A1000-liter (l) tank contains 500 l of water with a salt concentration of 10 g/l. water with a salt concentration of 50 g/l flows into the tank at a rate of rin=80 l/minutes (min). the fluid mixes instantaneously and is pumped out at a specified rate rout. let y(t) denote the quantity of salt in the tank at time t. assume that rout=40l/min. (a) set up and solve the differential equation for y(t). (b) what is the salt concentration when the tank overflows?

a)

b)

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is , where is the rate of salt entering and is the rate of salt going outside.

Then we have, , and

So we obtain.  , then

, and using the integrating factor , therefore  , we get   , after integrating both sides , therefore , to find we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions , so

Finally we can write an expression for the amount of salt in the tank at any time t, it is

b) The tank will overflow due Rin>Rout, at a rate of , due we have 500 L to overflow , so we can evualuate the expression of a) , is the salt concentration when the tank overflows

9 x 8 divided by (5-3)=36

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step-by-step explanation: