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Problem 2300
2300
A tank containing water is in the form of a cone with vertex C. The axis is vertical and the semi-vertical angle is 60°, as shown in the diagram. At time t = 0, the tank is full and the depth of water is H. At this instant, a tap at C is opened and water begins to flow out. The volume of water in the tank decreases at a rate proportional to \(\sqrt{h}\), where h is the depth of water at time t. The tank becomes empty when \(t = 60\).
(i) Show that h and t satisfy a differential equation of the form \(\frac{dh}{dt} = -Ah^{-\frac{3}{2}}\), where A is a positive constant.
(ii) Solve the differential equation given in part (i) and obtain an expression for t in terms of h and H.
(iii) Find the time at which the depth reaches \(\frac{1}{2}H\).
[The volume V of a cone of vertical height h and base radius r is given by \(V = \frac{1}{3} \pi r^2 h\).]
Solution
(i) The volume of the cone is given by \(V = \frac{1}{3} \pi r^2 h\). Given the semi-vertical angle is 60°, \(r = h \tan(60^{\circ}) = \sqrt{3}h\). Thus, \(V = \pi h^3\). The rate of change of volume is \(\frac{dV}{dt} = -k\sqrt{h}\). Using \(V = \pi h^3\), \(\frac{dV}{dh} = 3\pi h^2\). Therefore, \(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt} = 3\pi h^2 \cdot \frac{dh}{dt}\). Equating gives \(3\pi h^2 \cdot \frac{dh}{dt} = -k\sqrt{h}\), leading to \(\frac{dh}{dt} = -\frac{k}{3\pi} h^{-\frac{3}{2}}\). Let \(A = \frac{k}{3\pi}\), then \(\frac{dh}{dt} = -Ah^{-\frac{3}{2}}\).
(ii) Separate variables: \(\int h^{\frac{3}{2}} \, dh = -A \int \, dt\). Integrating gives \(\frac{2}{5} h^{\frac{5}{2}} = -At + C\). At \(t = 0, h = H\), so \(C = \frac{2}{5} H^{\frac{5}{2}}\). At \(t = 60, h = 0\), so \(0 = -60A + \frac{2}{5} H^{\frac{5}{2}}\), giving \(A = \frac{1}{150} H^{\frac{5}{2}}\). Substitute back to find \(t = 60 \left( 1 - \left( \frac{h}{H} \right)^{\frac{5}{2}} \right)\).
(iii) Substitute \(h = \frac{1}{2}H\) into the expression for t: \(t = 60 \left( 1 - \left( \frac{1}{2} \right)^{\frac{5}{2}} \right) = 49.4\).
