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.

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.