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# water tanks..4 part question.... watch

1. 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..
2. 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,

2nd Part

Substituting these values into the expression from part 1,

Using the Newton-Raphson method,

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
==========
3. (1)
G(t) = G + 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.
4. 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,

or

and t = 1.589 satisfies the area under that graph.

But the other eqns,

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 ??
5. (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

s/(0.5 + s)^2 ds
= 1/(0.5 + s) - 0.5/(0.5 + s)^2 ds
= ln(0.5 + s) + 0.5/(0.5 + s)
= 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.
6. Thanks.

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Updated: December 25, 2005
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