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Previous Problem List Next (1 point) Water leaks from a vertical cylindrical tank through a small hole in its base at a rate proportional to the square root of the volume of water remaining. The tank initially contains 100 liters and 20 liters leak out during the first day A. When will the tank be half empty? t 3 B. How much water will remain in the tank after 2 days? volume62.04 days Liters

Solution

From the given data,

the differential equation is dV/dt = k (V)^1/2

a)   By separating the variables, we have

     V ^1/2dV = k dt

Integrate on both sides,

V ^1/2dV = k dt

It follows that

2V^1/2 = kt + C

or V = (At + B)² for some constants A and B.

Since V (0) = 100   (at t=0),

   B = (100)^1/2 = 10

After 1 day, V (1) = 80

So,

80 = [ A.1 + 10] ²

A = (80)^1/2 -10

Now, when V (t) = 50 (half empty),

50 = (At + B)²

At+B = 50^1/2

   t= [50^1/2 - B ] / A

     = [50^1/2 - 10 ]/ [ (80)^1/2 -10]

    = 2.77 days

Therefore,

tank will be empty after 2.77 days

B) How much water will there be after 2 days?

V(t) = (At + B)²

V(2) = {[(80)^1/2 -10].2 + 10 ]}²

         = 62.23 litres