Question

A tank contains 70 kg of salt and 1000 L of water. Water containing 0.4kgL of salt enters the tank at the rate 16 Lmin. The solution is mixed and drains from the tank at the rate 4 Lmin. A(t) is the amount of salt in the tank at time t measured in kilograms.

Answer 1

`A(t) = 1600 - e^{-((t)/(250) - 7.33)}`

Step-by-step explanation:

A(t) is the amount of salt in the tank at time t, measured in kilograms.

A(0) = 70

`(dA)/(dt) =`rate in - rate out

=0.4* 16 - (A/1000)*4

=
`(1600 - A)/(250)` kg/min

`\int\limits {(1)/(1600-A) } \, dA = \int\ {(1)/(250) } \, dt\\\\ -ln(1600-A) = (t)/(250) + C\\\\`

A(0) = 70

-ln (1600 - 70) = 0/250 + C

-7.33 = C

`-ln(1600-A) = (t)/(250) - 7.33\\\\`

`1600 - A = e^{-((t)/(250) - 7.33)} \\\\A = 1600 - e^{-((t)/(250) - 7.33)}`
`1600-A = e^{-((t)/(250) - 7.33)} \\\\ A = 1600 - e^{-((t)/(250) - 7.33)}`