We appreciate your visit to Same situation as the previous problem but now the solution drains at 7 liters per minute A tank contains 40 liters of a saltwater solution. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
(a) The amount of salt added in the tank will be equal to the amount of salt drained from the tank.
(b) The concentration of salt in the solution in the tank is a decreasing function of time, and it has a vertical asymptote at t = 40.
(a) Given that a tank contains 40 liters of a saltwater solution. The solution contains 4 kg of salt. A second saltwater solution containing 0.5 kg of salt per liter is added to the tank at 6 liters per minute. The solution in the tank is kept thoroughly mixed and drains from the tank at 7 liters per minute.
Let's determine the amount of salt added in the tank per minute
= 0.5 kg/liter × 6 liters/minute
= 3 kg/minute
Let's determine the amount of salt drained from the tank per minute
= 4 kg/40 liters × 7 liters/minute
= 0.7 kg/minute
The amount of salt added in the tank per minute is more than the amount of salt drained from the tank per minute
i.e 3 kg/minute - 0.7 kg/minute = 2.3 kg/minute.
Therefore, the amount of salt in the tank after t minutes is given by,
`S(t) = S(0) + 2.3t`
where S(0) is the initial amount of salt present in the tank,
S(t) is the amount of salt present after t minutes.
Since there are 4 kg of salt in the solution initially,
`S(0) = 4 kg`
The domain of the function `S(t) = 4 + 2.3t` is 0 ≤ t ≤ 17.39
This is because after 17.39 minutes, the amount of salt added in the tank will be equal to the amount of salt drained from the tank. And after that time, the amount of salt in the tank will decrease.
(b) Let C(t) be the concentration of salt in the tank after t minutes.
Then we have,
`C(t) = S(t)/V(t)`
where V(t) is the volume of the solution in the tank after t minutes.
Since the solution is draining from the tank at 7 liters per minute, we have,
V(t) = 40 + (6 - 7)t= 40 - t liters
Therefore, the concentration of the solution in the tank after t minutes is given by,
`C(t) = S(t)/(40 - t)`
Substituting the value of S(t) from part (a), we get,
`C(t) = (4 + 2.3t)/(40 - t)`
The domain of the function `C(t) = (4 + 2.3t)/(40 - t)` is 0 ≤ t < 40.
This is because as t approaches 40, the volume of the solution in the tank approaches 0, and the concentration of salt in the solution becomes undefined.
The concentration of salt in the solution in the tank is given by the function
`C(t) = (4 + 2.3t)/(40 - t)`.
The domain of this function is 0 ≤ t < 40, and the range is the set of all positive real numbers. As t increases, the concentration of salt in the solution in the tank increases, but as t approaches 40, the concentration becomes undefined. This is because as t approaches 40, the volume of the solution in the tank approaches 0, and the concentration of salt in the solution becomes undefined.
Therefore, we can conclude that the concentration of salt in the solution in the tank is a decreasing function of time, and it has a vertical asymptote at t = 40.
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