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Thread: mixing problem

  1. #1
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    mixing problem

    tank contains 1000L of water with 15kg of salt dissolved in it. Pure water enters tank at 10L/min and the mixture leaves at the same rate. How much salt is in the tank after t minutes?

    Work:
    let y(t)=mass of salt in water
    y(0)=15kg
    dy/dt = rate in - rate out
    rate in = 0 because no salt is entering
    rate out = \frac{y(t)}{1000L}
    \longrightarrow \frac{dy}{dt}=0-{y(t)}{1000L}\times\frac{10L}{min}=\frac{-y(t)}{100min}
    This becomes a seperatable equation so I get \int\frac{-1}{y(t)}dy=\int\frac{1}{100t} dt = -ln|y| = \frac{ln|t|}{100} solving for y I get \frac{1}{t^{\frac{1}{100}}}

    The answer given is 15e^{\frac{-t}{100}}kg
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  2. #2
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    Quote Originally Posted by superdude View Post
    tank contains 1000L of water with 15kg of salt dissolved in it. Pure water enters tank at 10L/min and the mixture leaves at the same rate. How much salt is in the tank after t minutes?

    Work:
    let y(t)=mass of salt in water
    y(0)=15kg
    dy/dt = rate in - rate out
    rate in = 0 because no salt is entering
    rate out = \frac{y(t)}{1000L}
    \longrightarrow \frac{dy}{dt}=0-{y(t)}{1000L}\times\frac{10L}{min}=\frac{-y(t)}{100min}
    This becomes a seperatable equation so I get \int\frac{-1}{y(t)}dy=\int\frac{1}{100t} dt = -ln|y| = \frac{ln|t|}{100} solving for y I get \frac{1}{t^{\frac{1}{100}}}

    The answer given is 15e^{\frac{-t}{100}}kg

    \frac{dy}{dt} = -\frac{1}{100} y<br />

    separate variables ...

    \frac{dy}{y} = -\frac{1}{100} dt

    integrate ...

    \ln{y} = -\frac{t}{100} + C

    change to an exponential equation ...

    y = Ae^{-\frac{t}{100}}

    utilize the given initial condition, at t = 0 ... y = 15

    15 = Ae^0

    y = 15e^{-\frac{t}{100}}
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  3. #3
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    I get lost after "change to an exponential equation". Are you doing e^{\ln y} = e^{\frac{-t}{100}} + e^C ? If yes, where does the A come from?
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  4. #4
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    Quote Originally Posted by superdude View Post
    I get lost after "change to an exponential equation". Are you doing e^{\ln y} = e^{\frac{-t}{100}} + e^C ? If yes, where does the A come from?
    \ln{y} = -\frac{t}{100} + C

    \Rightarrow e^{\ln{y}} = e^{-\frac{t}{100} + C} = e^{-\frac{t}{100}} e^C

    using the usual index law

    \Rightarrow y =  e^{-\frac{t}{100}} e^C.

    Since C is arbitrary e^C is also arbitrary and so can be given a new symbol, like A. Therefore:

    y = A e^{-\frac{t}{100}}.
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