(1 point) A tank contains 80 kg of salt and 2000 L of water....

(1 point) A tank contains 80 kg of salt and 2000 L of water. A solution of a concentration 0.02 kg of salt per liter enters a tank at the rate 6 L/min. The solution is mixed and drains from the tank at the same rate. (a) What is the concentration of our solution in the tank initially? concentration = (-4x)/(1000+4t) (kg/L) (b) Set up an initial value problem for the quantity y, in kg, of salt in the tank at time t minutes. dy/dt = 80000/(1000+4x) (kg/min) y(0) = 80 (kg) (c) Solve the initial value problem in part (b). y(t) = 64.52 (d) Find the amount of salt in the tank after 1 hours. amount = 0 (kg) (e) Find the concentration of salt in the solution in the tank as time approaches infinity. concentration = 0 (kg/L)

Transcript

00:01 Hello everyone, in this problem we are given that a tank contains 80 kg of salt and 2000 liters of water and it is given that the solution is of the concentration of 0 .02 kg of salt per liter enters a tank at the rate of 6 liter per minute.

00:29 So in the first part of the question, we need to find the concentration of our solution in the tank initially.

00:35 So, initial concentration is of 80 divided by 2000 on dividing this we have the value to be 0 .04 kg per liter.

01:05 So that is 80 kg of salt in 2000 liter of water was there in the tank initially.

01:14 Now let us move on to the next part of the problem where we need to set up an initial value problem for the quantity of y in kg of salt in the tank at the time t minutes.

01:27 So here let the amount of salt be y.

01:45 So then y dash to be equal to rate in minus rate out.

01:53 So here the rate in is 6 multiplied by 0 .02 and rate out is 6y divided by 2000.

02:02 So here the solution enters and leaves at the same rate of 6 liters per minute.

02:09 So now simplifying this we have this value to be 0 .12 minus 6y divided by 2000.

02:20 So simplifying this we have the value of dy by dt to be equal to 0 .12 minus 6y divided by 2000.

02:32 So from this we get dy multiplied by 1 divided by 6x minus 240 to be equal to 1 divided by 2000 dt.

02:46 On integrating both sides we get log of 6y minus sorry here is also y.

02:58 So log of 6y minus 240 to be equal to minus 60 divided by 2000 plus log of c.

03:10 So from this we get the value of y of t to be equal to 40 plus 1 by 6 c e power minus 6 by 2000.

03:23 So this is the required initial solution and here the initial value is of y of 0 to be equal to 80.

03:36 Now substituting this here as in the given so it would be 80 substituting value of t to be equal to 0.

03:45 So we have 80 to be equal to 40 plus c by 6...

Question

Please give Ace some feedback