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Igoryamba
8 days ago
12

Initially 5 grams of salt are dissolved into 10 liters of water. Brine with concentration of salt 5 grams per liter is added at

a rate of 3 liters per minute. The tank is well mixed and drained at 3 liters per minute.Let xbe the amount of salt, in grams, in the solution 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.dxdt=equation editorEquation Editorgrams/minuteFind a formula for the amount of salt, in grams, after tminutes have elapsed.x(t)=equation editorEquation EditorgramsHow long must the process continue until there are exactly 20 grams of salt in the tank?
Mathematics
1 answer:
Svet_ta [4.3K]8 days ago
3 0

The salt enters at a rate of (5 g/L)*(3 L/min) = 15 g/min.

The salt exits at a rate of (x/10 g/L)*(3 L/min) = 3x/10 g/min.

Thus, the total rate of salt flow, represented by x(t) in grams, is defined by the differential equation,

x'(t)=15-\dfrac{3x(t)}{10}

which is linear. Shift the x term to the right side, then multiply both sides by e^{3t/10}:

e^{3t/10}x'+\dfrac{3e^{t/10}}{10}x=15e^{3t/10}

\implies\left(e^{3t/10}x\right)'=15e^{3t/10}

Next, integrate both sides and solve for x:

e^{3t/10}x=50e^{3t/10}+C

\implies x(t)=50+Ce^{-3t/10}

Initially, the tank contains 5 g of salt at time t=0, so we have

5=50+C\implies C=-45

\implies\boxed{x(t)=50-45e^{-3t/10}}

The duration required for the tank to contain 20 g of salt is t, such that

20=50-45e^{-3t/10}\implies t=\dfrac{20}3\ln\dfrac32\approx2.7031\,\mathrm{min}

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