Question

Analyzing models The following models were discussed in Section 9.1 and reappear in later sections of this chapter. In each case carry out the indicated analysis using direction fields. Drug infusion 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. Let \(I=10 \mathrm{mg} / \mathrm{hr}\) and \(k=0.05 \mathrm{hr}^{-1}\). a. Draw the direction field, for \(0 \leq t \leq 100,0 \leq y \leq 600\) b. For what initial values \(m(0)=A\) are solutions increasing? Decreasing? c. What is the equilibrium solution?

Short answer

Answer: The solutions are increasing if the initial drug amount, \(m(0)=A\), is less than 200 mg. The solutions are decreasing if A is greater than 200 mg. The equilibrium solution is at 200 mg, where the drug amount remains constant over time.

Step-by-step solution

Draw the direction field

In this step, we will draw the direction field for \(0\leq t \leq 100\) and \(0\leq m(t) \leq 600\). We have the differential equation: \(m'(t)+km(t)=I\), with \(I=10\,\text{mg/hr}\) and \(k=0.05\,\text{hr}^{-1}\). Replace the given values of \(I\) and \(k\) into the equation: \(m'(t) + 0.05m(t) = 10\) Now, this is a first-order linear differential equation, and we can generate its direction field using software or by hand using a consistent increment of \(t\) and \(m(t)\) values.

Identify increasing and decreasing initial values

To find out for which initial values \(m(0)=A\) the solutions are increasing or decreasing, let's examine the value of \(m'(t)\) at various initial conditions. If \(m'(t)>0\), the solution is increasing, and if \(m'(t)<0\), the solution is decreasing. From the equation \(m'(t) + 0.05m(t) = 10\), let's solve for \(m'(t)\): \(m'(t) = 10 - 0.05m(t)\) Now we can analyze \(m'(t)\) at \(t=0\): \(m'(0) = 10 - 0.05m(0) = 10 - 0.05A\) If \(A<200\), then \(m'(0)>0\), and the solution is increasing. If \(A>200\), then \(m'(0)<0\), and the solution is decreasing. If \(A=200\), then \(m'(0)=0\), and the solution remains constant.

Find the equilibrium solution

An equilibrium solution is a constant solution that doesn't change with time. In this case, the equilibrium solution occurs when \(m'(t)=0\). Let's find the equilibrium solution using the equation we derived in step 2: \(m'(t) = 10 - 0.05m(t)\) Set \(m'(t)\) to zero: \(0 = 10 - 0.05m(t)\) Now, solve for \(m(t)\): \(m(t) = \frac{10}{0.05} = 200\text{ mg}\) Thus, the equilibrium solution is \(m(t) = 200\text{ mg}\).

Key concepts

Differential Equations

Differential equations play an integral role in modeling real-world processes in science and engineering. They establish a relationship between a function and its derivatives, representing how a particular quantity changes over time.

For example, the differential equation mentioned in our exercise, \(m'(t) + km(t) = I\), is a first-order linear differential equation. It is first-order because it involves the first derivative of the function \(m(t)\), and it is linear because the function and its derivative appear to the first power and are not multiplied together.

To visualize the behavior of solutions to differential equations, direction fields (also known as slope fields) are used. A direction field is a grid of small line segments on the \(t\)-\(m(t)\) plane, each segment indicating the slope of the solution curve passing through its midpoint. By observing the pattern formed by these segments, you can predict the behavior of solution curves and understand how different initial conditions affect the system's dynamics.

Equilibrium Solution

An equilibrium solution is a special type of solution to a differential equation that remains constant over time and does not change. In other words, for any point that lies on the equilibrium solution, the derivative (rate of change) is zero.

In our exercise, we found that \(m(t) = 200 \text{ mg}\) is an equilibrium solution. This means that once the drug in the bloodstream reaches a mass of 200 mg, it will remain at this level indefinitely (assuming the infusion rate and absorption rate remain constant). In the context of the model, it indicates a steady state where the rate of drug infusion exactly balances the rate of drug absorption, leading to no net change in the mass of the drug.

To determine equilibrium solutions, we generally set the derivative to zero (\(m'(t)=0\)) and solve for the function. Equilibrium solutions can often be graphically represented as horizontal lines on the direction field, where the slopes of the solution curves are zero.

Initial Values

The term 'initial values' refers to the conditions of the system at the start of our observation (usually given as a certain value at time \(t=0\)). In the context of differential equations, initial values (also known as initial conditions) help us determine the particular solution that fits our scenario.

With the given equation \(m'(t) + 0.05m(t) = 10\) and by examining different initial values \(m(0) = A\), we determine whether the mass of the drug in the bloodstream is increasing, decreasing, or remains constant over time. It's the starting point that influences the trajectory of the solution curve on the direction field. For instance, if \(A < 200\) mg, the drug mass in the bloodstream is increasing; if \(A > 200\) mg, it is decreasing; and if \(A = 200\) mg, it represents the equilibrium solution where the mass does not change.

Understanding initial values is crucial because it allows us to apply general solutions of differential equations to specific situations, which can be vital for tasks such as predicting medication levels in medical treatments or determining the initial concentration of a chemical reaction.