Respuesta :
Answer:
The differential equation for the amount of salt A(t) in the tank at a time t > 0 is [tex]\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}[/tex].
Step-by-step explanation:
We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.
The concentration of the solution entering is 4 lb/gal.
Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;
[tex]\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t[/tex]
where, [tex]\text{R}_i_n[/tex] = concentration of salt in the inflow [tex]\times[/tex] input rate of brine solution
and [tex]\text{R}_o_u_t[/tex] = concentration of salt in the outflow [tex]\times[/tex] outflow rate of brine solution
So, [tex]\text{R}_i_n[/tex] = 4 lb/gal [tex]\times[/tex] 3 gal/min = 12 lb/gal
Now, the rate of accumulation = Rate of input of solution - Rate of output of solution
= 3 gal/min - 2 gal/min
= 1 gal/min.
It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.
So, [tex]\text{R}_o_u_t[/tex] = concentration of salt in the outflow [tex]\times[/tex] outflow rate of brine solution
= [tex]\frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min}[/tex] = [tex]\frac{2A(t)}{500+t} \text{ lb/min }[/tex]
Now, the differential equation for the amount of salt A(t) in the tank at a time t > 0 is given by;
= [tex]\frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }[/tex]
or [tex]\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}[/tex].
