Problem 4 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Continuous infusion of penicillin. Suppose a patient recovering from surgery is to be administered penicillin intravenously at a constant rate of 5 grams per hour. The patient's kidneys will remove penicillin at a rate proportional to the serum penicillin concentration. Let \(P(t)\) be the penicillin concentration \(t\) hours after infusion is begun. Then a simple model of penicillin pharmacokinetics is \(\begin{array}{l}\text { Net Rate of Increase } & \text { Clearance } & \text { Infusion }\end{array}\) $$P^{\prime}(t)=-K \times P(t)+5$$ $$\frac{\mathrm{gm}}{\mathrm{hr}} \quad \frac{1}{\mathrm{hr}} \times \operatorname{gm} \quad \frac{\mathrm{gm}}{\mathrm{hr}}$$ The proportionality constant, \(K,\) must have units \(\frac{1}{\mathrm{hr}}\) in order for the units on the equation to balance. We initially assume that \(K=2.5 \frac{1}{\mathrm{hr}}\) which is in the range of physiological reality. It is reasonable to assume that there was no penicillin in the patient at time \(t=0,\) so that \(P(0)=0\). a. Draw the phase plane for the differential equation $$P(0)=0 \quad P^{\prime}(t)=-2.5 P(t)+5$$ b. Find the equilibrium point of \(P^{\prime}=-2.5 P+5\). c. Is the equilibrium point stable? d. Show that the units of the equilibrium point are grams. e. Suppose the patient's kidneys are impaired and only operating at \(60 \%\) of normal. Then \(K=1.5\) instead of \(2.5 .\) What effect does this have on the equilibrium point.

Step-by-Step Solution

Answer

The equilibrium point with \( K = 2.5 \) is 2 grams and is stable. With \( K = 1.5 \), it increases to approximately 3.33 grams.

Step 1: Write Down the Differential Equation

The given differential equation is \( P'(t) = -2.5P(t) + 5 \). This describes the rate of change of penicillin concentration in the blood with respect to time.

Step 2: Draw the Phase Plane

Determine the dynamics by examining the signs of \( P'(t) \) based on \( P(t) \). The line \( P(t) = 0 \) produces \( P'(t) = 5 > 0 \), indicating increase. When \( P(t) = 2 \), \( P'(t) = -2.5 \times 2 + 5 = 0 \), so \( P(t) = 2 \) is the equilibrium.

Step 3: Find the Equilibrium Point

To find the equilibrium point, set \( P'(t) = 0 \):\[ 0 = -2.5P + 5 \]Solve for \( P \):\[ 2.5P = 5 \]\[ P = 2 \].

Step 4: Analyze the Stability of the Equilibrium

The derivative \( P'(t) = -2.5P + 5 \) shows that for \( P > 2 \), \( P' < 0 \); for \( P < 2 \), \( P' > 0 \). Thus, the equilibrium at \( P = 2 \) is stable since the system tends towards it from both directions.

Step 5: Identify the Units of the Equilibrium Point

The units of each term in the equation must be grams per hour. Solving for the equilibrium point \( P = 2 \) confirms that this is a concentration in grams since the units are consistent throughout the equation.

Step 6: Adjust for Impaired Kidney Function

With \( K = 1.5 \), the equation becomes \( P'(t) = -1.5P(t) + 5 \). To find the new equilibrium point, set \( P'(t) = 0 \):\[ 0 = -1.5P + 5 \]\[ 1.5P = 5 \]\[ P = \frac{5}{1.5} = \frac{10}{3} \approx 3.33 \]. The impaired function results in a higher equilibrium concentration of penicillin.

Key Concepts

Continuous Infusion

Continuous infusion is a method used in pharmacology and medicine where a drug is administered into a patient's system at a consistent and constant rate. In this particular problem, penicillin is being given at 5 grams per hour continuously. The idea behind continuous infusion is to maintain a steady drug concentration in the blood, avoiding peaks and troughs that could lead to under-dosing or toxicity.
This process is especially important for drugs, like antibiotics, that require precise dosing to effectively treat infections. With continuous infusion:
* A constant rate ensures drug levels remain therapeutic. * It avoids the complications associated with bolus or "one-shot" dosing.
In our model, the infusion rate remains constant while the rate of drug clearance is proportional to the concentration of penicillin in the blood, leading to a dynamic balance that stabilizes over time.

Equilibrium Point

The equilibrium point in a differential equation is the condition where the rate of change of the quantity in question becomes zero. When a system reaches equilibrium, it means that whatever is going in is perfectly balanced by what is going out, leading to a stable state.
In the exercise example, the equilibrium is found by setting the rate of change, given by the differential equation, to zero. For penicillin, when we solve \[ P'(t) = -2.5P(t) + 5 = 0 \]we find that the equilibrium concentration of penicillin is 2 grams. This means that when the concentration of penicillin reaches this level, the amount being infused is exactly counterbalanced by the amount being cleared by the kidneys.
Reaching this point is crucial because it signifies that the ideal therapeutic level is maintained. This prevents overdosing while ensuring that the concentration remains sufficient to be effective.

Stability of Equilibrium

The stability of an equilibrium point determines how a system behaves when it is disturbed from this state. A stable equilibrium means that if the system is slightly perturbed, it will return to equilibrium over time.
In our scenario, the equation\[ P'(t) = -2.5P(t) + 5 \]is analyzed for stability. For values of penicillin concentration, \(P\), greater than 2 grams, \( P' < 0 \) meaning the system will decrease back toward the equilibrium.
Similarly, for \(P\) values less than 2 grams, \( P' > 0 \), causing the concentration to rise towards equilibrium. This analysis reveals that the equilibrium at \(P=2\) is stable. This ensures that despite fluctuations in concentration, such as those resulting from changes in patient metabolism, the system will correct itself and stabilize around the equilibrium.
This understanding is essential in pharmacotherapy, underlining the importance of designing drug dosing regimens that ensure stability for patient safety.