The quantity of salt in the tank as it’s about to overflow is 47.5 pounds
The initial volume of the water in the tank is:
V = 100
The solution enters at a rate of 4 gal/min, and leaves at a rate of 2 gal/min.
So, the amount of water in the tank at time t is:
V(t) = 100 + (4 - 2)*t
V(t) = 100 + 2t
Represent the quantity of salt with S.
So, the rate of change of the amount of salt in the tank with time is:
[tex]\frac{dS}{dt} = \frac14 * R_{in} - \frac{S}{V(t)}* R_{out}[/tex]
This gives
[tex]\frac{dS}{dt} = \frac14 * 4 - \frac{S}{100 + 2t}* 2[/tex]
[tex]\frac{dS}{dt} = 1 - \frac{S}{50 + t}[/tex]
Collect like terms
[tex]\frac{dS}{dt} + \frac{S}{50 + t}= 1[/tex]
Multiply through by 50 + t
[tex]\frac{dS}{dt}(50 + t) + S= 50 + t[/tex]
Let 50 + t be the integrating factor.
So, we have:
[tex]S'(50 + t) = 50 + t[/tex]
Integrate
[tex]S(50 + t) = 50t + \frac{t^2}{2} + c[/tex]
Initially there are 20 pounds of salt in the tank.
So, we have:
[tex]S(0) = 20[/tex]
[tex]c = 47.5[/tex]
The equation becomes
[tex]S(50 + t) = 50t + \frac{t^2}{2} + 47.5[/tex]
Recall that:
V(t) = 100 + 2t
When the tank is full, we have: i.e. 200 gallons
100 + 2t = 200
Solve for t
t = 50
Calculate S(50)
[tex]S(50 + 0) = 50*0 + \frac{0^2}{2} + 47.5[/tex]
[tex]S(50) = 0 + 0 + 47.5[/tex]
[tex]S(50) = 47.5[/tex]
Hence, the quantity of salt in the tank as it’s about to overflow is 47.5 pounds
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