/*! 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 A drug has first order eliminati... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 23

A drug has first order elimination kinetics, meaning that a fixed fraction of drug is eliminated from the body in each unit of time. So if no further drug is absorbed into the patient's blood after time \(t=0\), the amount of drug in their blood will decay with time according to: $$ \frac{d M}{d t}=-k_{1} M $$ where \(k_{1}>0\) is the fraction of drug eliminated in one unit of time. (a) Assuming \(M(0)=M_{0}\), solve the differential equation. (b) According to your model, does \(M(t)\) ever reach 0 ? (c) Given that \(M_{0}=10\) and \(k_{1}=2\), calculate the time at which \(M(t)\) drops to \(M=1\)

Short Answer

Expert verified
(a) \( M(t) = M_0 e^{-k_1 t} \); (b) No, \( M(t) \) never reaches 0; (c) \( t \approx 1.151 \) units of time.

Step by step solution

01

Solve the Differential Equation

Given the differential equation \( \frac{dM}{dt} = -k_1 M \) with the initial condition \( M(0) = M_0 \), we recognize this as a first-order linear differential equation in the form \( \frac{dM}{dt} + k_1 M = 0 \). This can be solved using separation of variables. We separate the variables as follows: \( \frac{dM}{M} = -k_1 dt \). Integrating both sides, we have \( \int \frac{dM}{M} = \int -k_1 \, dt \), which leads to \( \ln|M| = -k_1 t + C \), where \( C \) is the constant of integration. Exponentiating both sides gives \( M = e^{-k_1 t + C} = e^C e^{-k_1 t} \). Define \( e^C = M_0 \), thus \( M(t) = M_0 e^{-k_1 t} \).
02

Analyze the behavior as t approaches infinity

The model \( M(t) = M_0 e^{-k_1 t} \) describes exponential decay. As \( t \) approaches infinity, \( e^{-k_1 t} \) approaches zero, meaning \( M(t) \) asymptotically approaches zero. However, \( M(t) \) never exactly reaches zero, because the exponential function approaches zero but doesn't touch it or become zero in finite time.
03

Calculate the time when M(t) = 1

We need to find \( t \) such that \( M(t) = 1 \). Using the equation \( M(t) = M_0 e^{-k_1 t} \), we set \( 1 = 10 e^{-2t} \). Solving for \( t \) involves isolating the exponent:1. Rearrange to get: \( e^{-2t} = \frac{1}{10} \).2. Take the natural logarithm of both sides: \( -2t = \ln(\frac{1}{10}) \).3. Solve for \( t \): \( t = -\frac{1}{2} \ln(\frac{1}{10}) \).Using \( \ln(0.1) \approx -2.302 \), \( t = -\frac{1}{2} \times -2.302 \approx 1.151 \).

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Key Concepts

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

First-Order Linear Differential Equation
A first-order linear differential equation involves the derivatives of a function with respect to one variable, commonly time. In this context, it involves the rate at which a drug is eliminated from the body. The general form is given by - \(\frac{dM}{dt} + k_1 M = 0\), where \(M\) is the amount of drug, and \(k_1\) is a constant. - Often, it's solved using separation of variables or integrating factors. These equations are quite prevalent in modeling natural processes, like cooling rates or chemical reactions. By solving them, we understand how a process evolves over time, given initial conditions like the example of initial drug concentration \(M(0) = M_0\). The specific solution \(M(t) = M_0 e^{-k_1 t}\) shows how the amount changes, emphasizing a declining trend due to elimination.
Exponential Decay
Exponential decay occurs when a quantity decreases at a rate proportional to its current value. In the case of drug elimination from the body, the formula \(M(t) = M_0 e^{-k_1 t}\) describes this process. - \(M_0\) represents the initial drug concentration. - \(k_1\) indicates the fractional rate of elimination. Exponential functions are unique because they decay rapidly at first, slowing over time, yet they never fully reach zero. They asymptotically approach zero, showcasing a prolonged decline. This behavior crucially informs us about the long-term presence of substances, explaining why they can linger in biological systems even with high elimination rates.
Separation of Variables
Separation of variables is a mathematical technique to solve differential equations. It's particularly effective for first-order linear equations where variables can be isolated on either side of the equation. In our specific differential equation, \(\frac{dM}{M} = -k_1 \, dt\), variables \(M\) and \(t\) are separated, enabling us to integrate both sides. - Integrating \(\int \frac{dM}{M}\), leads to \(\ln|M|\). - Meanwhile, integrating \(\int -k_1 \, dt\) results in \(-k_1 t\). This integration yields an expression involving a constant of integration, leading to exponential expressions like \(M(t) = M_0 e^{-k_1 t}\). It's a powerful approach that translates complex, dynamic systems into manageable, solvable equations, providing insights into how quantities evolve according to time.

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Most popular questions from this chapter

Compartment models are used to model the flow of traffic between different roads, by treating each road as a compartment. As an example, consider how the number of cars on a freeway on-ramp, \(N(t)\), changes with time. For a simplified model let's assume that cars join the on-ramp at a constant rate \(q\) (that is, \(q\) cars join the on-ramp in one unit of time). Cars then leave the on-ramp by entering the freeway itself. Assume that a fraction \(f\) of the cars on the on-ramp enter the freeway in one unit of time. (a) Derive a differential equation for \(N(t) .\) Your differential equation will include the unknown constants \(f\) and \(q\). (b) Analyze your model from part (a) to find the equilibrium number of cars on the on-ramp, and determine whether this equilibrium is stable or unstable. (c) Suppose that the maximum capacity of the on-ramp is 90 cars, and the rate at which cars flow onto the on-ramp is \(q=60\) cars per min. Find the value of \(f\) that is needed to keep \(N\) below the on-ramp's capacity.

Bite strength varies as animals grow, which may mean that the animal's diet must change. Christiansen and Adolfsson (2005) studied the relationship between the strength of animal teeth with skull size in carnivores from the cat and dog families. They found that tooth strength \(S\), and skull length \(L\), were related in a power law: $$ S=C L^{2.85} $$ where \(C\) is some constant. Find the relationship between the relative rates of growth of \(S\) and \(L\) (i.e., between \(\frac{1}{S} \frac{d S}{d t}\) and \(\left.\frac{1}{L} \frac{d L}{d t}\right)\).

Draw the vector field plot of the differential equation. Then, using the given initial conditions, sketch the solutions (i.e., draw a graph showing the dependent variable as a function of the independent variable). \(\frac{d y}{d x}=(y-1)(y-2)(y-5)\) (a) \(y(0)=0\), (b) \(y(0)=4\) (c) \(y(0)=3 / 2\), (d) \(y(0)=6\).

Find all equilibria, and, by calculating the eigenvalue of the differential equation, determine which equilibria are stable and which are unstable. $$ \frac{d y}{d t}=y(2-y)(y-3) $$

For make vector field plots of each of the differential equations. Find any equilibria of each differential equation and use your vector field plot to classify whether each equilibrium is stable or unstable. $$ \frac{d x}{d t}=\frac{x}{x+1}, x \neq-1 $$
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