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{\bf Question}

The rate at which a liquid runs from a container is proportional
to the square root of the depth $h$ of the opening below the
surface of the liquid. A cylindrical petrol storage tank is sunk
in the ground with its axis vertical. There is a leak in the tank
at an unknown depth. The level of the petrol in the tank,
originally full, is found to drop by 20 cm in 1 hour, and by 19 cm
in the next hour. Show that the differential equation modelling
the leak is of the form

$$-\ds\frac{dh}{dt}=kh^{\frac{1}{2}},\ \ k=\rm{constant}$$

Solve this with the given data to find the depth at which the leak
is located.
\medskip

{\bf Answer}

PICTURE \vspace{0.5in}

Let height of petrol level above hole be $h$. Rate of flowing out
$=-\ds\frac{dh}{dt}$ (negative because $h$ is decreasing with $t$)

$\Rightarrow \propto \sqrt{h}$

Proportional to \lq\lq$\propto$" is the same as
\lq\lq$=k\times\cdots$" where $k$ is unknown at this stage

$\begin{array} {rrcl} \rm{Thus} & \ds\frac{dh}{dt} & = &
-k\sqrt{h}\\ \rm{or} & \ds\frac{dh}{dt} & = & -kh^{\frac{1}{2}}\ \
\rm{variables\ separable};\\ &
\ds\int\ds\frac{dh}{h^{\frac{1}{2}}}=-\ds\int k \,dt\\ \Rightarrow
& 2h^{\frac{1}{2}} & = & -kt+c\ (\star) \end{array}$

Now when $h=0$ the petrol is at the level of the hole. Thus when
$h=0$,

$$t=\ds\frac{c}{k}$$

Thus we must find $c,\ k$. To do this use the two bits of
information given. Let the height of the petrol above the hole at
$t=0$ be $H$.

\newpage
Thus $(\star)$ gives

$$2H^{\frac{1}{2}}=c$$

(1)

Now when $t=1$ hr, $h=(H-20)$cm

$$\Rightarrow 2(H-20)^{\frac{1}{2}}=-k+c$$

(2)

and when $t=2$hrs, $h=H-20-19=(H-39)$cm

$$\Rightarrow 2(H-39)^{\frac{1}{2}}=-2k+c$$

(3)

We must find $H$. If we know $c$, we know $H$ from (1). Thus
consider $2 \times (2)-(3)$:

$$4(H-20)^{\frac{1}{2}}=-2k+2c$$

$$\un{2(H-39)^{\frac{1}{2}}=-2k+c}$$

$$4(H-20)^{\frac{1}{2}}-2(H-39)^{\frac{1}{2}}=c$$

But from (1) we have $c=2H^{\frac{1}{2}}$, thus

$$4(H-20)^{\frac{1}{2}}-2(H-39)^{\frac{1}{2}}=2H^{\frac{1}{2}}$$

This looks difficult to solve, but isn't really.

$$2(H-20)^{\frac{1}{2}}-(H-39)^{\frac{1}{2}}=H^{\frac{1}{2}}$$

Square both sides:

$\left[2(H-20)^{\frac{1}{2}}-(H-39)^{\frac{1}{2}}\right]^2=
4(H-20)+(H-39)-4(H-20)^{\frac{1}{2}}(H-39)^{\frac{1}{2}}=H$

Thus

$4H-80+H-39-H=4(H-20)^{\frac{1}{2}}(H-39)^{\frac{1}{2}}$

$\ds\frac{4H-119}{4}=(H-20)^{\frac{1}{2}}(H-39)^{\frac{1}{2}}$

\newpage
Square again:

$\left(\ds\frac{4H-119}{4}\right)^2=(H-20)(H-39)$

$\Rightarrow H^2-\ds\frac{2 \times
119}{4}H+\left(\ds\frac{119}{4}\right)^2=H^2-20H-39H+(20 \times
39)$

$\Rightarrow
-\ds\frac{119}{2}H+\left(\ds\frac{119}{4}\right)^2=-59H+780$

$\Rightarrow
\left(59-\ds\frac{119}{2}\right)H=780-\left(\ds\frac{119}{4}\right)^2$

$\Rightarrow -\ds\frac{H}{2}=780-\left(|ds\frac{119}{4}\right)^2$

Thus $H=\left[\left(\ds\frac{119}{4}\right)^2-780\right]\times 2
cm = 210.125 cm$

$$\un{H \approx 2.1m}$$
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