Carbon Dating

Subsection 3.3.1 Carbon Dating

Scientists can determine the age of objects containing organic material by a method called carbon dating or radiocarbon dating ¹Willard Libby, of Chicago University was awarded the Nobel Prize in Chemistry in 1960, for developing radiocarbon dating.. Cosmic rays hitting the atmosphere convert nitrogen into a radioactive isotope of carbon, ${}^{14}C\text{,}$ with a half–life of about 5730 years ²A good question to ask yourself is “How can a scientist (who presumably doesn't live 60 centuries) measure this quantity?” One way exploits the little piece of calculus we are about to discuss.. Vegetation absorbs carbon dioxide from the atmosphere through photosynthesis and animals acquire ${}^{14}C$ by eating plants. When a plant or animal dies, it stops replacing its carbon and the amount of ${}^{14}C$ begins to decrease through radioactive decay. More precisely, let $Q(t)$ denote the amount of ${}^{14}C$ in the plant or animal $t$ years after it dies. The number of radioactive decays per unit time, at time $t\text{,}$ is proportional to the amount of ${}^{14}C$ present at time $t\text{,}$ which is $Q(t)\text{.}$ Thus

Equation 3.3.1 Radioactive decay

\begin{align*} \diff{Q}{t}(t) &=-k Q(t) \end{align*}

Here $k$ is a constant of proportionality that is determined by the half–life. We shall explain what half-life is and also determine the value of $k$ in Example 3.3.3, below. Before we do so, let's think about the sign in equation 3.3.1.

Recall that $Q(t)$ denotes a quantity, namely the amount of ${}^{14}C$ present at time $t\text{.}$ There cannot be a negative amount of ${}^{14}C\text{,}$ nor can this quantity be zero (otherwise we wouldn't use carbon dating, so we must have $Q(t) \gt 0\text{.}$
As the time $t$ increases, $Q(t)$ decreases, because ${}^{14}C$ is being continuously converted into ${}^{14}N$ by radioactive decay ³The precise transition is ${}^{14}C\rightarrow {}^{14}N+ e^- + \bar{\nu}_e$ where $e^-$ is an electron and $\bar{\nu}_e $ is an electron neutrino.. Thus $\diff{Q}{t}(t) \lt 0\text{.}$
The signs $Q(t) \gt 0$ and $\diff{Q}{t}(t) \lt 0$ are consistent with equation 3.3.1 provided the constant of proportionality $k \gt 0\text{.}$
In equation 3.3.1, we chose to call the constant of proportionality “$-k$”. We did so in order to make $k \gt 0\text{.}$ We could just as well have chosen to call the constant of proportionality “$K$”. That is, we could have replaced equation 3.3.1 by $\diff{Q}{t}(t)=K Q(t)\text{.}$ The constant of proportionality $K$ would have to be negative, (and $K$ and $k$ would be related by $K=-k$).

Now, let's guess some solutions to equation 3.3.1. We wish to guess a function $Q(t)$ whose derivative is just a constant times itself. Here is a short table of derivatives. It is certainly not complete, but it contains the most important derivatives that we know.

$F(t)$	$1$	$t^a$	$\sin t$	$\cos t$	$\tan t$	$e^t$	$\log t$	$\arcsin t$	$\arctan t$
$\diff{}{t}F(t)$	$0$	$at^{a-1}$	$\cos t$	$-\sin t$	$\sec^2 t$	$e^t$	$\frac{1}{t}$	$\frac{1}{\sqrt{1-t^2}}$	$\frac{1}{1+t^2}$

There is exactly one function in this table whose derivative is just a (nonzero) constant times itself. Namely, the derivative of $e^t$ is exactly $e^t = 1\times e^t\text{.}$ This is almost, but not quite what we want. We want the derivative of $Q(t)$ to be the constant $-k$ (rather than the constant $1$) times $Q(t)\text{.}$ We want the derivative to “pull a constant” out of our guess. That is exactly what happens when we differentiate $e^{at}\text{,}$ where $a$ is a constant. Differentiating gives

\begin{gather*} \diff{}{t}e^{at} = a e^{at} \end{gather*}

i.e. “pulls the constant $a$ out of $e^{at}$”.

We have succeeded in guessing a single function, namely $e^{-kt}\text{,}$ that obeys equation 3.3.1. Can we guess any other solutions? Yes. If $C$ is any constant, $Ce^{-kt}$ also obeys equation 3.3.1:

\begin{gather*} \diff{}{t}(Ce^{-kt}) = C\diff{}{t}e^{-kt} = Ce^{-kt}(-k) = -k (Ce^{-kt}) \end{gather*}

You can try guessing some more solutions, but you won't find any, because with a little trickery we can prove that a function $Q(t)$ obeys equation 3.3.1 if and only if $Q(t)$ is of the form $Ce^{-kt}\text{,}$ where $C$ is some constant.

The trick ⁴Notice that is very similar to what we needed in Example 3.1.2, except that here the constant is multiplicative rather than additive. That is $const \times f(t)$ rather than $const + f(t)\text{.}$ is to imagine that $Q(t)$ is any (at this stage, unknown) solution to equation 3.3.1 and to compare $Q(t)$ and our known solution $e^{-kt}$ by studying the ratio $Q(t)/e^{-kt}\text{.}$ We will show that $Q(t)$ obeys equation 3.3.1 if and only if the ratio $Q(t)/e^{-kt}$ is a constant, i.e. if and only if the derivative of the ratio is zero. By the product rule

\begin{gather*} \diff{}{t}\big[Q(t)/e^{-kt}\big]= \diff{}{t}\big[e^{kt}Q(t)\big] =ke^{kt} Q(t)+e^{kt}Q'(t) \end{gather*}

Since $e^{kt}$ is never $0\text{,}$ the right hand side is zero if and only if $k Q(t)+Q'(t)=0\text{;}$ that is $Q'(t)=-kQ(t)\text{.}$ Thus

\begin{gather*} \diff{}{t}Q(t) = -k Q(t) \iff \diff{}{t}\big[Q(t)/e^{-kt}\big] =0 \end{gather*}

as required.

We have succeed in finding all functions that obey 3.3.1. That is we have found the general solution to 3.3.1. This is worth stating as a theorem.

Theorem 3.3.2

A differentiable function $Q(t)$ obeys the differential equation

\begin{gather*} \diff{Q}{t}(t)=-k Q(t) \end{gather*}

if and only if there is a constant $C$ such that

\begin{gather*} Q(t)= C e^{-kt} \end{gather*}

Before we start to apply the above theorem, we take this opportunity to remind the reader that in this text we will use $\log x$ with no base to indicate the natural logarithm. That is

\begin{gather*} \log x = \log_e x = \ln x \end{gather*}

Both of the notations $\log(x)$ and $\ln(x)$ are used widely and the reader should be comfortable with both.

Example 3.3.3 Carbon dating and half–life

In this example, we determine the value of the constant of proportionality $k$ in equation 3.3.1 that corresponds to the half–life of ${}^{14}C\text{,}$ which is 5730 years.

Imagine that some plant or animal contains a quantity $Q_0$ of ${}^{14}C$ at its time of death. Let's choose the zero point of time $t=0$ to be the instant that the plant or animal died.
Denote by $Q(t)$ the amount of ${}^{14}C$ in the plant or animal $t$ years after it died. Then $Q(t)$ must obey both equation 3.3.1 and $Q(0)=Q_0\text{.}$
Since $Q(t)$ must obey equation 3.3.1, Theorem 3.3.2 tells us that there must be a constant $C$ such that $Q(t)= C e^{-kt}\text{.}$ To also have $Q_0=Q(0) =Ce^{-k\times 0}\text{,}$ the constant $C$ must be $Q_0\text{.}$ That is, $Q(t) = Q_0 e^{-kt}$ for all $t\ge 0\text{.}$
By definition, the half–life of ${}^{14}C$ is the length of time that it takes for half of the ${}^{14}C$ to decay. That is, the half–life $t_{1/2}$ is determined by
\begin{align*} Q(t_{1/2})=\half Q(0)&=\half Q_0 & \text{but we know }Q(t) = Q_0 e^{-kt}\\ Q_0 e^{-kt_{1/2}}&=\half Q_0 & \text{now cancel } Q_0\\ e^{-kt_{1/2}}&=\half\\ \end{align*}
Taking the logarithm of both sides gives
\begin{align*} -k t_{1/2} &=\log \frac{1}{2} = -\log 2 & \text{ and so}\\ k &=\frac{\log 2}{t_{1/2}}. \end{align*}
We are told that, for ${}^{14}C\text{,}$ the half–life $t_{1/2}=5730\text{,}$ so
\begin{align*} k&=\frac{\log 2}{5730} = 0.000121 &\text{ to 6 digits} \end{align*}

From the work in the above example we have accumulated enough new facts to make a corollary to Theorem 3.3.2.

Corollary 3.3.4

The function $Q(t)$ satisfies the equation

\begin{align*} \diff{Q}{t} &= -k Q(t) \end{align*}

if and only if

\begin{align*} Q(t) &= Q(0) \cdot e^{-kt}. \end{align*}

The half-life is defined to be the time $t_{1/2}$ which obeys

\begin{align*} Q(t_{1/2}) &= \frac{1}{2} \cdot Q(0). \end{align*}

The half-life is related to the constant $k$ by

\begin{align*} t_{1/2} &= \frac{\log 2}{k} \end{align*}

Now here is a typical problem that is solved using Corollary 3.3.4.

Example 3.3.5 Determining the age of an artefact

A particular piece of parchment contains about 64$\%$ as much ${}^{14}C$ as plants do today. Estimate the age of the parchment.

Solution Let $Q(t)$ denote the amount of ${}^{14}C$ in the parchment $t$ years after it was first created.

By equation 3.3.1 and Example 3.3.3,

\begin{gather*} \diff{Q}{t}=-k Q(t)\qquad\text{with }k = \frac{\log 2}{5730} = 0.000121. \end{gather*}

By Corollary 3.3.4

\begin{align*} Q(t) &= Q(0) \cdot e^{-kt} \end{align*}

The time at which $Q(t)$ reaches $0.64 Q(0)$ is determined by

\begin{align*} Q(t) &=0.64 Q(0) & \text{ but } Q(t) = Q(0) e^{-kt}\\ Q(0)e^{-kt} &=0.64 Q(0) & \text{cancel $Q(0)$}\\ e^{-kt} &=0.64 & \text{take logarithms}\\ -kt &=\log 0.64\\ t &=\frac{\log 0.64}{-k} =\frac{\log 0.64}{-0.000121} = 3700 & \text{to 2} \text{significant digits.} \end{align*}

That is, the parchment ⁵The British Museum has an Egyptian mathematical text from the seventeenth century B.C. is about 37 centuries old.

We have stated that the half-life of ${}^{14}C$ is 5730 years. How can this be determined? We can explain this using the following example.

Example 3.3.6 Computing a half-life

A scientist in a B-grade science fiction film is studying a sample of the rare and fictitious element, implausium ⁶Implausium leads to even weaker plots than unobtainium.. With great effort he has produced a sample of pure implausium. The next day — 17 hours later — he comes back to his lab and discovers that his sample is now only 37% pure. What is the half-life of the element?

Solution We can again set up our problem using Corollary 3.3.4. Let $Q(t)$ denote the quantity of implausium at time $t\text{,}$ measured in hours. Then we know

\begin{align*} Q(t)&= Q(0) \cdot e^{-kt} \end{align*}

We also know that

\begin{align*} Q(17) &= 0.37 Q(0). \end{align*}

That enables us to determine $k$ via

\begin{align*} Q(17) = 0.37 Q(0) &= Q(0) e^{-17k} & \text{ divide both sides by $Q(0)$}\\ 0.37 &= e^{-17k}\\ \end{align*}

and so

\begin{align*} k &= -\frac{\log 0.37}{17} = 0.05849 \end{align*}

We can then convert this to the half life using Corollary 3.3.4:

\begin{align*} t_{1/2} &= \frac{\log 2}{k} \approx 11.85 \text{ hours} \end{align*}

While this example is entirely fictitious, one really can use this approach to measure the half-life of materials.

\(F(t)\)	\(1\)	\(t^a\)	\(\sin t\)	\(\cos t\)	\(\tan t\)	\(e^t\)	\(\log t\)	\(\arcsin t\)	\(\arctan t\)
\(\diff{}{t}F(t)\)	\(0\)	\(at^{a-1}\)	\(\cos t\)	\(-\sin t\)	\(\sec^2 t\)	\(e^t\)	\(\frac{1}{t}\)	\(\frac{1}{\sqrt{1-t^2}}\)	\(\frac{1}{1+t^2}\)