/*! 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 27 Infusion of a Drug A drug is inf... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Infusion of a Drug A drug is infused into a patient's bloodstream at a constant rate of \(r\) grams per second. Simultaneously, the drug is removed at a rate proportional to the amount \(x(t)\) of the drug present at time \(t\). Determine a differential equation governing the amount \(x(t)\).

Short Answer

Expert verified
The differential equation for the drug amount is \(\frac{dx}{dt} = r - kx(t)\).

Step by step solution

01

Understand the Problem

We need to create a differential equation describing how the amount of drug, \(x(t)\), changes over time. The drug is infused at a constant rate \(r\), and it is removed at a rate proportional to \(x(t)\).
02

Express the Proportional Rate of Removal

The removal rate is proportional to the drug amount, which can be represented as \(kx(t)\), where \(k\) is the proportionality constant.
03

Set Up the Differential Equation

Combine the rates of infusion and removal to describe the net change in drug amount over time. This results in the differential equation: \[ \frac{dx}{dt} = r - kx(t). \]
04

Interpret the Differential Equation

The equation \(\frac{dx}{dt} = r - kx(t)\) shows that the rate of change of the drug amount is the difference between the infusion rate \(r\) and the removal rate \(kx(t)\).

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.

Proportional Rate
A proportional rate, in the context of differential equations, refers to a scenario where a certain change within a system is directly proportional to another variable. In our example of drug infusion into a patient's bloodstream, the rate at which the drug is removed follows this principle.

The amount of drug removed over time is proportional to the quantity already present, denoted as \(x(t)\), the drug amount at time \(t\). We represent this removal rate mathematically as \(kx(t)\). Here, \(k\) is a constant that determines the degree of proportion. If \(k\) is large, the drug will be removed quickly because the rate of removal is more significant in relation to \(x(t)\).

Understanding this concept is crucial because it helps us grasp how biological systems, such as the human body's drug clearance, operate using feedback mechanisms to maintain balance or homeostasis. Changes in dosage or infusion might alter \(x(t)\), but the proportional rate mechanism ensures the system adapts accordingly.
Constant Infusion
Constant infusion is a process where a drug is added into the bloodstream at a steady, unchanging rate. In the given problem, this rate is defined as \(r\). Here, \(r\) signifies how much of the drug is delivered every second, consistently over time.

The significance of constant infusion lies in maintaining a desired concentration of medication within the body. A steady rate helps avoid peaks and troughs in drug levels, promoting a uniform therapeutic effect.

For example, if \(r\) is set to administer a particular dose that meets the exact needs of the patient without excess or deficiency, the patient receives optimum treatment benefits. In the differential equation, the constant infusion is highlighted by the term \(r\) in \(\frac{dx}{dt} = r - kx(t)\). This reflects the steady input of the drug that counters the proportional removal rate, aiming for equilibrium in the drug concentration.
Rate of Change
The rate of change in a differential equation illustrates how a particular quantity varies over time. In our problem, it depicts how the amount of drug \(x(t)\) in the bloodstream changes due to infusion and removal processes.

Our equation \(\frac{dx}{dt} = r - kx(t)\) combines both the constant infusion of the drug and the proportional removal. The left side, \(\frac{dx}{dt}\), represents the derivative of \(x(t)\) with respect to time, indicating how fast \(x(t)\) is changing.

When interpreting this, if the infusion rate \(r\) equals the proportional removal rate \(kx(t)\), then \( \frac{dx}{dt} = 0\), showing a steady state where the drug amount remains constant. If \(r\) exceeds \(kx(t)\), \(x(t)\) increases, signifying more drug remains in the bloodstream. Conversely, if \(kx(t)\) surpasses \(r\), \(x(t)\) diminishes. Understanding these dynamics is critical for pharmacologists to predict drug behavior in vivo, ensuring consistent medication delivery.

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

At a time \(t=0\), a technological innovation is introduced into a community with a fixed population of \(n\) people. Determine a differential equation governing the number of people \(x(t)\) who have adopted the innovation at time \(t\) if it is assumed that the rate at which the innovation spreads through the community is jointly proportional to the number of people who have adopted it and the number of people who have not adopted it.

What function (or functions) do you know from calculus is such that its second derivative is itself? Its second derivative is the negative of itself? Write each answer in the form of a second-order differential equation with a solution.

Suppose that $$ A^{\prime}(t)=-0.0004332 A(t) $$ represents a mathematical model for the decay of radium 226 , where \(A(t)\) is the amount of radium (measured in grams) remaining at time \(t\) (measured in years). How much of the radium sample remains at time \(t\) when the sample is decaying at a rate of \(0.002\) grams per year?

In Problems 19 and 20, verify that the indicated expression is an implicit solution of the given first-order differential equation. Find at least one explicit solution \(y=\phi(x)\) in each case. Use a graphing utility to obtain the graph of an explicit solution. Give an interval \(I\) of the definition of each solution \(\phi\). $$ \frac{d X}{d t}=(X-1)(1-2 X) ; \quad \ln \left(\frac{2 X-1}{X-1}\right)=t $$

State the order of the given ordinary differential equation. Determine whether the equation is linear or nonlinear by matching it with. $$ \frac{d^{2} u}{d r^{2}}+\frac{d u}{d r}+u=\cos (r+u) $$

See all solutions

Recommended explanations on Physics 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.