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Cal Prob

cheungTim

2009-10-23 2:36 pm

Water is leaking out of an inverted conical tank at a rate of 9000 cm^3/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.

回答 (1)

Prof. Physics

2009-10-23 5:12 pm

✔ 最佳答案

Let V be the volume of water in the tank,

h be the height of the water,

r be the radius of the surface of water.

Let x be the rate at which water is being pumped into the tank.

By similar figures,

(400/2)/600 = r/h

r = h/3

dV/dt = [x - (9 X 10^-3)] m^3/min

dh/dt = 0.2 m/min (Be aware of the units)

V = 1/3 pir^2h = 1/3 pi(h/3)^2h = (pi/27)h^3

Differentiate both sides with respect to time, t

dV/dt = (pi/27)(3h^2)dh/dt

x - (9 X 10^-3) = 0.2pi/9 h^2

Put h = 2,

x - (9 X 10^-3) = 0.2pi/9 (2)^2

x = 4pi /45 + 9 X 10^-3

x = 0.288 m^3/min (3 sig. fig.)

So, the rate of water being pumped in is 0.288 m^3/min (2.88 X 10^5 m^3/min)

參考: Physics king

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