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

1. Show that \(h\) satisfies the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.01 ( 3 - \sqrt { } h )$$ It is given that, after making the substitution \(x = 3 - \sqrt { } h\), the equation in part (i) becomes $$( x - 3 ) \frac { \mathrm { d } x } { \mathrm {~d} t } = 0.005 x$$
2. Using the fact that \(x = 3\) when \(t = 0\), solve this differential equation, obtaining an expression for \(t\) in terms of \(x\).
3. Find the time at which the depth of water reaches 4 m .