A fish tank initially contains 40 liters of pure water. Brine of constant, but u

Question

A fish tank initially contains 40 liters of pure water. Brine of constant, but unknown, concentration of salt is flowing in at 2 liters per minute. The solution is mixed well and drained at 2 liters per minute. a Let x be the amount of salt, in grams, in the fish tank after t minutes have elapsed. Find a formula for the rate of change in the amount of salt, dx/dt, in terms of the amount of salt in the solution x and the unknown concentration of incoming brine c. dx/dt = grams/minute b. Find a formula for the amount of salt. in grams, after t minutes have elapsed. Your answer should be in terms of c and t. x(t) = c. In 25 minutes there are 20 grams of salt in the fish tank. What is the concentration of salt in the incoming brine? c = g/L

Explanation / Answer

(a) The Tank have initially 40 liters of water

Concentration of salt flowing in=2 ltr/min

Solution is drained out = 2 ltr/min

let x= Amount of salt after t min

COncentration of slution =c

therfore dx/dt = 2c - (2/40)x=2c-(1/20)c

(b) => dx/(2c - (1/20)x) = dt

=> 20dx/(40c - x) = dt

=> 20ln(40c - x) = -t + C1

=> ln(40c - x) = (-t+ C1)/20

=> 40c-x=Ce^(-t/20) where C is an arbitrary constant

Given, When t = 0, x = 0

=> C = 40c

=>40c - x = 40ce^(-t/20)

=>x(t) = 40c - 40ce^(-t/20)

(c)t= 25 min, x=20 gm

then

20=40c-40ce^(-25/20)=40c(1-e^(-5/4))=40c*0.713=28.539

=> c=20/28.539=0.7007g/L

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