Consider the flow of a liquid through a hole in the bottom of a container. If h(

Question

Consider the flow of a liquid through a hole in the bottom of a container. If h(t) is the height of the liquid above the hole, then the velocity of the liquid emerging from the hole is given by V=(2gh)^1/2
where g is the acceleration of gravity and c is an empirical constant.

a) show that dh/dt = -ac(2gh)^1/2 / A(h)
where a is the cross sectional area of the hole and A(h) is the cross sectional area of the tank at a height h.

b)How does h vary with time if the tank shape is a regular cone of cone angle

Explanation / Answer

a) Say v' is the rate of change of the volume of the liquid in the container. Then v' = h' * A(h), since this gives the rate of change at that height times of the cross-sectional area at that height, which equals the rate of change of the volume. Also, v' equals the negative of the velocity V of the liquid through the whole times the cross-sectional area of the whole, since this is the rate with which liquid leaves the container. SO v' = -ac(2gh)1/2. Setting these two expressions for v' equal gives us the desired result:

h' * A(h) = v' = -ac(2gh)1/2   -->   dh/dt = -ac(2gh)1/2 / A(h).

b) If the tank is a regular cone with angle , then A(h) = h2tan2. So dh/dt = -ac(2gh)1/2 / (h2tan2) = -(ac/)h-2(2gh)1/2cot2.

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