IDNLearn.com provides a collaborative environment for finding and sharing answers. Join our knowledgeable community and get detailed, reliable answers to all your questions.
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