Application to Carbon-14 Dating

All carbon on Earth contains some carbon-14, which is radioactive. When a living organism dies, the carbon-14 begins to decay at a fixed rate. The formula \(P( t) =100e^{-0.000121t}\) gives the percentage of carbon-14 present at time \(t\) years. Notice that when \(t=0\), the percentage of carbon-14 present is 100%. When the preserved bodies of \(15\)-year-old La Doncella and her two children were found in Peru in 2005, 93.5% of the carbon-14 remained in their bodies, indicating that the three had died about \(550\) years earlier.

1. What is the rate of change of the percentage of carbon-14 present in a \(550\)-year-old fossil?
2. What is the rate of change of the percentage of carbon-14 present in a \(2000\)-year-old fossil?

Solution (a) The rate of change of \(P\) is given by its derivative \[ \begin{eqnarray*} && P^\prime ( t) =\dfrac{d}{dt}(100e^{-0.000121t}) \underset{\underset{\color{#0066A7} { {\hbox{\( \tfrac{d}{dt}e^{u}=e^{u}\tfrac{du}{dt} \)}} }}{\color{#0066A7}{{\uparrow}}}}{=} 100(-0.000121e^{-0.000121t}) = -0.0121e^{-0.000121t}\\ \end{eqnarray*} \]

At \(t=550\) years, \[ P^\prime (550) =-0.0121e^{-0.000121(550)} \approx -0.0113 \]

The percentage of carbon-14 present in a \(550\)-year-old fossil is decreasing at the rate of 1.13% per year.

(b) When \(t=2000\) years, the rate of change is \[ P^\prime ( 2000) =-0.0121e^{-0.000121( 2000) }\approx -0.0095 \]

The percentage of carbon-14 present in a \(2000\)-year-old fossil is decreasing at the rate of 0.95% per year.