Answer:
[tex]16e^{-\frac{t}{100}}[/tex]kg
Step-by-step explanation:
We are given that
Volume,V=9000 L
[tex]\frac{dV}{dt}=90L/min[/tex]
We have to find the salt in the tank after t minutes.
Let y(t) be the amount of salt dissolved in the tank
y(0)=16
[tex]\frac{dy}{dt}=[/tex]rate in-rate out
Rate in=0
Rate out=[tex]\frac{y}{9000}\times \frac{dV}{dt}=\frac{y(t)}{9000}\times 90=\frac{y}{100}[/tex]
[tex]\frac{dy}{dt}=0-\frac{y}{100}[/tex]
[tex]\int dy=-\int \frac{y}{100}dt[/tex]
[tex]\int \frac{dy}{y}=-\int\frac{1}{100}dt[/tex]
[tex]ln y=-\frac{t}{100}+C[/tex]
[tex]y=e^{-\frac{t}{100}+C}=e^{-\frac{t}{100}}\cdot e^C=Ke^{-\frac{t}{100}}[/tex]
Where [tex]k=e^{C}[/tex]
Substitute t=0 and y=16
[tex]16=C[/tex]
[tex]y(t)=16e^{-\frac{t}{100}}[/tex]kg
The required time is [tex]y(t)=16e^{-\frac{t}{100}[/tex]
Differential calculus deals with the rate of change of one quantity with respect to another.
Let [tex]y(t)[/tex] be the amount of salt in the tank then we get,
[tex]y(0)=16[/tex]
[tex]\frac{dy}{dt} =Rate\ in\ -\ Rate\ out[/tex]
Since pure water enters the tank the rate of in is zero and the rate of out is,
[tex]\frac{y(t)}{9000} \times \frac{90l}{min}=\frac{y(t)}{100} l/min[/tex]
So, we have,
[tex]\frac{dy}{dt}=-\frac{y}{100}\\ \frac{dy}{y}=-\frac{1}{100} dt[/tex]
Integrating both sides we get,
[tex]ln y=-\frac{t}{100}+c\\ y=Ae^{-\frac{t}{100} }[/tex]
Using the initial condition [tex]y(0)=16, A=16[/tex] then,
[tex]y(t)=16e^{-\frac{t}{100}[/tex]
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