Brine mixture

[macca1990]

This may be in the wrong section. If so please could someone move it to the right one.

I am struggling with that attached questions. I have found the answers online but the explanation is not that good so I can’t follow it through to understand how they arrived at the answer.

 

[MarkFL]

I am proceeding based on the assumption that the first problem is what you're asking about. The rate of change of the amount \(A\) of salt (in kg) in the tank at time \(t\) (in hrs) is equivalent to the amount of salt coming in less the amount going out. Adding in the given initial condition gives rise to the following IVP:

[MATH]\d{A}{t}=34\cdot\frac{1}{11}(1+\cos(t))-22.7\cdot\frac{A}{V}[/MATH] where \(A(0)=2.273\)

\(V\) represents the volume of brine present in the tank at time \(t\) and it is itself the solution to the simple IVP:

[MATH]\d{V}{t}=34-22.7=11.3[/MATH] where \(V(0)=2270\)

Let's solve this second IVP first, and obtain \(V(t)\) which we can then plug into the first IVP. What do you get for \(V(t)\)?

 

[macca1990]

I am unsure how to solve this second IVP

 

[HallsofIvy]

What course is this for? Even in Calculus I or II you should have learned that if dV/dt= A, a constant, then V(t)= At+ C .

 

[MarkFL]

What course is this for? Even in Calculus I or II you should have learned that if dV/dt= A, a constant, then V(t)= At+ C .

Yes, that's why I have not replied further because if the OP cannot solve the second IVP, then the first will be impossible.