12.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{29b56d51-120a-4275-a761-8b8aed7bca54-40_471_949_219_493}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{figure}
Figure 4 shows a right cylindrical water tank. The diameter of the circular cross section of the tank is 4 m and the height is 2.25 m . Water is flowing into the tank at a constant rate of \(0.4 \pi \mathrm {~m} ^ { 3 } \mathrm {~min} ^ { - 1 }\). There is a tap at a point \(T\) at the base of the tank. When the tap is open, water leaves the tank at a rate of \(0.2 \pi \sqrt { h } \mathrm {~m} ^ { 3 } \mathrm {~min} ^ { - 1 }\), where \(h\) is the height of the water in metres.
- Show that at time \(t\) minutes after the tap has been opened, the height \(h \mathrm {~m}\) of the water in the tank satisfies the differential equation
$$20 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 2 - \sqrt { h }$$
At the instant when the tap is opened, \(t = 0\) and \(h = 0.16\)
- Use the differential equation to show that the time taken to fill the tank to a height of 2.25 m is given by
$$\int _ { 0.16 } ^ { 2.25 } \frac { 20 } { 2 - \sqrt { h } } \mathrm {~d} h$$
Using the substitution \(h = ( 2 - x ) ^ { 2 }\), or otherwise,
- find the time taken to fill the tank to a height of 2.25 m .
Give your answer in minutes to the nearest minute.
(Solutions based entirely on graphical or numerical methods are not acceptable.)