/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Blood alcohol content (BAC) is t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Blood alcohol content (BAC) is the amount of alcohol present in your blood as you drink. It is calculated by determining how many grams of alcohol are present in 100 milliliters (1 deciliter) of blood. So if a person has 0.08 grams of alcohol per 100 milliliters of blood, the \(\mathrm{BAC}\) is \(0.08 \mathrm{~g} / \mathrm{dl} .\) After a person has stopped drinking, the BAC declines over time as his or her liver metabolizes the alcohol. Metabolism proceeds at a steady rate and is impossible to speed up. For instance, an average ( \(150-1 \mathrm{~b}\) ) male metabolizes about 8 to 12 grams of alcohol an hour (the amount in one bottle of beer^{3} ). The behavioral effects of alcohol are closely related to the blood alcohol content. For example, if an average 150 -lb male drank two bottles of beer within an hour, he would have a BAC level of 0.05 and could suffer from euphoria, inhibition, loss of motor coordination, and overfriendliness. The same male after drinking four bottles of beer in an hour would be legally drunk with a BAC of \(0.10 .\) He would likely suffer from impaired motor function and decision making, drowsiness, and slurred speech. After drinking twelve beers in one hour, he will have attained the dosage for stupor (0.30) and possibly death (0.40) . The following table gives the BAC of an initially legally drunk person over time (assuming he doesn't drink any additional alcohol). $$ \begin{array}{lccccc} \hline \text { Time (hours) } & 0 & 1 & 2 & 3 & 4 \\ \text { BAC (g/dI) } & 0.100 & 0.067 & 0.045 & 0.030 & 0.020 \\ \hline \end{array} $$ a. Graph the data from the table (be sure to carefully label the axes). b. Justify the use of an exponential function to model the data. Then construct the function where \(B(t)\) is the \(\mathrm{BAC}\) for time \(t\) in hours. c. By what percentage does the BAC decrease every hour? d. What would be a reasonable domain for your function? What would be a reasonable range? e. Assuming the person drinks no more alcohol, when does the BAC reach \(0.005 \mathrm{~g} / \mathrm{dl}\) ?

Short Answer

Expert verified
a) Graph time vs. BAC. b) Use exponential decay function: \( B(t) = 0.100 e^{-0.4055t} \). c) BAC decreases by about 33.3% per hour. d) Domain: 0 to 10 hours. Range: 0.100 to 0 g/dl. e) BAC reaches 0.005 g/dl at 7.39 hours.

Step by step solution

01

Graphing the Data

Plot the given BAC data against time on a graph. Time (hours) will be on the x-axis and BAC (g/dl) will be on the y-axis. Plot points: (0, 0.100), (1, 0.067), (2, 0.045), (3, 0.030), (4, 0.020). Draw a smooth curve through these points.
02

Justifying the Use of an Exponential Function

Exponential functions are suitable for modeling situations where the rate of change is proportional to the current value. Observing the BAC data, each successive value is a percentage of the previous value, indicating exponential decay.
03

Constructing the Exponential Function

Use the general form of an exponential decay function: \[B(t) = B_0 e^{-kt}\] where \( B_0 \) is the initial BAC, k is the decay constant, and t is time in hours. Given \( B(0) = 0.100 \), use the data points to find k:\[ B(1) = 0.067 \rightarrow 0.067 = 0.100 e^{-k} \rightarrow k \approx 0.4055 \]. The model function is: \[ B(t) = 0.100 e^{-0.4055t} \].
04

Calculating the Percentage Decrease in BAC per Hour

Solve for the percentage decrease using the decay factor. \[ B(t) = 0.100 e^{-0.4055t} \ implies \ B(1) = 0.100 \times e^{-0.4055} \approx 0.100 \times 0.667 \approx 0.067 \]. The BAC decreases by approximately \(100\text{ - }66.7\text{ = }33.3\) percent per hour.
05

Determining a Reasonable Domain and Range

The domain (t) should start from 0 (time of initial measurement) and extend to where BAC reaches negligible levels (practically around 10 hours). The range should start from 0.100 g/dl to levels lower than legal impairment (around 0.020 g/dl and decreases to almost 0 g/dl).
06

Finding When BAC Reaches 0.005 g/dl

Set \( B(t) = 0.005 \) and solve for t: \[ 0.005 = 0.100 e^{-0.4055t} \rightarrow e^{-0.4055t} = \frac{0.005}{0.100}\rightarrow e^{-0.4055t} = 0.05 \rightarrow -0.4055t = \text{ln}(0.05) \rightarrow t \approx 7.39 \text{ hours} \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Decay
Understanding how BAC (Blood Alcohol Content) decreases over time can be modeled using the concept of exponential decay. Exponential decay describes a process where the rate of change of a quantity is proportional to its current value. This means that as time progresses, the BAC decreases by a consistent percentage. In our example, the BAC declines over time after a person stops drinking because their liver metabolizes the alcohol at a constant rate. Here, this constant rate defines the exponential decay. To put it simply, if you start with a certain amount of alcohol in your bloodstream, your body will slowly and steadily work to remove it, reflecting a percentage decrease over regular intervals.
Graphing Data
Graphing the BAC data over time helps visualize how the BAC decreases. When we plot the given data points on a graph with time (in hours) on the x-axis and BAC (in g/dl) on the y-axis, we see the points: (0, 0.100), (1, 0.067), (2, 0.045), (3, 0.030), and (4, 0.020). Connecting these points with a smooth curve reveals the characteristic shape of exponential decay. This visual representation confirms that the BAC is not decreasing at a constant amount per hour but rather at a constant percentage, supporting the use of an exponential function to model this behavior.
Mathematical Modeling
To create a mathematical model of the decrease in BAC, we use an exponential decay function. The formula for an exponential decay model is given by \[ B(t) = B_0 e^{-kt} \] where \( B_0 \) is the initial BAC, \( k \) is the decay constant, and \( t \) is time in hours. In our example, the initial BAC is 0.100 g/dl, and we use the given data points to find the decay constant \( k \). With this model, we can predict the BAC at any given time. For instance, using \( k \approx 0.4055 \,\), the function becomes \[ B(t) = 0.100 e^{-0.4055t} \]. This function allows us to calculate the BAC at different times and understand how quickly alcohol is metabolized.
Alcohol Metabolism
Alcohol metabolism is the process by which the liver breaks down alcohol in the bloodstream. The rate at which this occurs is generally constant, meaning the liver metabolizes a set amount of alcohol per hour. For an average 150-pound male, this rate is about 8 to 12 grams per hour, equivalent to one bottle of beer. This steady rate of metabolism is why the BAC declines consistently over time after a person stops drinking. Behavioral effects, legal implications, and health risks associated with different BAC levels are linked to how quickly the body can process the alcohol. For example, as BAC decreases from 0.10 g/dl to lower levels, the associated impairments reduce accordingly.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

It takes 3 months for a malignant lung tumor to double in size. At the time a lung tumor was detected in a patient, its mass was 10 grams. a. If untreated, determine the size in grams of the tumor at each of the listed times in the table. Find a formula to express the tumor mass \(M\) (in grams) at any time \(t\) (in months). $$ \begin{array}{cc} \hline t, \text { Time (months) } & M, \text { Mass }(\mathrm{g}) \\ \hline 0 & 10 \\ 3 & \\ 6 & \\ 9 & \\ 12 & \\ \hline \end{array} $$ b. Lung cancer is fatal when a tumor reaches a mass of 2000 grams. If a patient diagnosed with lung cancer went untreated, estimate how long he or she would survive after the diagnosis. c. By what percentage of its original size has the 10 -gram tumor grown when it reaches 2000 grams?

Each of the following tables contains values representing either linear or exponential functions. Find the equation for each function. $$ \begin{aligned} &\text { a. }\\\ &\begin{array}{cccccc} \hline x & -2 & -1 & 0 & 1 & 2 \\ f(x) & 1.12 & 2.8 & 7 & 17.5 & 43.75 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { b. }\\\ &\begin{array}{cccccc} \hline x & -2 & -1 & 0 & 1 & 2 \\ g(x) & 0.1 & 0.3 & 0.5 & 0.7 & 0.9 \\ \hline \end{array} \end{aligned} $$

Construct both a linear and an exponential function that go through the points (0,6) and (1,9) .

Plutonium- 238 is used in bombs and power plants but is dangerously radioactive. It decays very slowly into nonradioactive materials. If you started with 100 grams today, a year from now you would have 99.2 grams. a. Construct an exponential function to describe the decay of plutonium- 238 over time. b. How much of the original 100 grams of plutonium- 238 would be left after 50 years? After 500 years?

(Requires technology to find a best-fit function.) The accompanying table shows the U.S. international trade in goods and services. $$ \begin{aligned} &\text { U.S }\\\ &\text { International Trade (Billions of Dollars) }\\\ &\begin{array}{crr} \hline & \text { Total } & \text { Total } \\ \text { Year } & \text { Exports } & \text { Imports } \\ \hline 1960 & 25.9 & 22.4 \\ 1965 & 35.3 & 30.6 \\ 1970 & 56.6 & 54.4 \\ 1975 & 132.6 & 120.2 \\ 1980 & 271.8 & 291.2 \\ 1985 & 288.8 & 410.9 \\ 1990 & 537.2 & 618.4 \\ 1995 & 793.5 & 891.0 \\ 2000 & 1070.6 & 1448.2 \\ 2005 & 1275.2 & 1992.0 \\ \hline \end{array} \end{aligned} $$ a. U.S. imports and exports both expanded rapidly between 1960 and \(2005 .\) Use technology to plot the total U.S. exports and total U.S. imports over time on the same graph. b. Now change the vertical axis to a logarithmic scale and generate a semi-log plot of the same data as in part (a). What is the shape of the data now, and what does this suggest would be an appropriate function type to model U.S. exports and imports? c. Construct appropriate function models for total U.S. imports and for total exports. d. The difference between the values of exports and imports is called the trade balance. If the balance is negative, it is called a trade deficit. The balance of trade has been an object of much concern lately. Calculate the trade balance for each year and plot it over time. Describe the overall pattern. e. We have a trade deficit that has been increasing rapidly in recent years. But for quantities that are growing exponentially, the "relative difference" is much more meaningful than the simple difference. In this case the relative difference is \(\frac{\text { exports }-\text { imports }}{\text { exports }}\) This gives the trade balance as a fraction (or if you multiply by 100 , as a percentage) of exports. Calculate the relative difference for each year in the above table and graph it as a function of time. Does this present a more or less worrisome picture? That is, in particular over the last decade, has the relative difference remained stable or is it also rapidly increasing in magnitude?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.