Consider a tank which initially holds `V_(0)` liter of brine that contains a lb of salt. Another brine solution, containing b lb of salt per liter is poured into the tank at the rate of `eL//"min"`. The problem is to find the amount of salt in the tank at any time t.
Let Q denote the amount of salt in the tank at any time. The time rate of change of Q, `(dQ)/(dt)`, equals the rate at which salt enters the tank at the rate at which salt leaves the tank. Salt enters the tank at the rate of be lb/min. To determine the rate at which salt leaves the tank, we first calculate the volume of brine in the tank at any time t, which is the initial volume `V_(0)` plus the volume of brine added et minus the volume of brine removed ft. Thus, the volume of brine at any time is
`V_(0)+et-ft`
The concentration of salt in the tank at any time is `Q//(V_(0)+et-ft)`, from which it follows that salt leaves the tank at the rate of `f(Q/(V_(0)+et-ft))`lb/min. Thus,
`(dQ)/(dt)=be-f(Q/(V_(0)+et-ft))Q=be`
A tank initially holds 100 L of a brine solution containing 20 lb of salt. At t=0, fresh water is poured into the tank at the rate of 5 L/min, while the well-stirred mixture leaves the tank at the same rate. Then the amount of salt in the tank after 20 min.

Options

A. 20/e
B. 10/e
C. `40//e^(2)`
D. 5/e L

Answer

A

Explanation

Here, `V_(0)=100, a=20, b=0` and `e=f=5`. Hence
`(dQ)/(dt)+1/20Q=0`
The solution of this linear equation is `Q=ce^(-t//20)`…………….(1)
At t=0, we are given that Q=a=20.
Substituting these values into equations (1), we find that `c=20,` so that equation (1), can be rewritten as `Q=20e^(-t//20)`.
For `t=20, Q=20/e`