A pond initially contains 1,000,000 gallons of water and some amount of an undes

Question

A pond initially contains 1,000,000 gallons of water and some amount of an undesirable chemical. Water containing 0.01 g of this chemical per gallon flows into the pond at a rate of 300 gal/hr. The water in the pond is well mixed and the mixture flows out at the same rate, so the amount of water in the pond remains constant. a) Write a differential equation for the amount of chemical in the pond at any time. b) How much of the chemical will be in the pond after a very long time? Does this limiting amount depend on the amount that was present initially?

Explanation / Answer

(a) Let a(t) denote the amount of chemical (in gms) at any point of time, t.

Since the amount of chemical depends on the chemical mixed water flow at time t, we need to write an equation for the rate of change of amount of chemical in the pond that is given by da/dt gm/hr.

Here, there are two things to consider: first the inflow of chemical mixed water which increases the amount of chemical. Second, the outflow of water which decreases the amout of chemical. Hence the equation for da/dt can be computed as:

da/dt= rate at which amount of chemical increases due to inflow - rate at which amount of chemical increases due to inflow

Inflow: The mixed water flowing into the pond contains 0.01 g of the chemical per gallon and it enters into pond at the rate of 300 gal/h. Therefore the amount of chemical increases at the rate = number of gallons of water x amount of chemical per gallon of water = 300 x 0.01 = 3 gm/hr.

Outflow: Given the water flows out of the pond at the same rate at which it flows in (300 gal/h). As above, we need to multiply 300 by the amount of chemical each gallon contains. The amount of chemical in the entire pond depends on t and is equal to a(t). The number of gallons in the entire pond is 1,000,000. So the amount of chemical in each gallon is a(t)/1, 000, 000.

Hence the amount of chemical leaving the pond every hour is

300 x a(t)/1, 000, 000 = 3/10, 000a(t) gm/hr.

Therefore, the final value for the rate of change of the amout of the chemical da/dt=3 - 3/10, 000a(t) gm/hr

(b)To decide how much chemical will be in the pond after a long time, we have to find the equilibrium solution. To find the equilibrium solution, we set the rate of change equal to 0.

0 = 3 3/10, 000a(t) 3 = 3/10, 000a(t) or   a(t) = 10, 000 is the only equillibrium solution.

Hence after a very long time there will be 10, 000 g of chemical in the pond.

(c) We can see that when a(t) < 10000, da/dt > 0; while when a(t) > 10000, da/dt < 0. Thus a(t) = 10000 is asymptotically stable. Therefore limt Q(t) = 10000, irrespective the intial amout of chemical.

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