Question

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. The delivery of a drug (such as an antibiotic) through an intravenous line may be modeled by the differential equation $m^{\prime }(t)+km(t)=I,$ where $m(t)$ is the mass of the drug in the blood at time $t\ge 0,k$ is a constant that describes the rate at which the drug is absorbed, and $I$ is the infusion rate. a. Show by substitution that if the initial mass of drug in the blood is zero $(m(0)=0),$ then the solution of the initial value problem is $m(t)=\frac{I}{k}(1-e^{-kt})$ b. Graph the solution for $I=10\mathrm{mg}/\mathrm{hr}$ and $k=0.05{\mathrm{hr}}^{-1}$ c. Evaluate $\underset{t\to \mathrm{\infty }}{lim}m(t),$ the steady-state drug level, and verify the result using the graph in part (b).

Short answer

Answer: The steady-state drug level is 200 mg.

Step-by-step solution

Verify given solution

Firstly, we are given a solution to the differential equation and initial value problem $m(t)=\frac{I}{k}(1-e^{-kt})$ when $m(0)=0$. To verify this solution, we have to plug it into the given differential equation, and check if it satisfies the equation. The original differential equation is $m^{\prime }(t)+km(t)=I$. Now, we find the derivative of the given solution with respect to time, $m^{\prime }(t)$. $m^{\prime }(t)=\frac{d}{dt}\left(\frac{I}{k}(1-e^{-kt})\right)=\frac{I}{k}\frac{d}{dt}(1-e^{-kt})=\frac{I}{k}\cdot k{e}^{-kt}$ Now, we plug the given solution $m(t)$ and its derivative $m^{\prime }(t)$ into the differential equation: $m^{\prime }(t)+km(t)=\frac{I}{k}\cdot k{e}^{-kt}+k\left(\frac{I}{k}(1-e^{-kt})\right)=I\cdot e^{-kt}+I-Ik\cdot e^{-kt}$ As we can see, the terms $I\cdot e^{-kt}$ and $-Ik\cdot e^{-kt}$ cancel out, and we are left with just: $I=I$ This means that the given solution $m(t)$ is, in fact, correct. Now, we want to verify the initial condition $m(0)=0$: $m(0)=\frac{I}{k}(1-e^{-k(0)})=\frac{I}{k}(1-e^0)=\frac{I}{k}(1-1)=0$ The initial condition is also satisfied.

Graph the solution

We are asked to graph the solution $m(t)=\frac{I}{k}(1-e^{-kt})$ with the given values $I=10\mathrm{mg}/\mathrm{hr}$ and $k=0.05{\mathrm{hr}}^{-1}$. Replacing these values gives: $m(t)=\frac{10}{0.05}(1-e^{-0.05t})=200(1-e^{-0.05t})$ To graph the solution, plot the function $m(t)=200(1-e^{-0.05t})$ over a suitable range of time values (e.g. $t=0$ to $t=50\mathrm{hr}$). You should see the graph starting from zero and increasing until it reaches a steady-state value (which we'll find in the next step).

Find the steady-state drug level

We are asked to find the steady-state drug level, which is the limit of the mass of the drug as time goes to infinity: $\underset{t\to \mathrm{\infty }}{lim}m(t)=\underset{t\to \mathrm{\infty }}{lim}200(1-e^{-0.05t})$ As $t\to \mathrm{\infty }$, the exponent $-0.05t\to -\mathrm{\infty }$, so $e^{-0.05t}\to 0$: $\underset{t\to \mathrm{\infty }}{lim}200(1-e^{-0.05t})=200(1-0)=200\mathrm{mg}$ This means the steady-state drug level is $200\mathrm{mg}$. To verify this result using the graph from step 2, observe that the graph approaches a horizontal asymptote at $m(t)=200\mathrm{mg}$ as time goes to infinity.

Key concepts

Initial Value Problem

An initial value problem in mathematics, particularly in differential equations, involves finding a function that satisfies a given differential equation and that also meets specific initial conditions. In pharmacokinetics and other applications, this allows modelers to predict how a system behaves over time, starting from an initial state.

For instance, when a drug is administrated intravenously, the initial value problem can describe how the concentration of the drug changes in the bloodstream over time. Given the differential equation m'(t) + k m(t) = I, where m(t) represents the mass of the drug at time t, and certain parameters are known (like the rate of absorption k and the infusion rate I), providing an initial condition (such as m(0) = 0 when no drug is initially present) allows us to predict the drug concentration at any subsequent time.

This kind of problem is a staple in mathematics because it can be applied to many real-world phenomena, where we know the rate of change of something and its initial state, and we want to find out how it evolves over time.

Steady-State Drug Level

The term steady-state drug level refers to the concentration of a medication in the bloodstream that does not change significantly over time, assuming a constant rate of drug administration and elimination. This is a crucial concept in pharmacology as it helps in determining the proper dosing regimen to achieve therapeutic drug concentrations without reaching toxic levels.

With respect to differential equations, the steady-state level corresponds to what happens when the net change over time balances out. Mathematically, we reach this value when the derivative of our concentration function, m'(t), approaches zero as time goes towards infinity. It's important for healthcare providers to understand this concept to ensure that patients receive the correct amount of a medication necessary to maintain therapeutic levels without overdosing.

Limit of a Function

In calculus, the limit of a function is a fundamental concept that describes the behavior of that function as it approaches a particular point or as the input (or