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Thread: Mixtures

  1. #1
    No one in Particular VonNemo19's Avatar
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    Mixtures

    A large tank is filled to capacity with 500 gallons of pure water. Brine (a mixture of salt and water) containing 2 pounds of salt per gallon is pumped into the tank at a rate of 5 gallons per minute. The well-mixed solution is pumped out at a rate of 10 gallons per minute. Find the number A(t) of pounds of salt in the tank at time t.

    OK. I know that the rate at which the amount of salt is changing inside the tank with respect to time - \frac{dA}{dt} - is simply the net rate of the amount being pumped in and the amount being pumped out, that is

    \frac{dA}{dt}=R_{in}-R_{out}.

    Now, R_{in}=(2\frac{lbs}{gal})\cdot(5\frac{gal}{min})=1  0\frac{lbs}{min}

    and I think that R_{out}=\frac{A(t)lbs}{(500-5t)gal}\cdot10\frac{gal}{min}

    Assuming my R_{out} is correct, this would imply the DE

    \frac{dA}{dt}+\frac{5}{100-t}A=10 which is linear with the integrating factor

    e^{\int\frac{5}{100-t}dt}=e^{-5\ln|100-t|}=e^{\ln(100-t)^{-5}}=(100-t)^{-5}.

    So, I have

    \int\frac{d}{dt}\left[(100-t)^{-5}A\right]dt=\int10(100-t)^{-5}dt

    Am I on the right track?
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  2. #2
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    Quote Originally Posted by VonNemo19 View Post
    A large tank is filled to capacity with 500 gallons of pure water. Brine (a mixture of salt and water) containing 2 pounds of salt per gallon is pumped into the tank at a rate of 5 gallons per minute. The well-mixed solution is pumped out at a rate of 10 gallons per minute. Find the number A(t) of pounds of salt in the tank at time t.

    OK. I know that the rate at which the amount of salt is changing inside the tank with respect to time - \frac{dA}{dt} - is simply the net rate of the amount being pumped in and the amount being pumped out, that is

    \frac{dA}{dt}=R_{in}-R_{out}.

    Now, R_{in}=(2\frac{lbs}{gal})\cdot(5\frac{gal}{min})=1  0\frac{lbs}{min}

    and I think that R_{out}=\frac{A(t)lbs}{(500-5t)gal}\cdot10\frac{gal}{min}

    Assuming my R_{out} is correct, this would imply the DE

    \frac{dA}{dt}+\frac{5}{100-t}A=10 which is linear with the integrating factor

    e^{\int\frac{5}{100-t}dt}=e^{-5\ln|100-t|}=e^{\ln(100-t)^{-5}}=(100-t)^{-5}.

    So, I have

    \int\frac{d}{dt}\left[(100-t)^{-5}A\right]dt=\int10(100-t)^{-5}dt

    Am I on the right track?
    You should have

    \displaystyle A'+\frac{2A}{100-t}=10
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Oh, duh. But everything else is OK?
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  4. #4
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    Quote Originally Posted by VonNemo19 View Post
    Oh, duh. But everything else is OK?
    Yeah once you change your P(t) numbers.
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