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September 2, 2014

September 2, 2014

Posted by **POD** on Wednesday, April 30, 2008 at 8:26pm.

How many gallons of water leak out of the tank from time t=0 to t=3 minutes?

- calculus -
**Reiny**, Wednesday, April 30, 2008 at 9:29pmthe effective rate of change of the water is dV/dt = 8 - (t+1)^½ + c

so integrating we get

V = 8t - (2/3)(t+1)^(3/2) + c , where c is a constant

but when t=0 , V = 30

30 = 0 - 2/3(1)^(3/2) + c

c = 92/3

so V = 8t - (2/3)(t+1)^(3/2) + 92/3

when t = 3 minutes

V = 24 - (2/3)(4)^(3/2) + 92/3

= 24 - 16/3 + 92/3

= 148/3

so the increase is 19.3333333

but we poured in 8 gallons/min for 3 minutes which is 24 gallons.

so the leakage must be 24 - 19.3333

or 4.66666 gallons

- calculus -
**JOE**, Thursday, May 1, 2008 at 9:58amya

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