calculus

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Water is Pumped into an underground tank at a constant rate of 8 gallons per minute.Water leaks out of the tank at the rate of (t+1)^½ gallons per minute, for 0<t<120 minutes. At time t=0, the tank contains 30 gallons of water.
How many gallons of water leak out of the tank from time t=0 to t=3 minutes?

  • calculus -

    the 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 -

    ya

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