lulu5563

11.10.2021 •

Mathematics

Homecoming 2021 at surry county highschool #6

Daniel Talkington

#7

Nathen Turner

#55

Maliki

Homecoming 2021 at surry county highschool

#6Daniel Talkington#7 Nathen Turner#55Maliki

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Ответ:

bellagracebulle8018

29.07.2021

Mathematics

$A(t)=300-290e^{-\frac{t}{60}}$

Step-by-step explanation:

The volume of fluid in the tank = 300 Liters

Initial amount of Salt in the tank, A(0)=10 grams

Change in the Amount of Salt in the Tank

$\dfrac{dA}{dt}=R_{in}-R_{out}$

Rate In =(concentration of salt in inflow)(input rate of brine)

$R_{in}=(1\frac{gram}{L})( 5\frac{L}{min})=5\frac{grams}{min}$

Rate Out =(concentration of salt in outflow)(output rate of brine)

$R_{out}=(\frac{A(t)}{300})( 5\frac{L}{min})=\frac{A(t)}{60}$

Therefore:

$\dfrac{dA}{dt}=5-\dfrac{A(t)}{60}\\$Rearranging, we have:\\\dfrac{dA}{dt}+\dfrac{A(t)}{60}=5$

We solve the resulting linear differential equation for A(t)

$\text{The integrating factor: } e^{\int \frac{1}{60}dt} =e^{\frac{t}{60}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{60}}+\dfrac{A}{60}e^{\frac{t}{60}}=5e^{\frac{t}{60}}\\(Ae^{\frac{t}{60}})'=5e^{\frac{t}{60}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{60}})'=\int 5e^{\frac{t}{60}} dt\\Ae^{\frac{t}{60}}=5*60e^{\frac{t}{60}}+C, $(C a constant of integration)\\Ae^{\frac{t}{60}}=300e^{\frac{t}{60}}+C\\$Divide all through by e^{\frac{t}{60}}\\A(t)=300+Ce^{-\frac{t}{60}}$

Recall that when t=0, A(t)=10 grams (our initial condition)

$10=300+Ce^{-\frac{0}{60}}\\10-300=Ce^0\\C=-290$

Therefore, the number A(t) of grams of salt in the tank at time t is:

$A(t)=300-290e^{-\frac{t}{60}}$

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Homecoming 2021 at surry county highschool #6Daniel Talkington#7 Nathen Turner#55Maliki