/*! 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 30 Hydrocodone bitartrate is used a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Hydrocodone bitartrate is used as a cough suppressant. After the drug is fully absorbed, the quantity of drug in the body decreases at a rate proportional to the amount left in the body. The half-life of hydrocodone bitartrate in the body is 3.8 hours, and the usual oral dose is 10 mg. (a) Write a differential equation for the quantity, \(Q,\) of hydrocodone bitartrate in the body at time \(t,\) in hours since the drug was fully absorbed. (b) Find the equilibrium solution of the differential equation. Based on the context, do you expect the equilibrium to be stable or unstable? (c) Solve the differential equation given in part (a). (d) Use the half-life to find the constant of proportionality, \(k\) (e) How much of the 10 mg dose is still in the body after 12 hours?

Short Answer

Expert verified
After 12 hours, about 0.83 mg of the drug remains in the body.

Step by step solution

01

Understanding the Proportional Decrease

The rate of decrease of the quantity of hydrocodone, \(Q\), is proportional to the amount left in the body, \(Q\). Mathematically, this is expressed by the differential equation \(\frac{dQ}{dt} = -kQ\), where \(k\) is the constant of proportionality.
02

Finding Equilibrium Solution

The equilibrium solution to a differential equation is when \(\frac{dQ}{dt} = 0\). From \(-kQ = 0\), solving for \(Q\) gives \(Q = 0\). In the context of the drug, this means that eventually all the drug is eliminated from the body, which is stable.
03

Solving the Differential Equation

The differential equation \(\frac{dQ}{dt} = -kQ\) is separable. Separating the variables and integrating, we get \(\int \frac{1}{Q} dQ = -k \int dt\), which results in \(\ln |Q| = -kt + C\). Exponentiating both sides, we find \(Q(t) = Ce^{-kt}\), where \(C\) is the initial amount of drug.
04

Finding the Constant of Proportionality

Using the half-life, we know \(Q(3.8) = \frac{1}{2}C\). Substituting into \(Ce^{-k \, 3.8} = \frac{1}{2}C\), canceling \(C\), and solving, we get \(e^{-3.8k} = \frac{1}{2}\), so \(-3.8k = \ln\left( \frac{1}{2} \right)\). Thus, \(k = \frac{\ln(2)}{3.8}\approx 0.1824\).
05

Calculating Drug Quantity After 12 Hours

We use \(Q(t) = 10e^{-0.1824t}\) with \(t = 12\) hours. So, \(Q(12) = 10e^{-0.1824 \cdot 12}\approx 0.8309 \, \text{mg}\). After 12 hours, about 0.83 mg of the drug remains in the body.

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
Exponential decay is a commonly encountered concept in differential equations and is particularly relevant in processes like the reduction of drug concentrations in the body over time. In this context, we deal with the drug hydrocodone bitartrate which decreases at a rate proportional to its current amount. This characteristic implies that at any given time, the rate of decrease is a constant fraction of the drug left. This phenomenon can be mathematically described using the differential equation \(\frac{dQ}{dt} = -kQ\), where \(Q\) is the quantity of the drug, and \(k\) is a positive constant known as the decay constant.
This equation tells us that the change in \(Q\) is negative, indicating a decrease, and is proportional to \(Q\) itself. As the process continues, less and less drug remains, making the decrease slower over time. This is a signature of exponential decay, where the concentration drops rapidly at first and then slowly approaches zero.
Half-Life
Half-life is a critical concept tied closely to the idea of exponential decay. It is defined as the time required for a quantity to reduce to half its initial amount. In pharmacology, the half-life is used to describe how quickly a drug is metabolized and eliminated from the body. For hydrocodone bitartrate, the half-life is 3.8 hours, meaning every 3.8 hours, the body's drug concentration reduces by half.
The half-life is crucial for determining the constant of proportionality, \(k\), in our exponential decay equation. This value is found by using the formula for half-life in terms of the decay constant: \(t_{1/2} = \frac{\ln(2)}{k}\). Solving for \(k\) with the given half-life, we obtain \(k = \frac{\ln(2)}{3.8} \approx 0.1824\). Knowing \(k\) allows us to predict how the concentration of a drug decreases over any period.
Drug Absorption
Drug absorption is the process by which a drug is taken into the body and reaches the bloodstream. In this exercise, we assume that the drug is already fully absorbed when we begin to analyze its decay. This simplification allows us to focus entirely on how the drug's concentration decreases over time, ignoring other factors that involve initial absorption processes.
Understanding drug absorption and the time frame of decay helps in designing dosage regimens. Accurate knowledge about how quickly a drug is absorbed and its half-life informs how often doses need to be administered to maintain a therapeutic level in the body, without reaching toxic levels or becoming ineffective.
Proportional Rate
The concept of a proportional rate is central to understanding the nature of the differential equation \(\frac{dQ}{dt} = -kQ\). The term "proportional" indicates that the rate of decline in the drug's concentration is directly related to its current amount. Simply put, as the amount of the drug decreases, so does the speed at which it continues to decline.
This relationship forms the basis for solutions to many real-world problems, where dynamics are governed by proportional relationships. In our drug model, the proportional rate ensures that the drug's concentration decreases in a way that follows exponential decay, reflecting a continuous, naturally occurring process, such as radioactive decay or cooling of a hot object. Understanding this helps predict how much of the drug will remain at any point in time, as demonstrated when calculating the remaining drug 12 hours after administration.

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

Policy makers are interested in modeling the spread of information through a population. For example, agricultural ministries use models to understand the spread of technical innovations or new seed types through their countries. Two models, based on how the information is spread, follow. Assume the population is of a constant size \(M\) (a) If the information is spread by mass media (TV, radio, newspapers), the rate at which information is spread is believed to be proportional to the number of people not having the information at that time. Write a differential equation for the number of people having the information by time \(t .\) Sketch a solution assuming that no one (except the mass media) has the information initially. (b) If the information is spread by word of mouth, the rate of spread of information is believed to be proportional to the product of the number of people who know and the number who don't. Write a differential equation for the number of people having the information by time \(t .\) Sketch the solution for the cases in which (i) No one \(\quad\) (ii) \(5 \%\) of the population (iii) \(75 \%\) of the population knows initially. In each case, when is the information spreading fastest?

Give an example of: A differential equation for any quantity which grows in two ways simultaneously: on its own at a rate proportional to the cube root of the amount present and from an external contribution at a constant rate.

In Problems \(55-58,\) give an example of: A differential equation all of whose solutions form the family of functions \(f(x)=x^{2}+C\).

Consider a conflict between two armies of \(x\) and \(y\) soldiers, respectively. During World War I, F. W. Lanchester assumed that if both armies are fighting a conventional battle within sight of one another, the rate at which soldiers in one army are put out of action (killed or wounded) is proportional to the amount of fire the other army can concentrate on them, which is in turn proportional to the number of soldiers in the opposing army. Thus Lanchester assumed that if there are no reinforcements and \(t\) represents time since the start of the battle, then \(x\) and \(y\) obey the differential equations $$\begin{array}{l} \frac{d x}{d t}=-a y \\ \frac{d y}{d t}=-b x \quad a, b>0 \end{array}$$. In this problem we adapt Lanchester's model for a conventional battle to the case in which one or both of the armies is a guerrilla force. We assume that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (a) Give a justification for the assumption that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (b) Write the differential equations which describe a conflict between a guerrilla army of strength \(x\) and a conventional army of strength \(y,\) assuming all the constants of proportionality are 1 (c) Find a differential equation involving \(d y / d x\) and solve it to find equations of phase trajectories. (d) Describe which side wins in terms of the constant of integration. What happens if the constant is zero? (e) Use your solution to part (d) to divide the phase plane into regions according to which side wins.

A bank account earns \(5 \%\) annual interest, compounded continuously. Money is deposited in a continuous cash flow at a rate of 1200 dollars per year into the account. (a) Write a differential equation that describes the rate at which the balance \(B=f(t)\) is changing. (b) Solve the differential equation given an initial balance \(B_{0}=0\) (c) Find the balance after 5 years.

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.