Water is leaking out of an inverted conical tank at a rate of 6000 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank.
2006-10-25 09:20:26 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The volume of water in the tank is (1/3)pi(r^2)h, and similarity will tell you that the diameter/height ratio is constant and equal to 4/6 = 2/3. So r = h/3, always, meaning V = (1/3)pi((h/3)^2)h = (1/27)pi*h^3. That means that dV/dh = 3*(1/27)pi*h^2 = (1/9)pi*h^2, which you can calculate for the given height. The rate of change of the water level is dh/dt, and is given. You need to find the total rate of volume change, dV/dt. The formula is dV/dt = (dV/dh)(dh/dt). Since you have dV/dh and dh/dt, the only unknown is dV/dt and you can calculate it easily. You know the rate at which water is leaking out, so you'll need to solve algebraically for the rate of water addition to yield the net rate of volume change that you calculate.
2006-10-25 09:27:01 · answer #1 · answered by DavidK93 7 · 0⤊ 0⤋