The Leaking Tank
A cylindrical tank with a radius of 2 meters and a height of 5 meters is filled with water. A small hole at the bottom causes water to leak out at a rate proportional to the square root of the depth of the water. Specifically, the rate of change of the water height h(t) (in meters) with respect to time t (in minutes) is given by dh/dt = -k√h, where k is a positive constant. After 10 minutes, the water height has decreased to 3.2 meters. Assume the tank remains cylindrical and the hole size does not change.
1. Determine the constant k and find the time when the tank will be empty.
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We solve the differential equation (\frac{dh}{dt} = -k\sqrt{h}) with initial condition (h(0) = 5). Separating variables: [ \frac{dh}{\sqrt{h}} = -k,dt \quad\Rightarrow\quad 2\sqrt{h} = -kt + C. ] Using (h(0)=5): (2\sqrt{5} = C). So (2\sqrt{h} = 2\sqrt{5} - kt). At (t=10), (h=3.2). Then (2\sqrt{3.2} = 2\sqrt{5} - 10k). Compute: (\sqrt{3.2} \approx 1.78885), so (2\sqrt{3.2} \approx 3.5777). (2\sqrt{5} \approx 4.4721). Thus (10k \approx 4.4721 - 3.5777 = 0.8944), so (k \approx 0.08944). The tank is empty when (h=0), i.e., (2\sqrt{0} = 2\sqrt{5} - kt), so (kt = 2\sqrt{5}), (t = \frac{2\sqrt{5}}{k} \approx \frac{4.4721}{0.08944} \approx 50) minutes. More precisely, (k = \frac{2(\sqrt{5}-\sqrt{3.2})}{10}), and (t = \frac{2\sqrt{5}}{k} = \frac{10\sqrt{5}}{\sqrt{5}-\sqrt{3.2}} \approx 50.0) minutes.
2. Find the volume of water that has leaked out after 20 minutes. Use the value of k you found (or an exact expression).
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The volume of water in the tank at height h is (V = \pi r^2 h = \pi (2)^2 h = 4\pi h). Initially, (V_0 = 4\pi \cdot 5 = 20\pi) cubic meters. From the solution (2\sqrt{h} = 2\sqrt{5} - kt), we have (\sqrt{h} = \sqrt{5} - \frac{k}{2}t), so (h(t) = \left(\sqrt{5} - \frac{k}{2}t\right)^2). At (t=20), using (k \approx 0.08944): (\frac{k}{2} \approx 0.04472), so (\sqrt{h} \approx \sqrt{5} - 0.04472 \cdot 20 = 2.23607 - 0.8944 = 1.34167), hence (h \approx 1.800) m (more precisely, (h = (\sqrt{5} - 0.8944)^2)). Then (V(20) = 4\pi \cdot 1.800 \approx 7.2\pi \approx 22.62) m³. The leaked volume is (V_0 - V(20) = 20\pi - 7.2\pi = 12.8\pi \approx 40.21) m³. Exact: (h(20) = (\sqrt{5} - 0.8944)^2), but using exact (k = \frac{2(\sqrt{5}-\sqrt{3.2})}{10} = \frac{\sqrt{5}-\sqrt{3.2}}{5}), then (\frac{k}{2} = \frac{\sqrt{5}-\sqrt{3.2}}{10}). So (\sqrt{h} = \sqrt{5} - \frac{\sqrt{5}-\sqrt{3.2}}{10} \cdot 20 = \sqrt{5} - 2(\sqrt{5}-\sqrt{3.2}) = 2\sqrt{3.2} - \sqrt{5}). Thus (h = (2\sqrt{3.2} - \sqrt{5})^2 \approx (3.5777 - 2.2361)^2 = (1.3416)^2 \approx 1.800). So leaked volume (= 4\pi(5 - 1.800) = 4\pi \cdot 3.200 = 12.8\pi) m³ exactly? Wait, check: (h = (2\sqrt{3.2} - \sqrt{5})^2 = 4 \cdot 3.2 - 4\sqrt{16} + 5 = 12.8 - 16 + 5 = 1.8). Yes, exactly 1.8 m. So leaked volume = (4\pi (5 - 1.8) = 4\pi \cdot 3.2 = 12.8\pi) m³.