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Your swimming pool containing 60,000 gal of water has been contaminated by 6 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 250 gal/min.
(a) What is the initial value problem for the filtering process? Let q be the amount of dye in grams in the pool at any time t. dq dt = g/min, q(0) = grams
(b) Solve the problem in part (a). q(t) = g
(c) 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? (For this part you will need your calculators; compute at least 5 decimal places.) Compute the concentration of the dye after 4 hours of filtering.
Note: Units must be consistent; consider g vs kg and min vs hrs.

Respuesta :

LammettHash

[a] Dye is removed from the pool at a rate of

(250 gal/min) * (q/60,000 g/gal) = -q/240 g/gal

where q denotes the amount of dye in the pool at time t. Clean water is pumped back into the pool, so no dye is being re-added.

So the net rate of change of the amount of dye in the pool is given by the differential equation,

[tex]\dfrac{\mathrm dq}{\mathrm dt}=-\dfrac{q(t)}{240}[/tex]

with the intial value, q(0) = 6000 g (or 6 kg).

[b] The ODE above is separable as

[tex]\dfrac{\mathrm dq}q=-\dfrac{\mathrm dt}{240}[/tex]

Integrate both sides to get

[tex]\ln|q|=-\dfrac t{240}+C[/tex]

[tex]e^{\ln|q|}=e^{-t/240+C}[/tex]

[tex]\implies q(t)=e^{-t/240+C}=e^{-t/240}e^C=Ce^{-t/240}[/tex]

Now plug in the initial condition:

[tex]6000=Ce^0\implies C=6000[/tex]

so the particular solution to the IVP is

[tex]q(t)=6000e^{-t/240}[/tex]

[c] The acceptable concentration of the dy is 0.03 g/gal, which in a pool containing 60,000 gal of water corresponds to

(0.03 g/gal) * (60,000 gal) = 1800 g = 1.8 kg

of dye. Find the time t when this occurs:

[tex]1800=6000e^{-t/240}\implies0.3=e^{-t/240}[/tex]

[tex]\implies\ln0.3=-\dfrac t{240}[/tex]

[tex]\implies t=-240\ln0.3\approx288.953[/tex]

so the amount of dye in the pool is within the acceptable tolerance after about 289 min have passes, or about 4.82 hrs. So no, the filtration system is not up to the task.

sqdancefan

Answer:

  a) q'(t) = -q/240; q(0) = 6000

  b) q(t) = 6000e^(-t/240)

  c) no

Step-by-step explanation:

a) The initial value in grams is given as ...

  q(0) = 6000 . . . . . that is, 6 kg

The rate of removal is the fraction 250/60,000 per minute:

  q'(t) = -q/240

These two equations comprise the initial value problem.

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b) q(t) = 6000·e^(-t/240) . . . . where time is in minutes

(The solution to such an initial value problem is well-known. The multiplier of the exponent is the coefficient in the differential equation. If you like, you can solve by separation of variables: dq/q = -dt/240; ln(q) = -t/240+C)

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c) 4 hours of filtering is 240 minutes, so the remaining grams will be ...

  q(240) = 6000·e^(-240/240) ≈ 2207.3 g

In the volume of water, this is 2207.3 g/(60000 gal) ≈ 0.0368 g/gal

The filtering system cannot reduce the concentration to the required level in 4 hours. (It will take 4 hours, 49 minutes.)

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