Problem 17 - Calculus Concepts

Question

Medicine The rate of change with respect to time of the quantity \(q\) of pain reliever in a person's body \(t\) hours after the individual takes the medication is proportional to the quantity of medication remaining. Assume that 2 hours after a person takes 200 milligrams of a pain reliever, onehalf of the original dose remains. a. Write a differential equation for the rate of change of the quantity of pain reliever in the body. b. Find a particular solution for this differential equation. c. How much pain reliever will remain after 4 hours; after 8 hours?

Step-by-Step Solution

Answer

After 4 hours, 50 mg remains; after 8 hours, 12.5 mg remains.

Step 1: Formulate the Differential Equation

The problem states that the rate of change of the quantity of pain reliever, \( q(t) \), with respect to time, \( t \), is proportional to the quantity remaining. Mathematically, this is expressed as \( \frac{dq}{dt} = -kq \), where \( k > 0 \) is the constant of proportionality.

Step 2: Solve the Differential Equation

This is a separable differential equation. We rewrite it as \( \frac{1}{q} \frac{dq}{dt} = -k \), integrate both sides: \(\int \frac{1}{q} \, dq = -k \int dt \), resulting in \( \ln |q| = -kt + C \). Solving for \( q \), we have \( q(t) = Ce^{-kt} \), where \( C \) is the integration constant.

Step 3: Determine the Constant C Using Initial Conditions

We know that initially, \( q(0) = 200 \). Thus, \( 200 = C \cdot e^{0} \) implies \( C = 200 \). So the solution becomes \( q(t) = 200e^{-kt} \).

Step 4: Use Given Condition to Find the Rate Constant k

It's given that after 2 hours, half of the original dose remains, so \( q(2) = 100 \). Substituting, \( 100 = 200e^{-2k} \), solving for \( k \), we get \( e^{-2k} = 0.5 \). Taking the natural logarithm, \( -2k = \ln(0.5) \), resulting in \( k = -\frac{1}{2} \ln(0.5) \).

Step 5: Evaluate for 4 and 8 Hours

Substitute \( k = -\frac{1}{2} \ln(0.5) \) into the model \( q(t) = 200e^{-kt} \). For \( t = 4 \) hours, \( q(4) = 200 \times e^{-2k} \), which simplifies to 50 mg. For \( t = 8 \) hours, \( q(8) = 200 \times e^{-4k} \), which simplifies to 12.5 mg.

Key Concepts

Exponential Decay

Exponential decay is a mathematical process where quantities decrease over time at a rate proportional to their current value. This is seen in the way the pain reliever dissolves in the body. As described in the medicine exercise, the reliever's concentration diminishes in a similar fashion. The rate of change is captured by the equation \( \frac{dq}{dt} = -kq \), reflecting how each moment, the remaining quantity decreases in proportion to its current amount.
Exponential decay occurs often in nature and various scientific fields:

• Radioactive decay, where unstable atoms lose particles.
• Cooling of a hot object, where heat escapes to the cooler surroundings.
• The discharge of a capacitor, showing how voltage decreases over time.

By understanding exponential decay, we learn how substances like medicine release their effect over time, and why careful timing of dosages is crucial.

Pharmacokinetics

Pharmacokinetics is the study of how drugs move through the body over time. It covers how substances enter, circulate, and leave the system. This exercise mimics a real-world pharmacokinetic scenario with a pain reliever's concentration decreasing exponentially.
Key processes involved in pharmacokinetics include:

• Absorption: How the drug enters the bloodstream.
• Distribution: How it spreads through body tissues.
• Metabolism: The chemical alteration of the drug.
• Excretion: The elimination of the drug from the body.

In the exercise, the focus is on the rate at which the medication depletes, showcasing the elimination aspect. Understanding pharmacokinetics can inform medical professionals about dose adjustments necessary for safe and effective treatment.

Separable Differential Equations

A separable differential equation is a common type of differential equation that can be split into two functions, each in one variable only. Solving involves integrating these two separate parts.
In the case presented, the differential equation \( \frac{dq}{dt} = -kq \) is separable. The steps to solve it included separating the variables, integrating, and then rearranging to find the solution. Steps are as follows:

• Rearrange: \( \frac{1}{q} \frac{dq}{dt} = -k \)
• Integrate both sides: \( \int \frac{1}{q} \, dq = -k \int dt \)
• Solve: Resulting in \( \ln|q| = -kt + C \)
• Exponentiate to remove \( \ln \): \( q(t) = Ce^{-kt} \)

Separable differential equations like this one allow for simple solutions in exponential models, exemplifying how basic calculus techniques apply to real-world problems.