2015 NJC P1 Q10

The population (in thousands) of fish present in a lake at time \(t\) years is denoted by \(x\). It is found that the growth rate of \(x\) is proportional to \(\left( 200-2t-x \right)\).

It is given that the initial population of the fish in the lake is \(8000\) and the population grows at a rate of \(16000\) per year initially. Show that the growth rate of \(x\) at time \(t\) years can be modelled by the differential equation

\(\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{200-2t-x}{12}\).[2]

Find \(x\) in terms of \(t\) by using the substitution \(u=2t+x\). Deduce, to the nearest number of years, the time taken for the fish to die out in this lake.[6]

It is given that the solution curve that describes that population size of the fish at time \(t\) years intersects the graph of \(x=200-2t\) at the point \(\left( {{t}_{1}},\mathrm{ }{{x}_{1}} \right)\). Describe, in context, what \({{t}_{1}}\) and \({{x}_{1}}\) represent.[2]