{"id":1192,"date":"2021-06-30T17:02:08","date_gmt":"2021-06-30T17:02:08","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/exponential-decay-model\/"},"modified":"2022-03-19T03:45:39","modified_gmt":"2022-03-19T03:45:39","slug":"exponential-decay-model","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/exponential-decay-model\/","title":{"raw":"Exponential Decay Model","rendered":"Exponential Decay Model"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use the exponential decay model in applications, including radioactive decay and Newton\u2019s law of cooling<\/li>\r\n \t<li>Explain the concept of half-life<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1167793931577\">Exponential functions can also be used to model populations that shrink (from disease, for example), or chemical compounds that break down over time. We say that such systems exhibit <strong>exponential decay<\/strong>, rather than exponential growth. The model is nearly the same, except there is a negative sign in the exponent. Thus, for some positive constant [latex]k,[\/latex] we have [latex]y={y}_{0}{e}^{\\text{\u2212}kt}.[\/latex]<\/p>\r\n<p id=\"fs-id1167793948836\">As with exponential growth, there is a differential equation associated with exponential decay. We have<\/p>\r\n\r\n<div id=\"fs-id1167793948839\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{y}^{\\prime }=\\text{\u2212}k{y}_{0}{e}^{\\text{\u2212}kt}=\\text{\u2212}ky.[\/latex]<\/div>\r\n&nbsp;\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Exponential Decay Model<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793245247\">Systems that exhibit <strong>exponential decay<\/strong> behave according to the model<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y={y}_{0}{e}^{\\text{\u2212}kt},[\/latex]<\/div>\r\nwhere [latex]{y}_{0}[\/latex] represents the initial state of the system and [latex]k&gt;0[\/latex] is a constant, called the <em>decay constant<\/em>.\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793770726\">The following figure shows a graph of a representative exponential decay function.<\/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\/11213337\/CNX_Calc_Figure_06_08_002.jpg\" alt=\"This figure is a graph in the first quadrant. It is a decreasing exponential curve. It begins on the y-axis at 2000 and decreases towards the t-axis.\" width=\"293\" height=\"353\" \/> Figure 2. An example of exponential decay.[\/caption]\r\n<p id=\"fs-id1167793692898\">Let\u2019s look at a physical application of exponential decay. <span class=\"no-emphasis\">Newton\u2019s law of cooling<\/span> says that an object cools at a rate proportional to the difference between the temperature of the object and the temperature of the surroundings. In other words, if [latex]T[\/latex] represents the temperature of the object and [latex]{T}_{a}[\/latex] represents the ambient temperature in a room, then<\/p>\r\n\r\n<div id=\"fs-id1167794122396\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{T}^{\\prime }=\\text{\u2212}k(T-{T}_{a}).[\/latex]<\/div>\r\nNote that this is not quite the right model for exponential decay. We want the derivative to be proportional to the function, and this expression has the additional [latex]{T}_{a}[\/latex] term. Fortunately, we can make a change of variables that resolves this issue. Let [latex]y(t)=T(t)-{T}_{a}.[\/latex] Then [latex]{y}^{\\prime }(t)={T}^{\\prime }(t)-0={T}^{\\prime }(t),[\/latex] and our equation becomes\r\n<div id=\"fs-id1167793510327\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{y}^{\\prime }=\\text{\u2212}ky.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793770718\">From our previous work, we know this relationship between [latex]y[\/latex] and its derivative leads to exponential decay. Thus,<\/p>\r\n\r\n<div id=\"fs-id1167793655314\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y={y}_{0}{e}^{\\text{\u2212}kt},[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793655284\">and we see that<\/p>\r\n\r\n<div id=\"fs-id1167793655287\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill T-{T}_{a}&amp; =\\hfill &amp; ({T}_{0}-{T}_{a}){e}^{\\text{\u2212}kt}\\hfill \\\\ \\hfill T&amp; =\\hfill &amp; ({T}_{0}-{T}_{a}){e}^{\\text{\u2212}kt}+{T}_{a}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793943909\">where [latex]{T}_{0}[\/latex] represents the initial temperature. Let\u2019s apply this formula in the following example.<\/p>\r\n\r\n<div id=\"fs-id1167793936902\" class=\"textbook exercises\">\r\n<h3>Example: Newton\u2019s Law of Cooling<\/h3>\r\nAccording to experienced baristas, the optimal temperature to serve coffee is between [latex]155\\text{\u00b0}\\text{F}[\/latex] and [latex]175\\text{\u00b0}\\text{F}.[\/latex] Suppose coffee is poured at a temperature of [latex]200\\text{\u00b0}\\text{F},[\/latex] and after 2 minutes in a [latex]70\\text{\u00b0}\\text{F}[\/latex] room it has cooled to [latex]180\\text{\u00b0}\\text{F}.[\/latex] When is the coffee first cool enough to serve? When is the coffee too cold to serve? Round answers to the nearest half minute.\r\n<div id=\"fs-id1167793936904\" class=\"exercise\">[reveal-answer q=\"fs-id1167793498796\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793498796\"]\r\n<p id=\"fs-id1167793498796\">We have<\/p>\r\n\r\n<div id=\"fs-id1167793498799\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill T&amp; =\\hfill &amp; ({T}_{0}-{T}_{a}){e}^{\\text{\u2212}kt}+{T}_{a}\\hfill \\\\ \\hfill 180&amp; =\\hfill &amp; (200-70){e}^{\\text{\u2212}k(2)}+70\\hfill \\\\ \\hfill 110&amp; =\\hfill &amp; 130{e}^{-2k}\\hfill \\\\ \\hfill \\frac{11}{13}&amp; =\\hfill &amp; {e}^{-2k}\\hfill \\\\ \\hfill \\text{ln}\\frac{11}{13}&amp; =\\hfill &amp; -2k\\hfill \\\\ \\hfill \\text{ln}11-\\text{ln}13&amp; =\\hfill &amp; -2k\\hfill \\\\ \\hfill k&amp; =\\hfill &amp; \\frac{\\text{ln}13-\\text{ln}11}{2}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167794049090\">Then, the model is<\/p>\r\n\r\n<div id=\"fs-id1167794049093\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T=130{e}^{(\\text{ln}11-\\text{ln}13\\text{\/}2)t}+70.[\/latex]<\/div>\r\n<p id=\"fs-id1167793605057\">The coffee reaches [latex]175\\text{\u00b0}\\text{F}[\/latex] when<\/p>\r\n\r\n<div id=\"fs-id1167793580253\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill 175&amp; =\\hfill &amp; 130{e}^{(\\text{ln}11-\\text{ln}13\\text{\/}2)t}+70\\hfill \\\\ \\hfill 105&amp; =\\hfill &amp; 130{e}^{(\\text{ln}11-\\text{ln}13\\text{\/}2)t}\\hfill \\\\ \\hfill \\frac{21}{26}&amp; =\\hfill &amp; {e}^{(\\text{ln}11-\\text{ln}13\\text{\/}2)t}\\hfill \\\\ \\hfill \\text{ln}\\frac{21}{26}&amp; =\\hfill &amp; \\frac{\\text{ln}11-\\text{ln}13}{2}t\\hfill \\\\ \\hfill \\text{ln}21-\\text{ln}26&amp; =\\hfill &amp; \\frac{\\text{ln}11-\\text{ln}13}{2}t\\hfill \\\\ \\hfill t&amp; =\\hfill &amp; \\frac{2(\\text{ln}21-\\text{ln}26)}{\\text{ln}11-\\text{ln}13}\\approx 2.56.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167793926217\">The coffee can be served about 2.5 minutes after it is poured. The coffee reaches [latex]155\\text{\u00b0}\\text{F}[\/latex] at<\/p>\r\n\r\n<div id=\"fs-id1167793541952\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill 155&amp; =\\hfill &amp; 130{e}^{(\\text{ln}11-\\text{ln}13\\text{\/}2)t}+70\\hfill \\\\ \\hfill 85&amp; =\\hfill &amp; 130{e}^{(\\text{ln}11-\\text{ln}13)t}\\hfill \\\\ \\hfill \\frac{17}{26}&amp; =\\hfill &amp; {e}^{(\\text{ln}11-\\text{ln}13)t}\\hfill \\\\ \\hfill \\text{ln}17-\\text{ln}26&amp; =\\hfill &amp; (\\frac{\\text{ln}11-\\text{ln}13}{2})t\\hfill \\\\ \\hfill t&amp; =\\hfill &amp; \\frac{2(\\text{ln}17-\\text{ln}26)}{\\text{ln}11-\\text{ln}13}\\approx 5.09.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167793615264\">The coffee is too cold to be served about 5 minutes after it is poured.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793360848\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSuppose the room is warmer [latex](75\\text{\u00b0}\\text{F})[\/latex] and, after 2 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.\r\n<div>[reveal-answer q=\"708032\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"708032\"]Use the process from the previous example.[\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1167793360852\" class=\"exercise\">[reveal-answer q=\"fs-id1167793282623\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793282623\"]\r\n<p id=\"fs-id1167793282623\">The coffee is first cool enough to serve about 3.5 minutes after it is poured. The coffee is too cold to serve about 7 minutes after it is poured.<\/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=280&amp;end=693&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\/6.8TryItProblems280to693_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]59457[\/ohm_question]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793956504\">Just as systems exhibiting exponential growth have a constant doubling time, systems exhibiting exponential decay have a constant<strong> half-life<\/strong>. To calculate the half-life, we want to know when the quantity reaches half its original size. Therefore, we have<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\frac{{y}_{0}}{2}&amp; =\\hfill &amp; {y}_{0}{e}^{\\text{\u2212}kt}\\hfill \\\\ \\hfill \\frac{1}{2}&amp; =\\hfill &amp; {e}^{\\text{\u2212}kt}\\hfill \\\\ \\hfill -\\text{ln}2&amp; =\\hfill &amp; \\text{\u2212}kt\\hfill \\\\ \\hfill t&amp; =\\hfill &amp; \\frac{\\text{ln}2}{k}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793570058\"><em>Note<\/em>: This is the same expression we came up with for doubling time.<\/p>\r\n\r\n<div id=\"fs-id1167793570061\" class=\"textbox shaded\">\r\n<div class=\"title\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\n<\/div>\r\nIf a quantity decays exponentially, the <strong>half-life<\/strong> is the amount of time it takes the quantity to be reduced by half. It is given by\r\n<div id=\"fs-id1167793546938\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Half-life}=\\frac{\\text{ln}2}{k}[\/latex]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793607668\" class=\"textbook exercises\">\r\n<h3>Example: Radiocarbon Dating<\/h3>\r\nOne of the most common applications of an exponential decay model is <span class=\"no-emphasis\">carbon dating<\/span>. [latex]\\text{Carbon-}14[\/latex] decays (emits a radioactive particle) at a regular and consistent exponential rate. Therefore, if we know how much carbon was originally present in an object and how much carbon remains, we can determine the age of the object. The half-life of [latex]\\text{carbon-}14[\/latex] is approximately 5730 years\u2014meaning, after that many years, half the material has converted from the original [latex]\\text{carbon-}14[\/latex] to the new nonradioactive [latex]\\text{nitrogen-}14.[\/latex] If we have 100 g [latex]\\text{carbon-}14[\/latex] today, how much is left in 50 years? If an artifact that originally contained 100 g of carbon now contains 10 g of carbon, how old is it? Round the answer to the nearest hundred years.\r\n<div id=\"fs-id1167793607670\" class=\"exercise\">[reveal-answer q=\"fs-id1167793318477\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793318477\"]\r\n<p id=\"fs-id1167793318477\">We have<\/p>\r\n\r\n<div id=\"fs-id1167793318481\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill 5730&amp; =\\hfill &amp; \\frac{\\text{ln}2}{k}\\hfill \\\\ \\hfill k&amp; =\\hfill &amp; \\frac{\\text{ln}2}{5730}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167793929641\">So, the model says<\/p>\r\n\r\n<div id=\"fs-id1167793929644\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=100{e}^{\\text{\u2212}(\\text{ln}2\\text{\/}5730)t}.[\/latex]<\/div>\r\n<p id=\"fs-id1167793504048\">In 50 years, we have<\/p>\r\n\r\n<div id=\"fs-id1167793504057\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill y&amp; =\\hfill &amp; 100{e}^{\\text{\u2212}(\\text{ln}2\\text{\/}5730)(50)}\\hfill \\\\ &amp; \\approx \\hfill &amp; 99.40.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167793446651\">Therefore, in 50 years, 99.40 g of [latex]\\text{carbon-}14[\/latex] remains.<\/p>\r\n<p id=\"fs-id1167793628300\">To determine the age of the artifact, we must solve<\/p>\r\n\r\n<div id=\"fs-id1167793628303\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill 10&amp; =\\hfill &amp; 100{e}^{\\text{\u2212}(\\text{ln}2\\text{\/}5730)t}\\hfill \\\\ \\hfill \\frac{1}{10}&amp; =\\hfill &amp; {e}^{\\text{\u2212}(\\text{ln}2\\text{\/}5730)t}\\hfill \\\\ \\hfill t&amp; \\approx \\hfill &amp; 19035.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167794296558\">The artifact is about 19,000 years old.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793932261\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nIf 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.\r\n<div>[reveal-answer q=\"270940\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"270940\"]Use the process from the previous example.[\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1167793932265\" class=\"exercise\">[reveal-answer q=\"fs-id1167793590438\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793590438\"]\r\n<p id=\"fs-id1167793590438\">A total of 94.13 g of carbon remains. The artifact is approximately 13,300 years old.<\/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=700&amp;end=1025&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\/6.8TryItProblems700to1025_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]5787[\/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 decay model in applications, including radioactive decay and Newton\u2019s law of cooling<\/li>\n<li>Explain the concept of half-life<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1167793931577\">Exponential functions can also be used to model populations that shrink (from disease, for example), or chemical compounds that break down over time. We say that such systems exhibit <strong>exponential decay<\/strong>, rather than exponential growth. The model is nearly the same, except there is a negative sign in the exponent. Thus, for some positive constant [latex]k,[\/latex] we have [latex]y={y}_{0}{e}^{\\text{\u2212}kt}.[\/latex]<\/p>\n<p id=\"fs-id1167793948836\">As with exponential growth, there is a differential equation associated with exponential decay. We have<\/p>\n<div id=\"fs-id1167793948839\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{y}^{\\prime }=\\text{\u2212}k{y}_{0}{e}^{\\text{\u2212}kt}=\\text{\u2212}ky.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Exponential Decay Model<\/h3>\n<hr \/>\n<p id=\"fs-id1167793245247\">Systems that exhibit <strong>exponential decay<\/strong> behave according to the model<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y={y}_{0}{e}^{\\text{\u2212}kt},[\/latex]<\/div>\n<p>where [latex]{y}_{0}[\/latex] represents the initial state of the system and [latex]k>0[\/latex] is a constant, called the <em>decay constant<\/em>.<\/p>\n<\/div>\n<p id=\"fs-id1167793770726\">The following figure shows a graph of a representative exponential decay function.<\/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\/11213337\/CNX_Calc_Figure_06_08_002.jpg\" alt=\"This figure is a graph in the first quadrant. It is a decreasing exponential curve. It begins on the y-axis at 2000 and decreases towards the t-axis.\" width=\"293\" height=\"353\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. An example of exponential decay.<\/p>\n<\/div>\n<p id=\"fs-id1167793692898\">Let\u2019s look at a physical application of exponential decay. <span class=\"no-emphasis\">Newton\u2019s law of cooling<\/span> says that an object cools at a rate proportional to the difference between the temperature of the object and the temperature of the surroundings. In other words, if [latex]T[\/latex] represents the temperature of the object and [latex]{T}_{a}[\/latex] represents the ambient temperature in a room, then<\/p>\n<div id=\"fs-id1167794122396\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{T}^{\\prime }=\\text{\u2212}k(T-{T}_{a}).[\/latex]<\/div>\n<p>Note that this is not quite the right model for exponential decay. We want the derivative to be proportional to the function, and this expression has the additional [latex]{T}_{a}[\/latex] term. Fortunately, we can make a change of variables that resolves this issue. Let [latex]y(t)=T(t)-{T}_{a}.[\/latex] Then [latex]{y}^{\\prime }(t)={T}^{\\prime }(t)-0={T}^{\\prime }(t),[\/latex] and our equation becomes<\/p>\n<div id=\"fs-id1167793510327\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{y}^{\\prime }=\\text{\u2212}ky.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793770718\">From our previous work, we know this relationship between [latex]y[\/latex] and its derivative leads to exponential decay. Thus,<\/p>\n<div id=\"fs-id1167793655314\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y={y}_{0}{e}^{\\text{\u2212}kt},[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793655284\">and we see that<\/p>\n<div id=\"fs-id1167793655287\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill T-{T}_{a}& =\\hfill & ({T}_{0}-{T}_{a}){e}^{\\text{\u2212}kt}\\hfill \\\\ \\hfill T& =\\hfill & ({T}_{0}-{T}_{a}){e}^{\\text{\u2212}kt}+{T}_{a}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793943909\">where [latex]{T}_{0}[\/latex] represents the initial temperature. Let\u2019s apply this formula in the following example.<\/p>\n<div id=\"fs-id1167793936902\" class=\"textbook exercises\">\n<h3>Example: Newton\u2019s Law of Cooling<\/h3>\n<p>According to experienced baristas, the optimal temperature to serve coffee is between [latex]155\\text{\u00b0}\\text{F}[\/latex] and [latex]175\\text{\u00b0}\\text{F}.[\/latex] Suppose coffee is poured at a temperature of [latex]200\\text{\u00b0}\\text{F},[\/latex] and after 2 minutes in a [latex]70\\text{\u00b0}\\text{F}[\/latex] room it has cooled to [latex]180\\text{\u00b0}\\text{F}.[\/latex] When is the coffee first cool enough to serve? When is the coffee too cold to serve? Round answers to the nearest half minute.<\/p>\n<div id=\"fs-id1167793936904\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793498796\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793498796\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793498796\">We have<\/p>\n<div id=\"fs-id1167793498799\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill T& =\\hfill & ({T}_{0}-{T}_{a}){e}^{\\text{\u2212}kt}+{T}_{a}\\hfill \\\\ \\hfill 180& =\\hfill & (200-70){e}^{\\text{\u2212}k(2)}+70\\hfill \\\\ \\hfill 110& =\\hfill & 130{e}^{-2k}\\hfill \\\\ \\hfill \\frac{11}{13}& =\\hfill & {e}^{-2k}\\hfill \\\\ \\hfill \\text{ln}\\frac{11}{13}& =\\hfill & -2k\\hfill \\\\ \\hfill \\text{ln}11-\\text{ln}13& =\\hfill & -2k\\hfill \\\\ \\hfill k& =\\hfill & \\frac{\\text{ln}13-\\text{ln}11}{2}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167794049090\">Then, the model is<\/p>\n<div id=\"fs-id1167794049093\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T=130{e}^{(\\text{ln}11-\\text{ln}13\\text{\/}2)t}+70.[\/latex]<\/div>\n<p id=\"fs-id1167793605057\">The coffee reaches [latex]175\\text{\u00b0}\\text{F}[\/latex] when<\/p>\n<div id=\"fs-id1167793580253\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill 175& =\\hfill & 130{e}^{(\\text{ln}11-\\text{ln}13\\text{\/}2)t}+70\\hfill \\\\ \\hfill 105& =\\hfill & 130{e}^{(\\text{ln}11-\\text{ln}13\\text{\/}2)t}\\hfill \\\\ \\hfill \\frac{21}{26}& =\\hfill & {e}^{(\\text{ln}11-\\text{ln}13\\text{\/}2)t}\\hfill \\\\ \\hfill \\text{ln}\\frac{21}{26}& =\\hfill & \\frac{\\text{ln}11-\\text{ln}13}{2}t\\hfill \\\\ \\hfill \\text{ln}21-\\text{ln}26& =\\hfill & \\frac{\\text{ln}11-\\text{ln}13}{2}t\\hfill \\\\ \\hfill t& =\\hfill & \\frac{2(\\text{ln}21-\\text{ln}26)}{\\text{ln}11-\\text{ln}13}\\approx 2.56.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793926217\">The coffee can be served about 2.5 minutes after it is poured. The coffee reaches [latex]155\\text{\u00b0}\\text{F}[\/latex] at<\/p>\n<div id=\"fs-id1167793541952\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill 155& =\\hfill & 130{e}^{(\\text{ln}11-\\text{ln}13\\text{\/}2)t}+70\\hfill \\\\ \\hfill 85& =\\hfill & 130{e}^{(\\text{ln}11-\\text{ln}13)t}\\hfill \\\\ \\hfill \\frac{17}{26}& =\\hfill & {e}^{(\\text{ln}11-\\text{ln}13)t}\\hfill \\\\ \\hfill \\text{ln}17-\\text{ln}26& =\\hfill & (\\frac{\\text{ln}11-\\text{ln}13}{2})t\\hfill \\\\ \\hfill t& =\\hfill & \\frac{2(\\text{ln}17-\\text{ln}26)}{\\text{ln}11-\\text{ln}13}\\approx 5.09.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793615264\">The coffee is too cold to be served about 5 minutes after it is poured.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793360848\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Suppose the room is warmer [latex](75\\text{\u00b0}\\text{F})[\/latex] and, after 2 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>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q708032\">Hint<\/span><\/p>\n<div id=\"q708032\" class=\"hidden-answer\" style=\"display: none\">Use the process from the previous example.<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1167793360852\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793282623\">Show Solution<\/span><\/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 3.5 minutes after it is poured. The coffee is too cold to serve about 7 minutes after it is poured.<\/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=280&amp;end=693&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\/6.8TryItProblems280to693_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 class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm59457\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=59457&theme=oea&iframe_resize_id=ohm59457&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p id=\"fs-id1167793956504\">Just as systems exhibiting exponential growth have a constant doubling time, systems exhibiting exponential decay have a constant<strong> half-life<\/strong>. To calculate the half-life, we want to know when the quantity reaches half its original size. Therefore, we have<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\frac{{y}_{0}}{2}& =\\hfill & {y}_{0}{e}^{\\text{\u2212}kt}\\hfill \\\\ \\hfill \\frac{1}{2}& =\\hfill & {e}^{\\text{\u2212}kt}\\hfill \\\\ \\hfill -\\text{ln}2& =\\hfill & \\text{\u2212}kt\\hfill \\\\ \\hfill t& =\\hfill & \\frac{\\text{ln}2}{k}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793570058\"><em>Note<\/em>: This is the same expression we came up with for doubling time.<\/p>\n<div id=\"fs-id1167793570061\" class=\"textbox shaded\">\n<div class=\"title\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<\/div>\n<p>If a quantity decays exponentially, the <strong>half-life<\/strong> is the amount of time it takes the quantity to be reduced by half. It is given by<\/p>\n<div id=\"fs-id1167793546938\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Half-life}=\\frac{\\text{ln}2}{k}[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1167793607668\" class=\"textbook exercises\">\n<h3>Example: Radiocarbon Dating<\/h3>\n<p>One of the most common applications of an exponential decay model is <span class=\"no-emphasis\">carbon dating<\/span>. [latex]\\text{Carbon-}14[\/latex] decays (emits a radioactive particle) at a regular and consistent exponential rate. Therefore, if we know how much carbon was originally present in an object and how much carbon remains, we can determine the age of the object. The half-life of [latex]\\text{carbon-}14[\/latex] is approximately 5730 years\u2014meaning, after that many years, half the material has converted from the original [latex]\\text{carbon-}14[\/latex] to the new nonradioactive [latex]\\text{nitrogen-}14.[\/latex] If we have 100 g [latex]\\text{carbon-}14[\/latex] today, how much is left in 50 years? If an artifact that originally contained 100 g of carbon now contains 10 g of carbon, how old is it? Round the answer to the nearest hundred years.<\/p>\n<div id=\"fs-id1167793607670\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793318477\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793318477\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793318477\">We have<\/p>\n<div id=\"fs-id1167793318481\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill 5730& =\\hfill & \\frac{\\text{ln}2}{k}\\hfill \\\\ \\hfill k& =\\hfill & \\frac{\\text{ln}2}{5730}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793929641\">So, the model says<\/p>\n<div id=\"fs-id1167793929644\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=100{e}^{\\text{\u2212}(\\text{ln}2\\text{\/}5730)t}.[\/latex]<\/div>\n<p id=\"fs-id1167793504048\">In 50 years, we have<\/p>\n<div id=\"fs-id1167793504057\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill y& =\\hfill & 100{e}^{\\text{\u2212}(\\text{ln}2\\text{\/}5730)(50)}\\hfill \\\\ & \\approx \\hfill & 99.40.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793446651\">Therefore, in 50 years, 99.40 g of [latex]\\text{carbon-}14[\/latex] remains.<\/p>\n<p id=\"fs-id1167793628300\">To determine the age of the artifact, we must solve<\/p>\n<div id=\"fs-id1167793628303\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill 10& =\\hfill & 100{e}^{\\text{\u2212}(\\text{ln}2\\text{\/}5730)t}\\hfill \\\\ \\hfill \\frac{1}{10}& =\\hfill & {e}^{\\text{\u2212}(\\text{ln}2\\text{\/}5730)t}\\hfill \\\\ \\hfill t& \\approx \\hfill & 19035.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167794296558\">The artifact is about 19,000 years old.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793932261\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\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>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q270940\">Hint<\/span><\/p>\n<div id=\"q270940\" class=\"hidden-answer\" style=\"display: none\">Use the process from the previous example.<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1167793932265\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793590438\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793590438\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793590438\">A total of 94.13 g of carbon remains. The artifact is approximately 13,300 years old.<\/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=700&amp;end=1025&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\/6.8TryItProblems700to1025_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 class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm5787\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5787&theme=oea&iframe_resize_id=ohm5787&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-1192\">\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 2. <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\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/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-2\/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":32,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/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-1192","chapter","type-chapter","status-publish","hentry"],"part":1160,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1192","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":4,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1192\/revisions"}],"predecessor-version":[{"id":1657,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1192\/revisions\/1657"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/1160"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1192\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1192"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1192"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1192"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1192"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}