Respuesta :
Answer:
47.25 pounds
Step-by-step explanation:
[tex]\dfrac{dA}{dt}=R_{in}-R_{out}[/tex]
First, we determine the Rate In
Rate In=(concentration of salt in inflow)(input rate of brine)
[tex]=(0.5\frac{lbs}{gal})( 6\frac{gal}{min})\\R_{in}=3\frac{lbs}{min}[/tex]
Change In Volume of the tank, [tex]\frac{dV}{dt}=6\frac{gal}{min}-4\frac{gal}{min}=2\frac{gal}{min}[/tex]
Therefore, after t minutes, the volume of fluid in the tank will be: 100+2t
Rate Out
Rate Out=(concentration of salt in outflow)(output rate of brine)
[tex]R_{out}=(\frac{A(t)}{100+2t})( 4\frac{gal}{min})\\\\R_{out}=\frac{4A(t)}{100+2t}[/tex]
Therefore:
[tex]\dfrac{dA}{dt}=3-\dfrac{4A(t)}{100+2t}\\\\\dfrac{dA}{dt}=3-\dfrac{4A(t)}{2(50+t)}\\\\\dfrac{dA}{dt}=3-\dfrac{2A(t)}{50+t}\\\\\dfrac{dA}{dt}+\dfrac{2A(t)}{50+t}=3[/tex]
This is a linear differential equation in standard form, therefore the integrating factor:
[tex]e^{\int \frac{2}{50+t}dt}=e^{2\ln|50+t|}=e^{\ln(50+t)^2}=(50+t)^2[/tex]
Multiplying the DE by the integrating factor, we have:
[tex](50+t)^2\dfrac{dA}{dt}+(50+t)^2\dfrac{2A(t)}{50+t}=3(50+t)^2\\\{(50+t)^2A(t)\}'=3(50+t)^2\\$Taking the integral of both sides\\\int \{(50+t)^2A(t)\}'= \int 3(50+t)^2\\(50+t)^2A(t)=(50+t)^3+C $ (C a constant of integration)\\Therefore:\\A(t)=(50+t)+C(50+t)^{-2}[/tex]
Initially, 20 pounds of salt was dissolved in the tank, therefore: A(0)=20
[tex]20=(50+0)+C(50+0)^{-2}\\20-50=C(50)^{-2}\\C=-\dfrac{30}{(50)^{-2}} =-30X50^2=-75000[/tex]
Therefore, the amount of salt in the tank at any time t is:
[tex]A(t)=(50+t)-75000(50+t)^{-2}[/tex]
After 15 minutes, the amount of salt in the tank is:
[tex]A(15)=(50+15)-75000(50+15)^{-2}\\=47.25$ pounds[/tex]
Following are the solution to the given question:
Using formula:
[tex]\to \frac{dA}{dt} =R_{in}-R_{out}\\\\[/tex]
Find:
[tex]\to R_{in}\ and \ R_{out}\ =? \\\\[/tex]
Solution:
[tex]\to R_{ in} = \text{(concentration of salt in inflow)}[/tex][tex]\times \text{(i\1nput rate of brine)}\\[/tex]
[tex]\therefore\\ R_{in} = (\frac{1}{2}\frac{lb}{gal})\cdot (3\frac{gal}{min}) = 3\frac{lb}{min}\\[/tex]
Since in this question the solution is pumped out at a slower rate, therefore the water amount in the tank accumulates at the rate of[tex](6- 4) = 2 \ \frac{gal}{min} Thus,[/tex]
after t minutes there will be [tex]100+ 2t[/tex]gallons in the tank.
[tex]\because R_{out}=\text{(concentration of salt in outflow)}\cdot \text{(o\1utput rate of brine)}[/tex]
[tex]\therefore R_{out} =\frac{A(t)}{100+2t} \frac{lb}{gal} \cdot (4\frac{gal}{min})= \frac{4A(t)}{100+2t}\ \frac{lb}{min}[/tex]
Now, we substitute these results in the DE to get
[tex]\to \frac{dA}{dt}=3-\frac{4A(t)}{100+2t}\\\\\to \frac{dA}{dt}= 3 -\frac{2A(t)}{50+t} \\\\\to \frac{dA}{dt}+\frac{2}{50+t}\ A(t) -3\\\\[/tex]
Which is a linear DE in the standard form Thus, the integrating factor is
[tex]e ^{\frac{2}{ 50+t} \ dt} + e^{2\ln|(50+t)|} =e^{\ln(50 + t)^2}=(50+t)^2[/tex]
Multiplying the DE by integrated factor:
[tex](50+t)^2 \frac{dA}{dt} +2(50 + t) A(t) = 3(50 +t)^2\\\\ \therefore \frac{d}{dt}(50+t)^2 \frac{dA}{dt}-3(50+t)^2\\\\\therefore (50+t)^2\ A(t) -3(50+t)^2 \ dt \\\\\therefore (50+t)^2\ A(t) -\frac{3}{3}(50+t)^3 + c \\\\\therefore \ A(t) -(50+t) + c(50+t)^{-2} \\\\[/tex]
Now, applying the initial condition [tex]A(0) = 20[/tex]to get
[tex]\to 20 = (50) +\frac{c}{2500}\\\\\to 20 - 50=\frac{c}{2500}\\\\[/tex]
[tex]\therefore \\\\\to \frac{c}{2500} = -30\\\\ \to c= -30 \times 2500 \\\\\ \to c= -75000\\\\[/tex]
Now, substitute by this result in the solution to get
[tex]\to A(t) = (50 +t) - \frac{-75000}{(50 +t)^2}[/tex]
Now, after 15 minutes we find that
[tex]\to A(30) = 50 + 15-\frac{-75000}{(50+15)^2 } = 65+\frac{75000}{65^2} \\\\[/tex]
therefore
[tex]\to A(30)=65 +\frac{75000}{4225} \\\\[/tex]
[tex]=65 +\frac{15000}{845} \\\\ =65+\frac{3000}{169}\\\\= 65+17.75 \\\\= 82.75 \ \frac{lb}{gal} \\\\[/tex]
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