A 120-gallon vat initially contains 90lb of salt dissolved in 90 gallons of water. Brinecontaining 2 lb/gal flows into the vat at a rate of 4gal/min. The water in the vat is kept well mixed, and the mixture is removed at the rate of 3 gal/min. Derive a DE for he total amount S(t) of salt in the vat at time t.

dS/dt = 8 - (3S(t)/(90 + t))

Step-by-step explanation:

The differential equation of this problem is;

dS/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 = 2 lb/gal

Input rate of brine = 4 gal/min

Thus;

R_in = 2 × 4 = 8 lb/min

Due to the fact that mixture is removed out at a slower rate, thus it is accumulating at the rate of (4 - 3)gal/min = 1 gal/min

Now, R_out = 3(concentration of solution outflow)

Now, concentration of solution outflow is given by the formula;

c_out = S(t)/v(t)

Where v is volume of fluid in tank at time t.

Since initial volume is 90 and we have a net total of 1 gallon added every minute. Thus, v(t) = 90 + t

So, c_out = S(t)/(90 + t)

R_out = 3S(t)/(90 + t)

So, final final differential equation is now;

dS/dt = R_in - R_out = 8 - 3S(t)/(90 + t)

false

step-by-step explanation:

4y - 6 + 4y^2 - 9

before simplifying this contains 4 terms

after simplifying

4y -15 + 4y^2

it contains 3 terms