of diameter 60cm with water at a rate of 2 L/min. Find the rate at which the water is rising in the basin when it is HALF full. [Use the following facts: 1 L is 1000 cm3. V = pi (rh^2 -1/3 h^3]I need help with this asap. Thank you!!
Equation of the hemispherical surface x? + y? + z? = R?, (-R z 0) The volume of water in the basin with the height h: . . . (-R+h) V = ∫ π(x? + y?) dz . . . -R . . . (h-R) V = π ∫ (R? + z?) dz . . . -R V = πR?[(h - R) - (-R)] + (π/3)[(h - R)? - (-R)?] V = πR?h + (π/3)(h? - 3h?R + 3hR? - R? + R?) V = πR?h + (π/3)(h? - 3h?R + 3hR?) dV/dh = πR? + (π/3)(3h? - 6hR + 3R?) dV/dh = πR? + π(h? - 2Rh + R?) dV/dh = π(h? - 2Rh + 2R?) dV/dh = (dV/dt)/(dh/dt) dh/dt = (dV/dt)/(dV/dh) dh/dt = (dV/dt)/π(h? - 2Rh + 2R?) There is an ambiguity when you say HALF full. It can be interpreted as the water was filled up to half its maximum height (R) or filled up to half the volume of the basin (2πR?/3) I will take the case in which the water was filled up to half its maximum height: h = R/2 dh/dt = (dV/dt)/π(R?/4 - 2R*R/2 + 2R?) dh/dt = (dV/dt)/π(R?/4 - R? + 2R?) dh/dt = 4(dV/dt)/5πR? Plugging in dV/dt = 2000 cm?/min R = 30 cm dh/dt = 4*2000/4500π dh/dt = 16/9π cm/min. dh/dt ≈ 0.57 cm/min.
