\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\dy}{\frac{dy}{dx}}
\parindent=0pt
\begin{document}
{\bf Question}
$(*)$ A wooden post is excavated from some ruins in the centre of
Southampton and the University is asked to determine it's probable
age. It is known that all living matter has a certain fraction of
its carbon as Carbon 14 and the remaining fraction as Carbon 12.
Once the matter has died the Carbon 14 decays radioactively to
Carbon 12 at a rate proportional to the concentration of Carbon 14
and such that exactly half the Carbon 14 decays to Carbon 12 in
5568 years (called the half-life). Write down the ODE for the
concentration of Carbon 14 in dead wood and determine the
constants in this equation. Measurements on the post show that it
has lost 22\% of the amount of Carbon 14 that the living post
would have. How old is the post?
\vspace{0.25in}
{\bf Answer}
Concentration of Carbon 14 = $\ds C$, the decay is generated by
$\ds \frac{dC}{dt}=-kC$, and the solution is $\ds C=Ae^{-kt}$.
If $\ds C=C_0 \rm{\, at \,\,} t=0 \rm{\, then \,\,}
C=\frac{C_0}{2} \rm{\, at \,\,} t=5568 \rm{\, years}$
$\ds \Rightarrow A=C_0 \Rightarrow \frac{C_0}{2}=C_0e^{-k(5568)}$
$\ds \Rightarrow k=\frac{1}{5568}\ln2\approx0.0001245$
For the age of the wooden post we need to find $\ds T$ where
$\ds C=C_0 \rm{\, at \,\,} t=T, \rm{\, and \,\,} C=(1-0.22)C_0
\rm{\, at \,\,} t=2000.$
So $\ds C=Ae^{-\left(\frac{1}{5568}\ln2\right)t}$
and $\ds C_0=Ae^{-\left(\frac{1}{5568}\ln2\right)T}$ and $\ds
0.78C_0=Ae^{-\left(\frac{1}{5568}\ln2\right)(2000)}$
eliminating $\ds C_0$ and $\ds A \Rightarrow
0.78=e^{\left(\frac{1}{5568}\ln2\right)(T-2000)}$
$\ds T=2000+5568\frac{\ln0.78}{\ln2}\approx 4$ AD.
\end{document}