[latex]y={y}_{0}{e}^{\\text{\u2212}kt}.[\/latex]<\/p>\n
As with exponential growth, there is a differential equation associated with exponential decay. We have<\/p>\n
exponential decay model<\/h3>\n
Systems that exhibit exponential decay behave according to the model<\/p>\n
where [latex]{y}_{0}[\/latex] represents the initial state of the system and [latex]k>0[\/latex] is a constant, called the decay constant.<\/p>\n<\/section>\n<\/div>\n
The following figure shows a graph of a representative exponential decay function.<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"293\"]![\"This](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213337\/CNX_Calc_Figure_06_08_002.jpg\")
Figure 2. An example of exponential decay.[\/caption]\n\n
Let\u2019s look at a physical application of exponential decay.<\/p>\n
Newton\u2019s Law of Cooling<\/span><\/h3>\nNewton\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.<\/p>\nIn 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
[latex]{T}^{\\prime }=\\text{\u2212}k(T-{T}_{a}).[\/latex]<\/div>\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.<\/p>\n
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
[latex]{y}^{\\prime }=\\text{\u2212}ky.[\/latex]<\/div>\nFrom our previous work, we know this relationship between [latex]y[\/latex] and its derivative leads to exponential decay. Thus,<\/p>\n
[latex]y={y}_{0}{e}^{\\text{\u2212}kt},[\/latex]<\/div>\nand we see that<\/p>\n
[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>\nwhere [latex]{T}_{0}[\/latex] represents the initial temperature.<\/p>\n
Let\u2019s apply this formula in the following example.<\/p>\n\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.<\/p>\n
[reveal-answer q=\"fs-id1167793498796\"]Show Solution[\/reveal-answer]
\n[hidden-answer a=\"fs-id1167793498796\"]\n\nWe have<\/p>\n
[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>\nThen, the model is<\/p>\n
[latex]T=130{e}^{(\\text{ln}11-\\text{ln}13\\text{\/}2)t}+70.[\/latex]<\/div>\nThe coffee reaches [latex]175\\text{\u00b0}\\text{F}[\/latex] when<\/p>\n
[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>\nThe 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
[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>\nThe coffee is too cold to be served about 5 minutes after it is poured.<\/p>\n
[\/hidden-answer]<\/p>\n<\/div>\n<\/section>\n\n[ohm_question hide_question_numbers=1]288450[\/ohm_question]<\/p>\n<\/section>\n
Half-Life and Radioactive Decay<\/h3>\n
Just as systems exhibiting exponential growth have a constant doubling time, systems exhibiting exponential decay have a constant half-life<\/strong>.<\/p>\nTo calculate the half-life, we want to know when the quantity reaches half its original size. Therefore, we have<\/p>\n
[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>\nNote<\/em>: This is the same expression we came up with for doubling time.<\/p>\n\n\nhalf-life<\/h3>\n
If a quantity decays exponentially, the half-life is the amount of time it takes the quantity to be reduced by half. It is given by<\/p>\n
[latex]\\text{Half-life}=\\frac{\\text{ln}2}{k}[\/latex]<\/div>\n<\/div>\n<\/section>\nOne of the most common applications of an exponential decay model is 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.<\/p>\nThe half-life of [latex]\\text{carbon-}14[\/latex] is approximately [latex]5730[\/latex] 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]<\/p>\n\nAnother example of radioactive decay is uranium-238, which decays into lead-206 over a much longer period, with a half-life of about [latex]4.5[\/latex] billion years. This property is often used to date rocks and fossils, providing important insights into the history of the Earth and its geological events.<\/p>\n<\/section>\n\nIf we have [latex]100[\/latex] g [latex]\\text{carbon-}14[\/latex] today, how much is left in [latex]50[\/latex] years? <\/p>\n
[reveal-answer q=\"fs-id1167793318477\"]Show Solution[\/reveal-answer]
\n[hidden-answer a=\"fs-id1167793318477\"]\n\nWe have<\/p>\n
[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>\nSo, the model says<\/p>\n
[latex]y=100{e}^{\\text{\u2212}(\\text{ln}2\\text{\/}5730)t}.[\/latex]<\/div>\nIn 50 years, we have<\/p>\n
[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>\nTherefore, in 50 years, 99.40 g of [latex]\\text{carbon-}14[\/latex] remains.<\/p>\n
[\/hidden-answer]<\/p>\n<\/div>\n<\/section>\n\nIf an artifact that originally contained [latex]100[\/latex] g of carbon now contains [latex]10[\/latex] g of carbon, how old is it? Round the answer to the nearest hundred years.<\/p>\n
[reveal-answer q=\"fs-id1167793318478\"]Show Solution[\/reveal-answer]
\n[hidden-answer a=\"fs-id1167793318478\"]<\/p>\n
To determine the age of the artifact, we must solve<\/p>\n
[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>\nThe artifact is about 19,000 years old.<\/p>\n
[\/hidden-answer]<\/p>\n<\/section>\n\n[ohm_question hide_question_numbers=1]5787[\/ohm_question]<\/p>\n<\/section>\n","rendered":"
Exponential Decay Model<\/h2>\n
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 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<\/p>\n[latex]y={y}_{0}{e}^{\\text{\u2212}kt}.[\/latex]<\/p>\n
As with exponential growth, there is a differential equation associated with exponential decay. We have<\/p>\n
[latex]{y}^{\\prime }=\\text{\u2212}k{y}_{0}{e}^{\\text{\u2212}kt}=\\text{\u2212}ky.[\/latex]<\/div>\n\n\nexponential decay model<\/h3>\n
Systems that exhibit exponential decay behave according to the model<\/p>\n
[latex]y={y}_{0}{e}^{\\text{\u2212}kt},[\/latex]<\/div>\nwhere [latex]{y}_{0}[\/latex] represents the initial state of the system and [latex]k>0[\/latex] is a constant, called the decay constant.<\/p>\n<\/section>\n<\/div>\n
The following figure shows a graph of a representative exponential decay function.<\/p>\nFigure 2. An example of exponential decay.<\/figcaption><\/figure>\nLet\u2019s look at a physical application of exponential decay.<\/p>\n
Newton\u2019s Law of Cooling<\/span><\/h3>\nNewton\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.<\/p>\nIn 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
[latex]{T}^{\\prime }=\\text{\u2212}k(T-{T}_{a}).[\/latex]<\/div>\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.<\/p>\n
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
[latex]{y}^{\\prime }=\\text{\u2212}ky.[\/latex]<\/div>\nFrom our previous work, we know this relationship between [latex]y[\/latex] and its derivative leads to exponential decay. Thus,<\/p>\n
[latex]y={y}_{0}{e}^{\\text{\u2212}kt},[\/latex]<\/div>\nand we see that<\/p>\n
[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>\nwhere [latex]{T}_{0}[\/latex] represents the initial temperature.<\/p>\n
Let\u2019s apply this formula in the following example.<\/p>\n\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.<\/p>\n
\n
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
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.<\/p>\n
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
From our previous work, we know this relationship between [latex]y[\/latex] and its derivative leads to exponential decay. Thus,<\/p>\n
and we see that<\/p>\n
where [latex]{T}_{0}[\/latex] represents the initial temperature.<\/p>\n
Let\u2019s apply this formula in the following example.<\/p>\n 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 We have<\/p>\n Then, the model is<\/p>\n The coffee reaches [latex]175\\text{\u00b0}\\text{F}[\/latex] when<\/p>\n 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 The coffee is too cold to be served about 5 minutes after it is poured.<\/p>\n [\/hidden-answer]<\/p>\n<\/div>\n<\/section>\n [ohm_question hide_question_numbers=1]288450[\/ohm_question]<\/p>\n<\/section>\n Just as systems exhibiting exponential growth have a constant doubling time, systems exhibiting exponential decay have a constant half-life<\/strong>.<\/p>\n To calculate the half-life, we want to know when the quantity reaches half its original size. Therefore, we have<\/p>\n Note<\/em>: This is the same expression we came up with for doubling time.<\/p>\n If a quantity decays exponentially, the half-life is the amount of time it takes the quantity to be reduced by half. It is given by<\/p>\n One of the most common applications of an exponential decay model is 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.<\/p>\n The half-life of [latex]\\text{carbon-}14[\/latex] is approximately [latex]5730[\/latex] 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]<\/p>\n Another example of radioactive decay is uranium-238, which decays into lead-206 over a much longer period, with a half-life of about [latex]4.5[\/latex] billion years. This property is often used to date rocks and fossils, providing important insights into the history of the Earth and its geological events.<\/p>\n<\/section>\n If we have [latex]100[\/latex] g [latex]\\text{carbon-}14[\/latex] today, how much is left in [latex]50[\/latex] years? <\/p>\n We have<\/p>\n So, the model says<\/p>\n In 50 years, we have<\/p>\n Therefore, in 50 years, 99.40 g of [latex]\\text{carbon-}14[\/latex] remains.<\/p>\n [\/hidden-answer]<\/p>\n<\/div>\n<\/section>\n If an artifact that originally contained [latex]100[\/latex] g of carbon now contains [latex]10[\/latex] g of carbon, how old is it? Round the answer to the nearest hundred years.<\/p>\n [reveal-answer q=\"fs-id1167793318478\"]Show Solution[\/reveal-answer] To determine the age of the artifact, we must solve<\/p>\n The artifact is about 19,000 years old.<\/p>\n [\/hidden-answer]<\/p>\n<\/section>\n [ohm_question hide_question_numbers=1]5787[\/ohm_question]<\/p>\n<\/section>\n","rendered":" 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 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<\/p>\n [latex]y={y}_{0}{e}^{\\text{\u2212}kt}.[\/latex]<\/p>\n As with exponential growth, there is a differential equation associated with exponential decay. We have<\/p>\n Systems that exhibit exponential decay behave according to the model<\/p>\n where [latex]{y}_{0}[\/latex] represents the initial state of the system and [latex]k>0[\/latex] is a constant, called the decay constant.<\/p>\n<\/section>\n<\/div>\n The following figure shows a graph of a representative exponential decay function.<\/p>\n Let\u2019s look at a physical application of exponential decay.<\/p>\n 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.<\/p>\n 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 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.<\/p>\n 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 From our previous work, we know this relationship between [latex]y[\/latex] and its derivative leads to exponential decay. Thus,<\/p>\n and we see that<\/p>\n where [latex]{T}_{0}[\/latex] represents the initial temperature.<\/p>\n Let\u2019s apply this formula in the following example.<\/p>\n 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
\n[hidden-answer a=\"fs-id1167793498796\"]\n\nHalf-Life and Radioactive Decay<\/h3>\n
half-life<\/h3>\n
\n[hidden-answer a=\"fs-id1167793318477\"]\n\n
\n[hidden-answer a=\"fs-id1167793318478\"]<\/p>\nExponential Decay Model<\/h2>\n
exponential decay model<\/h3>\n
Newton\u2019s Law of Cooling<\/span><\/h3>\n