population reaches 100,000 bacteria after 310.73 minutes.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<p>Let\u2019s now turn our attention to a financial application: compound interest.<\/p>\n<h3>Compound Interest<\/h3>\n<p>Interest that is not compounded is called <span class=\"no-emphasis\" style=\"font-size: 16px; font-weight: 400;\"><em>simple interest<\/em><\/span><span style=\"font-size: 16px; font-weight: 400;\">. Simple interest is paid once, at the end of the specified time period (usually 1 year). <\/span><\/p>\n<section class=\"textbox example\">\n<p><span style=\"font-size: 16px; font-weight: 400;\">If we put [latex]$1000[\/latex] in a savings account earning [latex]2\\%[\/latex] simple interest per year, then at the end of the year we have,<\/span><\/p>\n<div id=\"fs-id1167793294235\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]1000(1+0.02)=$1020.[\/latex]<\/div>\n<\/section>\n<p id=\"fs-id1167793414250\">Compound interest is paid multiple times per year, depending on the compounding period.<\/p>\n<section class=\"textbox example\">\n<p>If a bank compounds the interest every [latex]6[\/latex] months, it credits half of the year\u2019s interest to the account after [latex]6[\/latex] months.<\/p>\n<p>During the second half of the year, the account earns interest not only on the initial [latex]$1000,[\/latex] but also on the interest earned during the first half of the year. Mathematically speaking, at the end of the year, we have<\/p>\n<div id=\"fs-id1167794188340\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]1000{(1+\\frac{0.02}{2})}^{2}=$1020.10.[\/latex]<\/div>\n<p id=\"fs-id1167794136411\">Similarly, if the interest is compounded every [latex]4[\/latex] months, we have<\/p>\n<div id=\"fs-id1167794337261\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]1000{(1+\\frac{0.02}{3})}^{3}=$1020.13,[\/latex]<\/div>\n<p id=\"fs-id1167794002738\">and if the interest is compounded daily [latex](365[\/latex] times per year), we have [latex]$1020.20.[\/latex]<\/p>\n<p>If we extend this concept, so that the interest is compounded continuously, after [latex]t[\/latex] years we have<\/p>\n<div id=\"fs-id1167794059054\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]1000\\underset{n\\to \\infty }{\\text{lim}}{(1+\\frac{0.02}{n})}^{nt}.[\/latex]<\/div>\n<div>\n<p id=\"fs-id1167794172746\">Now let\u2019s manipulate this expression so that we have an exponential growth function. Recall that the number [latex]e[\/latex] can be expressed as a limit:<\/p>\n<div id=\"fs-id1167794075520\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e=\\underset{m\\to \\infty }{\\text{lim}}{(1+\\frac{1}{m})}^{m}.[\/latex]<\/div>\n<p id=\"fs-id1167793281296\">Based on this, we want the expression inside the parentheses to have the form [latex](1+1\\text{\/}m).[\/latex] Let [latex]n=0.02m.[\/latex] Note that as [latex]n\\to \\infty ,[\/latex] [latex]m\\to \\infty[\/latex] as well. Then we get<\/p>\n<div id=\"fs-id1167794119172\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]1000\\underset{n\\to \\infty }{\\text{lim}}{(1+\\frac{0.02}{n})}^{nt}=1000\\underset{m\\to \\infty }{\\text{lim}}{(1+\\frac{0.02}{0.02m})}^{0.02mt}=1000{\\left[\\underset{m\\to \\infty }{\\text{lim}}{(1+\\frac{1}{m})}^{m}\\right]}^{0.02t}.[\/latex]<\/div>\n<p id=\"fs-id1167793444559\">We recognize the limit inside the brackets as the number [latex]e.[\/latex] So, the balance in our bank account after [latex]t[\/latex] years is given by [latex]1000{e}^{0.02t}.[\/latex]<\/p>\n<div id=\"fs-id1167793257856\" class=\"equation unnumbered\" style=\"text-align: center;\">&nbsp;<\/div>\n<\/div>\n<\/section>\n<p>Generalizing this concept, we see that if a bank account with an initial balance of [latex]$P[\/latex] earns interest at a rate of [latex]r\\text{%},[\/latex] compounded continuously, then the balance of the account after [latex]t[\/latex] years is<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Balance}=P{e}^{rt}.[\/latex]<\/div>\n<div>\n<section class=\"textbox example\">\n<p>A 25-year-old student is offered an opportunity to invest some money in a retirement account that pays [latex]5\\%[\/latex] annual interest compounded continuously. How much does the student need to invest today to have [latex]$1[\/latex] million when she retires at age [latex]65?[\/latex]<\/p>\n<p>What if she could earn [latex]6\\%[\/latex] annual interest compounded continuously instead?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793245842\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793245842\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793245842\">We have<\/p>\n<div id=\"fs-id1167793245845\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill 1,000,000& =\\hfill & P{e}^{0.05(40)}\\hfill \\\\ \\hfill P& =\\hfill & 135,335.28.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793832121\">She must invest [latex]$135,335.28[\/latex] at [latex]5\\%[\/latex] interest.<\/p>\n<p id=\"fs-id1167793883723\">If, instead, she is able to earn [latex]6\\text{%},[\/latex] then the equation becomes<\/p>\n<div id=\"fs-id1167793952541\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill 1,000,000& =\\hfill & P{e}^{0.06(40)}\\hfill \\\\ \\hfill P& =\\hfill & 90,717.95.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793956677\">In this case, she needs to invest only [latex]$90,717.95.[\/latex] This is roughly two-thirds the amount she needs to invest at [latex]5\\text{%}.[\/latex] The fact that the interest is compounded continuously greatly magnifies the effect of the 1% increase in interest rate.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm64715\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=64715&theme=lumen&iframe_resize_id=ohm64715&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<h3>Doubling Time<\/h3>\n<p id=\"fs-id1167793977352\">If a quantity grows exponentially, the time it takes for the quantity to double remains constant. In other words, it takes the same amount of time for a population of bacteria to grow from [latex]100[\/latex] to [latex]200[\/latex] bacteria as it does to grow from [latex]10,000[\/latex] to [latex]20,000[\/latex] bacteria. This time is called the <strong>doubling time<\/strong>.<\/p>\n<p>To calculate the doubling time, we want to know when the quantity reaches twice its original size. So we have<\/p>\n<div id=\"fs-id1167793719089\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill 2{y}_{0}& =\\hfill & {y}_{0}{e}^{kt}\\hfill \\\\ \\hfill 2& =\\hfill & {e}^{kt}\\hfill \\\\ \\hfill \\text{ln}2& =\\hfill & kt\\hfill \\\\ \\hfill t& =\\hfill & \\frac{\\text{ln}2}{k}.\\hfill \\end{array}[\/latex]<\/div>\n<section class=\"textbox keyTakeaway\">\n<h3>doubling time<\/h3>\n<p id=\"fs-id1167794171070\">If a quantity grows exponentially, the doubling time is the amount of time it takes the quantity to double. It is given by<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Doubling time}=\\frac{\\text{ln}2}{k}.[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Assume a population of fish grows exponentially. A pond is stocked initially with [latex]500[\/latex] fish. After [latex]6[\/latex] months, there are [latex]1000[\/latex] fish in the pond. The owner will allow his friends and neighbors to fish on his pond after the fish population reaches [latex]10,000[\/latex]. When will the owner\u2019s friends be allowed to fish?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793367799\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793367799\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793367799\">We know it takes the population of fish [latex]6[\/latex] months to double in size. So, if [latex]t[\/latex] represents time in months, by the doubling-time formula, we have [latex]6=(\\text{ln}2)\\text{\/}k.[\/latex] Then, [latex]k=(\\text{ln}2)\\text{\/}6.[\/latex] Thus, the population is given by [latex]y=500{e}^{((\\text{ln}2)\\text{\/}6)t}.[\/latex] To figure out when the population reaches [latex]10,000[\/latex] fish, we must solve the following equation:<\/p>\n<div id=\"fs-id1167793937195\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill 10,000& =\\hfill & 500{e}^{(\\text{ln}2\\text{\/}6)t}\\hfill \\\\ \\hfill 20& =\\hfill & {e}^{(\\text{ln}2\\text{\/}6)t}\\hfill \\\\ \\hfill \\text{ln}20& =\\hfill & (\\frac{\\text{ln}2}{6})t\\hfill \\\\ \\hfill t& =\\hfill & \\frac{6(\\text{ln}20)}{\\text{ln}2}\\approx 25.93.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793495113\">The owner\u2019s friends have to wait [latex]25.93[\/latex] months (a little more than [latex]2[\/latex] years) to fish in the pond.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm100768\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=100768&theme=lumen&iframe_resize_id=ohm100768&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n<\/div>\n","protected":false},"author":6,"menu_order":10,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"6.8 Try It Problems\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":479,"module-header":"","content_attributions":[{"type":"cc","description":"Calculus Volume 1","author":"Gilbert Strang, Edwin (Jed) Herman","organization":"OpenStax","url":"https:\/\/openstax.org\/details\/books\/calculus-volume-1","project":"","license":"cc-by-nc-sa","license_terms":"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction"},{"type":"original","description":"6.8 Try It Problems","author":"Ryan Melton","organization":"","url":"","project":"","license":"cc-by","license_terms":""}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/489"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":0,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/489\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/479"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/489\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=489"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=489"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=489"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=489"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}