[Cambridge/9709/2004] A rectangular reservoir has a horizontal base of area 1000 $\text{m}^2$. At time $t = 0$, it is empty and water begins to flow into it at a constant rate of 30 $\text{m}^3\text{ s}^{-1}$. At the same time, water begins to flow out at a rate proportional to $\sqrt{h}$, where $h$ m is the depth of the water at time $t$ s. When $h = 1$, $\frac{\text{d}h}{\text{d}t} = 0.02$.

1. Express the rate of change of the volume of water with respect to time.

Let $V$ be the volume of water in the reservoir. The net rate of change of volume is the inflow rate minus the outflow rate:
$\frac{\text{d}V}{\text{d}t} = 30 - k\sqrt{h}$
where $k$ is a constant of proportionality.

2. Relate the rate of change of volume to the rate of change of depth.

Since the reservoir has a constant horizontal base area of 1000 $\text{m}^2$, the volume is $V = 1000h$.
Differentiating with respect to $t$ gives:
$\frac{\text{d}V}{\text{d}t} = 1000\frac{\text{d}h}{\text{d}t}$
Substituting this into our initial equation:
$1000\frac{\text{d}h}{\text{d}t} = 30 - k\sqrt{h}$

3. Use the given conditions to find the constant $k$.

We are given that when $h = 1$, $\frac{\text{d}h}{\text{d}t} = 0.02$. Substituting these values into the equation:
$1000(0.02) = 30 - k\sqrt{1}$
$20 = 30 - k$
$k = 10$

4. Substitute $k$ back and rearrange to show the target differential equation.

Substitute $k = 10$ into the differential equation:
$1000\frac{\text{d}h}{\text{d}t} = 30 - 10\sqrt{h}$
Divide both sides by 1000:
$\frac{\text{d}h}{\text{d}t} = \frac{30}{1000} - \frac{10}{1000}\sqrt{h}$
$\frac{\text{d}h}{\text{d}t} = 0.03 - 0.01\sqrt{h}$
Factorising out $0.01$ gives the required result:
$\frac{\text{d}h}{\text{d}t} = 0.01(3 - \sqrt{h}) \quad \text{(shown)}$

1. Separate the variables to prepare for integration.

Divide both sides by $x$ and rearrange to isolate $x$ on one side and $t$ on the other:
$\frac{x - 3}{x} \frac{\text{d}x}{\text{d}t} = 0.005$
$\int \left( 1 - \frac{3}{x} \right) \text{d}x = \int 0.005 \, \text{d}t$

2. Integrate both sides.

$x - 3\ln x = 0.005t + C$
(Note: $x > 0$ since $h$ starts at $0$ making $x=3$, so the absolute value in $\ln|x|$ is not necessary.)

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

Substitute $x = 3$ when $t = 0$ into the integrated equation:
$3 - 3\ln 3 = 0.005(0) + C$
$C = 3 - 3\ln 3$

4. Rearrange to express $t$ in terms of $x$.

Substitute $C$ back into the equation:
$x - 3\ln x = 0.005t + 3 - 3\ln 3$
$0.005t = x - 3 - 3\ln x + 3\ln 3$
Since $\frac{1}{0.005} = 200$, multiply the entire equation by $200$:
$t = 200(x - 3 - 3\ln x + 3\ln 3)$
$t = 200 \left( x - 3 + 3\ln \left(\frac{3}{x}\right) \right)$

1. Solve the differential equation from part (ii) to find $t$ in terms of $x$ (since this result is required for the target).

$\int \frac{x - 3}{x} \, \text{d}x = \int 0.005 \, \text{d}t$
$\int \left(1 - \frac{3}{x}\right) \, \text{d}x = 0.005t + C$
$x - 3 \ln x = 0.005t + C \quad (\text{since } x > 0)$
Substituting the initial conditions $t = 0$ and $x = 3$:
$3 - 3 \ln 3 = C$
$0.005t = x - 3 \ln x - 3 + 3 \ln 3$

2. Determine the value of $x$ when the water depth reaches 4 m.

Given the substitution $x = 3 - \sqrt{h}$, when $h = 4$:
$x = 3 - \sqrt{4}$
$x = 1$

3. Substitute $x = 1$ into the solution to find $t$.

$0.005t = 1 - 3 \ln 1 - 3 + 3 \ln 3$
$0.005t = -2 + 3 \ln 3$
$t = \frac{3 \ln 3 - 2}{0.005}$
$t = 200(3 \ln 3 - 2)$
$t \approx 259.17$
$t \approx 259 \text{ s} \quad (\text{to 3 s.f.})$