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Mathematics 20 Online
OpenStudy (anonymous):

How to write a half-life function? I am working on a project for math and one of the things I have to do is write a half-life function with the data I came up with. I looked all through my book and can't even find a mention of the term.

OpenStudy (anonymous):

Here is the data I have to write the function based off of

OpenStudy (whpalmer4):

The half-life is the period of time it takes to fall to half of its value. But looking at your table, I don't quite see how that applies here...

OpenStudy (whpalmer4):

Are your various trials supposed to represent measurements of the remaining quantity in identical samples after an identical amount of time?

OpenStudy (anonymous):

Oh i see what I did wrong!!! Let me attach another table...

OpenStudy (anonymous):

OpenStudy (whpalmer4):

Ah, that makes more sense! :-)

OpenStudy (anonymous):

Yes! I re-read the instructions and saw I was supposed to eliminate objects...

OpenStudy (anonymous):

@whpalmer4

OpenStudy (whpalmer4):

Sorry, openstudy is being a bit flaky for me :-( So your data looks like this when plotted (I'm plotting the number of coins still face up vs. the trial number, assuming the trial number is equivalent to elapsed time):

OpenStudy (whpalmer4):

To find the half-life, you need to fit the right sort of curve through those points, and determine how much time has to elapse to cut the initial quantity in half

OpenStudy (anonymous):

@whpalmer4

OpenStudy (anonymous):

simplest way is using the function. it is in the form: \[f(x) = f(0)e^{-ax}\] f(0) is the initial amount (when x = 0). so the half-life is when f(x) = f(0)/2 re-written: \[\frac{ f(0) }{ 2 }= f(0)e^{-0.61x}\] re-written: \[\frac{ 1 }{ 2 } = e^{-0.61x}\] \[\ln(1/2) = -0.61x\] x is the half-life amount

OpenStudy (anonymous):

@Euler271 Do i simplify further or leave it like that?

OpenStudy (whpalmer4):

Actually, I think your time series should be 20, 13, 8, 5, 3, 1 as 20 is the initial amount — trial 0 is the amount remaining after 1 sampling period has elapsed

OpenStudy (anonymous):

you would simplify to find x

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