{"id":602,"date":"2021-02-04T16:36:13","date_gmt":"2021-02-04T16:36:13","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=602"},"modified":"2022-03-16T22:53:11","modified_gmt":"2022-03-16T22:53:11","slug":"exponential-growth-model","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/exponential-growth-model\/","title":{"raw":"Exponential Growth Model","rendered":"Exponential Growth Model"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use the exponential growth model in applications, including population growth and compound interest<\/li>\r\n \t<li>Explain the concept of doubling time<\/li>\r\n<\/ul>\r\n<\/div>\r\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 <em>growth constant<\/em>. Notice that in an exponential growth model, we have<\/p>\r\n\r\n<div id=\"fs-id1167793543548\" class=\"equation\" style=\"text-align: center;\">[latex]{y}^{\\prime }=k{y}_{0}{e}^{kt}=ky[\/latex]<\/div>\r\n&nbsp;\r\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. The equation above involves derivatives and is called a <em>differential equation.<\/em> We learn more about differential equations in <a class=\"target-chapter\" href=\"https:\/\/cnx.org\/contents\/HTmjSAcf@2.46:KzA8da4p@2\/Introduction\">Introduction to Differential Equations<\/a> in the second volume of this text.<\/p>\r\n\r\n<div id=\"fs-id1167793498261\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Exponential Growth Model<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793936451\">Systems that exhibit exponential growth increase according to the mathematical model<\/p>\r\n\r\n<div id=\"fs-id1167793951818\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y={y}_{0}{e}^{kt},[\/latex]<\/div>\r\n&nbsp;\r\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 <em>growth constant<\/em>.<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793392254\"><span class=\"no-emphasis\">Population growth<\/span> is a common example of exponential growth. 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. Figure 1 and the table below represent the growth of a population of bacteria with an initial population of 200 bacteria and a growth constant of 0.02. Notice that after only 2 hours [latex](120[\/latex] minutes), the population is 10 times its original size!<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" 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]\r\n<table id=\"fs-id1167793939650\" summary=\"This table has two columns. The columns are labeled Time (minutes) and Population size (no. of bacteria). The times start at 10 and increase in increments of 10 minutes up until 120 minutes. The number of bacteria start at 244 and increase to 2205.\"><caption>Exponential Growth of a Bacterial Population<\/caption>\r\n<thead>\r\n<tr valign=\"top\">\r\n<th><strong>Time (min)<\/strong><\/th>\r\n<th><strong>Population Size (no. of bacteria)<\/strong><\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>10<\/td>\r\n<td>244<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>20<\/td>\r\n<td>298<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>30<\/td>\r\n<td>364<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>40<\/td>\r\n<td>445<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>50<\/td>\r\n<td>544<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>60<\/td>\r\n<td>664<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>70<\/td>\r\n<td>811<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>80<\/td>\r\n<td>991<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>90<\/td>\r\n<td>1210<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>100<\/td>\r\n<td>1478<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>110<\/td>\r\n<td>1805<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>120<\/td>\r\n<td>2205<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNote 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.\r\n<div id=\"fs-id1167794022966\" class=\"textbook exercises\">\r\n<h3>Example: Population Growth<\/h3>\r\nConsider 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?\r\n<div id=\"fs-id1167793957317\" class=\"exercise\">[reveal-answer q=\"fs-id1167794170510\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794170510\"]\r\n<p id=\"fs-id1167794170510\">We have [latex]f(t)=200{e}^{0.02t}.[\/latex] Then<\/p>\r\n\r\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>\r\n<p id=\"fs-id1167793814662\">There are 80,686 bacteria in the population after 5 hours.<\/p>\r\n<p id=\"fs-id1167793607551\">To find when the population reaches 100,000 bacteria, we solve the equation<\/p>\r\n\r\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>\r\n<p id=\"fs-id1167794186931\">The population reaches 100,000 bacteria after 310.73 minutes.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167794050614\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nConsider a population of bacteria that grows according to the function [latex]f(t)=500{e}^{0.05t},[\/latex] where [latex]t[\/latex] is measured in minutes. How many bacteria are present in the population after 4 hours? When does the population reach 100 million bacteria?\r\n<div id=\"fs-id1167793245191\" class=\"exercise\">[reveal-answer q=\"fs-id1167793427517\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793427517\"]\r\n<p id=\"fs-id1167793427517\">There are 81,377,396 bacteria in the population after 4 hours. The population reaches 100 million bacteria after 244.12 minutes.<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1167793450732\">Use the process from the previous example.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167793967055\">Let\u2019s now turn our attention to a financial application: <span class=\"no-emphasis\">compound interest<\/span>. Interest that is not compounded is called <span class=\"no-emphasis\"><em>simple interest<\/em><\/span>. Simple interest is paid once, at the end of the specified time period (usually 1 year). So, if we put [latex]$1000[\/latex] in a savings account earning 2% simple interest per year, then at the end of the year we have<\/p>\r\n\r\n<div id=\"fs-id1167793294235\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]1000(1+0.02)=$1020.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793414250\">Compound interest is paid multiple times per year, depending on the compounding period. Therefore, if the bank compounds the interest every 6 months, it credits half of the year\u2019s interest to the account after 6 months. 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>\r\n\r\n<div id=\"fs-id1167794188340\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]1000{(1+\\frac{0.02}{2})}^{2}=$1020.10.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794136411\">Similarly, if the interest is compounded every 4 months, we have<\/p>\r\n\r\n<div id=\"fs-id1167794337261\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]1000{(1+\\frac{0.02}{3})}^{3}=$1020.13,[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794002738\">and if the interest is compounded daily [latex](365[\/latex] times per year), we have [latex]$1020.20.[\/latex] If we extend this concept, so that the interest is compounded continuously, after [latex]t[\/latex] years we have<\/p>\r\n\r\n<div id=\"fs-id1167794059054\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]1000\\underset{n\\to \\infty }{\\text{lim}}{(1+\\frac{0.02}{n})}^{nt}.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794172746\">Now let\u2019s manipulate this expression so that we have an exponential growth function. Recall that the number [latex]e[\/latex] can be expressed as a limit:<\/p>\r\n\r\n<div id=\"fs-id1167794075520\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e=\\underset{m\\to \\infty }{\\text{lim}}{(1+\\frac{1}{m})}^{m}.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793281296\">Based on this, we want the expression inside the parentheses to have the form [latex](1+1\\text{\/}m).[\/latex] Let [latex]n=0.02m.[\/latex] Note that as [latex]n\\to \\infty ,[\/latex] [latex]m\\to \\infty [\/latex] as well. Then we get<\/p>\r\n\r\n<div id=\"fs-id1167794119172\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]1000\\underset{n\\to \\infty }{\\text{lim}}{(1+\\frac{0.02}{n})}^{nt}=1000\\underset{m\\to \\infty }{\\text{lim}}{(1+\\frac{0.02}{0.02m})}^{0.02mt}=1000{\\left[\\underset{m\\to \\infty }{\\text{lim}}{(1+\\frac{1}{m})}^{m}\\right]}^{0.02t}.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793444559\">We recognize the limit inside the brackets as the number [latex]e.[\/latex] So, the balance in our bank account after [latex]t[\/latex] years is given by [latex]1000{e}^{0.02t}.[\/latex] Generalizing this concept, we see that if a bank account with an initial balance of [latex]$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>\r\n\r\n<div id=\"fs-id1167793257856\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Balance}=P{e}^{rt}.[\/latex]<\/div>\r\n&nbsp;\r\n<div id=\"fs-id1167793879685\" class=\"textbook exercises\">\r\n<h3>Example: Compound Interest<\/h3>\r\nA 25-year-old student is offered an opportunity to invest some money in a retirement account that pays 5% 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] What if she could earn 6% annual interest compounded continuously instead?\r\n<div id=\"fs-id1167793879688\" class=\"exercise\">[reveal-answer q=\"fs-id1167793245842\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793245842\"]\r\n<p id=\"fs-id1167793245842\">We have<\/p>\r\n\r\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>\r\n<p id=\"fs-id1167793832121\">She must invest [latex]$135,335.28[\/latex] at 5% interest.<\/p>\r\n<p id=\"fs-id1167793883723\">If, instead, she is able to earn [latex]6\\text{%},[\/latex] then the equation becomes<\/p>\r\n\r\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>\r\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>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793638827\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSuppose instead of investing at age [latex]25\\sqrt{{b}^{2}-4ac},[\/latex] the student waits until age 35. How much would she have to invest at [latex]5\\text{%}?[\/latex] At [latex]6\\text{%}?[\/latex]\r\n<div id=\"fs-id1167793541183\" class=\"exercise\">[reveal-answer q=\"fs-id1167793627769\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793627769\"]\r\n<p id=\"fs-id1167793627769\">At 5% interest, she must invest [latex]$223,130.16.[\/latex] At 6% interest, she must invest [latex]$165,298.89.[\/latex]<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1167793515383\">Use the process from the previous example.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]64715[\/ohm_question]\r\n\r\n<\/div>\r\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 100 to 200 bacteria as it does to grow from 10,000 to 20,000 bacteria. This time is called the <strong>doubling time<\/strong>. To calculate the doubling time, we want to know when the quantity reaches twice its original size. So we have<\/p>\r\n\r\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>\r\n&nbsp;\r\n<div id=\"fs-id1167793770692\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\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>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Doubling time}=\\frac{\\text{ln}2}{k}.[\/latex]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793294341\" class=\"textbook exercises\">\r\n<h3>Example: Using the Doubling Time<\/h3>\r\nAssume a population of fish grows exponentially. A pond is stocked initially with 500 fish. After 6 months, there are 1000 fish in the pond. The owner will allow his friends and neighbors to fish on his pond after the fish population reaches 10,000. When will the owner\u2019s friends be allowed to fish?\r\n<div id=\"fs-id1167793294343\" class=\"exercise\">[reveal-answer q=\"fs-id1167793367799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793367799\"]\r\n<p id=\"fs-id1167793367799\">We know it takes the population of fish 6 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 10,000 fish, we must solve the following equation:<\/p>\r\n\r\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>\r\n<p id=\"fs-id1167793495113\">The owner\u2019s friends have to wait 25.93 months (a little more than 2 years) to fish in the pond.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793276903\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSuppose it takes 9 months for the fish population in the last example to reach 1000 fish. Under these circumstances, how long do the owner\u2019s friends have to wait?\r\n<div id=\"fs-id1167793926203\" class=\"exercise\">[reveal-answer q=\"fs-id1167794067551\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794067551\"]\r\n<p id=\"fs-id1167794067551\">38.90 months<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1167793545796\">Use the process from the previous example.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/Z6SUpWJXWSo?controls=0&amp;start=0&amp;end=279&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/6.8TryItProblems0to279_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"6.8 Try It Problems\" here (opens in new window)<\/a>.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]100768[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use the exponential growth model in applications, including population growth and compound interest<\/li>\n<li>Explain the concept of doubling time<\/li>\n<\/ul>\n<\/div>\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 <em>growth constant<\/em>. 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>&nbsp;<\/p>\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. The equation above involves derivatives and is called a <em>differential equation.<\/em> We learn more about differential equations in <a class=\"target-chapter\" href=\"https:\/\/cnx.org\/contents\/HTmjSAcf@2.46:KzA8da4p@2\/Introduction\">Introduction to Differential Equations<\/a> in the second volume of this text.<\/p>\n<div id=\"fs-id1167793498261\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Exponential Growth Model<\/h3>\n<hr \/>\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 <em>growth constant<\/em>.<\/p>\n<\/div>\n<p id=\"fs-id1167793392254\"><span class=\"no-emphasis\">Population growth<\/span> is a common example of exponential growth. 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. Figure 1 and the table below represent the growth of a population of bacteria with an initial population of 200 bacteria and a growth constant of 0.02. Notice that after only 2 hours [latex](120[\/latex] minutes), the population is 10 times its original size!<\/p>\n<div style=\"width: 303px\" class=\"wp-caption aligncenter\"><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\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. An example of exponential growth for bacteria.<\/p>\n<\/div>\n<table id=\"fs-id1167793939650\" summary=\"This table has two columns. The columns are labeled Time (minutes) and Population size (no. of bacteria). The times start at 10 and increase in increments of 10 minutes up until 120 minutes. The number of bacteria start at 244 and increase to 2205.\">\n<caption>Exponential Growth of a Bacterial Population<\/caption>\n<thead>\n<tr valign=\"top\">\n<th><strong>Time (min)<\/strong><\/th>\n<th><strong>Population Size (no. of bacteria)<\/strong><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>10<\/td>\n<td>244<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>20<\/td>\n<td>298<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>30<\/td>\n<td>364<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>40<\/td>\n<td>445<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>50<\/td>\n<td>544<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>60<\/td>\n<td>664<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>70<\/td>\n<td>811<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>80<\/td>\n<td>991<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>90<\/td>\n<td>1210<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>100<\/td>\n<td>1478<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>110<\/td>\n<td>1805<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>120<\/td>\n<td>2205<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\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<div id=\"fs-id1167794022966\" class=\"textbook exercises\">\n<h3>Example: Population Growth<\/h3>\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 id=\"fs-id1167793957317\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794170510\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<div id=\"fs-id1167794050614\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Consider a population of bacteria that grows according to the function [latex]f(t)=500{e}^{0.05t},[\/latex] where [latex]t[\/latex] is measured in minutes. How many bacteria are present in the population after 4 hours? When does the population reach 100 million bacteria?<\/p>\n<div id=\"fs-id1167793245191\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793427517\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793427517\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793427517\">There are 81,377,396 bacteria in the population after 4 hours. The population reaches 100 million bacteria after 244.12 minutes.<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1167793450732\">Use the process from the previous example.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167793967055\">Let\u2019s now turn our attention to a financial application: <span class=\"no-emphasis\">compound interest<\/span>. Interest that is not compounded is called <span class=\"no-emphasis\"><em>simple interest<\/em><\/span>. Simple interest is paid once, at the end of the specified time period (usually 1 year). So, if we put [latex]$1000[\/latex] in a savings account earning 2% simple interest per year, then at the end of the year we have<\/p>\n<div id=\"fs-id1167793294235\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]1000(1+0.02)=$1020.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793414250\">Compound interest is paid multiple times per year, depending on the compounding period. Therefore, if the bank compounds the interest every 6 months, it credits half of the year\u2019s interest to the account after 6 months. 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>&nbsp;<\/p>\n<p id=\"fs-id1167794136411\">Similarly, if the interest is compounded every 4 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>&nbsp;<\/p>\n<p id=\"fs-id1167794002738\">and if the interest is compounded daily [latex](365[\/latex] times per year), we have [latex]$1020.20.[\/latex] If we extend this concept, so that the interest is compounded continuously, after [latex]t[\/latex] years we have<\/p>\n<div id=\"fs-id1167794059054\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]1000\\underset{n\\to \\infty }{\\text{lim}}{(1+\\frac{0.02}{n})}^{nt}.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794172746\">Now let\u2019s manipulate this expression so that we have an exponential growth function. Recall that the number [latex]e[\/latex] can be expressed as a limit:<\/p>\n<div id=\"fs-id1167794075520\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]e=\\underset{m\\to \\infty }{\\text{lim}}{(1+\\frac{1}{m})}^{m}.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793281296\">Based on this, we want the expression inside the parentheses to have the form [latex](1+1\\text{\/}m).[\/latex] Let [latex]n=0.02m.[\/latex] Note that as [latex]n\\to \\infty ,[\/latex] [latex]m\\to \\infty[\/latex] as well. Then we get<\/p>\n<div id=\"fs-id1167794119172\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]1000\\underset{n\\to \\infty }{\\text{lim}}{(1+\\frac{0.02}{n})}^{nt}=1000\\underset{m\\to \\infty }{\\text{lim}}{(1+\\frac{0.02}{0.02m})}^{0.02mt}=1000{\\left[\\underset{m\\to \\infty }{\\text{lim}}{(1+\\frac{1}{m})}^{m}\\right]}^{0.02t}.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793444559\">We recognize the limit inside the brackets as the number [latex]e.[\/latex] So, the balance in our bank account after [latex]t[\/latex] years is given by [latex]1000{e}^{0.02t}.[\/latex] Generalizing this concept, we see that if a bank account with an initial balance of [latex]$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<p>&nbsp;<\/p>\n<div id=\"fs-id1167793879685\" class=\"textbook exercises\">\n<h3>Example: Compound Interest<\/h3>\n<p>A 25-year-old student is offered an opportunity to invest some money in a retirement account that pays 5% 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] What if she could earn 6% annual interest compounded continuously instead?<\/p>\n<div id=\"fs-id1167793879688\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793245842\">Show Solution<\/span><\/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 5% 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<\/div>\n<\/div>\n<div id=\"fs-id1167793638827\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Suppose instead of investing at age [latex]25\\sqrt{{b}^{2}-4ac},[\/latex] the student waits until age 35. How much would she have to invest at [latex]5\\text{%}?[\/latex] At [latex]6\\text{%}?[\/latex]<\/p>\n<div id=\"fs-id1167793541183\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793627769\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793627769\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793627769\">At 5% interest, she must invest [latex]$223,130.16.[\/latex] At 6% interest, she must invest [latex]$165,298.89.[\/latex]<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1167793515383\">Use the process from the previous example.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm64715\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=64715&theme=oea&iframe_resize_id=ohm64715&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\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 100 to 200 bacteria as it does to grow from 10,000 to 20,000 bacteria. This time is called the <strong>doubling time<\/strong>. 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<p>&nbsp;<\/p>\n<div id=\"fs-id1167793770692\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\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<\/div>\n<div id=\"fs-id1167793294341\" class=\"textbook exercises\">\n<h3>Example: Using the Doubling Time<\/h3>\n<p>Assume a population of fish grows exponentially. A pond is stocked initially with 500 fish. After 6 months, there are 1000 fish in the pond. The owner will allow his friends and neighbors to fish on his pond after the fish population reaches 10,000. When will the owner\u2019s friends be allowed to fish?<\/p>\n<div id=\"fs-id1167793294343\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793367799\">Show Solution<\/span><\/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 6 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 10,000 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 25.93 months (a little more than 2 years) to fish in the pond.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793276903\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Suppose it takes 9 months for the fish population in the last example to reach 1000 fish. Under these circumstances, how long do the owner\u2019s friends have to wait?<\/p>\n<div id=\"fs-id1167793926203\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794067551\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794067551\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794067551\">38.90 months<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1167793545796\">Use the process from the previous example.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/Z6SUpWJXWSo?controls=0&amp;start=0&amp;end=279&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/6.8TryItProblems0to279_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;6.8 Try It Problems&#8221; here (opens in new window)<\/a>.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm100768\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=100768&theme=oea&iframe_resize_id=ohm100768&show_question_numbers\" width=\"100%\" height=\"150\"><\/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-602\">\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>6.8 Try It Problems. <strong>Authored by<\/strong>: Ryan Melton. <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>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/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":17533,"menu_order":31,"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\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-602","chapter","type-chapter","status-publish","hentry"],"part":65,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/602","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":9,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/602\/revisions"}],"predecessor-version":[{"id":4920,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/602\/revisions\/4920"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/65"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/602\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=602"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=602"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=602"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=602"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}