A square-root removal law has an implicit solution

In a tank at an aquarium, fish pollute the water at a constant rate. A filter is installed into a new clean tank to remove the pollutant. The rate at which it removes the pollutant from the tank is proportional to the square root of the amount of pollutant present in the tank at that instant. Immediately after being cleaned of all pollutant, the amount of pollutant in the tank is increasing at a rate of 5 units per day. It is known that eventually, with the filter installed in a new clean tank, the amount of pollutant in the tank would stabilize at 75 units.

(i) Show that dQdt=5−Q/3\frac{dQ}{dt}=5-\sqrt{Q/3}, where QQ is the amount of pollutant present at time tt. marks

(ii) Using the substitution x=Q/3x=\sqrt{Q/3}, show that the differential equation in part (i) can be written as 6xdxdt=5−x6x\frac{dx}{dt}=5-x. marks

(iii) Hence or otherwise, show that the particular solution of the differential equation in part (i) is (a−Q/3)beQ/3=mept(a-\sqrt{Q/3})^be^{\sqrt{Q/3}}=me^{pt} where a,b,p,ma,b,p,m are constants to be determined. marks

The water in the tank becomes toxic if the pollutant level gets too high. To avoid this, the tank is cleaned when the pollutant level reaches 48 units.

(iv) By comparing the time taken for the pollutant level to reach 48 units with and without the filter, comment on the effectiveness of the filter. marks

[2023/RVHS/Prelim/I/12]

Total marks: 13

Solution

(i)

Forming the equation

Show that dQdt=5−Q/3\dfrac{dQ}{dt}=5-\sqrt{Q/3}.

Hint

  1. dQdt=input−removal=A−cQ\frac{dQ}{dt}=\text{input}-\text{removal}=A-c\sqrt Q.

  2. Use the two given conditions.

Answer · (i)

Q′=5−Q/3Q\prime=5-\sqrt{Q/3}.

(ii)

Substitution

With x=Q/3x=\sqrt{Q/3}, show 6xdxdt=5−x6x\dfrac{dx}{dt}=5-x.

Hint

  1. Q=3x2Q=3x^2.

Answer · (ii)

6xx′=5−x6xx\prime=5-x.

(iii)

Implicit solution

Show that the particular solution is (a−Q/3)beQ/3=mept(a-\sqrt{Q/3})^be^{\sqrt{Q/3}}=me^{pt}, and find a,b,p,ma,b,p,m.

Explore this idea

Hint

  1. ∫6x5−xdx=∫dt\int\frac{6x}{5-x}dx=\int dt.

  2. Write 6x5−x=−6+305−x\frac{6x}{5-x}=-6+\frac{30}{5-x}.

Answer · (iii)

a=5,b=5,m=3125,p=−1/6a=5,b=5,m=3125,p=-1/6; (5−x)5ex=3125e−t/6(5-x)^5e^x=3125e^{-t/6}.

x=Q/3∈[0,5)x=\sqrt{Q/3}\in[0,5).

Explore the idea

The pollutant level

Each dash shows the slope the differential equation gives at that point. Change the starting value and the solution follows the dashes.

Starting value of Q at t = 00
Read the solution at t24.3
Try an example
Q at t = 24.3
48.05
Rate dQ/dt there
0.998
  • Solution that starts at the purple dot
Key idea

The level rises towards the equilibrium 75 more and more slowly.

Ideas to try
  1. Start the solution above 75: the level falls towards 75.
  2. Read the time when the curve reaches 48.
(iv)

Comparing times

Compare the time to reach 48 units with and without the filter.

Hint

  1. With filter: x=48/3=4x=\sqrt{48/3}=4.

  2. Without filter: Q=5tQ=5t.

Answer · (iv)

With filter: 30ln⁡5−24≈24.28330\ln5-24\approx24.283 days; without: 9.69.6 days. Effective: delays the toxic threshold by about 14.6814.68 days.