Question

The delivery of a drug (such as an antibiotic) through an intravenous line may be modeled by the differential equation \(m^{\prime}(t)+k m(t)=I,\) where \(m(t)\) is the mass of the drug in the blood at time \(t \geq 0, k\) 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 \(m(t)=\frac{I}{k}\left(1-e^{-k t}\right)\). b. Graph the solution for \(I=10 \mathrm{mg} / \mathrm{hr}\) and \(k=0.05 \mathrm{hr}^{-1}\). c. Evaluate \(\lim m(t),\) the steady-state drug level, and verify the result using the graph in part (b).

Short answer

2. What function represents the mass of a drug in the blood over time, when the values of \(I\) and \(k\) are \(10 \, \mathrm{mg/hr}\) and \(0.05 \, \mathrm{hr}^{-1}\), respectively? 3. What is the steady-state drug level after computing the limit of \(m(t)\) as \(t\) approaches infinity?

Step-by-step solution

a. Show by substitution that the given solution is correct.

To show that \(m(t)=\frac{I}{k}\left(1-e^{-kt}\right)\) is the solution to the initial value problem, we can first check if it satisfies the differential equation \(m^{\prime}(t)+km(t)=I\), and then check if it meets the initial condition \(m(0)=0\). First, let's find the derivative of the given solution: \(m(t) = \frac{I}{k}(1-e^{-kt})\) \(m^{\prime}(t) = \frac{I}{k}(-e^{-kt}( -k))\) \(m^{\prime}(t) = Ie^{-kt}\) Now, let's substitute the function and its derivative into the differential equation and see if the equation holds true: \(m^{\prime}(t) + km(t) = Ie^{-kt} + k\left(\frac{I}{k}(1-e^{-kt})\right)\) \(Ie^{-kt} + I(1-e^{-kt}) = Ie^{-kt} + I - Ie^{-kt} = I\) As the equation holds true, the provided solution is correct. Finally, we need to check if it meets the initial condition: \(m(0) = \frac{I}{k}(1 - e^{-k(0)}) = \frac{I}{k}(1 - 1) = 0\) So, the given solution meets the initial condition as well.

b. Graph the solution for given values of \(I\) and \(k\).

The given values for \(I\) and \(k\) are \(10 \, \mathrm{mg/hr}\) and \(0.05 \, \mathrm{hr}^{-1}\), respectively. Substitute these into the solution we derived earlier: \(m(t) = \frac{10}{0.05}(1 - e^{-0.05t})\) \(m(t) = 200(1 - e^{-0.05t})\) Now, plot this function on the graph with the time \(t\) on the x-axis and \(m(t)\) on the y-axis. This will show how the mass of drug in the blood changes over time.

c. Evaluate the limit of \(m(t)\) and verify the result using the graph.

To find the steady-state drug level, we need to compute the limit of \(m(t)\) as \(t\) approaches infinity: \(\lim_{t\to\infty} m(t) = \lim_{t\to\infty} 200(1 - e^{-0.05t})\) As \(t\) approaches infinity, \(e^{-0.05t}\) approaches zero: \(\lim_{t\to\infty} m(t) = 200(1 - 0) = 200\) So, the steady-state drug level is \(200 \, \mathrm{mg}\). To verify this result using the graph, check that the graph approaches a horizontal line at \(200 \, \mathrm{mg}\) as \(t \to \infty\). If the graph plateaus around this value, it confirms our computed limit.

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) is a type of differential equation that comes with specific conditions called initial conditions. These are the values of the unknown function and possibly some of its derivatives at a given point. In our exercise, we are solving the differential equation \(m'(t) + km(t) = I\) under the initial condition \(m(0)=0\). This means that at time \(t=0\), the mass of the drug in the bloodstream is zero.
Handling IVPs typically involves finding a function that not only satisfies the differential equation for all time \(t\), but also adheres to the initial conditions specified. For instance, in this problem, the initial condition ensures we find the correct function representing the mass of the drug over time, starting from zero and depending on the infusion rate \(I\) and removal constant \(k\).

Steady-State Solution

The Steady-State Solution represents a condition where the variables in a system reach a constant level over time, remaining unchanged as time progresses. In the context of our drug delivery problem, the steady-state occurs when the amount of drug entering the bloodstream equals the amount being absorbed or removed, resulting in a stable mass of drug at times approaching infinity.
Mathematically, we find the steady-state solution by evaluating the limit of the function \(m(t)\) as \(t\) approaches infinity: \( \lim_{t \to \infty} m(t)\). For the given solution \(m(t) = \frac{I}{k}(1 - e^{-kt})\), as \(t\) goes to infinity, \(e^{-kt}\) approaches zero, leaving the steady-state mass at \(\frac{I}{k}\). In this exercise, with \(I = 10 \text{ mg/hr}\) and \(k = 0.05 \text{ hr}^{-1}\), the steady-state level is \(200 \text{ mg}\).

Infusion Rate

The Infusion Rate, denoted as \(I\), is a critical component in the study of drug delivery through an intravenous line. It refers to the rate at which the drug is introduced into the patient's bloodstream. Measured in units such as \(\text{mg/hr}\), this rate essentially determines how quickly the drug is administered.
In our equation \(m'(t) + km(t) = I\), the infusion rate directly affects how quickly the mass of the drug in the body builds up over time. A higher infusion rate \(I\) will lead to a quicker rise in drug mass until it reaches steady-state. This problem uses an infusion rate of \(10 \text{ mg/hr}\), showing how the drug level stabilizes over time and the continual balance required to maintain desired drug concentration in the bloodstream.

Rate of Absorption

The Rate of Absorption, represented by the constant \(k\), describes how quickly the drug is absorbed, distributed, and eventually removed from the bloodstream. In our differential equation, \(k\) is a critical factor because it directly influences how fast this process occurs.
Understanding \(k\) helps in predicting how the drug concentration changes over time. A higher rate of absorption will lead to the drug being removed from the bloodstream more quickly, affecting how long it takes to reach the steady-state level. Conversely, a smaller \(k\) means the drug remains in the bloodstream longer, potentially leading to a slower approach to steady-state. In our exercise, the given value for \(k = 0.05 \text{ hr}^{-1}\) influences the shape and timewise behavior of the solution \(m(t)\).
Understanding both \(k\) and \(I\) is essential for designing effective drug delivery systems that achieve optimal therapeutic levels without reaching toxicity.