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Andrej [43]
3 years ago
15

A tank contains 100 L of pure water. Brine that contains 0.1 kg of salt per liter enters the tank at a rate of 10 L/min. The sol

ution is kept thoroughly mixed and drains from the tank at the same rate. How much salt is in the tank after 6 minutes?
Mathematics
1 answer:
garri49 [273]3 years ago
3 0
Let y(t) be the amount of salt (in kg) after t minutes. Then y(0) = 0. The amount of liquid in the tank is 100 L at all times, so the concentration at time t (in minutes) is y(t) / 1000\ \mathrm{kg/L) and

\dfrac{dy}{dt} = \left( 0.1\dfrac{\mathrm{kg}}{\mathrm{L}}\right) \left( 10 \dfrac{\text{L} }{\text{ min}}\right)- \left[ \dfrac{y(t)}{100} \dfrac{\mathrm{kg}}{\mathrm{L}}\right] \left( 10 \dfrac{\text{L} }{\text{ min}}\right) \\ \\ 
\dfrac{dy}{dt}  = 1 \dfrac{\mathrm{kg}}{\mathrm{min}}-\dfrac{y(t)}{10}\dfrac{\mathrm{kg}}{\mathrm{min}} \\ \\
\dfrac{dy}{dt} = 1 - \dfrac{y}{10} = \dfrac{10 - y}{10} \\
\displaystyle\int \dfrac{dy}{10 - y} = \int \frac{1}{10} dt \\
-\ln|10-y| = \frac{t}{10} + C

and y(0) = 0 ⇒ -\ln|10-0| = \frac{0}{10} + C\ \Rightarrow\ C = -\ln 10

-\ln|10-y| = \frac{t}{10} - \ln 10 \\
\Rightarrow \ln|10-y| = \ln 10 - \frac{t}{10}  \\
\Rightarrow  |10 - y| = e^{\ln 10 - t/10} \\
\Rightarrow  |10 - y| = 10e^{- t/10} \\ 
\Rightarrow  10 - y = \pm 10e^{- t/10} \\ 
\Rightarrow  y = 10 \pm 10e^{- t/10}

Choose (-) because that satisfies y(0) = 0

y(t) = 10 - 10e^{- t/10} \\
y(6) = 10 - 10e^{-6/10} \approx 4.512\ \mathrm{kg}

4.512 kg
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