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The next day — 17 hours later — he comes back to his lab and discovers that his sample is now only 37% pure. Solution We can again set up our problem using Corollary 3.3.4.
Let \(Q(t)\) denote the quantity of implausium at time \(t\text\) measured in hours.
Vegetation absorbs carbon dioxide from the atmosphere through photosynthesis and animals acquire \(^C\) by eating plants.
With great effort he has produced a sample of pure implausium.
Differentiating gives \begin \diffe^ = a e^ \end i.e. We have succeeded in guessing a single function, namely \(e^\text\) that obeys equation 3.3.1. The trick is to imagine that \(Q(t)\) is any (at this stage, unknown) solution to equation 3.3.1 and to compare \(Q(t)\) and our known solution \(e^\) by studying the ratio \(Q(t)/e^\text\) We will show that \(Q(t)\) obeys equation 3.3.1 if and only if the ratio \(Q(t)/e^\) is a constant, i.e.
if and only if the derivative of the ratio is zero.
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View the full list Radiocarbon dating has transformed our understanding of the past 50,000 years.