(1) Let A(h) be the cross-sectional area of the water int he tank at height h an

Question

(1) Let A(h) be the cross-sectional area of the water int he tank at height h and a the areas of drain hole. the rate at which water if flowing out of the tank at time t can be expressed as the cross-sectional area at height h times the rate at which the height of the water is changing. Alternatively, the rate at which water flows out of the hole can be expressed as the area of the hole times the velocity of the draining water. Set these two equal to each other and insert Torricelli's Law to derive the differential euqation

A(h)*dh/dt= -a*Sqrt(2gh)

(2) The conical tank has a radius of 30 cm when it is filled to an inital depth of 50cm. A small round hole at the bottom has a diameter of 1 cm. Determine A(h) and a and then solve the differential euqaition A(h) dh/dt=-a*Sqrt(2gh), thus derviing a formla ralting time and the hegiht of the water in this tank.

Explanation / Answer

V = the cross-sectional area at height h times the rate at which the height of the water is changing = A(h)*dh/dt

V = the rate at which water flows out of the hole can be expressed as the area of the hole times the velocity of the draining water = a*velocity(v)

v^2 -0 = 2gh

=>

v = Sqrt(2gh)

=>

V = -a*Sqrt(2gh)

=>

A(h)*dh/dt= -a*Sqrt(2gh)
(thus proved)

(b)

A(h) = pi*(rh/H)^2 = pi*(0.3*h/0.5) = 0.36*pi*h^2

=>

0.36*pi*h^2*dh/dt= -pi*0.01*0.01*Sqrt(2*9.8h)

=>

h = 9.8 cm

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