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Math Help - Half-life.

  1. #1
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    Half-life.

    A radioactive substance has a half-life of 20 days.

    a) How much time is required so that only \frac{1}{32} of the original amount remains?
    b) Find the rate of decay at this time.

    My attempts:
    a) Since after 20 days the substance decays by half of its mass, after 40 days it decays to {1}{4}, and after 60 days it decays to \frac{1}{8}. After 80 days it decays to \frac {1}{16} of its original mass and after 100 days it decays to {1}{32} of its original mass. Therefore, it will take 100 days for the substance to decay to [tex]\frac{1}{32}/MATH] of its original mass.
    b) f(t) = \frac{1}{32}(\frac{1}{2})^{\frac{t}{20}
    f'(t) =\frac{1}{32}(\frac{1}{2})^{\frac{t}{20}}  (ln\frac{1}{2})  (\frac{1}{20})
    f'(100) =\frac{1}{32}(\frac{1}{2})^{\frac{100}{20}}  (ln\frac{1}{2})  (\frac{1}{20})
    f'(100) = -3.384  10^{-5} is the rate of decay when t = 100 days.

    I just wanted to know if my answers are correct? I checked with derivative calculators and they have given me very different answers for b. Thanks in advance.
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  2. #2
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    Quote Originally Posted by Pupil View Post
    A radioactive substance has a half-life of 20 days.

    a) How much time is required so that only \frac{1}{32} of the original amount remains?
    b) Find the rate of decay at this time.

    My attempts:
    a) Since after 20 days the substance decays by half of its mass, after 40 days it decays to {1}{4}, and after 60 days it decays to \frac{1}{8}. After 80 days it decays to \frac {1}{16} of its original mass and after 100 days it decays to {1}{32} of its original mass. Therefore, it will take 100 days for the substance to decay to [tex]\frac{1}{32}/MATH] of its original mass.
    b) f(t) = \frac{1}{32}(\frac{1}{2})^{\frac{t}{20}
    f'(t) =\frac{1}{32}(\frac{1}{2})^{\frac{t}{20}}  (ln\frac{1}{2})  (\frac{1}{20})
    f'(100) =\frac{1}{32}(\frac{1}{2})^{\frac{100}{20}}  (ln\frac{1}{2})  (\frac{1}{20})
    f'(100) = -3.384  10^{-5} is the rate of decay when t = 100 days.

    I just wanted to know if my answers are correct? I checked with derivative calculators and they have given me very different answers for b. Thanks in advance.
    The first part is correct However you could have solved it this way

    \displaystyle \left( \frac{1}{2}\right)^\frac{x}{20}=\frac{1}{32}=\left  ( \frac{1}{2}\right)^5 \iff \frac{x}{20}=5 \implies x=100

    For the 2nd part all we know is

    \displaystyle f(t)=A_0\left( \frac{1}{2}\right)^{\frac{x}{20}}=A_0 e^{\frac{x}{20}\cdot \ln(\frac{1}{2})}

    Now taking the derivative gives

    \displaystyle f'(t)=\left( {\frac{1}{20}\cdot \ln(\frac{1}{2})}\right)A_0 e^{\frac{x}{20}\cdot \ln(\frac{1}{2})}=A_0\left( {\frac{1}{20}\cdot \ln(\frac{1}{2})}\right)\left( \frac{1}{2}\right)^{\frac{x}{20}}
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  3. #3
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    Ah, I forgot to take the log of both sides and solve it much easier.

    Okay, so I derived the equation correctly? And the question asks me to find the rate of decay when t = 100 days. Knowing the derivative and the quantity at the time, wouldn't I just need to plug in 100 to find the rate of decay?
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  4. #4
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    Yes, f'(100) should tell you the rate of decay on the 100th day.
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  5. #5
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    Quote Originally Posted by NOX Andrew View Post
    Yes, f'(100) should tell you the rate of decay on the 100th day.
    So, is the rate of decay -3.384  10^{-5} when t = 100?
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  6. #6
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    What is the original amount? It should be the original amount times -log(2)/640.
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  7. #7
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    Quote Originally Posted by NOX Andrew View Post
    What is the original amount? It should be the original amount times -log(2)/640.
    It was not given. I only have \frac{1}{32} of the original substance to work with.
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