MediumMathematicsHigh School (11–12)Functions and CalculusNumerical

A team of archaeologists finds charcoal in an old hearth…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A team of archaeologists finds charcoal in an old hearth. Living wood has a steady fraction of carbon-14, and after death the carbon-14 decays with a half-life of $5730$ years. This half-life is the standard value, given here. The charcoal has $62\%$ of the carbon-14 of living wood. Find the decay constant $\lambda$ in per year, the age of the charcoal in years, and the time, from death, in which the carbon-14 falls to $25\%$ .

Need a reminder? The key ideaShow

A rate proportional to the amount gives $\dfrac{dN}{dt} = kN$ and $N = N_0 e^{kt}$ . The doubling time or the half-life is $\dfrac{\ln 2}{|k|}$ . The logistic equation $\dfrac{dP}{dt} = kP\left(1 - \dfrac{P}{M}\right)$ has the solution $P = \dfrac{M}{1 + Ae^{-kt}}$ with $A = \dfrac{M - P_0}{P_0}$ .

\displaystyle N = N_0 e^{kt}.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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