A square-root removal law has an implicit solution · source reference

Printed question

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 $\frac{dQ}{dt}=5-\sqrt{Q/3}$, where $Q$ is the amount of pollutant present at time $t$. [3]

(ii) Using the substitution $x=\sqrt{Q/3}$, show that the differential equation in part (i) can be written as $6x\frac{dx}{dt}=5-x$. [2]

(iii) Hence or otherwise, show that the particular solution of the differential equation in part (i) is $(a-\sqrt{Q/3})^be^{\sqrt{Q/3}}=me^{pt}$ where $a,b,p,m$ are constants to be determined. [6]

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. [2]

[2023/RVHS/Prelim/I/12]

Printed working

$dQ_{\rm in}/dt=k$; $dQ_{\rm out}/dt=c\sqrt Q$, $c>0$.

$Q'=\text{rate increase}-\text{rate removal}=k-c\sqrt Q$.

New clean tank: $t=0,Q=0,Q'=5$, so $5=k-c\sqrt0\Rightarrow k=5$.

As $t\to\infty$, $Q\to75$, $Q'\to0$: $0=5-c\sqrt{75}\Rightarrow c=1/\sqrt3$. Thus $Q'=5-\sqrt{Q/3}$.

Given $x=\sqrt{Q/3}$, $Q=3x^2$ and $Q'=6xx'$. Substitute: $6xx'=5-x$.

$\int6x/(5-x)dx=\int1dt$; $6\int[-1+5/(5-x)]dx=\int1dt$.

$-x-5\ln|5-x|=t/6+c$; $x+\ln|5-x|^5=-t/6-c$.

$\ln|5-x|^5=-x-t/6-c$; $|5-x|^5=e^{-x-t/6-c}$; $(5-x)^5=\pm e^{-c}e^{-x-t/6}$.

$(5-x)^5e^x=Ae^{-t/6}$, $A=\pm e^{-c}$.

$(5-\sqrt{Q/3})^5e^{\sqrt{Q/3}}=Ae^{-t/6}$. At 0, $Q=0$: $A=5^5=3125$.

$a=5,b=5,m=3125,p=-1/6$.

At $Q=48$, $(5-\sqrt{48/3})^5e^{\sqrt{48/3}}=3125e^{-t/6}$, so $t=-6\ln(e^4/3125)=24.3$ days (3 s.f.).

Without filter, $t=48/5=9.6$ days. Therefore filter is effective.

Printed answer key

(iii) $a=5,b=5,m=3125,p=-1/6$. (iv) Filter is effective.

Teacher note 1

$dQ/dt$ is the rate of change of pollutants in the tank at time $t$.

Teacher note 2

Given a DE of form $dx/dt=f(x)$, rearrange before integrating with respect to $t$: $\frac1{f(x)}\frac{dx}{dt}=1$, hence $\int\frac1{f(x)}dx=\int1dt$. Used result: $\int f'(x)/f(x)dx=\ln|f(x)|+c$.

Teacher note 3

Fish pollute water in the tank at a constant rate of 5 units per day.

Mathematical review

(i) Write $Q'=A-c\sqrt Q$. At Q=0 the rate is 5, so A=5; equilibrium at Q=75 gives $c=5/\sqrt{75}=1/\sqrt3$.

(ii) Q=3x² gives $Q'=6xx'=5-x$.

(iii) $\int6x/(5-x)\,dx=t+C$, hence $-6x-30\ln(5-x)=t+C$. Rearrangement gives $(5-x)^5e^x=me^{-t/6}$. At t=x=0, m=5⁵=3125. Thus a=5,b=5,p=−1/6. This implicit relation determines a unique 0≤x<5 for t≥0; x′ is singular at x=0 but Q′ tends to 5.

(iv) Q=48 means x=4. Then $e^4=3125e^{-t/6}$, so $t=30\ln5-24\approx24.283$ days. Without filtering Q=5t, so the time is 48/5=9.6 days. Filtering delays this threshold by roughly 14.68 days, though eventual Q=75 still exceeds 48.

Provenance

TJC JC2 2025 H2 Maths (9758) MSM. Native transcription checked against original visible question, working and sidebars. Question page(s): 5. Working page(s): 18, 19. Original records remain private for QA; this native article has no PDF or raster worksheet dependency.