There are 20 animals in the barn. There are chickens and sheep in the barn. If there are
2 answers:
You might be interested in
[a] Dye is removed from the pool at a rate of
(250 gal/min) * (<em>q</em>/60,000 g/gal) = -<em>q</em>/240 g/gal
where <em>q</em> denotes the amount of dye in the pool at time <em>t</em>. 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,
with the intial value, <em>q</em>(0) = 6000 g (or 6 kg).
[b] The ODE above is separable as
Integrate both sides to get
Now plug in the initial condition:
so the particular solution to the IVP is
[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 <em>t</em> when this occurs:
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.