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Nov 2010 p33 q9
2306
A biologist is investigating the spread of a weed in a particular region. At time \(t\) weeks after the start of the investigation, the area covered by the weed is \(A \text{ m}^2\). The biologist claims that the rate of increase of \(A\) is proportional to \(\sqrt{2A - 5}\).
(i) Write down a differential equation representing the biologist’s claim.
(ii) At the start of the investigation, the area covered by the weed was \(7 \text{ m}^2\) and, 10 weeks later, the area covered was \(27 \text{ m}^2\). Assuming that the biologist’s claim is correct, find the area covered 20 weeks after the start of the investigation.
Solution
(i) The differential equation representing the biologist's claim is:
\(\frac{dA}{dt} = k\sqrt{2A - 5}\)
(ii) To solve the differential equation, separate variables and integrate:
\(\int \frac{1}{\sqrt{2A - 5}} \, dA = \int k \, dt\)
Integrating both sides, we obtain:
\((2A - 5)^{1/2} = kt + C\)
Using the initial condition \(t = 0\) and \(A = 7\), we find \(C\):
\((2 \times 7 - 5)^{1/2} = C\)
\(C = 3\)
Using \(t = 10\) and \(A = 27\) to find \(k\):
\((2 \times 27 - 5)^{1/2} = 10k + 3\)
\(7 = 10k + 3\)
\(k = 0.4\)
Substitute \(t = 20\), \(C = 3\), and \(k = 0.4\) to find \(A\):