-2[ - 3(6[ - 11) = -7
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Let A(t) denote the amount of salt in the tank at time t.
Salt flows in at a rate of
(1 lb/gal) * (3 gal/min) = 3 lb/min
and flows out at a rate of
(A(t)/(200 + t) lb/gal) * (2 gal/min) = 2 A(t)/(500 + t)
(in case you're unsure about the denominator: the tank starts off with 200 gal of solution, and each minute solution flows in at a rate of 3 gal/min and thus the tank gains (3 gal/min) * (1 min) = 3 gal. At the same time, solution flows out at a rate of 2 gal/min and thus the tank loses 2 gal, giving a net change in volume of (3 - 2)*t = t gal)
Then the net rate of salt flow is given by the ODE,
Multiply both sides by :
Integrating both sides and solving for gives
The tank starts off with 100 lb of salt in solution, so and we find
and so
The tank will begin to overflow once the volume of solution reaches 500 gal; this happens when
or 300 minutes or 5 hours after solution starts flowing. At this point, the tank will contain
or 2125 lb of salt.
Theoretically, the amount of salt in the tank will increase forever, since as
.