Math Problem Statement

A patient's blood alcohol concentration (BAC) at t = 0 hrs is 1.2 g/liter. Find a recursion relation for the blood alcohol concentration measured 1 hour later, 2 hours later, and so on, and solve it to deduce an explicit formula for BAC as a function of time. When does the blood alcohol concentration reach 0 g/liter?

Solution

The problem you're dealing with revolves around the elimination of alcohol from the body, following zeroth-order elimination kinetics. This means that alcohol is eliminated at a constant rate over time, regardless of its concentration in the blood.

From the information in the image:

  • The rate of alcohol elimination is given as 0.186g/liter/h0.186 \, \text{g/liter/h}.
  • A patient's blood alcohol concentration (BAC) at t=0hourst = 0 \, \text{hours} is 1.2g/liter1.2 \, \text{g/liter}.
  • The BAC is measured at regular intervals, and we are tasked with finding a formula for the BAC as a function of time and determining when the BAC reaches 0 g/liter.

Part (a):

To model the BAC over time, we can use the zeroth-order elimination equation: BAC(t)=BAC0kt\text{BAC}(t) = \text{BAC}_0 - k \cdot t where:

  • BAC0=1.2g/liter\text{BAC}_0 = 1.2 \, \text{g/liter} is the initial BAC,
  • k=0.186g/liter/hk = 0.186 \, \text{g/liter/h} is the elimination rate,
  • tt is the time in hours.

The equation becomes: BAC(t)=1.20.186t\text{BAC}(t) = 1.2 - 0.186 \cdot t

To find when the BAC reaches 0 g/liter, set BAC(t)=0\text{BAC}(t) = 0: 0=1.20.186t0 = 1.2 - 0.186 \cdot t Solving for tt: t=1.20.1866.45hourst = \frac{1.2}{0.186} \approx 6.45 \, \text{hours} So, the BAC will reach 0 after approximately 6.45 hours.

Part (b):

In this part, a driver was measured with a BAC of 0.61g/liter0.61 \, \text{g/liter} two hours after being pulled over. Using the equation from part (a), we need to estimate the BAC when the driver was first pulled over.

Let t=2hourst = 2 \, \text{hours} and BAC(2)=0.61g/liter\text{BAC}(2) = 0.61 \, \text{g/liter}. Substituting this into the formula: BAC(2)=BAC00.1862\text{BAC}(2) = \text{BAC}_0 - 0.186 \cdot 2 0.61=BAC00.3720.61 = \text{BAC}_0 - 0.372 Solving for BAC0\text{BAC}_0: BAC0=0.61+0.372=0.982g/liter\text{BAC}_0 = 0.61 + 0.372 = 0.982 \, \text{g/liter} Thus, the driver's BAC was approximately 0.982g/liter0.982 \, \text{g/liter} when they were first pulled over.


Let me know if you'd like further details or clarifications.

Here are 5 follow-up questions you might find useful:

  1. How does the concept of zeroth-order kinetics apply to other substances besides alcohol?
  2. What factors could alter the elimination rate of alcohol in the body?
  3. How would the equation change if the elimination followed first-order kinetics instead?
  4. Could external factors like body weight or food intake affect the initial BAC?
  5. How can the BAC equation be used in real-life situations, like estimating the time needed to be sober?

Tip: Always double-check the units used in any rate equations to ensure consistent and accurate results!

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Math Problem Analysis

Mathematical Concepts

Zeroth-order kinetics
Linear equations
Recursion

Formulas

BAC(t) = BAC_0 - k * t
t = BAC_0 / k

Theorems

Zeroth-order reaction kinetics

Suitable Grade Level

College/University Level (Introductory Chemistry or Biochemistry)