Problem 17 - Calculus

Question

Morphine is often used as a pain-relieving drug. The half-life of morphine in the body is 2 hours. Suppose morphine is administered to a patient intravenously at a rate of 2.5 mg per hour, and the rate at which the morphine is eliminated is proportional to the amount present. (a) Use the half-life to show that, to three decimal places, the constant of proportionality for the rate at which morphine leaves the body (in mg/hour) is \(k=-0.347\). (b) Write a differential equation for the quantity, \(Q,\) of morphine in the blood after \(t\) hours. (c) Use the differential equation to find the equilibrium solution. (This is the long-term amount of morphine in the body, once the system has stabilized.)

Step-by-Step Solution

Answer

(a) The constant \( k \approx -0.347 \). (b) Differential equation: \( \frac{dQ}{dt} = 2.5 - 0.347Q \). (c) Equilibrium solution: \( Q \approx 7.207 \) mg.

Step 1: Understand the Half-life Formula

The half-life of a substance is the time taken for its concentration to reduce to half its initial value. The formula is given by \( t_{1/2} = \frac{\ln(2)}{k} \), where \( k \) is the decay constant. For morphine, the half-life \( t_{1/2} = 2 \) hours.

Step 2: Determine the Decay Constant

Rearrange the half-life formula to solve for \( k \): \( k = \frac{\ln(2)}{t_{1/2}} \). By substituting \( t_{1/2} = 2 \), we get \( k = \frac{\ln(2)}{2} \). Compute this to find \( k \approx -0.347 \).

Step 3: Formulate the Differential Equation

The change in morphine, \( \frac{dQ}{dt} \), is the difference between the administration rate and the elimination rate. This can be written as: \( \frac{dQ}{dt} = 2.5 - kQ \), substituting \( k \approx -0.347 \), the equation becomes \( \frac{dQ}{dt} = 2.5 + 0.347Q \).

Step 4: Solve for the Equilibrium Solution

The equilibrium solution occurs when \( \frac{dQ}{dt} = 0 \). So, set \( 2.5 + 0.347Q = 0 \) and solve for \( Q \). This gives \( Q = -\frac{2.5}{0.347} \approx -7.206 \) mg. This does not make sense as the quantity of a substance cannot be negative, indicating an error. Reconsider the signs used in elimination and administration equations to ensure positivity of solutions.

Step 5: Correct the Differential Equation with Proper Signs

Reevaluate step 3, acknowledging that \( k \) should be positive in context so corpus decay is expressed similarly. Set the equation right: \( \frac{dQ}{dt} = 2.5 - 0.347Q \) by confirming signs by context of elimination rate implying \( Q \rightarrow + \)

Step 6: Compute a Valid Equilibrium Solution

Using the correct differential equation \( \frac{dQ}{dt} = 2.5 - 0.347Q \), set \( 0 = 2.5 - 0.347Q \). Solving for \( Q \), we get \( Q = \frac{2.5}{0.347} \approx 7.207 \) mg, ensuring a plausible physiologic condition.

Key Concepts

Half-life Calculation

Half-life is a fundamental concept in understanding the behavior of substances as they undergo decay or elimination. In the context of the given exercise, the half-life of morphine is 2 hours. This means that every 2 hours, the concentration of morphine in the body reduces by half. The half-life can be crucial in determining how quickly a drug exits the system.
To find the decay constant, which helps measure the rate of elimination, we use the half-life formula:

• The formula: \( t_{1/2} = \frac{\ln(2)}{k} \).
• Where \( t_{1/2} \) is the half-life, and \( k \) is the decay constant.

To find \( k \), rearrange the formula: \( k = \frac{\ln(2)}{t_{1/2}} \).
Substitute the half-life of morphine (2 hours) into the equation: \( k = \frac{\ln(2)}{2} \). This calculates to approximately \( k = -0.347 \), indicating the rate at which morphine is removed from the body. Understanding this decay constant helps predict the pharmacokinetics of the drug in real-world scenarios.

Decay Constant

The decay constant, \( k \), plays a pivotal role in quantifying the rate at which morphine is metabolized and cleared from the body. It is expressed in terms of inverse time since its value indicates how quickly morphine concentrations decrease over time.
In the context of decay, the decay constant represents the proportionality factor in the differential equation of morphine's elimination:

• \( \frac{dQ}{dt} = -kQ \), where \( Q \) is the quantity of morphine present and \( \frac{dQ}{dt} \) is the rate of change over time.

Calculating the decay constant using the half-life formula gives \( k \approx -0.347 \). This means that every hour, approximately 34.7% of the morphine remaining in the body is eliminated.
It is important to note that in practical applications, the decay constant provides insight into dosing regimens and helps determine appropriate intervals for administering medication to maintain effective therapeutic levels without causing toxicity.

Equilibrium Solution

An equilibrium solution in a differential equation context refers to a stable state where the system no longer changes because the input and output rates are balanced. For morphine administration, the equilibrium is reached when the rate at which the drug is administered equals the rate of elimination.
To find the equilibrium solution, we use the differential equation \( \frac{dQ}{dt} = 2.5 - 0.347Q \). Setting this equal to zero (as change ceases at equilibrium), we solve:

• \( 0 = 2.5 - 0.347Q \)
• Solve for \( Q \): \( Q = \frac{2.5}{0.347} \approx 7.207 \) mg.

This indicates that over time, the morphine levels will stabilize around 7.207 mg under continuous intravenous administration at a rate of 2.5 mg per hour.
This equilibrium value is essential for medical professionals to ensure it aligns with therapeutic thresholds, preventing overdose while maintaining analgesic efficacy.