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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?

Respuesta :

LammettHash

Salt flows in at a rate of (5 g/L)*(3 L/min) = 15 g/min.

Salt flows out at a rate of (x/10 g/L)*(3 L/min) = 3x/10 g/min.

So the net flow rate of salt, given by [tex]x(t)[/tex] in grams, is governed by the differential equation,

[tex]x'(t)=15-\dfrac{3x(t)}{10}[/tex]

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

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

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

Integrate both sides, then solve for [tex]x[/tex]:

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

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

Since the tank starts with 5 g of salt at time [tex]t=0[/tex], we have

[tex]5=50+C\implies C=-45[/tex]

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

The time it takes for the tank to hold 20 g of salt is [tex]t[/tex] such that

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