{"id":2634,"date":"2024-05-29T18:02:05","date_gmt":"2024-05-29T18:02:05","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=2634"},"modified":"2025-08-18T02:09:50","modified_gmt":"2025-08-18T02:09:50","slug":"exponential-growth-and-decay-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/exponential-growth-and-decay-fresh-take\/","title":{"raw":"Exponential Growth and Decay: Fresh Take","rendered":"Exponential Growth and Decay: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Apply the exponential growth formula to real-world cases like increasing populations or investments<\/li>\r\n\t<li>Describe how long it takes for quantities to double or reduce by half<\/li>\r\n\t<li>Implement the exponential decay formula for scenarios like radioactive substances decaying or objects cooling down<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Exponential Growth Model<\/h2>\r\n<div class=\"textbox shaded\">\r\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\r\n<ul>\r\n\t<li class=\"whitespace-normal break-words\">Exponential Growth Model:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">General form: [latex]y = y_0 e^{kt}[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">[latex]y_0[\/latex]: Initial state of the system<\/li>\r\n\t<li class=\"whitespace-normal break-words\">[latex]k[\/latex]: Positive growth constant<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Key Feature:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Rate of growth is proportional to current function value<\/li>\r\n\t<li class=\"whitespace-normal break-words\">[latex]y' = ky[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Applications:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Population growth<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Compound interest<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Compound Interest Formula:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">[latex]\\text{Balance} = P e^{rt}[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">[latex]P[\/latex]: Principal (initial investment)<\/li>\r\n\t<li class=\"whitespace-normal break-words\">[latex]r[\/latex]: Interest rate (as a decimal)<\/li>\r\n\t<li class=\"whitespace-normal break-words\">[latex]t[\/latex]: Time (in years)<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Doubling Time:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Time for a quantity to double in exponential growth<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Formula: [latex]\\text{Doubling time} = \\frac{\\ln 2}{k}[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section class=\"textbox example\">\r\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 [latex]4[\/latex] hours? When does the population reach [latex]100[\/latex] million bacteria?<\/p>\r\n\r\n[reveal-answer q=\"fs-id1167793427517\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1167793427517\"]\r\n\r\n<p id=\"fs-id1167793427517\">There are [latex]81,377,396[\/latex] bacteria in the population after [latex]4[\/latex] hours. The population reaches [latex]100[\/latex] million bacteria after [latex]244.12[\/latex] minutes.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\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>\r\n\r\n[reveal-answer q=\"fs-id1167793627769\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1167793627769\"]\r\n\r\n<p id=\"fs-id1167793627769\">At [latex]5\\% [\/latex] interest, she must invest [latex]$223,130.16.[\/latex] At [latex]6\\%[\/latex] interest, she must invest [latex]$165,298.89.[\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>Suppose it takes [latex]9[\/latex] months for the fish population in the last example to reach [latex]1000 [\/latex] fish. Under these circumstances, how long do the owner\u2019s friends have to wait?<\/p>\r\n\r\n[reveal-answer q=\"fs-id1167794067551\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1167794067551\"]\r\n\r\n<p id=\"fs-id1167794067551\">[latex]38.90[\/latex] months<\/p>\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\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\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\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>\r\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 \"6.8 Try It Problems\" here (opens in new window)<\/a>.<\/p>\r\n\r\n[\/hidden-answer]<\/section>\r\n<section class=\"textbox example\">\r\n<p>A bacteria culture starts with [latex]1000[\/latex] cells and grows exponentially. After [latex]2[\/latex] hours, there are [latex]1500[\/latex] cells. How many cells will be present after [latex]5[\/latex] hours?<\/p>\r\n<p><br \/>\r\n[reveal-answer q=\"279401\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"279401\"]<\/p>\r\n<p>Use the exponential growth model: [latex]y = y_0 e^{kt}[\/latex]<\/p>\r\n<p>We know:<\/p>\r\n<ul>\r\n\t<li>[latex]y_0 = 1000[\/latex] (initial number of cells)<\/li>\r\n\t<li>After [latex]2[\/latex] hours, [latex]y = 1500[\/latex]<\/li>\r\n<\/ul>\r\n<p>Find [latex]k[\/latex] using the [latex]2[\/latex]-hour data point:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br \/>\r\n1500 &amp;=&amp; 1000 e^{k(2)} \\\\<br \/>\r\n1.5 &amp;=&amp; e^{2k} \\\\<br \/>\r\n\\ln 1.5 &amp;=&amp; 2k \\\\<br \/>\r\nk &amp;=&amp; \\frac{\\ln 1.5}{2} \\approx 0.2027<br \/>\r\n\\end{array}[\/latex]<\/p>\r\n<p>Now use this [latex]k[\/latex] value to find the number of cells after [latex]5[\/latex] hours:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br \/>\r\ny &amp;=&amp; 1000 e^{0.2027(5)} \\\\<br \/>\r\n&amp;\\approx&amp; 2740 \\text{ cells}<br \/>\r\n\\end{array}[\/latex]<\/p>\r\n<p>Therefore, after [latex]5[\/latex] hours, there will be approximately [latex]2740[\/latex] bacteria cells in the culture.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<h2 class=\"entry-title\">Exponential Decay Model<\/h2>\r\n<div class=\"textbox shaded\">\r\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\r\n<ul>\r\n\t<li class=\"whitespace-normal break-words\">Exponential Decay Model:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">General form: [latex]y = y_0 e^{-kt}[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">[latex]y_0[\/latex]: Initial state of the system<\/li>\r\n\t<li class=\"whitespace-normal break-words\">[latex]k[\/latex]: Positive decay constant<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Key Feature:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Rate of decay is proportional to current function value<\/li>\r\n\t<li class=\"whitespace-normal break-words\">[latex]y' = -ky[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Applications:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Newton's Law of Cooling<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Radioactive decay<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Population decline<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Newton's Law of Cooling:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">[latex]T = (T_0 - T_a)e^{-kt} + T_a[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">[latex]T[\/latex]: Temperature of object<\/li>\r\n\t<li class=\"whitespace-normal break-words\">[latex]T_0[\/latex]: Initial temperature<\/li>\r\n\t<li class=\"whitespace-normal break-words\">[latex]T_a[\/latex]: Ambient temperature<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Half-Life:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Time for a quantity to reduce by half in exponential decay<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Formula: [latex]\\text{Half-life} = \\frac{\\ln 2}{k}[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section class=\"textbox example\">\r\n<p>Suppose the room is warmer [latex](75\\text{\u00b0}\\text{F})[\/latex] and, after [latex]2[\/latex] minutes, the coffee has cooled only to [latex]185\\text{\u00b0}\\text{F}.[\/latex] When is the coffee first cool enough to serve? When is the coffee be too cold to serve? Round answers to the nearest half minute.<\/p>\r\n\r\n[reveal-answer q=\"fs-id1167793282623\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1167793282623\"]\r\n\r\n<p id=\"fs-id1167793282623\">The coffee is first cool enough to serve about [latex]3.5[\/latex] minutes after it is poured. The coffee is too cold to serve about [latex]7[\/latex] minutes after it is poured.<\/p>\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/Z6SUpWJXWSo?controls=0&amp;start=280&amp;end=693&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\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>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/6.8TryItProblems280to693_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"6.8 Try It Problems\" here (opens in new window)<\/a>.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>If we have [latex]100[\/latex] g of [latex]\\text{carbon-}14,[\/latex] how much is left after [latex]t[\/latex] years? If an artifact that originally contained [latex]100[\/latex] g of carbon now contains [latex]20g[\/latex] of carbon, how old is it? Round the answer to the nearest hundred years.<\/p>\r\n\r\n[reveal-answer q=\"fs-id1167793590438\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1167793590438\"]\r\n\r\n<p id=\"fs-id1167793590438\">A total of [latex]94.13[\/latex] g of carbon remains. The artifact is approximately [latex]13,300[\/latex] years old.<\/p>\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<p><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/Z6SUpWJXWSo?controls=0&amp;start=700&amp;end=1025&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\r\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>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/6.8TryItProblems700to1025_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"6.8 Try It Problems\" here (opens in new window)<\/a>.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p>A radioactive sample initially contains [latex]100[\/latex] grams of a certain isotope. After [latex]30[\/latex] days, the sample contains [latex]80[\/latex] grams of the isotope. What is the half-life of this isotope?<\/p>\r\n<p><br \/>\r\n[reveal-answer q=\"706718\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"706718\"]<\/p>\r\n<p>Use the exponential decay model: [latex]y = y_0 e^{-kt}[\/latex]<\/p>\r\n<p>We know:<\/p>\r\n<ul>\r\n\t<li>[latex]y_0 = 100[\/latex] (initial amount)<\/li>\r\n\t<li>After [latex]30[\/latex] days, [latex]y = 80[\/latex]<\/li>\r\n<\/ul>\r\n<p>Find [latex]k[\/latex] using the given data:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br \/>\r\n80 &amp;=&amp; 100 e^{-k(30)} \\\\<br \/>\r\n0.8 &amp;=&amp; e^{-30k} \\\\<br \/>\r\n\\ln 0.8 &amp;=&amp; -30k \\\\<br \/>\r\nk &amp;=&amp; -\\frac{\\ln 0.8}{30} \\approx 0.00743<br \/>\r\n\\end{array}[\/latex]<\/p>\r\n<p>Use the half-life formula:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br \/>\r\n\\text{Half-life} &amp;=&amp; \\frac{\\ln 2}{k} \\\\<br \/>\r\n&amp;=&amp; \\frac{\\ln 2}{0.00743} \\\\<br \/>\r\n&amp;\\approx&amp; 93.3 \\text{ days}<br \/>\r\n\\end{array}[\/latex]<\/p>\r\n<p>Therefore, the half-life of this isotope is approximately [latex]93.3[\/latex] days.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>","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<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Exponential Growth Model:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">General form: [latex]y = y_0 e^{kt}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]y_0[\/latex]: Initial state of the system<\/li>\n<li class=\"whitespace-normal break-words\">[latex]k[\/latex]: Positive growth constant<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Key Feature:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Rate of growth is proportional to current function value<\/li>\n<li class=\"whitespace-normal break-words\">[latex]y' = ky[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Applications:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Population growth<\/li>\n<li class=\"whitespace-normal break-words\">Compound interest<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Compound Interest Formula:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]\\text{Balance} = P e^{rt}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]P[\/latex]: Principal (initial investment)<\/li>\n<li class=\"whitespace-normal break-words\">[latex]r[\/latex]: Interest rate (as a decimal)<\/li>\n<li class=\"whitespace-normal break-words\">[latex]t[\/latex]: Time (in years)<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Doubling Time:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Time for a quantity to double in exponential growth<\/li>\n<li class=\"whitespace-normal break-words\">Formula: [latex]\\text{Doubling time} = \\frac{\\ln 2}{k}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\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 [latex]4[\/latex] hours? When does the population reach [latex]100[\/latex] million bacteria?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793427517\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793427517\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793427517\">There are [latex]81,377,396[\/latex] bacteria in the population after [latex]4[\/latex] hours. The population reaches [latex]100[\/latex] million bacteria after [latex]244.12[\/latex] minutes.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\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 class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793627769\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793627769\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793627769\">At [latex]5\\%[\/latex] interest, she must invest [latex]$223,130.16.[\/latex] At [latex]6\\%[\/latex] interest, she must invest [latex]$165,298.89.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Suppose it takes [latex]9[\/latex] months for the fish population in the last example to reach [latex]1000[\/latex] fish. Under these circumstances, how long do the owner\u2019s friends have to wait?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794067551\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167794067551\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794067551\">[latex]38.90[\/latex] months<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/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\" data-mce-fragment=\"1\"><\/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>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>A bacteria culture starts with [latex]1000[\/latex] cells and grows exponentially. After [latex]2[\/latex] hours, there are [latex]1500[\/latex] cells. How many cells will be present after [latex]5[\/latex] hours?<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q279401\">Show Answer<\/button><\/p>\n<div id=\"q279401\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the exponential growth model: [latex]y = y_0 e^{kt}[\/latex]<\/p>\n<p>We know:<\/p>\n<ul>\n<li>[latex]y_0 = 1000[\/latex] (initial number of cells)<\/li>\n<li>After [latex]2[\/latex] hours, [latex]y = 1500[\/latex]<\/li>\n<\/ul>\n<p>Find [latex]k[\/latex] using the [latex]2[\/latex]-hour data point:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br \/>  1500 &=& 1000 e^{k(2)} \\\\<br \/>  1.5 &=& e^{2k} \\\\<br \/>  \\ln 1.5 &=& 2k \\\\<br \/>  k &=& \\frac{\\ln 1.5}{2} \\approx 0.2027<br \/>  \\end{array}[\/latex]<\/p>\n<p>Now use this [latex]k[\/latex] value to find the number of cells after [latex]5[\/latex] hours:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br \/>  y &=& 1000 e^{0.2027(5)} \\\\<br \/>  &\\approx& 2740 \\text{ cells}<br \/>  \\end{array}[\/latex]<\/p>\n<p>Therefore, after [latex]5[\/latex] hours, there will be approximately [latex]2740[\/latex] bacteria cells in the culture.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h2 class=\"entry-title\">Exponential Decay Model<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Exponential Decay Model:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">General form: [latex]y = y_0 e^{-kt}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]y_0[\/latex]: Initial state of the system<\/li>\n<li class=\"whitespace-normal break-words\">[latex]k[\/latex]: Positive decay constant<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Key Feature:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Rate of decay is proportional to current function value<\/li>\n<li class=\"whitespace-normal break-words\">[latex]y' = -ky[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Applications:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Newton&#8217;s Law of Cooling<\/li>\n<li class=\"whitespace-normal break-words\">Radioactive decay<\/li>\n<li class=\"whitespace-normal break-words\">Population decline<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Newton&#8217;s Law of Cooling:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]T = (T_0 - T_a)e^{-kt} + T_a[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]T[\/latex]: Temperature of object<\/li>\n<li class=\"whitespace-normal break-words\">[latex]T_0[\/latex]: Initial temperature<\/li>\n<li class=\"whitespace-normal break-words\">[latex]T_a[\/latex]: Ambient temperature<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Half-Life:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Time for a quantity to reduce by half in exponential decay<\/li>\n<li class=\"whitespace-normal break-words\">Formula: [latex]\\text{Half-life} = \\frac{\\ln 2}{k}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p>Suppose the room is warmer [latex](75\\text{\u00b0}\\text{F})[\/latex] and, after [latex]2[\/latex] minutes, the coffee has cooled only to [latex]185\\text{\u00b0}\\text{F}.[\/latex] When is the coffee first cool enough to serve? When is the coffee be too cold to serve? Round answers to the nearest half minute.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793282623\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793282623\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793282623\">The coffee is first cool enough to serve about [latex]3.5[\/latex] minutes after it is poured. The coffee is too cold to serve about [latex]7[\/latex] minutes after it is poured.<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/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=280&amp;end=693&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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\/6.8TryItProblems280to693_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>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>If we have [latex]100[\/latex] g of [latex]\\text{carbon-}14,[\/latex] how much is left after [latex]t[\/latex] years? If an artifact that originally contained [latex]100[\/latex] g of carbon now contains [latex]20g[\/latex] of carbon, how old is it? Round the answer to the nearest hundred years.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793590438\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793590438\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793590438\">A total of [latex]94.13[\/latex] g of carbon remains. The artifact is approximately [latex]13,300[\/latex] years old.<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<p><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/Z6SUpWJXWSo?controls=0&amp;start=700&amp;end=1025&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\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\/6.8TryItProblems700to1025_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>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>A radioactive sample initially contains [latex]100[\/latex] grams of a certain isotope. After [latex]30[\/latex] days, the sample contains [latex]80[\/latex] grams of the isotope. What is the half-life of this isotope?<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q706718\">Show Answer<\/button><\/p>\n<div id=\"q706718\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the exponential decay model: [latex]y = y_0 e^{-kt}[\/latex]<\/p>\n<p>We know:<\/p>\n<ul>\n<li>[latex]y_0 = 100[\/latex] (initial amount)<\/li>\n<li>After [latex]30[\/latex] days, [latex]y = 80[\/latex]<\/li>\n<\/ul>\n<p>Find [latex]k[\/latex] using the given data:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br \/>  80 &=& 100 e^{-k(30)} \\\\<br \/>  0.8 &=& e^{-30k} \\\\<br \/>  \\ln 0.8 &=& -30k \\\\<br \/>  k &=& -\\frac{\\ln 0.8}{30} \\approx 0.00743<br \/>  \\end{array}[\/latex]<\/p>\n<p>Use the half-life formula:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br \/>  \\text{Half-life} &=& \\frac{\\ln 2}{k} \\\\<br \/>  &=& \\frac{\\ln 2}{0.00743} \\\\<br \/>  &\\approx& 93.3 \\text{ days}<br \/>  \\end{array}[\/latex]<\/p>\n<p>Therefore, the half-life of this isotope is approximately [latex]93.3[\/latex] days.<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":13,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":782,"module-header":"fresh_take","content_attributions":[],"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\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2634"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2634\/revisions"}],"predecessor-version":[{"id":3771,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2634\/revisions\/3771"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/782"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2634\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=2634"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=2634"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=2634"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=2634"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}