A tank with capacity of 300 gal of water originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1 lb of salt per gallon is entering at a rate of 3 gal/min, and the mixture is allowed to flow out of the tank at a rate of 2 gal/min. Let Q(t) lb be the amount of salt in the tank, V(t) gal be the volume of water in the tank. Find the amount of salt in the tank at any time prior to the instant when the solution begins to overflow. Q(t)

The amounts of salt in the tank at any time prior to the instant when the solution begins to overflow is.

Step-by-step explanation:

Let Q(t) be the  amount of salt in the tank at any time .

Then, its time of change, by (Balance law).

Since three gallons of salt water runs in the tank per minute, containing lb of salt, the salt inflow rate is

The  amount of water in the tank at any time is,

Now, the outflow is gal of the solution in a minute. That is of the total solution content in the tank, hence of the salt salt content , that is ,

Initially, the tank contains lb of salt,

Therefore, we obtain the initial condition  

Thus, the model is

            here, Linear ODE

So, the integrating factor is

and the general solution is

using initial condition and find the value of constant c.

Hence, is the amount of salt in the tank at any moment t.

answer: 80,000

step-by-step explanation:

the population doubles every day.

10,000*2=20,000 day 4

20,000*2=40,000 day 5

40,000*2=80,000 day 6