<p>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, $V$cm$^{3}$, is given by $V=\frac{1}{3}\pi {{x}^{2}}\left( 3a-x \right)$. Water flows from the tank, through an outlet at its lowest point, at a rate $\pi k\sqrt{x}$cm$^{3}$ s$^{-1}$, where&nbsp;$k$ is a positive constant.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Show that&nbsp;$\left( 2ax-{{x}^{2}} \right)\frac{\mathrm{d}x}{\mathrm{d}t}=-k\sqrt{x}$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Find the general solution for&nbsp;$t$ in terms of $x$,$a$ and $k$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Find the ratio ${{T}_{1}}:{{T}_{2}}$, where&nbsp;${{T}_{1}}$ 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.<span class="inline-marks">[4]</span></div></li></ol>

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, $V$cm$^{3}$, is given by $V=\frac{1}{3}\pi {{x}^{2}}\left( 3a-x \right)$. Water flows from the tank, through an outlet at its lowest point, at a rate $\pi k\sqrt{x}$cm$^{3}$ s$^{-1}$, where $k$ is a positive constant.

Show that $\left( 2ax-{{x}^{2}} \right)\frac{\mathrm{d}x}{\mathrm{d}t}=-k\sqrt{x}$.

Find the general solution for $t$ in terms of $x$,$a$ and $k$.

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.

Differential Equations - General

Differential Equations

International Baccalaureate Higher Level Mathematics

IB HL Math

2019 NYJC P1 Q7

Given: $\dfrac{\mathrm{d}V}{\mathrm{d}t} = -\pi k \sqrt{x}$

$V = \dfrac{1}{3}\pi x^2(3a - x) = \pi a x^2 - \dfrac{1}{3}\pi x^3$

$\dfrac{\mathrm{d}V}{\mathrm{d}t} = 2\pi ax \dfrac{\mathrm{d}x}{\mathrm{d}t} - \pi x^2 \dfrac{\mathrm{d}x}{\mathrm{d}t}$

$-\pi k \sqrt{x} = (2ax - x^2)\pi \dfrac{\mathrm{d}x}{\mathrm{d}t}$

$(2ax - x^2)\dfrac{\mathrm{d}x}{\mathrm{d}t} = -k\sqrt{x}$

$\displaystyle\displaystyle\int \dfrac{2ax - x^2}{\sqrt{x}} \, \mathrm{d}x = -k \displaystyle\int 1 \, dt$

$\displaystyle\displaystyle\int 2ax^{\frac{1}{2}} - x^{\frac{3}{2}} \, \mathrm{d}x = -kt$

$\dfrac{2ax^{\frac{3}{2}}}{\frac{3}{2}} - \dfrac{x^{\frac{5}{2}}}{\frac{5}{2}} + C = -kt$

$-kt = \dfrac{4}{3}ax^{\frac{3}{2}} - \dfrac{2}{5}x^{\frac{5}{2}} + C$

$t = -\dfrac{1}{k}\left[\dfrac{4}{3}ax^{\frac{3}{2}} - \dfrac{2}{5}x^{\frac{5}{2}} + C\right]$

Case of $T_1$: When $t = 0$, $x = 2a$

$0 = \dfrac{4}{3}a(2a)^{\frac{3}{2}} - \dfrac{2}{5}(2a)^{\frac{5}{2}} + C_1$

$C_1 = -\dfrac{4}{3}(2)^{\frac{3}{2}}a^{\frac{5}{2}} + \dfrac{2}{5}(2)^{\frac{5}{2}}a^{\frac{5}{2}}$

$= -\dfrac{4}{3}\sqrt{2}^3 a^{\frac{5}{2}} + \dfrac{2}{5}\sqrt{2}^5 a^{\frac{5}{2}}$

$= \left(-\dfrac{8}{3}\sqrt{2} + \dfrac{2}{5}(4\sqrt{2})\right)a^{\frac{5}{2}} = -\dfrac{16\sqrt{2}}{15}a^{\frac{5}{2}}$

Case of $T_2$: When $t = 0$, $x = a$

$0 = \dfrac{4}{3}a(a)^{\frac{3}{2}} - \dfrac{2}{5}(a)^{\frac{5}{2}} + C_2$

$C_2 = -\dfrac{4}{3}a^{\frac{5}{2}} + \dfrac{2}{5}a^{\frac{5}{2}} = -\dfrac{14}{15}a^{\frac{5}{2}}$

For case of $T_1$: $T_1 = -\dfrac{1}{k}[0 + C_1]$

For case of $T_2$: $T_2 = -\dfrac{1}{k}[0 + C_2]$

$\therefore \dfrac{T_1}{T_2} = \dfrac{C_1}{C_2} = \dfrac{-\frac{16\sqrt{2}}{15}}{-\frac{14}{15}} = \dfrac{8\sqrt{2}}{7}$

2019 NYJC P1 Q7

2019 NYJC P1 Q7

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