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2 answers:
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Set up a differential equation using rate of salt going in minus rate going out.
Rate = dS/dt = salt/gal* gal/min
Also the number of gallons in tank is increasing at (4-2) gal/min = 2 gal/min
The volume of tank at any given time = 100+2t
Differential Equation:
Use integrating factor of t+50
![[S(t+50)]' = t+50](https://tex.z-dn.net/?f=%5BS%28t%2B50%29%5D%27%20%3D%20t%2B50)
Integrate both sides
Solve for constant c given S(0) = 20
c = 1000
Tank is full at 200 gallons when t =50
100+2t = 200 ---> t = 50
S(50) = 47.5
Final Answer: There are 47.5 pounds of salt
Rate = dS/dt = salt/gal* gal/min
Also the number of gallons in tank is increasing at (4-2) gal/min = 2 gal/min
The volume of tank at any given time = 100+2t
Differential Equation:
Use integrating factor of t+50
Integrate both sides
Solve for constant c given S(0) = 20
c = 1000
Tank is full at 200 gallons when t =50
100+2t = 200 ---> t = 50
S(50) = 47.5
Final Answer: There are 47.5 pounds of salt