2018 AJC MYE P1 Q11

As part of the décor of a seafood restaurant, the owner places a fish tank filled with freshwater fish near the entrance to the restaurant. Oxygenated water is pumped into the tank and flows out into a sump to ensure that the water in the tank does not overflow. At time \(t\) seconds, the depth of water in the tank is \(h\) meters. The relationship between \(t\) and \(h\) can be modelled by the differential equation \(\frac{\mathrm{d}h}{\mathrm{d}t}=\frac{{{\left( 2{{h}^{2}}-3 \right)}^{2}}}{h}\).

The depth of water in the tank is \(1\) metre initially.

1. Find \(h\) in terms of \(t\) and the exact time taken for the depth to reach \(1.2\) metres.[5]
2. Sketch a graph of \(h\) against \(t\). Based on this, suggest what the restaurant owner should take note of regarding the height of the fish tank.[2]

The owner decides to change the fish in the tank to saltwater fish. To do so, he adjusts the salinity of the water in the tank. The tank initially contains \(500\) litres of water with \(22\) kg of salt dissolved in it. Salt water with a concentration of \(0.025\) kg of salt per litre is pumped into the tank at a rate of \(8\) litres per minute. The thoroughly mixed water flows out at the same rate. If the amount of salt in the tank is \(M\)kg at time \(T\) minutes,

3. Show that \(\frac{\mathrm{d}M}{\mathrm{d}T}=0.2-0.016M\).[1]
4. By solving this differential equation, find the time taken for the concentration of the salt to reach the salinity level of \(0.035\) kg per litre.[4]