Turn on thread page Beta
    • Thread Starter

    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..

    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)}

    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

    G(t) = G + \int_0^t g(s) ds

    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.

    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
    \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 ??

    (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


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

    numerically gives t = 1.588.

Submit reply
Turn on thread page Beta
Updated: December 25, 2005

University open days

  • University of Lincoln
    Mini Open Day at the Brayford Campus Undergraduate
    Wed, 19 Dec '18
  • University of East Anglia
    UEA Mini Open Day Undergraduate
    Fri, 4 Jan '19
  • Bournemouth University
    Undergraduate Mini Open Day Undergraduate
    Wed, 9 Jan '19
Were you ever put in isolation at school?
Useful resources

Make your revision easier


Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here


How to use LaTex

Writing equations the easy way


Best calculators for A level Maths

Tips on which model to get

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.