A tank contains 9,000 L of brine with 18 kg of dissolved salt. Pure water enters the tank at a rate of 90 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate. (a) How much salt is in the tank after t minutes?
(b) How much salt is in the tank after 20 minutes?

Let x(t) denote the amount of salt (in kg) in the tank at time t. The tank starts with 18 kg of salt, so x (0) = 18.

The solution is drained from the tank at a rate of 90 L/min, so that the amount of salt in the tank changes according to the differential equation

dx(t)/dt = - (x(t) kg)/(9000 L) × (90 L/min) = -1/100 x(t) kg/min

or, more succintly,

x' = -1/100 x

This equation is separable as

dx/x = -1/100 dt

Integrating both sides gives

∫ dx/x = -1/100 ∫ dt

ln|x| = -1/100 t + C

x = exp(-1/100 t + C )

x = C exp(-t/100)

(a) Using the initial condition x (0) = 18, we find

18 = C exp(0)   ==>   C = 18

so that

x(t) = 18 exp(-t/100)

(b) After 20 minutes, we have

x (20) = 18 exp(-20/100) = 18 exp(-1/5) ≈ 14.74

so the tank contains approximately 14.74 kg of salt after this time.

step-by-step explanation:

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which expression represents the approximate length of bc?

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