Respuesta :
Answer:
(a)[tex]A(t)=18e^{ -\frac{t}{100}}[/tex]
(b)13.3 kg
Step-by-step explanation:
The volume of brine in the tank = 4000L
Initial Amount of salt, A(0)=18 kg
The rate of change in the amount of salt in the tank at any time t is represented by the equation:
[tex]\dfrac{dA}{dt}=$Rate In$-$Rate Out[/tex]
Rate In = (concentration of salt in inflow)(input rate of brine)
Since pure water enters the tank, concentration of salt in inflow =0
Rate In = 0
Rate Out=(concentration of salt in outflow)(output rate of brine)
[tex]=\frac{A(t)}{4000}\times 40\\ =\frac{A(t)}{100}[/tex]
Therefore:
[tex]\dfrac{dA}{dt}=-\dfrac{A(t)}{100}\\\dfrac{dA}{dt}+\dfrac{A(t)}{100}=0[/tex]
This is a linear D.E. which we can then solve for A(t).
Integrating Factor: [tex]e^{\int \frac{1}{100}d}t =e^{ \frac{t}{100}[/tex]
Multiplying all through by the I.F.
[tex]\dfrac{dA}{dt}e^{ \frac{t}{100}}+\dfrac{A(t)}{100}e^{ \frac{t}{100}}=0e^{ \frac{t}{100}}\\(Ae^{ \frac{t}{100}})'=0[/tex]
Taking integral of both sides
[tex]Ae^{ \frac{t}{100}}=C\\A(t)=Ce^{ -\frac{t}{100}}[/tex]
Recall our initial condition
A(0)=18 kg
[tex]18=Ce^{ -\frac{0}{100}}\\C=18[/tex]
Therefore, the amount of salt in the tank after t minutes is:
[tex]A(t)=18e^{ -\frac{t}{100}}[/tex]
(b)When t=30 mins
[tex]A(30)=18e^{ -\frac{30}{100}}\\=18e^{ -0.3}\\=13.3 $kg(correct to 1 decimal place)[/tex]
The amount of salt in the tank after 30 minutes is 13.3kg
In this exercise we have to use the integral to calculate the salt concentration:
(a)[tex]A(t)=18e^{-\frac{t}{100} }[/tex]
(b)[tex]13.3 kg[/tex]
Knowing that the volume of brine in the tank = 4000L, the initial Amount of salt, A(0)=18 kg. The rate of change in the amount of salt in the tank at any time t is represented by the equation:
[tex]\frac{dA}{dt} = Rate \ in - Rate \ out[/tex]
Rate In = (concentration of salt in inflow)(input rate of brine). Since pure water enters the tank, concentration of salt in inflow =0.
Rate In = 0
Rate Out=(concentration of salt in outflow)(output rate of brine)
[tex]\frac{A(t)}{4000}*(40)[/tex]
[tex]= \frac{A(t)}{100}[/tex]
Therefore:
[tex]\frac{dA}{dt} = \frac{A(t)}{100}\\\frac{dA}{dt} + \frac{A(t)}{100} = 0[/tex]
This is a linear D.E. which we can then solve for A(t). Integrating Factor: [tex]e^{\int\limits {\frac{t}{100} } \, dt\\e^{t/100}[/tex] . Multiplying all through by the Integrating Factor:
[tex]\frac{dA}{dt} = e^{t/100}+\frac{A(t)}{100}e^{t/100}\\(Ae^{1/100})'=0[/tex]
Taking integral of both sides:
[tex]Ae^{t/100}=C\\A(t)=Ce^{-t/100}[/tex]
Recall our initial condition:
[tex]A(0)=18 kg\\18=Ce^{0}\\C=18[/tex]
Therefore, the amount of salt in the tank after t minutes is:
[tex]A(t)=18e^{-t/100}[/tex]
(b)When t=30 mins
[tex]A(30)=18e^{-30/100}\\=18e^{-0.3}\\=13.3[/tex]
The amount of salt in the tank after 30 minutes is 13.3kg.
See more about concentration at brainly.com/question/12970323
