1. The problem statement, all variables and given/known data Tank is filled with water to a height h_0 = 1.00 meters Cross section area of tank (A_1 = .785 m^2) Hole at bottom of tank with an area, A_2 = .002 m^2 2. Relevant equations How much time does it tank to half empty the tank? find t_(1/2). Question also provides "a useful antiderivative." / = integral sign / x^(-1/2) dx = 2x^(1/2) 3. The attempt at a solution 1. V = A_2 * v_hole dV/dt = A_2 * v_hole 2. getting an equation for velocity coming out of hole v_hole = [2gh + (v_surf)^2] ; got that from using Bernoulli's principle. 3. equation of continuity to get v_surf v_surf = v_hole * A_hole / A_surf thus, v_hole = (2gh)^(1/2) , since [1 - (A_2/A_1)^2] = ~1 4. rate of change of the level of water in the tank dh/dt = (-v_hole * A_2)/(A_1) dh/dt = -.0113 5. h_2 - h_0 = -.0113 * t_(1/2) a half empty tank means, h goes down 1/2, 1/2 of h = 1, h=-.5 t_(1/2) = -.5m / -.0013 m/s = 44.3 s 44.3 s is wrong though. I don't get how come. I went through the hints for this problem. It mentions dy/dx = f(x) * g(y) ... f(x) and g(y) are known functions / dy/g(y) = / f(x)dx I don't get which known functions they are referring to. And it's also been probably 2 years since I ever used calculus, so those equations don't make much sense to me. Any help would be appreciated.