1. A cylindrical open tank needs cleaning. The tank is filled with water to a height meter, so you decide to empty it by letting the water flow steadily from an opening at the side of the tank, located near the bottom. The cross-sectional area of the tank is square meters, while that of the opening is square meters 2. I've got a bunch of calculations below. I need help with setting up and solving the final integral. 3. A_1 = area of surface, v_1 = velocity of flow at suraface A_2 = area of small tube, v_2 = velocity of flow out of tube A_1 >> A_2 V' = A_2 * v_2 = rate of discharge V' = A_1 * dh/dt, relates rate of discharge to drop in height of liquid v_2 = sqrt[2gh + v_1^2] v_1 = v_2 *(A_2/A_1) v_1 = sqrt[(2gh) / (1-(A_2/A_1)^2)] dh/dt = rate of change of height of water = -A_2/A_1 * sqrt(2gh) Ok, so now the strategy is to solve a separable first-order ODE like so: dy/dx = f(x)g(y) dy/g(y) = f(x)dx, and integrate both sides. So in my case, these are the integrals I obtain: dh/dt = -A_2/A_1 * sqrt[2gh] Plus in g = 9.81 and obtain: dh/dt = -4.23*(A_2/A_1)*sqrt(h) So I get the following integrals: h^(-0.5)dh = -4.23*(A_2/A_1)dt with h_0 and 0.5h_0 as the limits of integration. I have solved this a couple different ways and obtained incorrect answers. Can someone point out what the correct integral I need to solve is? Thanks!