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\includegraphics[max width=\textwidth, alt={}, center]{e26f21c5-3776-4c86-8440-6959c5e37486-12_444_382_258_886}
A water tank has vertical sides and a horizontal rectangular base, as shown in the diagram. The area of the base is \(2 \mathrm {~m} ^ { 2 }\). At time \(t = 0\) the tank is empty and water begins to flow into it at a rate of \(1 \mathrm {~m} ^ { 3 }\) per hour. At the same time water begins to flow out from the base at a rate of \(0.2 \sqrt { } h \mathrm {~m} ^ { 3 }\) per hour, where \(h \mathrm {~m}\) is the depth of water in the tank at time \(t\) hours.
- Form a differential equation satisfied by \(h\) and \(t\), and show that the time \(T\) hours taken for the depth of water to reach 4 m is given by
$$T = \int _ { 0 } ^ { 4 } \frac { 10 } { 5 - \sqrt { } h } \mathrm {~d} h$$
- Using the substitution \(u = 5 - \sqrt { } h\), find the value of \(T\).