Draining a sphere: height is not volume · source reference

Printed question

A spherical tank with negligible thickness and internal radius $a$ cm contains water. At time $t$ s, the water surface is at a height $x$ cm above the lowest point of the tank and the volume of water in the tank, $V$ cm³, is given by $V=\frac13\pi x^2(3a-x)$. Water flows from the tank, through an outlet at its lowest point, at a rate $\pi k\sqrt x$ cm³ s⁻¹, where $k$ is a positive constant.

(i) Show that $(2ax-x^2)\frac{dx}{dt}=-k\sqrt x$. [2]

(ii) Find the general solution for $t$ in terms of $x,a,k$. [3]

(iii) Find the ratio $T_1:T_2$, where $T_1$ is the time taken to empty the tank when initially it is completely full, and $T_2$ is the time taken to empty the tank when initially it is half full. [4]

[2019/NYJC/Prelim/I/7]

Printed working

$V=\pi x^2(3a-x)/3$; $dV/dt=(2\pi ax-\pi x^2)dx/dt$.

Since $dV/dt=-\pi k\sqrt x$, $(2\pi ax-\pi x^2)dx/dt=-\pi k\sqrt x\Rightarrow(2ax-x^2)dx/dt=-k\sqrt x$.

$\int(2ax-x^2)/\sqrt x\,dx=\int-kdt$; $\int(2a\sqrt x-x^{3/2})dx=-kt+c$; $4ax^{3/2}/3-2x^{5/2}/5=-kt+c$.

$t=[c-4ax^{3/2}/3+2x^{5/2}/5]/k$.

Initially full: $x=2a$ when $t=0$, so $c=(4/3)a(2a)^{3/2}-(2/5)(2a)^{5/2}=(16/15)a^2\sqrt{2a}$. Thus $T_1=c/k=(16/(15k))a^2\sqrt{2a}$.

Initially half full: $x=a$ when $t=0$, so $c=(4/3)a(a)^{3/2}-(2/5)a^{5/2}=(14/15)a^2\sqrt a$. Thus $T_2=(14/(15k))a^2\sqrt a$.

$T_1/T_2=16a^2\sqrt{2a}/(14a^2\sqrt a)=8\sqrt2/7$; required ratio $8\sqrt2:7$.

Printed answer key

(ii) $t=[c-4ax^{3/2}/3+2x^{5/2}/5]/k$. (iii) $8\sqrt2:7$.

Teacher note 1

Since water flows from the tank, volume decreases as time increases, so $dV/dt<0$. Need to put in the negative sign for the rate.

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 $\int\frac1{f(x)}dx=\int1dt$.

Teacher note 3

“Initially” means when $t=0$.

Mathematical review

(i) $dV/dx=\pi(2ax-x^2)$ and $dV/dt=-\pi k\sqrt x$. The chain rule gives the equation for $0<x<2a$.

(ii) $dt/dx=-(2a\sqrt x-x^{3/2})/k$. Thus $t=[C-(4a/3)x^{3/2}+(2/5)x^{5/2}]/k$.

(iii) Taking the initial time as zero and final height zero, $T_1=[(4a/3)(2a)^{3/2}-(2/5)(2a)^{5/2}]/k=16\sqrt2a^{5/2}/(15k)$. Half volume occurs at x=a, so $T_2=14a^{5/2}/(15k)$. Therefore $T_1:T_2=8\sqrt2:7$. The initial full-height derivative is singular, but the integrated emptying time is finite.

The continuation row on solution page 4 is numbered “6 (iii)” in print; its tank-emptying calculation belongs to Q5(iii).

Provenance

TJC JC2 2025 H2 Maths (9758) MSM. Native transcription checked against original visible question, working and sidebars. Question page(s): 1. Working page(s): 4. Original records remain private for QA; this native article has no PDF or raster worksheet dependency.