Respuesta :
Let [tex]A(t)[/tex] be the amount of salt in the tank at time [tex]t[/tex]. We're given that [tex]A(0)=40[/tex]. The rate at which this amount changes is given by
[tex]A'(t)=\dfrac{5\text{ gal}}{1\text{ min}}\cdot\dfrac{4\text{ oz}}{1\text{ gal}}-\dfrac{5\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ oz}}{100+(5-5)t\text{ gal}}[/tex]
[tex]A'(t)+\dfrac{A(t)}{20}=20[/tex]
[tex]e^{t/20}A'(t)+e^{t/20}\dfrac{A(t)}{20}=20e^{t/20}[/tex]
[tex]\bigg(e^{t/20}A(t)\bigg)'=20e^{t/20}[/tex]
[tex]e^{t/20}A(t)=400e^{t/20}+C[/tex]
[tex]A(t)=400+Ce^{-t/20}[/tex]
Since [tex]A(0)=40[/tex], we get
[tex]40=400+C\implies C=-360[/tex]
so that the amount of salt at time [tex]t[/tex] is
[tex]A(t)=400-360e^{-t/20}[/tex]
After 20 minutes, the tank contains
[tex]A(20)=400-360e^{-20/20}\approx267.56\text{ oz}[/tex]
[tex]A'(t)=\dfrac{5\text{ gal}}{1\text{ min}}\cdot\dfrac{4\text{ oz}}{1\text{ gal}}-\dfrac{5\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ oz}}{100+(5-5)t\text{ gal}}[/tex]
[tex]A'(t)+\dfrac{A(t)}{20}=20[/tex]
[tex]e^{t/20}A'(t)+e^{t/20}\dfrac{A(t)}{20}=20e^{t/20}[/tex]
[tex]\bigg(e^{t/20}A(t)\bigg)'=20e^{t/20}[/tex]
[tex]e^{t/20}A(t)=400e^{t/20}+C[/tex]
[tex]A(t)=400+Ce^{-t/20}[/tex]
Since [tex]A(0)=40[/tex], we get
[tex]40=400+C\implies C=-360[/tex]
so that the amount of salt at time [tex]t[/tex] is
[tex]A(t)=400-360e^{-t/20}[/tex]
After 20 minutes, the tank contains
[tex]A(20)=400-360e^{-20/20}\approx267.56\text{ oz}[/tex]
The amount of salt that is in the tank after 20 minutes is;
A(20) = 267.56 Oz
- We are given;
Initial amount of salt in tank; A(0) = 40 Ounces
A solution containing 4 oz salt per gallon is pumped into the tank at a rate of 5 gal/min.
This means rate in oz/min = 4/1 × 5/1 = 20 oz/min
- Now, the rate at which the initial amount in the tank changes will be;
A'(t) = 20 - [(A(t)/(100)) × 5/1]
A'(t) = 20 - A(t)/20
Rearranging gives;
A'(t) + A(t)/20 = 20
- Since this is a linear equation, the integrating factor will be; [tex]e^{t/20}[/tex]
Multiplying through by the integrating factor gives;
A'(t) [tex]e^{t/20}[/tex] + (A(t)/20) [tex]e^{t/20}[/tex] = 20[tex]e^{t/20}[/tex]
- Thus, the solution will be;
A(t) [tex]e^{t/20}[/tex] = 400[tex]e^{t/20}[/tex] + C
Divide through by [tex]e^{t/20}[/tex] to get;
A(t) = 400 + C [tex]e^{-t/20}[/tex]
At initial condition of A(0) = 40, we have;
40 = 400 + C
C = -360
Thus; at time (t), the amount of salt left is given by;
A(t) = 400 - 360 [tex]e^{-t/20}[/tex]
- After 20 minutes;
A(20) = 400 - 360 [tex]e^{-20/20}[/tex]
A(20) = 400 - 132.44
A(20) = 267.56 Oz
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