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Chapter 9: Problem 36

A physiological model A common assumption in modeling drug assimilation is that the blood volume in a person is a single compartment that behaves like a stirred tank. Suppose the blood volume is a four-liter tank that initially has a zero concentration of a particular drug. At time \(t=0,\) an intravenous line is inserted into a vein (into the tank) that carries a drug solution with a concentration of \(500 \mathrm{mg} / \mathrm{L} .\) The inflow rate is \(0.06 \mathrm{L} / \mathrm{min} .\) Assume the drug is quickly mixed thoroughly in the blood and that the volume of blood remains constant. a. Write an initial value problem that models the mass of the drug in the blood, for \(t \geq 0\) b. Solve the initial value problem, and graph both the mass of the drug and the concentration of the drug. c. What is the steady-state mass of the drug in the blood? d. After how many minutes does the drug mass reach \(90 \%\) of its steady-state level?

Short Answer

Expert verified
Question: Set up a differential equation to model the mass of a drug in the bloodstream, with a given inflow rate and concentration in the solution. Answer: The differential equation representing the mass of the drug in the bloodstream is given by: $$\frac{dM}{dt} = 30$$, where \(M(t)\) is the mass of the drug at time \(t\).

Step by step solution

01

Part a: Set up the initial value problem

To model the mass of the drug in the blood, let \(M(t)\) represent the mass of the drug in the bloodstream at time \(t\). The rate at which the drug enters the bloodstream is the product of the inflow rate and the concentration of the drug in the solution, which is \(0.06 \mathrm{L/min} \cdot 500 \mathrm{mg/L} = 30 \mathrm{mg/min}\). The differential equation representing the mass of the drug in the bloodstream is then given by: $$\frac{dM}{dt} = 30$$ At \(t=0\), the initial mass of the drug in the bloodstream is 0. So we have the initial value problem: $$ \frac{dM}{dt} = 30, \quad M(0) = 0 $$
02

Part b: Solve the initial value problem and graph the drug mass and concentration

To solve the initial value problem, integrate both sides of the differential equation with respect to \(t\): $$\int \frac{dM}{dt} dt = \int 30 dt$$ This gives us: $$M(t) = 30t + C$$ Since \(M(0) = 0\), we have: $$0 = 30(0) + C \Rightarrow C = 0$$ So the mass of the drug in the bloodstream at time \(t\), \(M(t)\), is given by: $$M(t) = 30t$$ Graphing the mass of the drug, we obtain a linear function with a slope of 30. To find the concentration of the drug, we must divide the mass of the drug by the volume of the blood, which is 4 liters. Hence, the concentration \(C(t)\) is given by: $$C(t) = \frac{M(t)}{4} = \frac{30t}{4} = 7.5t$$ Graphing the concentration of the drug, we also obtain a linear function with a slope of 7.5.
03

Part c: Determine the steady-state mass of the drug in the blood

The steady-state mass is achieved when the mass of the drug in the bloodstream no longer changes with time. Since the inflow rate is constant, the mass of the drug will continue to increase linearly with time and will never reach a steady state.
04

Part d: Find the time it takes for the drug mass to reach 90% of its steady-state level

As noted in part c, there is no steady-state mass for the drug in the blood. Therefore, we cannot determine the time it takes for the drug mass to reach 90% of its steady-state level.

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Key Concepts

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

Differential Equations in Pharmacokinetics
Differential equations are mathematical equations that describe how a quantity changes over time. They are an essential tool in pharmacokinetics, the study of how drugs move through the body. In the case of our drug concentration model, we are looking at a first-order differential equation which models the accumulation of a drug within the bloodstream over time.

The equation \(\frac{dM}{dt} = 30\) is a simple representation stating that the change in mass (\(M\)) of the drug in the bloodstream with respect to time (\(t\)) is constant. To solve this, we integrate with respect to \(t\), which gives us \(M(t) = 30t\), indicating the mass increases uniformly over time. An initial condition is provided, giving us an initial value problem: this combines the differential equation with the specific condition \(M(0) = 0\) to tailor the solution to our scenario.
Drug Concentration Model
A drug concentration model helps us understand how much of a drug is present within a specific volume at any given time. For example, we imagine a stirred tank representing the bloodstream, which keeps the drug evenly distributed. The model takes into account the drug's entry rate into the bloodstream and the constant volume of the blood.

We calculate concentration by dividing mass by volume, which in our case manifests as the simple equation \(C(t) = \frac{M(t)}{4} = \frac{30t}{4} = 7.5t\). Graphing this relationship, we create a visual representation of how drug concentration in the bloodstream increases with time—a critical factor when considering dosages and potential side effects in medical treatments.
Steady-State Analysis
Steady-state analysis in pharmacokinetics is the process of determining when a drug's concentration becomes constant in the body; this happens when the rate of drug input equals the rate of drug elimination. However, in our scenario, since the rate of drug entry is constant and we have no elimination or other changes happening in the system, we will not reach a steady-state concentration. The drug mass will continue to increase linearly with time, as reflected by our graph which shows a straight line with a positive slope. Consequently, we cannot calculate a time at which the drug mass reaches 90% of its steady state, as such a steady-state does not exist in this case.

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Most popular questions from this chapter

In this section, several models are presented and the solution of the associated differential equation is given. Later in the chapter, we present methods for solving these differential equations. Logistic population growth Widely used models for population growth involve the logistic equation \(P^{\prime}(t)=r P\left(1-\frac{P}{K}\right)\) where \(P(t)\) is the population, for \(t \geq 0,\) and \(r>0\) and \(K>0\) are given constants. a. Verify by substitution that the general solution of the equation is \(P(t)=\frac{K}{1+C e^{-r t}},\) where \(C\) is an arbitrary constant. b. Find the value of \(C\) that corresponds to the initial condition \(P(0)=50\) c. Graph the solution for \(P(0)=50, r=0.1,\) and \(K=300\) d. Find \(\lim _{t \rightarrow \infty} P(t)\) and check that the result is consistent with the graph in part (c).

Solve the equation \(y^{\prime}(t)=k y+b\) in the case that \(k y+b<0\) and verify that the general solution is \(y(t)=C e^{k t}-\frac{b}{k}\)

Convergence of Euler's method Suppose Euler's method is applied to the initial value problem \(y^{\prime}(t)=a y, y(0)=1,\) which has the exact solution \(y(t)=e^{a t} .\) For this exercise, let \(h\) denote the time step (rather than \(\Delta t\) ). The grid points are then given by \(t_{k}=k h .\) We let \(u_{k}\) be the Euler approximation to the exact solution \(y\left(t_{k}\right),\) for \(k=0,1,2, \ldots\). a. Show that Euler's method applied to this problem can be written \(u_{0}=1, u_{k+1}=(1+a h) u_{k},\) for \(k=0,1,2, \ldots\). b. Show by substitution that \(u_{k}=(1+a h)^{k}\) is a solution of the equations in part (a), for \(k=0,1,2, \ldots\). c. Recall from Section 4.7 that \(\lim _{h \rightarrow 0}(1+a h)^{1 / h}=e^{a} .\) Use this fact to show that as the time step goes to zero \((h \rightarrow 0,\) with \(\left.t_{k}=k h \text { fixed }\right),\) the approximations given by Euler's method approach the exact solution of the initial value problem; that is, \(\lim _{h \rightarrow 0} u_{k}=\lim _{h \rightarrow 0}(1+a h)^{k}=y\left(t_{k}\right)=e^{a t_{k}}\).

Find the equilibrium solution of the following equations, make a sketch of the direction field, for \(t \geq 0,\) and determine whether the equilibrium solution is stable. The direction field needs to indicate only whether solutions are increasing or decreasing on either side of the equilibrium solution. $$y^{\prime}(t)=-6 y+12$$

Consider the differential equation \(y^{\prime \prime}(t)-k^{2} y(t)=0,\) where \(k>0\) is a real number. a. Verify by substitution that when \(k=1\), a solution of the equation is \(y(t)=C_{1} e^{t}+C_{2} e^{-t} .\) You may assume this function is the general solution. b. Verify by substitution that when \(k=2\), the general solution of the equation is \(y(t)=C_{1} e^{2 t}+C_{2} e^{-2 t}\) c. Give the general solution of the equation for arbitrary \(k>0\) and verify your conjecture. d. For a positive real number \(k\), verify that the general solution of the equation may also be expressed in the form \(y(t)=C_{1} \cosh k t+C_{2} \sinh k t,\) where cosh and sinh are the hyperbolic cosine and hyperbolic sine, respectively (Section \(7.3)\)
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