Question

Your swimming pool containing 65,000 gal of water has been contaminated by 7 kg of a nontoxic dye that leaves a swimmer's skin an unattractive green. The pool's filtering system can take water from the pool, remove the dye, and return the water to the pool at a flow rate of 200 gal/min. Assume that the dye is uniformly distributed throughout the pool.
(a) What is the initial value problem for the filtering process?
(b) You have invited several dozen friends to a pool party that is scheduled to begin in 4 h. You have determined that the effect of the dye is imperceptible if its concentration is less than 0.03 g/gal. Is your filtering system capable of reducing the dye
concentration to this level within 4 h?

Answer 1

Answer:

a) $a(t)=7e^{\frac{-t}{325} }$

b) it is noticeable

Step-by-step explanation:

a) Let a(t) be the amount of die at time t. a is in kg and t is in minutes. Therefore:

$\frac{da}{dt} =\frac{-a(t)}{65000}*\frac{200gal}{1 min}=\frac{-a}{325}\\\frac{da}{a} =\frac{-1}{325} dt\\\int\limits{\frac{1}{a} } \, da =\int\limits{\frac{-1}{325} } \, dt\\ln(a)=\frac{-1}{325} t+c\\a=ce^{\frac{-t}{325} }\\at\ t=0, a=7\\ 7=ce^0\\c=7\\a(t)=7e^{\frac{-t}{325} }$

b) 4 hrs = (4 * 60) min = 240 min

$a(t)=7e^{\frac{-t}{325} }\\a(240)=7e^\frac{-240}{325} = 3.345kg$

3.345 kg = (3.345 * 1000 / 60000) g/gal = 0.0557 g/gal

0.0557 g/gal > 0.03 g/gal, therefore it is noticeable