Practice section 6 of 6

A town has population $3000$. The number infected after $t$ days satisfies $dI/dt=kI(3000-I)$, $k>0$.

(i) Given $I(0)=30$, show $I=3000/(1+99e^{-3000kt})$. [6]

(ii) Given $I(7)=240$, find when $90\%$ of the population is infected. [3]

(iii) State one contextual assumption needed for the model. [1]

An insect population $N$ after $t$ weeks satisfies $dN/dt=N-kN^2$, $k>0$. Show, rather than merely verify, the general solution $N=1/(k+Ae^{-t})$. [4]

Initially there are $250$ insects and the population tends to $10000$ after a very long time. Find the time to reach three times the initial population, to the nearest day. [3]

Find the general solution of $dy/dt=ky(1-y)$, $0<y<1$, $k>0$, in the form $y=f(t)$. [3]

Find the particular solution with $y(0)=1/4$, and sketch it. [3]

Here $y$ models the proportion infected by a disease. Explain why the differential equation may be suitable. [2]

A tank initially contains $2\,\mathrm{m}^3$. Water enters at $4\,\mathrm{m}^3\mathrm{s}^{-1}$ and leaves at a rate proportional to the volume $V$. When $V=6\,\mathrm{m}^3$, the net rate is $1\,\mathrm{m}^3\mathrm{s}^{-1}$. Set up and solve a differential equation to show $V=8-6e^{-0.5t}$. [7]

(i) Describe what happens to the volume eventually. [1]

(ii) Sketch $V$ against $t$. [2]

The given diagram has quantity $Q$ tonnes on the horizontal axis and price $P$ dollars per kg on the vertical axis. Consumer curve $C_1$ and producer line $C_2$ meet at $A=(1,4)$. Consumer quantity is inversely proportional to price, while producer quantity is directly proportional to price.

(i) Interpret $A$ in context.

(ii) Find both equations, expressing $Q$ in terms of $P$.

A shortage occurs when consumer quantity exceeds producer quantity. The rate of increase of price after $t$ months is proportional to this shortage.

(iii) Given an initial price of $3 and a price of $3.65 after one month, find $P(t)$ and sketch it, showing its long-term behaviour. Suggest why producers might use $P=aQ+b$, $a,b>0$, instead of $C_2$.