[剑桥/9709/2009] 一个地下储水罐正在被注满液体（如图所示）。初始时储水罐是空的。在开始注水后 $t$ 小时，液体的体积为 $V\text{ m}^3$，液体的深度为 $h\text{ m}$。已知 $V = \frac{4}{3}h^3$。 液体以每小时 $20\text{ m}^3$ 的速率注入，但由于泄漏，液体流失的速率与 $h^2$ 成正比。当 $h = 1$ 时，$\frac{\text{d}h}{\text{d}t} = 4.95$。

1. Express the rate of change of volume $\frac{\text{d}V}{\text{d}t}$ in terms of $h$ and an unknown constant of proportionality $k$.

The net rate of change of volume is the rate at which liquid is poured in minus the rate of leakage.
$\frac{\text{d}V}{\text{d}t} = 20 - kh^2$ where $k$ is a constant of proportionality.

2. Find $\frac{\text{d}V}{\text{d}h}$ from the given volume formula.

Given $V = \frac{4}{3}h^3$, differentiating with respect to $h$ gives:
$\frac{\text{d}V}{\text{d}h} = 4h^2$

3. Relate $\frac{\text{d}h}{\text{d}t}$ to $\frac{\text{d}V}{\text{d}t}$ using the chain rule.

By the chain rule for connected rates of change:
$\frac{\text{d}V}{\text{d}t} = \frac{\text{d}V}{\text{d}h} \cdot \frac{\text{d}h}{\text{d}t}$
Substituting our expressions:
$20 - kh^2 = 4h^2 \frac{\text{d}h}{\text{d}t}$
Dividing both sides by $4h^2$:
$\frac{\text{d}h}{\text{d}t} = \frac{20 - kh^2}{4h^2} = \frac{5}{h^2} - \frac{k}{4}$

4. Use the initial conditions to find the value of the constant term and obtain the required differential equation.

We are given that when $h = 1$, $\frac{\text{d}h}{\text{d}t} = 4.95$. Substituting these values into the equation gives:
$4.95 = \frac{5}{1^2} - \frac{k}{4}$
$4.95 = 5 - \frac{k}{4}$
$\frac{k}{4} = 5 - 4.95 = 0.05 = \frac{1}{20}$
Substituting $\frac{k}{4} = \frac{1}{20}$ back into the differential equation gives:
$\frac{\text{d}h}{\text{d}t} = \frac{5}{h^2} - \frac{1}{20} \quad \text{(shown)}$

1. Express the right-hand side over a common denominator and simplify to match the left-hand side.

$\text{RHS} = -20 + \frac{2000}{(10 - h)(10 + h)}$
$= -20 + \frac{2000}{100 - h^2}$
$= \frac{-20(100 - h^2) + 2000}{100 - h^2}$
$= \frac{-2000 + 20h^2 + 2000}{100 - h^2}$
$= \frac{20h^2}{100 - h^2}$
$= \text{LHS}$

1. Separate the variables in the differential equation.

From part (i), we have the differential equation:
$\frac{\text{d}h}{\text{d}t} = \frac{5}{h^2} - \frac{1}{20}$
Combine the terms on the right-hand side over a common denominator:
$\frac{\text{d}h}{\text{d}t} = \frac{100 - h^2}{20h^2}$
Separate the variables $h$ and $t$:
$\int \frac{20h^2}{100 - h^2} \,\text{d}h = \int 1 \,\text{d}t$

2. Substitute the identity from part (ii) and integrate both sides.

Substitute the algebraic identity given in part (ii):
$\int \left( -20 + \frac{2000}{(10 - h)(10 + h)} \right) \,\text{d}h = \int 1 \,\text{d}t$
$\int \left( -20 + \frac{2000}{100 - h^2} \right) \,\text{d}h = \int 1 \,\text{d}t$
Integrate using the standard result $\int \frac{1}{a^2-x^2}\,\text{d}x = \frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right|$:
$-20h + 2000 \left( \frac{1}{2(10)} \ln\left|\frac{10+h}{10-h}\right| \right) = t + C$
$-20h + 100 \ln\left|\frac{10+h}{10-h}\right| = t + C$
(Note: This can equivalently be integrated by expanding $\frac{2000}{(10-h)(10+h)}$ into partial fractions $\frac{100}{10-h} + \frac{100}{10+h}$).

3. Use the initial condition to find the constant of integration.

The problem states that the tank is initially empty. Therefore, when $t = 0$, the depth $h = 0$.
Substitute these values into the integrated equation to find $C$:
$-20(0) + 100 \ln\left|\frac{10+0}{10-0}\right| = 0 + C$
$100 \ln(1) = C$
$C = 0$

4. Express $t$ in terms of $h$.

Substitute $C = 0$ back into the equation:
$t = 100 \ln\left|\frac{10+h}{10-h}\right| - 20h$
Since the tank is filling starting from $h = 0$, we have $0 \le h < 10$. This ensures that $\frac{10+h}{10-h} > 0$, so the absolute value signs can be dropped.
$t = 100 \ln\left(\frac{10+h}{10-h}\right) - 20h$