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Math Help - exp growth

  1. #1
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    exp growth

    need a help setting up this story problem.

    The mass of a tumor grows at a rate proportional to its size. The first measurement of its size was 4.0 grams. Four months later its mass was 6.76 grams. How large was the tumor six months before the first measurement?
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  2. #2
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    Hello, viet!

    The mass of a tumor grows at a rate proportional to its size.
    The first measurement of its size was 4.0 grams.
    Four months later its mass was 6.76 grams.
    How large was the tumor six months before the first measurement?

    The rate of growth of the mass of a tumor is proportional to its mass.
    This translates to: . \frac{dM}{dt} = k\!\cdot\!M, . where t is measured in months.

    We have: . \frac{dM}{M} = k\,dt

    Integrate: . \ln|M| \:=\:kt + C

    Then: . M \:=\:e^{kt + c} \:=\:e^{kt}\!\cdot\!e^c\quad\Rightarrow\quad M(t)\:=\:Ce^{kt}


    We are told that: . M(0) = 4
    . . We have: . Ce^0 = 4\quad\Rightarrow\quad C = 4
    The function (so far) is: . M(t)\:=\:4e^{kt}

    We are told that: . M(4) = 6.75
    . . We have: . 4e^{4k} = 6.76\quad\Rightarrow\quad e^{4k} = 1.69\quad\Rightarrow\quad 4k = \ln(1.69)
    . . . . . . k \:= \:\frac{1}{4}\ln(1.69) \:=\:0.1311821332 \:\approx\:0.13 1

    The function is: . M(t) \:=\:4e^{0.131t}


    Six months before (t = \text{-}6), we have: . M(\text{-}6) \:=\:4e^{(0.131)(\text{-}6)} \:\approx\:1.82 grams.

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