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Chapter 11: Problem 20

Suppose that a drug is administered to a person in a single dose, and assume that the drug does not accumulate in body tissue, but is excreted through urine. Denote the amount of drug in the body at time \(t\) by \(x_{1}(t)\) and in the urine at time \(t\) by \(x_{2}(t) .\) If \(x_{1}(0)=6 \mathrm{mg}\) and \(x_{2}(0)=0\), find a system of differential equations for \(x_{1}(t)\) and \(x_{2}(t)\) if it takes 20 minutes for the drug to be at onehalf of its initial amount in the body.

Short Answer

Expert verified
The system is \(\frac{dx_1}{dt} = -\frac{\ln(2)}{20} x_1(t)\) and \(\frac{dx_2}{dt} = \frac{\ln(2)}{20} x_1(t)\) with \(x_1(0)=6\), \(x_2(0)=0\).

Step by step solution

01

Understanding the Problem

We need to set up a system of differential equations to model the amount of drug in the body and the amount being excreted over time. Initially, there are 6 mg of the drug in the body and none in the urine. It is given that the drug decreases to half its amount every 20 minutes.
02

Determine the Rate of Excretion

Since the drug does not accumulate and is excreted over time, we assume it follows first-order kinetics. The rate of change of the drug in the body, \(x_1(t)\), can be expressed as \(\frac{dx_1}{dt} = -k x_1(t)\), where \(k\) is the rate constant. The half-life formula \(t_{1/2} = \frac{\ln(2)}{k}\) can be used to find \(k\): \(k = \frac{\ln(2)}{20}\).
03

Equation for Drug in the Body

Using the rate constant \(k\), we find that the rate of change of the drug in the body is \(\frac{dx_1}{dt} = -\frac{\ln(2)}{20} \cdot x_1(t)\). This describes how the drug reduces over time.
04

Equation for Drug in the Urine

All the excreted drug goes into the urine. Therefore, the rate at which the drug enters the urine is the same as the rate it leaves the body: \(\frac{dx_2}{dt} = \frac{\ln(2)}{20} \cdot x_1(t)\). This describes the accumulation of the drug in urine.
05

Formulate the System of Equations

Putting both expressions together, the system of differential equations is: \[ \begin{align*} \frac{dx_1}{dt} &= -\frac{\ln(2)}{20} x_1(t), \ \frac{dx_2}{dt} &= \frac{\ln(2)}{20} x_1(t). \end{align*} \]The initial conditions are \(x_1(0) = 6\) mg and \(x_2(0) = 0\) mg.

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Key Concepts

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

First-Order Kinetics
First-order kinetics is a common concept used to describe processes where the rate of change of a quantity is directly proportional to the amount present. In our context, this applies to the concentration of a drug in the body. The rate at which the drug concentration decreases is proportional to its current concentration. This means that if you have more of the drug to start with, it will decrease faster. However, as the substance diminishes, the rate of decrease also slows.

In mathematical terms, this is modeled using a differential equation of the form:
  • \( \frac{dx}{dt} = -kx \)
Here, \(x\) represents the amount of drug, while \(k\) is the rate constant. The negative sign implies the concentration is reducing over time, which aligns with the idea of excretion or decay. Understanding this model is crucial for predicting how the amount of drug in the body decreases over time.
Rate Constant
The rate constant, typically denoted as \(k\), is a critical factor in first-order kinetics. It is a measure of how quickly a reactant is being converted into a product, or, in this context, how fast a drug is being removed from the body. The larger the rate constant, the faster the process occurs.

Rate constants are specific to each substance and reaction conditions. They depend on various factors such as temperature, the medium of reaction, and the nature of the reactants. For our exercise, the rate constant \(k\) can be determined using the half-life formula when the half-life is known.

The formula connecting the rate constant to half-life is:
  • \( t_{1/2} = \frac{\ln(2)}{k} \)
This equation tells us that the half-life is inversely proportional to the rate constant: as the rate constant increases, the half-life decreases. In other words, a drug with a high rate constant will have a short half-life.
Half-Life Formula
The half-life, often symbolized as \(t_{1/2}\), is a concept that describes the time required for a quantity to reduce to half its initial value. This idea is not only significant in chemistry and pharmacology but also in physics and many other fields where decay processes are studied.

For the situation described in the exercise, the half-life of our drug is given as 20 minutes. Using the half-life formula, we can solve for the rate constant \(k\), which in turn helps us understand the rate at which the drug is excreted.
  • The formula is: \( k = \frac{\ln(2)}{t_{1/2}} \)
Plugging in the given half-life:
  • \( k = \frac{\ln(2)}{20} \)
This rate constant can then be used in our differential equations to accurately predict how the concentration of the drug changes over time. Understanding half-lives allows us to plan effective dosing schedules and manage drug excretion effectively.

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Most popular questions from this chapter

Find the general solution of each given system of differential equations and sketch the lines in the direction of the eigenvectors. Indicate on each line the direction in which the solution would move if it starts on that line. $$ \left[\begin{array}{c} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{ll} -5 & 3 \\ -2 & 0 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$

We consider differential equations of the form $$ \frac{d \mathbf{x}}{d t}=A \mathbf{x}(t) $$ where $$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] $$ The eigenvalues of \(A\) will be complex conjugates. Analyze the stability of the equilibrium \((0,0)\), and classify the equilibrium according to whether it is a stable spiral, an unstable spiral, or a center. $$ A=\left[\begin{array}{ll} -1 & 1 \\ -3 & 1 \end{array}\right] $$

Consider communities composed of two species. The abundance of species 1 at time \(t\) is given by \(N_{1}(t)\), the abundance of species 2 at time \(t\) by \(N_{2}(t) .\) Their dynamics are described by $$ \begin{array}{l} \frac{d N_{1}}{d t}=f_{1}\left(N_{1}, N_{2}\right) \\ \frac{d N_{2}}{d t}=f_{2}\left(N_{1}, N_{2}\right) \end{array} $$ Assume that when both species are at low abundances their abundances increase and that \(f_{1}\) and \(f_{2}\) change sign when crossing their zero isoclines. In each problem, determine the sign structure of the community matrix at the nontrivial equilibrium (indicated by a dot) on the basis of the graph of the zero isoclines. Determine the stability of the equilibria if possible.

The Michaelis-Menten law [Equation (11.76)] states that $$ \frac{d p}{d t}=\frac{v_{m} s}{K_{m}+s} $$ where \(p=p(t)\) is the concentration of the product of the enzymatic reaction at time \(t, s=s(t)\) is the concentration of the substrate at time \(t\), and \(v_{m}\) and \(K_{m}\) are positive constants. Set $$ f(s)=\frac{v_{m} s}{K_{m}+s} $$ where \(v_{m}\) and \(K_{m}\) are positive constants. (a) Show that $$ \lim _{s \rightarrow \infty} f(s)=v_{m} $$ (b) Show that $$ f\left(K_{m}\right)=\frac{v_{m}}{2} $$ (c) Show that, for \(s \geq 0, f(s)\) is (i) nonnegative, (ii) increasing, and (iii) concave down. Sketch a graph of \(f(s) .\) Label \(v_{m}\) and \(K_{m}\) on your graph. (d) Explain why we said that the reaction rate \(d p / d t\) is limited by the availability of the substrate.

Assume that $$ \begin{array}{l} \frac{d x_{1}}{d t}=x_{1}\left(10-x_{1}-2 x_{2}\right) \\ \frac{d x_{2}}{d t}=x_{2}\left(10-2 x_{1}-x_{2}\right) \end{array} $$ (a) Graph the zero isoclines. (b) Show that \(\left(\frac{10}{3}, \frac{10}{3}\right)\) is an equilibrium, and use the analytical approach to determine its stability.
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