Can someone explain this to me Please show all work i would

Can someone explain this to me? Please show all work i would apprciate it alot

A worker pours salt into a 100 gal tank at a rate of 2 lb/min. The target is a 0.01 lb/gal concentration of salty water. Derive an equation that determines the rate of change of concentration in time. Determine how long the worker must pour in the salt to reach the desired concentration. Answer: 30 sec.

Now suppose the worker pours in the salt from a big bag at a rate that dies away exponentially. Specifically the rate is 4 exp(-2t) lb/min. How long will it take to reach the desired concentration? Answer: 0.35 min.

Solution

If N(t) is the mass of salt in the tank at time t (in pounds mass), the concentration in the tank at time t is simply C(t) = N(t)/V (in lbs/gal), where V is the volume of water in the tank in gallons.

No salt ever leaves the tank (it doesn\'t overflow), so the rate of change in the amount of salt in the tank is then:

dN/dt = A(t),

where A(t) is either a constant 2lb/min for the first case, or (4lb/min)*exp(-2t/min) in the second case.

The first case yields a differential equation:

dN/dt = +2lb/min

or

dC/dt = (1/V)*dN/dt = (2/V)*(lb/min)

Because V = 100 gal,


dC/dt = (1/50)*(lb/(gal*min))

dC = (1/50)*(lb/(gal*min)) dt

C(t) = (t/50min)*(lb/gal) + Co

where Co is the initial concentration of salt in the tank, which is not specified in the question. Assuming the initial concentration at t = 0 is C(0) = 0, then Co = 0 and :


C(t) = (t/50min)*(lb/gal)

If the desired concentration is 0,01 lb/gal, then this concentration is reached when:

0.01 = (T/50min)

0.5min = T

The target concentration would be reached in half a minute.

Now, if the salt is added at an exponentially declining rate:

dN/dt = (4lb/min)*exp(-2t/min)

dC/dt = 1lb/(25 gal*min)*exp(-2t/min)

This is again a separable equation:

dC = 1lb/(25 gal*min)*exp(-2t/min) dt

C(t) = -(1min/2)*(1lb/(25gal*min)*exp(-2t/min) + D

where D is the constant of integration.

C(t) = D - (1lb/50gal)*exp(-2t/min)

Again, we need to know the inital concentration of salt in the tank. Assuming it\'s zero, then:

C(0) = 0 = D - (1lb/50gal)

D = (1lb/50gal)

so

C(t) = (1lb/50gal)*(1 - exp(-2t/min))

The time to reach the desired concentration of 0.01 lb/gal is then given by:

0.01 = (1/50)*(1 - exp(-2T/min))

0.5 = 1 - exp(-2T/min)

0.5 = exp(-2T/min)

ln(1/2) = -2T/min

ln(2)*min = 2T

ln(2)*min/2 = T

T = 0.347 min