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Chapter 7: Problem 15

The Weibull equation for the dynamics of the drug concentration is $$\frac{d c}{d t}=\frac{k}{t^{b}}\left(c_{s}-c\right)$$ where \(k, c_{s},\) and \(b\) are positive constants and \(b<1 .\) Notice hat this differential equation is undefined when \(t=0 .\) Is this differential equation pure-time, autonomous, or nonautonomous? State in words what this differential equation says about how drug dissolution occurs. Verify that $$c=c_{s}\left(1-e^{-\alpha t^{1-b}}\right)$$ is a solution for \(t \neq 0,\) where \(\alpha=k /(1-b)\).

Short Answer

Expert verified
Nonautonomous; describes faster dissolution at early times.

Step by step solution

01

Understand the Differential Equation Types

A differential equation is classified based on how its variables are involved. A **pure-time** equation involves only the time derivative and a function of time. An **autonomous** equation depends only on the dependent variable (here, concentration) and its derivatives, but not explicitly on the independent variable (time). A **nonautonomous** equation explicitly includes both the dependent variable and the independent variable.
02

Classify the Given Differential Equation

The given equation \( \frac{dc}{dt} = \frac{k}{t^b}(c_s - c) \) has explicit dependence on \(t\) (the term \( \frac{k}{t^b} \)), therefore, it is not autonomous. Because it involves time \(t\) explicitly, it is a nonautonomous differential equation.
03

Interpret the Differential Equation

The equation \( \frac{dc}{dt} = \frac{k}{t^b}(c_s - c) \) indicates that the change in drug concentration \( c \) over time \( t \) is governed by the difference \( c_s - c \) scaled by the factor \( \frac{k}{t^b} \), suggesting faster rates of change when \(c\) is far from saturation \(c_s\) and when \(t\) is small (since \( b < 1 \)). This implies that dissolution is more rapid at early times.
04

Check the Solution Substitution

Substitute \( c = c_s (1 - e^{-\alpha t^{1-b}}) \) into the left-hand side of the differential equation to find \( \frac{dc}{dt} \). First, differentiate:\[ \frac{dc}{dt} = \frac{d}{dt}\left(c_s (1 - e^{-\alpha t^{1-b}})\right) = c_s \alpha (1-b) t^{-b} e^{-\alpha t^{1-b}} \]
05

Substitute and Simplify

Substitute \( \frac{dc}{dt} \) into the right-hand side of the differential equation:\[ \frac{k}{t^b} (c_s - c) = \frac{k}{t^b} (c_s - c_s(1 - e^{-\alpha t^{1-b}})) = \frac{k c_s}{t^b} e^{-\alpha t^{1-b}} \]Since \( \alpha = \frac{k}{1-b} \),\( c_s \alpha (1-b) t^{-b} e^{-\alpha t^{1-b}} = \frac{k c_s}{t^b} e^{-\alpha t^{1-b}} \), confirming the expression.
06

Conclusion: Verify Solution

The substitution confirms that both sides of the original differential equation are equal when using the proposed solution. This verifies that \( c = c_s (1 - e^{-\alpha t^{1-b}}) \) is indeed a solution for \( t eq 0 \).

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Key Concepts

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

Drug Concentration Dynamics
Understanding the dynamics of drug concentration is crucial in pharmacokinetics, which is the study of how drugs move through the body. The Weibull differential equation models how the concentration of a drug changes over time.

This equation states that the rate of change in drug concentration, \( \frac{dc}{dt} \), is influenced by several factors:
  • It is proportional to the difference between the saturation concentration \( c_s \) and the current concentration \( c \).
  • It is inversely proportional to a power of time \( t^b \), which means that early times affect the concentration dynamics more significantly.
The constants \( k \), \( c_s \), and \( b \) are crucial as they define the rate and style of the drug's dissolution process.

The implication here is that drug dissolution occurs more rapidly when the current concentration is far from saturation and during the initial phase (small \( t \)). As time progresses, the change rate decreases as the concentration approaches its maximum (saturation level).
Nonautonomous Differential Equations
Differential equations play a pivotal role in modeling dynamic systems. The Weibull equation given in the problem is classified as a nonautonomous differential equation.

Nonautonomous equations are distinguished by the explicit inclusion of the independent variable, which in this case is time \( t \). Unlike autonomous equations, where changes depend solely on the dependent variable, nonautonomous equations calculate change by considering both the dependent variable and the time.
  • In the analyzed equation, the presence of \( \frac{k}{t^b} \) illustrates this dependency on \( t \).
  • This aspect suggests that the influence of time diminishes as \( t \) becomes larger because the term \( t^b \) in the denominator grows.
This particular trait of nonautonomous equations makes them more suited to applications where the time itself impacts the outcome, essential in many biological scenarios like drug concentration.

The classification of the equation helps understand how various states and transitions are regulated over time, a critical insight in drug dissolution models.
Differential Equations in Biology
Differential equations are extensively used in biology to describe systems that change over time, including populations, cellular processes, and pharmacokinetics.

The exploration of drug concentration dynamics through differential equations, such as the Weibull equation, sheds light on how substances interact with biological systems.
  • This equation is apt for modeling how drugs are absorbed, distributed, and eliminated in biological systems.
  • It reflects real-life kinetics where events such as uptake and dispersion do not occur at a constant rate over time.
Such equations help clinicians and scientists predict how long it will take for a drug to reach effective concentrations and how quickly it will diminish from the body.

This approach supports more effective medication dosing and timing strategies, demonstrating the significance of differential equations in biological research and healthcare applications. As seen in the exercise, solving these equations allows for making precise predictions about the behavior of biological systems undergoing dynamic change.

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Most popular questions from this chapter

A system of differential equations is given. (a) Use a phase plane analysis to determine the values of the constant \(a\) for which the sole equilibrium of the differential equations is locally stable. (b) Obtain an expression for each equilibrium (it may be a function of the constant \(a\) ). \(x^{\prime}=a y^{2}-x+1, \quad y^{\prime}=2(1-y)\)

(a) Sketch the curve by using the parametric equations to plot points. Indicate with an arrow the direction in which the curve is traced as \(t\) increases. (b) Eliminate the parameter to find a Cartesian equation of the curve. \(x=t^{2}, \quad y=t^{3}\)

(a) Use Euler's method with each of the following step sizes to estimate the value of \(y(0.4),\) where \(y\) is the solution of the initial-value problem \(y^{\prime}=y, y(0)=1\) (i) \(h=0.4 \quad\) (ii) \(h=0.2 \quad\) (iii) \(h=0.1\) (b) We know that the exact solution of the initial-value problem in part (a) is \(y=e^{x} .\) Draw, as accurately as you can, the graph of \(y=e^{x}, 0 \leqslant x \leqslant 0.4,\) together with the Euler approximations using the step sizes in part (a). (Your sketches should resemble Figures \(12,\) 13, and \(15 .\) Use your sketches to decide whether your estimates in part (a) are underestimates or overestimates. (c) The error in Euler's method is the difference between the exact value and the approximate value. Find the errors made in part (a) in using Euler's method to estimate the true value of \(y(0.4),\) namely, \(e^{0.4} .\) What happens to the error each time step size is halved?

Sketch the direction field of the differential equation. Then use it to sketch a solution curve that passes through the given point. \(y^{\prime}=x y-x^{2}, \quad(0,1)\)

(a) Sketch the curve by using the parametric equations to plot points. Indicate with an arrow the direction in which the curve is traced as \(t\) increases. (b) Eliminate the parameter to find a Cartesian equation of the curve. \(x=3 t-5, \quad y=2 t+1\)
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