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After several drinks, a person has a blood alcohol level of 200 mg/dL (milligrams per deciliter). If the amount of alcohol in the blood decays exponentially, with one fourth being removed every hour, find the person's blood alcohol level after four hours.

2007-01-18 20:53:24 · 3 answers · asked by Anonymous in Science & Mathematics Mathematics

3 answers

The typical property of exponantial decay is. that the time for the reduction to certain fraction, which is called the fractional life, is constant.
In this problem there is a reduction to one quarter after on hour. After another hour there remains one quarter of one quarter of the initial amount, that is one sixteenth part. After another hour there is one quarter of one sixteenth... etc.
So after four hours there is
(1/4)^4=1/256 of the initial 200mg/dl alcohol remaining in the blood.

The general formula for such a problem is
c(t) =c0 * f^(t/tf)
with
t time from start of decay
c(t) concentration (or amount) at time t
c0 initial concentration (or amount)
tf fractional life
f fraction

Derivation of the formula:
The standard formula of an exponential dacay function is
c(t) = c0*exp{-k*t) where k is the constant of decay
<=> ln{c(t)/c0}=-k*t
particularly for a given fractional life
ln{f}=-k*tf
=> k =-ln{f}/tf
substituting k in the first equation gives
c(t)=c0*exp{ln{f}*(t/tf)}
<=>
c(t)=c0*f^(t/tf)

2007-01-18 22:12:16 · answer #1 · answered by schmiso 7 · 0 0

The ratio that remains after one hour is 1 - 1/4 = 3/4.

The ratio remaining after 4 hours is (3/4)^4 = 81/256.

The amount after 4 hours is:

200(81/256) = 63.28125 mg/dL

2007-01-18 21:02:44 · answer #2 · answered by Northstar 7 · 0 0

a) set t = 0, the time of first counting and are available to a determination for p(0) b) set t = 60, the form of minutes elapsed one hour after counting, and are available to a determination for p(60) c) set p(t) = a million and are available to a determination for t that's the form of minutes after counting that the inhabitants reaches a million.

2016-12-16 08:09:07 · answer #3 · answered by Anonymous · 0 0

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