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Math Help - Word problem help

  1. #1
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    Word problem help

    One hundred grams of a certain chemical is removed from its airtight container and
    immediately placed in a 20 gram dish on a sensitive scale. The scale reads 120 grams. The
    chemical absorbs water from the air at a rate of 15e^-t grams per minute, where t is the
    time in minutes that the chemical has been exposed to air.

    a) What is the reading on the scale after t minutes?
    b) What is the reading on the scale after a very long time has passed?


    I am not asking you to do this for me, I just have no idea how to do it an help would be appreciated
    Thanks in advance
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Word problem help

    Let R(t) be the reading on the scale in grams after t minutes. From the information provided, we may state the IVP:

    \frac{dR}{dt}=15e^{-t} where R(0)=120

    Can you proceed?
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  3. #3
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    Re: Word problem help

    Thank you for the help
    I got R(t)= 15e^-t + 135 for part a and 135 for part b
    Can someone please confirm that this is the answer or help me figure out what I did wrong?
    Thanks
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Word problem help

    You can confirm whether or not your solution to the IVP is correct by differentiating it to see if you get back to the differential equation, or at least check to see if the initial condition is met.
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  5. #5
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    Re: Word problem help

    R(t) = 135 - 15e^{-t}

    \lim_{t \to \infty} R(t) = 135
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Word problem help

    Yes, using the boundaries and introducing dummy variables, we find:

    \int_{120}^{R(t)}\,du=15\int_0^te^{-v}\,dv

    R(t)-120=-15(e^{-t}-1)

    R(t)=135-15e^{-t}=15(9-e^{-t})
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