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Sagot :
Let s(t) be the amount of salt in the tank at time t. Then s(0) = 50 lb.
Salt flows into the tank at a rate of
(2 gal/min) (6 lb/gal) = 12 lb/min
and flows out at a rate of
(2 gal/min) (s(t)/300 lb/gal) = s(t)/150 lb/min
so that the net rate at which salt is exchanged through the tank is
ds(t)/dt = 12 - s(t)/150 … (lb/min)
Solve for s(t). The DE is separable, so we have
ds/dt = 12 - s/150
150 ds/dt = 1800 - s
150/(1800 - s) ds = dt
Integrate both sides to get
-150 ln|1800 - s| = t + C
Solve for s :
ln|1800 - s| = -t/150 + C
1800 - s = exp(-t/150 + C )
1800 - s = C exp(-t/150)
s = 1800 - C exp(-t/150)
Now given that s(0) = 50, we solve for C :
50 = 1800 - C exp(-0/150)
50 = 1800 - C
C = 1750
Then the amount of salt in the tank at any time t ≥ 0 is
s(t) = 1800 - 1750 exp(-t/150)
To find the time it takes for the tank to hold 100 lb of salt, solve for t in
100 = 1800 - 1750 exp(-t/150)
1700 = 1750 exp(-t/150)
34/35 = exp(-t/150)
ln(34/35) = -t/150
t = -150 ln(34/35) ≈ 4.348
So it would take approximately 4.348 minutes.
By the way, we didn't have to solve for s(t), we could have instead stopped with
-150 ln|1800 - s| = t + C
Solve for C - this C is not the same as the one we found using the other method. s(0) = 50, so
-150 ln|1800 - 50| = 0 + C
C = -150 ln|1750|
==> t = 150 ln(1750) - 150 ln|1800 - s|
Then s(t) = 100 lb when
t = 150 ln(1750) - 150 ln(1700)
t = 150 ln(1750/1700)
t = 150 ln(35/34)
t ≈ 4.348
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