Separate the formal model from inconsistent pump directions · source reference

Printed question

The cylindrical tank in a research laboratory has a cross-sectional area of 4 m². To cool the tank, water is pumped in and out of the tank simultaneously. The volume and height of the water in the tank at any time $t$ minutes is given by $V$ (m³) and $h$ (metres) respectively. Clean water is pumped into the tank at a rate that is proportional to $h^2$ and the water is pumped out from the tank at a rate that is proportional to $h$.

(i) Assume that the water does not overflow and that there is no change to the height of the water when $h=10$, show that $\frac{dh}{dt}=\frac{kh(h-10)}4$ where $k$ is a real constant. [4]

The tank was initially filled with clean water to a height of 2 metres. When the height of the water is 5 metres, the volume of water is increasing at a rate of 5.5 m³ per minute.

(ii) Find the exact value of $k$. Hence find $h$ in terms of $t$. [5]

(iii) Sketch a graph of $h$ against $t$. Hence write down the minimum height of the cylindrical tank that will not result in the overflow of the water. [3]

[2022/ASRJC/Prelim/I/11]

Printed working

$V'=\text{rate pumped in}-\text{rate pumped out}=V'_{\rm in}-V'_{\rm out}=Ah^2-Bh$, $A,B\in\mathbb R$ as printed.

At $h=10$, $V'=0\Rightarrow B=10A$. Given base area 4 m², $V=4h$, so $V'=4h'$ (1).

$4h'=Ah^2-10Ah\Rightarrow h'=kh(h-10)/4$, where $A=k$ (2).

At $h=5$, $V'=5.5$: $5.5=[5k(5-10)/4]\times4\Rightarrow k=-11/50$.

$h'=-11h(h-10)/200$; $\int dh/(h^2-10h)=\int-11/200dt$.

Method 1 (MF27): $\int dh/[(h-5)^2-5^2]=-11t/200+c$; $\frac1{10}\ln|(h-10)/h|=-11t/200+c$.

Method 2 (partial fractions): $\int dh/[h(h-10)]=-\frac1{10}\int[1/h-1/(h-10)]dh=-\frac1{10}[\ln|h|-\ln|h-10|]=\frac1{10}\ln|(h-10)/h|$.

$\ln|(h-10)/h|=-11t/20+10c$; $|1-10/h|=e^{-11t/20+10c}$; $1-10/h=\pm e^{-11t/20+10c}$.

$10/h=1+Ae^{-11t/20}$, $A=\pm e^{10c}$; $h=10/(1+Ae^{-11t/20})$.

At 0, $h=2$: $2=10/(1+A)\Rightarrow A=4$. Thus $h=10/(1+4e^{-11t/20})$.

As $t\to\infty$, $e^{-11t/20}\to0$, $h\to10$. The source gives minimum tank height 10 metres. This formal solution’s conflict with positive physical pump rates is reviewed separately.

Printed answer key

(ii) $-11/50$; $h=10/(1+4e^{-11t/20})$. (iii) 10 metres; native sketch below.

Teacher note 1

$dV/dt=dV_{\rm in}/dt-dV_{\rm out}/dt$.

1. You cannot express both the rate pumped in and rate pumped out as $dV/dt$; the net derivative is neither $Ah^2$ alone nor $-Bh$ alone.
2. Constants of proportionality for the two rates need not be the same.
3. Avoid using $k$ initially, since it represents the final constant in the DE to be shown. Use different letters, e.g. $A,B$.
4. Need to express the DE in terms of one constant: use the given condition to express $A$ in terms of $B$ or vice versa.
5. Need to show clearly how $A$ or $B$ relates to $k$.

Recall that a “Show” question needs clear explanation. No change in water height when $h=10$ implies $dV/dt=0$ there.

Teacher note 2

Given $dh/dt=f(h)$, integrate $\int dh/f(h)=\int1dt$. Used result: $\int f'(x)/(f(x)^2-a^2)dx=\frac1{2a}\ln|(f(x)-a)/(f(x)+a)|+c$.

Teacher note 3

As $t\to\infty$, $e^{-11t/20}\to0$, so $h\to10$. Horizontal asymptote $h=10$.

Diagram 10-17-height

Mathematical review

(i) Write inflow $Ah^2$ and outflow $Bh$. Since V=4h, $4h'=Ah^2-Bh$. Equilibrium at h=10 gives B=10A, hence the stated form with k=A.

(ii) Using the stated datum formally, $5.5=k\cdot5(5-10)=-25k$, so $k=-11/50$. The formal DE is $h'=(11/200)h(10-h)$. Separation gives $\ln[h/(10-h)]=(11/20)t+C$. Initial h=2 gives odds 1/4, hence $h=10/(1+4e^{-11t/20})$. Substitution verifies both the formal DE and V′=5.5 at h=5.

(iii) This formal graph starts at (0,2), rises sigmoidally through inflection h=5 and tends to 10 from below. It suggests a minimum tank height 10 m. However, it requires A=k<0 and B=10A<0 in the original inflow/outflow description: these are negative pumped volumes. No positive physical pump-rate constants satisfy all the supplied statements. Swapping the proportional laws (inflow ∝h, outflow ∝h²) would produce the formal logistic behavior, but that is a proposed repair, not the printed question.

Provenance

TJC JC2 2025 H2 Maths (9758) MSM. Native transcription checked against original visible question, working and sidebars. Question page(s): 6. Working page(s): 27, 28. Original records remain private for QA; this native article has no PDF or raster worksheet dependency.