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Nov 2004 p3 q10
2313
A rectangular reservoir has a horizontal base of area 1000 m2. At time \(t = 0\), it is empty and water begins to flow into it at a constant rate of 30 m3s-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 \(h = 1\), \(\frac{dh}{dt} = 0.02\).
(i) Show that \(h\) satisfies the differential equation \(\frac{dh}{dt} = 0.01(3 - \sqrt{h})\).
It is given that, after making the substitution \(x = 3 - \sqrt{h}\), the equation in part (i) becomes \((x - 3) \frac{dx}{dt} = 0.005x\).
(ii) Using the fact that \(x = 3\) when \(t = 0\), solve this differential equation, obtaining an expression for \(t\) in terms of \(x\).
(iii) Find the time at which the depth of water reaches 4 m.
Solution
(i) The rate of change of volume \(\frac{dV}{dt}\) is given by the inflow minus the outflow: \(\frac{dV}{dt} = 30 - k\sqrt{h}\). Since \(V = 1000h\), \(\frac{dV}{dt} = 1000 \frac{dh}{dt}\). Equating gives \(1000 \frac{dh}{dt} = 30 - k\sqrt{h}\). Given \(\frac{dh}{dt} = 0.02\) when \(h = 1\), we find \(k = 10\). Thus, \(\frac{dh}{dt} = 0.01(3 - \sqrt{h})\).
(ii) Substitute \(x = 3 - \sqrt{h}\), then \(\sqrt{h} = 3 - x\) and \(h = (3 - x)^2\). Differentiate to get \(\frac{dh}{dt} = -2(3-x) \frac{dx}{dt}\). Substitute into the differential equation to get \(-2(3-x) \frac{dx}{dt} = 0.01(3 - (3-x))\), simplifying to \((x-3) \frac{dx}{dt} = 0.005x\). Separate variables and integrate: \(\int \frac{x-3}{x} \, dx = \int 0.005 \, dt\). This gives \(x - 3 \ln x = 0.005t + C\). Using \(x = 3\) when \(t = 0\), find \(C = 0\). Thus, \(t = 200(x - 3 \ln x + 3 \ln 3)\).
(iii) When \(h = 4\), \(\sqrt{h} = 2\), so \(x = 3 - 2 = 1\). Substitute \(x = 1\) into the expression for \(t\): \(t = 200(1 - 3 \ln 1 + 3 \ln 3) = 200(1 + 3 \ln 3)\). Calculate to find \(t = 259\) seconds.
