Respuesta :
Answer:
Therefore, the number of grams of salt in the tank at time t is [tex]A(t) = 150-140 e^{-\frac{t}{30} }[/tex]
Explanation:
Given:
Tank A contain [tex]V_{1} = 150[/tex] lit
Rate [tex]\alpha = 5 \frac{L}{min}[/tex]
Dissolved salt [tex]A = 10[/tex] gm
Salt pumped in one minute is [tex]4 \frac{L}{min}[/tex]
Salt pumped out is [tex]\frac{5L}{150L} = \frac{1}{30}[/tex] of initial amount added salt.
To find [tex]A(t)[/tex]
[tex]\frac{dA}{dt} = Rate _{in} - Rate _{out}[/tex]
[tex]A' = 5 - \frac{A}{30}[/tex]
[tex]A' + \frac{A}{30} = 5[/tex]
Solving above equation,
[tex]I .F = e^{\int\limits {p} \, dt }[/tex]
[tex]y = e^{\int\limits {\frac{1}{30} } \, dt }[/tex]
[tex]y = e^{\frac{t}{30} }[/tex]
[tex](Ae^{\frac{t}{30} } )' = 5 e^{\frac{t}{30} } + c[/tex]
Integrating on both side,
[tex]Ae^{\frac{t}{30} } = 5 \times 30 e^{\frac{t}{30} } +c[/tex]
Add [tex]e^{-\frac{t}{30} }[/tex] on above equation,
[tex]A = 150 + ce^{-\frac{t}{30} }[/tex]
Here given in question,[tex]A(t=0) = 10[/tex]
[tex]10 =150 +c[/tex]
[tex]c = -140[/tex]
Put value of constant in above equation, and find the number of grams of salt in the tank at time t.
[tex]A(t) = 150-140 e^{-\frac{t}{30} }[/tex]
Therefore, the number of grams of salt in the tank at time t is [tex]A(t) = 150-140 e^{-\frac{t}{30} }[/tex]
