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Question

If the rate of alcohol in your bloodstream decreases continuously by 50% every hour, and you find that by 10 PM you have a blood alcohol level of .15 (called your BAC):  

  a.  Find a continuous exponential model of your BAC for the time after 10 PM.

 

 

  b.  What would be your BAC at midnight?

 

 

  c.  If the legal BAC is .02 for drivers under 21, when will you sober enough to legally drive home (round to the nearest minute)?

 

 

 

d.  For adults over 21, the legal limit is .08.  If you are over 21, what exact time would you be able to legally drive home using this same scenario?  

Answer 1

a) Let B(t) represent the BAC at time t, in hours. We are given that

[ B(t+1) - B(t) ] / B(t) = -0.50

and that B(0) = 0.15, where t = 0 means 10:00:00 pm

Solving for B(t+1), we get

B(t+1) = B(t) / 2

When B(t) is a geometric progression, then each succeeding term is a constant times the previous term, which is what we have here. In this case, the constant term is 0.50 or 1/2.

Because B(t) is a geometric series, its general equation is of the form:

B(t) = B₀*r^t

where B₀ represents B(0), in this case 0.15 and r = 0.5

Thus, B(t) = 0.15 * (0.5)^t

So far, B(t) is a discrete function, meaning that it is valid for integral, non-negative values of t. To make is apply to all t > 0, we interpret the word "continuously" to mean that B(t+1+x) = 0.5*B(t+x) for all x>0. If that interpretation is correct, then we have our desired exponential model:

B(t) = 0.15 * (0.5)^t for all t ≥ 0.

If the word "exponential" is meant in terms of e, the base of natural logarithms, then we need to change 0.5 to some power k of e:

e^k = 0.5

Taking the natural log of both sides yields:

k = ln(0.5) = -0.6932

Therefor,

0.5 = e^-0.6932

and

0.5^t = e^-0.6932t

So our final answer is

B(t) = B₀*e^-0.6932t

b) At midnight, t = 2

B(2) = 0.15*(0.5)² = 0.15*(1/4) = .0375

c) B(t) = .02 for some t

We solve for t in terms of B, or simply t(B):

B = 0.15 * (0.5)^t(B)

B / 0.15 = (0.5)^t(B)

ln( B / 0.15 ) = ln (0.5)^t(B) = t(B) * ln(0.5)

t(B) = ln (B / 0.15 ) / ln(0.5)

In this case, B=0.02, so

t(0.02) = ln( 0.02 / 0.15 ) / ln( 0.5) = 2.9069 hrs = 2h 54.414m = 12:54 am.

d) t(.08) = ln( .08 / 0.15 ) / ln(0.5) = 0.9068906 hrs = 54.413 min = 54m 24.8 secs = 10:54:25 pm.