<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.

Concavity of Graphs

Applications of Differentiation

H2 Math

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

Video Solution: