water tanks..4 part question....

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..

Reply 1

Fermat

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)

Unparseable latex formula:

\mbox{Let } f(t) = ln(0.5+t) + \frac{0.5}{0.5+t} - 0.975603

Unparseable latex formula:

\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
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Reply 2

Jonny W

(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

Reply 3

Fermat

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 ??

Reply 4

Jonny W

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.