Hard word problem

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lamaclass

Junior Member
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Oct 18, 2009
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I have no clue how to go about solving or setting this problem up with all of the information given in it:

It's the first of your summer job at the community center. You slowly begin adding 10,000 gallons of clean water to the pool. Your last duty for the day is to add the required 4 pounds of chlorine granules. Or was it supposed to be 2 pounds? You check the list of instructions left for you by the manager and realize you've added twice the legal amount of chlorine to the pool. You decide to open the drain to let the water flow out of the pool, while adding fresh water to keep the volume the same. The water drains at a rate of 12 gallons/minute. Assuming that the water in the pool stays well mixed and you start at 5:30 P.M., will you get the chlorine concentration down to the correct level before 8 A.M. the next day?
 
This is a classic tank mixture problem reworded to involve a swimming pool and all that.

Initially, there are 4 lbs of chlorine in the 10,000 gallon pool. y(0)=4

There is pure water flowing in, so we have 0 as the rate in of the chlorine mixture.

The rate out is \(\displaystyle \left(\frac{y(t)}{10,000} \;\ \text{lbs/gal}\right)\left(12 \;\ \text{gal/min}\right)=\frac{3y}{2500} \ \text{lbs/min}\)

So, we have \(\displaystyle \frac{dy}{dt}=-\frac{3y}{2500}\). It is negative because the chlorine is being diluted.

\(\displaystyle \frac{dy}{dt}+\frac{3}{2500}y=0\)

the integrating factor is \(\displaystyle e^{\int\frac{3}{2500}dt}=e^{\frac{3t}{2500}}\)


\(\displaystyle \frac{d}{dt}\left[ye^{3t}{2500}\right]=0\)

Integrate and note that when we integrate 0 we get C:

\(\displaystyle ye^{\frac{3t}{2500}}=C\)

\(\displaystyle y=Ce^{\frac{-3t}{2500}}\)

Now, use y(0)=4 to find C. Then, set the equation equal to 2 and solve for t. It'll be in minutes.
 
That is not the answer I got.

y = Ce^(-3t/2500)
Now, use y(0)=4 to find C
Click to expand...

Plug in:
t = 0
y = 4
and then solve for C, you should get C = 4.

Once you have the value of C, you know that you need y to equal 2 lbs.

Set y = 2 and solve for the new value of t:

2 = 4e^(-3t/2500)
 
No, I did not get that either. I got 5 hundred and something minutes.
 
Final answer

So I guess my guestion is does the water dilute enough in time before 8 AM? I got 500 someting minutes too and if he has 870 minutes for it to drain in time then does this mean that the water will drain in time before morning?


alsotheres a third part that confuses me:

(3) Suppose the "clean" water you are adding actually contains 0.00005 pounds of chlorine granules per gallon of water. How does this affect your answer to Question (2)? You can assume that the water originally in the pool (before you added anything) was completely chlorine-free.
 
so is that 500 somthing that I got as well the correct answer? he keeps his job right?

there is also this part that im confused about:

(3) Suppose the "clean" water you are adding actually contains 0.00005 pounds of chlorine granules per gallon of water. How does this affect your answer to Question (2)? You can assume that the water originally in the pool (before you added anything) was completely chlorine-free.
 
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