Answer:
dA/dt = 12 - 2A/(100 + t)
Step-by-step explanation:
The differential equation of this problem is;
dA/dt = R_in - R_out
Where;
R_in is the rate at which salt enters
R_out is the rate at which salt exits
R_in = (concentration of salt in inflow) × (input rate of brine)
We are given;
Concentration of salt in inflow = 4 lb/gal
Input rate of brine = 3 gal/min
Thus;
R_in = 4 × 3 = 12 lb/min
Due to the fact that solution is pumped out at a slower rate, thus it is accumulating at the rate of (3 - 2)gal/min = 1 gal/min
So, after t minutes, there will be (100 + t) gallons in the tank
Therefore;
R_out = (concentration of salt in outflow) × (output rate of brine)
R_out = [A(t)/(100 + t)]lb/gal × 2 gal/min
R_out = 2A(t)/(100 + t) lb/min
So, we substitute the values of R_in and R_out into the Differential equation to get;
dA/dt = 12 - 2A(t)/(100 + t)
Since we are to use A foe A(t), thus the Differential equation is now;
dA/dt = 12 - 2A/(100 + t)
