1. The problem statement, all variables and given/known data Water si emptying out of a tank at a rate of 4 cubic metres/min. The tank is shaped like an inverted circular cone with a diameter of 8 m at the top. The sides of the cone make a 30 degree with the vertical. Determine the exact value of the rate of change of the water level when the height of the water is 3/4 the height of the cone. 2. Relevant equations v=(1/3) (pi) (r^2) (h) 3. The attempt at a solution v = (1/3) (pi) (3/4(h))^2 (h) =(3/16)(pi)(h^3) v'=(9/16)(pi)(h^2)(h') i don't know where to go from here, as i could use trig to get the height of the tank, but i don't have the height at the time specified. help?