Browsing as Guest. Progress, bookmarks and attempts are disabled.
Log in to track your work.
June 2022 p33 q8
2287
At time t days after the start of observations, the number of insects in a population is N. The variation in the number of insects is modelled by a differential equation of the form \(\frac{dN}{dt} = kN^{\frac{3}{2}} \cos 0.02t\), where k is a constant and N is a continuous variable. It is given that when t = 0, N = 100.
(a) Solve the differential equation, obtaining a relation between N, k and t.
\((b) Given also that N = 625 when t = 50, find the value of k.\)
(c) Obtain an expression for N in terms of t, and find the greatest value of N predicted by this model.
Solution
(a) Separate variables: \(\frac{dN}{N^{\frac{3}{2}}} = k \cos 0.02t \, dt\).
Integrate both sides: \(-\frac{2}{\sqrt{N}} = 50k \sin 0.02t - 0.2\).
Use initial condition \(t = 0, N = 100\) to find constant: \(-\frac{2}{10} = -0.2\).
(b) Substitute \(N = 625\) and \(t = 50\) into the expression: \(-\frac{2}{25} = 50k \sin 1 - 0.2\).
Solve for \(k\): \(k = 0.00285\).
(c) Rearrange to find \(N\) in terms of \(t\): \(N = 4(0.2 - 0.142(607) \sin 0.02t)^{-2}\).
Find the greatest value of \(N\) by evaluating the expression at critical points.