A tank initially contains 50 gallons of sugar water having a...

A tank initially contains 50 gallons of sugar water having a concentration of 2 lbs. of sugar for each gallon of water. At time zero, pure water begins pouring into the tank at a rate of 2 gallons per minute. Simultaneously, a drain is opened a the bottom of the tank so that the volume of sugar water solution in the tank remains constant. a. How much sugar is in the tank after 10 minutes? b. How long will it take the sugar content in the tank to drop below 20 lbs.? c. What will the be the eventual sugar content in the tank?

Transcript

00:01 So this question is about a tank containing sugar water that's being kind of washed.

00:09 So we have that the volume, which stays constant throughout, of sugar water in the tank, is v equals 50 gallons.

00:21 We have that the initial concentration, so mass of sugar at times zero divided by v, is two.

00:34 Pounds per gallon which means that the initial mass is going to be a hundred pounds we also have that the water is flowing out of the bottom of the tank at a rate of two gallons per minute which means that the mass is changing by minus the concentration at a given time times that that flow rate in gallons per minute, which i'm going to call alpha.

01:22 And alpha is, what did we say, two gallons per minute.

01:39 And the concentration is going to be c is the mass divided by the volume.

01:48 And this is going to give us a differential equation.

01:53 Ms dot equals minus alpha over v times ms and let's just check that the units all add up on this so what we're going to have here is the rate of change of the mass is going to have units of pounds per minute alpha is in gallons per minute so minus two gallons per minute divided by v which is 50 gallons and then multiplied by the mass in pounds and this is going to give minus minus two or minus 1 over 25 mass in units of pounds per minute so we have the ms dot equals minus 1 over 25 ms that that means that if we can integrate this equation, we'll get the ms of t is equal to e to the minus t over 25 minutes times the mass at t equals 0, which is 100 pounds, which we worked out earlier.

03:59 So the equation for the mass is a function of time is 100 pounds times e to the minus t over 25...

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