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Math Help - Related Rates

  1. #1
    Junior Member
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    Related Rates

    Can someone help me with the following question?

    A tank contains 100 litres of brine whose concentration is 3grams/litre. Three litres of brine whose concentration is 2 grams/litre flow into the tank each minute and at the same time 3 litres of miture flow out each minute.

    (a). Show that the quantity of salt, Q grams, in the tank at any time t is given by:

    Q = 200 + 100e^-0.03t

    The problem is I get my constant as a negative log.

    Thanks.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    dQ/dt = Salt in -Salt out

    dQ/dt = 6 - 3Q/100 Q(0) = 300

    Note the equillibrium solution is Q= 200 Since we start with 300 Q will always be greater than 200

    dQ/dt = 6 - .03 Q

    -1/.03ln|6-.03Q| = t + C

    ln|6 - .03Q| = -.03t + C
    6-.03Q < 0 so ln|6 - .03Q| = ln(.03Q -6)

    ln(.03Q -6) = -.03*t + C

    at t = 0 ln(3) = C

    .03*Q - 6 = 3*e^(-.03t)

    Q = 200 +100e^(-.03*t)
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