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A tank contains 1000 gallons of water in which 200 lb of salt is dissolved. Fifty gallons of brine, each containing (1 cos t)lb of dissolved salt, runs into the tank per minute, the mixture kept uniform by stirring, runs out at the same rate. Find the amount of salt y(t) in the tank at any time t.

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Final answer:

The problem requires setting up and solving a differential equation to determine the amount of salt y(t) in a tank over time, accounting for the inflow and outflow of brine with varying salt concentrations.

Step-by-step explanation:

The question involves setting up and solving a differential equation to find the amount of salt y(t) in a tank at any time t. The tank initially contains 1000 gallons of water with 200 lb of salt dissolved in it. As brine with (1 - cos t) lb of salt per 50 gallons enters the tank at a rate of 50 gallons per minute and the well-stirred mixture leaves at the same rate, we need to calculate the changing concentration of salt in the tank.

To solve this problem, we can use the concept of a dilute solution, assuming that the solution's density is roughly the density of water due to the low solubility of the salt. The differential equation is based on the rate of change of the amount of salt in the tank due to the incoming and outgoing brine.

Since 50 gallons of brine containing (1 cos t)lb of salt is flowing into the tank per minute, the amount of salt flowing in is given by the integral of (1 cos t) over the interval [0, t]. This integral represents the total amount of salt that has flowed in up to time t.

Subtracting the amount of salt that has flowed out gives us the amount of salt y(t) in the tank at any time t.

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