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pochemuha
3 years ago
5

A large tank contains 200 liters of a salt solution which has a concentration of 0.35 kilograms per liter. Pure water is added t

o the solution at a rate of 40 liters per minute. At the same time, the solution drains from the tank at 40 liters per minute
Mathematics
1 answer:
Ymorist [56]3 years ago
6 0
The question is incomplete, but from the given information it's clear that this is a standard mixing problem. Let A(t) be the amount of salt (in kilograms) in the tank at time t.

Then the rate of change of the amount of salt in the tank is described by the differential equation

\dfrac{\mathrm dA(t)}{\mathrm dt}=\left(0\dfrac{\text{kg}}{\text{L}}\right)\left(40\dfrac{\text{L}}{\text{min}}\right)-\left(\dfrac{A(t)}{200}\dfrac{\text{kg}}{\text{L}}\right)\left(40\dfrac{\text{L}}{\text{min}}\right)
\drac{\mathrm dA}{\mathrm dt}=-\dfrac A5

Separating the variables, we get

\dfrac{\mathrm dA}A=-\dfrac{\mathrm dt}5

and integrating both sides yields

\ln|A|=-\dfrac t5+C
\implies A=Ce^{-t/5}

Initially, the tank contains \left(0.35\dfrac{\text{kg}}{\text{L}}\right)\left(200\text{ L}\right)=70\text{ kg} of salt, i.e. A(0)=70. This means you have

70=Ce^{-0/5}\implies C=70

and so the amount of salt in the tank as a function of time is

A(t)=70e^{-t/5}
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