Problem 21 - Calculus

Question

Warfarin is a drug used as an anticoagulant. After administration of the drug is stopped, the quantity remaining in a patient's body decreases at a rate proportional to the quantity remaining. The half-life of warfarin in the body is 37 hours. (a) Sketch the quantity, \(Q,\) of warfarin in a patient's body as a function of the time, \(t,\) since stopping administration of the drug. Mark the 37 hours on your graph. (b) Write a differential equation satisfied by \(Q\) (c) How many days does it take for the drug level in the body to be reduced to \(25 \%\) of the original level?

Step-by-Step Solution

Answer

(a) Exponential decay graph with half-life at 37 hrs. (b) \( \frac{dQ}{dt} = -kQ \). (c) About 3.08 days.

Step 1: Understanding the Problem

We know that the quantity of warfarin decreases exponentially, meaning its rate of change is proportional to the quantity remaining. The half-life is 37 hours, which means after 37 hours, the quantity of warfarin in the body will be half of what it was initially.

Step 2: Sketching the Graph for Part (a)

The graph of the quantity, \( Q(t) \), of warfarin over time will be an exponential decay curve. Plot \( Q(t) \) on the y-axis and time \( t \) on the x-axis. Start with \( Q(0) \) as the initial quantity. At \( t = 37 \) hours, mark the point where the quantity is half of the initial value \( Q(0)/2 \). The curve will approach zero as time increases.

Step 3: Writing the Differential Equation for Part (b)

Since the rate of change of \( Q \) is proportional to \( Q \), the differential equation can be written as: \[ \frac{dQ}{dt} = -kQ \] where \( k \) is the proportionality constant.

Step 4: Solve for the Constant \( k \)

To find \( k \), use the half-life information. From the half-life formula, \[ Q(t) = Q_0 e^{-kt} \]and the condition that \( Q(37) = \frac{Q_0}{2} \). Substituting into the equation gives: \[ \frac{Q_0}{2} = Q_0 e^{-37k} \]Solving for \( k \) yields:\[ e^{-37k} = \frac{1}{2} \] \[ -37k = \ln\left(\frac{1}{2}\right) \] \[ k = -\frac{\ln(0.5)}{37} \]

Step 5: Calculating Time for 25% Reduction for Part (c)

We want \( Q(t) = \frac{1}{4}Q_0 \). Using the formula \[ Q(t) = Q_0 e^{-kt} \] substitute \( Q(t) = \frac{1}{4}Q_0 \) and solve:\[ \frac{1}{4}Q_0 = Q_0 e^{-kt} \] Cancel \( Q_0 \):\[ \frac{1}{4} = e^{-kt} \] Take logarithm:\[ -kt = \ln\left(\frac{1}{4}\right) \] Substitute \( k = -\frac{\ln(0.5)}{37} \): \[ t = \frac{\ln(4)}{\ln(2)/37} \] Calculate \( t \) to get \( t = 74 \) hours. Convert hours to days: \[ \frac{74}{24} \approx 3.08 \text{ days} \].

Key Concepts

Differential Equations

A differential equation is a mathematical equation that relates a function with its derivatives. It describes how a quantity changes over time and is a powerful tool in modeling various phenomena. In this exercise, the differential equation relates to the decay of warfarin in the human body. This is expressed as:\[ \frac{dQ}{dt} = -kQ \]where:

• \( \frac{dQ}{dt} \) is the rate of change of the quantity \( Q \) of warfarin.
• \( k \) is the proportionality constant, determining how quickly the drug decays.
• \( -kQ \) implies the decay is exponential, meaning as time increases, the amount of drug decreases at a rate proportional to its current amount.

Such equations are typically solved by separating variables or using integration techniques. Understanding this concept is crucial as it applies to many physical and biological processes beyond pharmaceuticals, like population dynamics and radioactive decay.

Half-Life

Half-life is a term that describes the time it takes for a substance to reduce to half its initial amount. In this context, it reflects how quickly the drug warfarin is eliminated from the human body. For warfarin, the half-life is 37 hours. This means that every 37 hours, the quantity of warfarin is halved.The relationship between half-life and exponential decay can be expressed using the formula:\[ Q(t) = Q_0 e^{-kt} \]where \( Q(t) \) is the quantity at time \( t \), and \( Q_0 \) is the initial quantity. When \( t = 37 \), we observe that:\[ Q(37) = \frac{Q_0}{2} = Q_0 e^{-37k} \]Solving for \( k \), we use the relationship:\[ e^{-37k} = \frac{1}{2} \]Half-life is a fundamental concept in understanding how substances decay over time, and it is widely used in fields such as chemistry, geology, and medicine.

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. They are characterized by their rapid growth or decay. The function representing the decay of warfarin is an exponential function:\[ Q(t) = Q_0 e^{-kt} \]Key characteristics of exponential decay:

• The base of the exponential function \( e \) is a mathematical constant approximately equal to 2.718.
• Exponential decay implies that the rate of change decreases over time, depicted by a decaying curve in a graph.
• This curve starts at \( Q_0 \) and asymptotically approaches zero, never quite reaching it.

Exponential functions elegantly describe how quantities decrease in contexts like radioactive decay, cooling processes, and the breakdown of pharmaceutical drugs. Understanding these functions helps in visualizing natural phenomena and solving real-world problems efficiently.