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    This is the first part of this water tank problem
    Water flows into a huge tank at a rate of g[t] gallons/hour. At time t = 0, the tank contains G gallons of water. Give a formula for the number of gallons of water in the tank at any time t greater than or equal to 0 before the tank is full.



    Here comes the second part:
    At time t = 0, a 207-gallon tank contains 100 gallons of old water. New water flows in at a rate of g[t] = (160t)/(0.5+t)^2 gallons/hour.
    Approximately how many hours pass before the tank overflows?

    I'll stop here for now to see if i understand the answer suggested by u guys and then i'll post the rest of the question there is 2 more parts of questions to this problem..
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    1st Part

    Let N(t) be the number of gallons of water in the tank at time t.
    Let G be the initial volume of water in the tank before any more is added.
    Let g(t) be the rate at which water is added to the tank.

    If g(t) were a constant value, rather than a function of time, then the total volume of water added after a time t woud be V = g(t)*t.
    Since g(t) is a variable function, however, then the total volume of water added is the area under the curve of the function g(t) against time t.
    Imagine g(t) is the y-axis and t is on the x-axis. Then V is simply the area under the curve for t = 0 to some value of t.

    This means we can write N(t) as,

    N(t) = G_0 + \int_0^t g(t)\ dt

    2nd Part

    N(t) = 207,\ G_0 = 100,\ g(t) = \frac{160t}{(0.5+t)^2}

    Substituting these values into the expression from part 1,

    207 = 100 + \int_0^t \frac{160t}{(0.5+t)^2}\ dt
    207 = 100 + 160*\left[ln(0.5+t) + \frac{0.5}{0.5+t}\right]_0^t
    \frac{107}{160} = ln(0.5+t) + \frac{0.5}{0.5+t} - ln(0.5) -1
    ln(0.5+t) + \frac{0.5}{0.5+t} - \left(ln(0.5) +1 + \frac{107}{160}\right)

    \mbox{Let } f(t) = ln(0.5+t) + \frac{0.5}{0.5+t} - 0.975603
    \mbox{Then } f'(t) = \frac{t}{(0.5+t)^2}

    Using the Newton-Raphson method,

    t_{n+1} = t_n - \frac{f(t_n)}{f'(t_n)}

    Code:
    n  t_n         f(t_n)           f'(t_n)     t_{n+1}
    1   1         -0.2368           0.4444      1.5328
    2   1.5328    -0.02022          0.37093     1.58732
    3   1.58732   -1.8097x10^-4     0.36432     1.587815
    4   1.587815  -1.4749x10^-8     0.36426     1.587815
    t = 1.588 hrs
    ==========
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    (1)
    G(t) = G + \int_0^t g(s) ds

    (2)
    The results of the Mathematica commands

    G[t_] = 100 + Integrate[160s/((1/2) + s)^2, {s, 0, t}]
    Plot[207 - G[t], {t, 0, 5}]
    FindRoot[G[t] == 207, {t, 1.6}]

    show that the tank overflows after about 1.588 hours.
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    I used graphmatica and also got t = 1.589 hrs as the time for overflow.

    But I've checked my working and I can't find where my error is!!

    There is this expresion,

    207 = 100 + \int_0^t \frac{160t}{(0.5+t)^2}\ dt
    or
    \int_0^t \frac{160t}{(0.5+t)^2}\ dt = 107

    and t = 1.589 satisfies the area under that graph.

    But the other eqns,

    207 = 100 + 160*\left\{ln(0.5+t) + \frac{0.5}{0.5+t}\right\}
    \frac{107}{160} = ln(0.5+t) + \frac{0.5}{0.5+t}

    f(t) = ln(0.5+t) + \frac{0.5}{0.5+t} - \frac{107}{160} = 0


    These eqns are all satisfied by t = 0.8463.

    Which suggests that the expression for the integral of g(t) is at fault. But I've checked that a few times, and the wolfram integrator gives the same value.

    So, can anyone see where the mistake is ??
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    (Original post by Fermat)
    Which suggests that the expression for the integral of g(t) is at fault. But I've checked that a few times, and the wolfram integrator gives the same value.
    The integral should equal 0 when t = 0.

    Partial fractions:
    t/(0.5 + t)^2 = A/(0.5 + t) + B/(0.5 + t)^2

    t = A(0.5 + t) + B
    A = 1
    B = -0.5

    \int_0^t s/(0.5 + s)^2 ds
    = \int_0^t 1/(0.5 + s) - 0.5/(0.5 + s)^2 ds
    = \Big[ ln(0.5 + s) + 0.5/(0.5 + s) \Big]_0^t
    = ln(0.5 + t) - ln(0.5) + 0.5/(0.5 + t) - 1

    Solving

    107 = 160*[ln(0.5 + t) - ln(0.5) + 0.5/(0.5 + t) - 1]

    numerically gives t = 1.588.
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    Thanks.
 
 
 
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