Radiocarbon Dating

The formula for radioactive decay is important in radiocarbon dating, which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby, who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about 1% error for plants or animals that died within the last 60,000 years.

Carbon-14 is a radioactive isotope of carbon that has a half-life of 5,730 years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on Earth is carbon-12, which has an atomic weight of 12 and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last 60,000 years, using tree rings and other organic samples of known dates—although the ratio has changed slightly over the centuries.

As long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated.

Since the half-life of carbon-14 is 5,730 years, the formula for the amount of carbon-14 remaining after [latex]t[/latex] years is  [latex]A\approx {A}_{0}{e}^{\left(\frac{\mathrm{ln}\left(0.5\right)}{5730}\right)t}[/latex]

where

• [latex]A[/latex] is the amount of carbon-14 remaining
• [latex]{A}_{0}[/latex] is the amount of carbon-14 when the plant or animal began decaying.

This formula is derived as follows:

To find the age of an object, we solve this equation for [latex]t:[/latex]  [latex]t=\frac{\mathrm{ln}\left(\frac{A}{{A}_{0}}\right)}{-0.000121}[/latex]

Out of necessity, we neglect here the many details that a scientist takes into consideration when doing carbon-14 dating, and we only look at the basic formula. The ratio of carbon-14 to carbon-12 in the atmosphere is approximately 0.0000000001%. Let [latex]r[/latex] be the ratio of carbon-14 to carbon-12 in the organic artifact or fossil to be dated, determined by a method called liquid scintillation. From the equation [latex]A\approx {A}_{0}{e}^{-0.000121t}[/latex] we know the ratio of the percentage of carbon-14 in the object we are dating to the percentage of carbon-14 in the atmosphere is [latex]r=\frac{A}{{A}_{0}}\approx {e}^{-0.000121t}.[/latex] We solve this equation for [latex]t,[/latex] to get  [latex]t=\frac{\mathrm{ln}\left(r\right)}{-0.000121}[/latex]

Calculating Doubling Time

For decaying quantities, we determined how long it took for half of a substance to decay. For growing quantities, we might want to find out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the doubling time.

Given the basic exponential growth equation[latex]\,A={A}_{0}{e}^{kt},[/latex] doubling time can be found by solving for when the original quantity has doubled, that is, by solving[latex]\,2{A}_{0}={A}_{0}{e}^{kt}.[/latex]

The formula is derived as follows: [latex]\begin{array}{ll}2{A}_{0}={A}_{0}{e}^{kt}\hfill & \hfill \\ \,\,\,\,\,\,\,2={e}^{kt}\hfill & \text{Divide by }{A}_{0}.\hfill \\ \,\,\mathrm{ln}2=kt\hfill & \text{Take the natural logarithm}.\hfill \\ \,\,\,\,\,\,\,\,t=\frac{\mathrm{ln}2}{k}\hfill & \text{Divide by the coefficient of }t.\hfill \end{array}[/latex]

Thus the doubling time is [latex] t=\frac{\mathrm{ln}2}{k}[/latex]

Using Newton’s Law of Cooling

Exponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object’s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a vertical shift of the generic exponential decay function. This translation leads to Newton’s Law of Cooling, the scientific formula for temperature as a function of time as an object’s temperature is equalized with the ambient temperature [latex] T\left(t\right)=a{e}^{kt}+{T}_{s}[/latex]

This formula is derived as follows:

Using Logistic Growth Models

Exponential growth cannot continue forever. Exponential models, while they may be useful in the short term, tend to fall apart the longer they continue. Consider an aspiring writer who writes a single line on day one and plans to double the number of lines she writes each day for a month. By the end of the month, she must write over 17 billion lines, or one-half-billion pages. It is impractical, if not impossible, for anyone to write that much in such a short period of time. Eventually, an exponential model must begin to approach some limiting value, and then the growth is forced to slow. For this reason, it is often better to use a model with an upper bound instead of an exponential growth model, though the exponential growth model is still useful over a short term, before approaching the limiting value.

The logistic growth model is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model’s upper bound, called the carrying capacity. For constants [latex]\text{a, b,}[/latex] and [latex]\text{c,}[/latex] the logistic growth of a population over time[latex]\,x\,[/latex]is represented by the model [latex]f\left(x\right)=\frac{c}{1+a{e}^{-bx}}[/latex]

The graph in Figure 6 shows how the growth rate changes over time. The graph increases from left to right, but the growth rate only increases until it reaches its point of maximum growth rate, at which point the rate of increase decreases.

Using the Logistic-Growth Model

An influenza epidemic spreads through a population rapidly, at a rate that depends on two factors: The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model a good one to study the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.

For example, at time[latex]\,t=0\,[/latex]there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is[latex]\,b=0.6030.\,[/latex]Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed.

Show Solution

We substitute the given data into the logistic growth model [latex]f\left(x\right)=\frac{c}{1+a{e}^{-bx}}[/latex]

Because at most 1,000 people, the entire population of the community, can get the flu, we know the limiting value is[latex]\,c=1000.\,[/latex]To find[latex]\,a,[/latex]we use the formula that the number of cases at time[latex]\,t=0\,[/latex]is[latex]\,\frac{c}{1+a}=1,[/latex] from which it follows that[latex]\,a=999.\text{}[/latex]This model predicts that, after ten days, the number of people who have had the flu is[latex]\,f\left(x\right)=\frac{1000}{1+999{e}^{-0.6030x}}\approx 293.8.\,[/latex]Because the actual number must be a whole number (a person has either had the flu or not) we round to 294. In the long term, the number of people who will contract the flu is the limiting value,[latex]\,c=1000.[/latex]

Analysis

Remember that, because we are dealing with a virus, we cannot predict with certainty the number of people infected. The model only approximates the number of people infected and will not give us exact or actual values.

The graph in Figure 7 gives a good picture of how this model fits the data.

 

Choosing an Appropriate Model for Data

Now that we have discussed various mathematical models, we need to learn how to choose the appropriate model for the raw data we have. Many factors influence the choice of a mathematical model, among which are experience, scientific laws, and patterns in the data itself. Not all data can be described by elementary functions. Sometimes, a function is chosen that approximates the data over a given interval. For instance, suppose data were gathered on the number of homes bought in the United States from the years 1960 to 2013. After plotting these data in a scatter plot, we notice that the shape of the data from the years 2000 to 2013 follow a logarithmic curve. We could restrict the interval from 2000 to 2010, apply regression analysis using a logarithmic model, and use it to predict the number of home buyers for the year 2015.

Three kinds of functions that are often useful in mathematical models are linear functions, exponential functions, and logarithmic functions. If the data lies on a straight line, or seems to lie approximately along a straight line, a linear model may be best. If the data is non-linear, we often consider an exponential or logarithmic model, though other models, such as quadratic models, may also be considered.

In choosing between an exponential model and a logarithmic model, we look at the way the data curves. This is called the concavity. If we draw a line between two data points, and all (or most) of the data between those two points lies above that line, we say the curve is concave down. We can think of it as a bowl that bends downward and therefore cannot hold water. If all (or most) of the data between those two points lies below the line, we say the curve is concave up. In this case, we can think of a bowl that bends upward and can therefore hold water. An exponential curve, whether rising or falling, whether representing growth or decay, is always concave up away from its horizontal asymptote. A logarithmic curve is always concave away from its vertical asymptote. In the case of positive data, which is the most common case, an exponential curve is always concave up, and a logarithmic curve always concave down.

A logistic curve changes concavity. It starts out concave up and then changes to concave down beyond a certain point, called a point of inflection.

After using the graph to help us choose a type of function to use as a model, we substitute points, and solve to find the parameters. We reduce round-off error by choosing points as far apart as possible.

Choosing a Mathematical Model

Does a linear, exponential, logarithmic, or logistic model best fit the values listed in Table 1 ? Find the model, and use a graph to check your choice.

[latex]x[/latex]	1	2	3	4	5	6	7	8	9
[latex]y[/latex]	0	1.386	2.197	2.773	3.219	3.584	3.892	4.159	4.394

<p style=”text-align: center”>Table 1

Show Solution

First, plot the data on a graph as in Figure 8. For the purpose of graphing, round the data to two significant digits.

 

 

Clearly, the points do not lie on a straight line, so we reject a linear model. If we draw a line between any two of the points, most or all of the points between those two points lie above the line, so the graph is concave down, suggesting a logarithmic model. We can try[latex]\,y=a\mathrm{ln}\left(bx\right).\,[/latex]Plugging in the first point,[latex]\,\left(\text{1,0}\right)\text{,}\,[/latex]gives[latex]\,0=a\mathrm{ln}b.\,[/latex]We reject the case that[latex]\,a=0\,[/latex](if it were, all outputs would be 0), so we know[latex]\,\mathrm{ln}\left(b\right)=0.\,[/latex]Thus[latex]\,b=1\,[/latex]and[latex]\,y=a\mathrm{ln}\left(\text{x}\right).\,[/latex]Next we can use the point[latex]\,\left(\text{9,4}\text{.394}\right)\,[/latex]to solve for[latex]\,a:[/latex]

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,\,\,y=a\mathrm{ln}\left(x\right)\hfill \\ 4.394=a\mathrm{ln}\left(9\right)\hfill \\ \,\,\,\,\,\,\,\,\,\,\,a=\frac{4.394}{\mathrm{ln}\left(9\right)}\hfill \end{array}[/latex]

Because[latex]\,a=\frac{4.394}{\mathrm{ln}\left(9\right)}\approx 2,[/latex]an appropriate model for the data is[latex]\,y=2\mathrm{ln}\left(x\right).[/latex]

To check the accuracy of the model, we graph the function together with the given points as in Figure 9.

 

We can conclude that the model is a good fit to the data.

Compare Figure 9 to the graph of[latex]\,y=\mathrm{ln}\left({x}^{2}\right)\,[/latex]shown in Figure 10.

The graphs appear to be identical when[latex]\,x>0.\,[/latex]A quick check confirms this conclusion:[latex]\,y=\mathrm{ln}\left({x}^{2}\right)=2\mathrm{ln}\left(x\right)\,[/latex]for[latex]\,x>0.[/latex]

 

However, if[latex]\,x0,[/latex] the graph of[latex]\,y=\mathrm{ln}\left({x}^{2}\right)\,[/latex] includes an “extra” branch, as shown in Figure 11. This occurs because, while[latex]\,y=2\mathrm{ln}\left(x\right)\,[/latex]cannot have negative values in the domain (as such values would force the argument to be negative), the function[latex]\,y=\mathrm{ln}\left({x}^{2}\right)\,[/latex]can have negative domain values.

Enter the data from Table A  into a graphing calculator and graph the resulting scatter plot. Determine whether the data from the table would likely represent a linear, exponential, or logarithmic function.

<p style=”text-align: center”>Table A

Expressing an Exponential Model in Base e

While powers and logarithms of any base can be used in modeling, the two most common bases are[latex]\,10\,[/latex]and[latex]\,e.\,[/latex]In science and mathematics, the base[latex]\,e\,[/latex]is often preferred. We can use laws of exponents and laws of logarithms to change any base to base[latex]\,e.[/latex]