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Advanced Algebra II/Trig finals Friday

  1. Dec 17, 2003 #1
    I have Advanced Algebra II/Trig finals Friday and I'm cramming(sue me :P). How does half-life work?
  2. jcsd
  3. Dec 17, 2003 #2
    half-life implies a substance decaying at a rate proportional to (ie a multiple of) the amount present.

    double the amount present and the rate at which it decays doubles.

    half the amount present and the rate at which it decays halves.

    via calculus you don't have to know, this leads to the equation for exponential decay:
    1. [tex]A\left( t\right) =A_{0}e^{kt}[/tex].

    A is the amount present and A0 is the initial amount. e is a fixed constant, like pi, which is roughly 2.72. k is a constant that relates how quickly the substance is decaying and it will be negative. t is time. btw, the rate at which the substance is decaying is given by [tex]kA\left( t\right) [/tex], multiple of how much is present. if k=0 then the amount present would be constant over time and if k>0 then the amount is growing exponentially. for example, you can use this equation to model moore's law which says that the rate of cpu speed growth doubles every 18 months to predict where cpu speed might be in 10 years.

    there are two other formulations of this which are equivalent (ie lead to the same final answers) that (a) your book might use and (b) that are more convienient for half-life specifically:
    2. [tex]A\left( t\right) =A_{0}2^{-t/h}[/tex]
    3. [tex]A\left( t\right) =A_{0}a^{t}[/tex].

    (the harvard reformists tend to use #3 while the formula worshippers would love #2.)

    your approach changes slightly depending on which of the three decay formulas you use. if you're using #2, the half life is h. that's all there is to it in that case. for example, someone says the half life of a substance is 2700 years and they say that 37% is left and ask how long it has been decaying, you can do this:
    I. h=2700
    II. A0 doesn't matter (and isn't given)
    III. [tex]A\left( t\right) =0.37A_{0}[/tex] is what is given. that translates to, using formula 2 above, [tex]A_{0}2^{-t/2700}=0.37A_{0}[/tex] which you now have to solve for t. you would get 3872.89 years. that could mean that the robe supposedly belonging to jesus comes out as too old to have been his (unless it was a really old robe!).

    you have to be able to take any of the three formulas for half life, all of which have 4 variables, take three facts, and solve for the 4th variable not given. 3 equations and 4 variables to solve for gives 12 different algebraic techniques that might come up though there is similarity in several reducing this a bit from 12. if you stick to one formula, there are 4 things you might be asked to solve for given three or fewer facts.

    if you're using formula 3, you'd have to find a first. if 0<a<1, the formula describes growth. if a=1, you get a constant function describing a substance whose amount isn't changing and if a>1, the amount is growing exponentially. the further a is from 1, the more rapid the change.
    I. half life of 2700 years means that [tex]A\left( 2700\right) =\frac{1}{2}A_{0}[/tex] which translates to [tex]A_{0}a^{2700}=\frac{1}{2}A_{0}[/tex]. after cancelling the A0, you get [tex]a=\left( \frac{1}{2}\right) ^{1/2700}[/tex]. btw, since [tex]\frac{1}{2}=2^{-1}[/tex], that can be written as [tex]\left( \frac{1}{2}\right) ^{1/2700}=2^{-1/2700}[/tex]; it's not at all a coincidence that that looks like formula 2. i'll use a six-digit approximation for a: [tex]a\approx 0.999743[/tex]. now we know that [tex]A\left( t\right) \approx A_{0}\left( 0.999743\right) ^{t}[/tex].

    see how close a is to 1? that's because 2700 years is a long time. if it only took 10^-36 seconds to decay half-way, a would be ultra close to 0.

    II. we're trying to find out how old something is. we know that 37% is left and we're solving for t: [tex]A\left( t\right) =0.37A_{0}[/tex]. this translates to [tex]A_{0}\left( 0.999743\right) ^{t}=0.37A_{0}[/tex]. after cancelling the [tex]A_{0}[/tex], we can use logs to solve for t:
    [tex]\log \left( 0.999743\right) ^{t}=\log 0.37[/tex].
    (you can use any log base you like, basically.) by the properties of logs they crammed into your skull, the t comes down and then we can divide the constant next it to solve for it:
    [tex]t\log 0.999743=\log 0.37[/tex]
    [tex]t=\frac{\log 0.37}{\log 0.999743}[/tex]. when i approximate the right hand side, i get, lo and behold!,
    III. the item is 3868.2 years old.

    (i hope the God of Math won't send me to hell for not using approximately equals when i should have!)

    any questions?

    i got one. why did i get 3872.89 the first way and 3868.2 the second way? does this invalidate the whole mathematical ediface or is something else at work (or play) here?

    first of all, these are different by 4.69 which is 0.12% of the average of the two answers; so it ain't a big difference. if you work out the math, you'll note that in both cases, the exact value of t can be algebraically reduced to [tex]t=\allowbreak \frac{2700\log \left( 0.37\right) }{\log 2}[/tex] which would mean they're actually exactly the same. the error is due to rounding off in certain places.

    these examples are roughly 2/12 of all the types of problems you might encounter with half-life. the hope is that once you learn x/12 of them, you'll figure out how to do all types.
    Last edited: Dec 18, 2003
  4. Dec 18, 2003 #3


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    No, it's the holiday season and I'm in a lenient mood.
  5. Dec 18, 2003 #4
    the math God will give you a credit you can cash in when you die if you can spot a flaw in what i wrote. i could just edit it but maybe the person who asked the question would like to find it. it's not a major one. something involving a statement about a. there might be other flaws, too.

    when i read math, sometimes i add a line at the bottom:
    there is a flaw somehere in this document.

    and if i can't find it in what preceeds the final statement, then it is the final statement that is flawed.
  6. Dec 18, 2003 #5
    My hat is off phoenixthoth, to your oustanding post! This one should be saved? I liked the typesetting too.
    Last edited: Dec 18, 2003
  7. Dec 18, 2003 #6
    i taught about 5 precalculus courses back before i became a full-time surfer/bum so that has something to do with it so i've thought about teaching this a few times before. in fact, half-life is one of my favorite parts of teaching precalculus which was one of my favorite classes to teach. i actually decided that within the next year i'm going to try to go back to work as a teacher for a year before going for a phd. still don't know if i got what it takes to get a phd though i suspect that i do. only one way to find out. if i don't go, i won't know.
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