Solving a Differential Equation for Decay

The skull of an animal found in an archaeological dig contains about 20% of the original amount of carbon-14. If the half-life of carbon-14 is 5730 years, how long ago did the animal die?

Solution Let \(A=A(t) \) be the amount of carbon-14 present in the skull at time \(t\). Then \(A\) satisfies the differential equation \(\dfrac{{\it dA}}{dt}=kA\), whose solution is \[ A=A_{0}e^{kt} \]

where \(A_{0}\) is the amount of carbon-14 present at time \(t=0\). To determine the constant \(k\), we use the fact that when \(t=5730,\) half of the original amount \(A_{0}\) remains. \[ \begin{eqnarray*} \dfrac{1}{2}A_{0} &=&A_{0}e^{5730k} \\ \dfrac{1}{2} &=&e^{5730k} \\ 5730k &=&\ln \dfrac{1}{2} = -\ln 2 \\ k &=&-\dfrac{\ln 2}{5730} \end{eqnarray*} \]

The relationship between the amount \(A\) of carbon-14 present and the time \(t\) is \[ A(t) =A_{0}e^{\left( -\ln 2/5730\right) t} \]

In this skull, 20% of the original amount of carbon-14 remains, so \(A(t) =0.20A_{0}.\) \[ \begin{eqnarray*} 0.20A_{0} &=&A_{0}e^{( -\ln 2/5730) t} \\ 0.20 &=&e^{( -\ln 2/5730) t} \end{eqnarray*} \]

Now, we take the natural logarithm of both sides. \[ \begin{eqnarray*} \ln 0.20 &=&-\dfrac{\ln 2}{5730}\cdot t \\ t &=&-5730\cdot \dfrac{\ln 0.20}{\ln 2}\approx 13{,}300 \end{eqnarray*} \]

The animal died approximately 13,300 years ago.