Q:: A tank initially holds 80 liters of saltwater with 1 gram of salt per liter.
ID: 2943764 • Letter: Q
Question
Q:: A tank initially holds 80 liters of saltwater with 1 gram of salt per liter. starting at time t=0, a solution containing 3 gram of salt per liter is poured into the tank at the rate of 2 liter/min, and a second solution containing 6 gram of salt per liter is poured into the tank at the rate of 1 liter/min, while the well-stirred mixture leaves the tank at the rate of 3 liter/min.(a) Set up the initial value problem for finding the amount of salt in the tank at any time t.
(b) solve the initial value problem.
Explanation / Answer
The amount (let's call it S) of the salt in the tank will be measured in grams. Now dS/dt (the rate of change of the amount of salt in the tank) will be equal to the rate of change that salt enters the tank minus the rate that salt leaves the tank. So let's try to figure that out: Rate that it enters: 3 grams/ liter * 2 liter/ min = 6 grams/ min 6 grams/ liter * 1 liter / min = 6 grams / min So the total is 12 grams / min Rate that it leaves: (Notice that the volume of the tank will remain constant at 80 L) S grams/ 80 liters * 3 liters / min = 3S/ 80 grams/ min Now we know dS/dt = 12 - 3S/80 So now we can solve this using separation of variables: dS/dt = 12 - 3S/80 dS/dt = 3(4 - S/80) dS / (4 - S/80) = 3dt integrate both sides -80ln(4-S/80) = 3t + c 4-S/80 = Ke^(-3t/80) S = Ke^(-3t/80) + 320 Now, we now that S(0) = 80 grams which means that K = -240 So, S(t) = 320 - 240e^(-3t/80)
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