Problem 72 - Calculus for Biology and Medicine

Question

Radioactive Decay Suppose \(W(t)\) denotes the amount of a radioactive material left after time \(t .\) Assume that \(W(0)=10\) and \(W(1)=8\) (a) Find the differential equation that describes this situation. (b) How much material is left at time \(t=5 ?\) (c) What is the half-life of the material?

Step-by-Step Solution

Answer

(a) \( \frac{dW}{dt} = -0.223W \); (b) \( W(5) \approx 3.68 \); (c) Half-life is approximately 3.11.

Step 1: Understand the Problem

We need to model radioactive decay, which follows an exponential decay law. We are given that \( W(0) = 10 \) and \( W(1) = 8 \). Our goal is to find a differential equation, determine how much material is left at \( t = 5 \), and calculate the half-life.

Step 2: Formulate the Differential Equation

The differential equation for radioactive decay is \( \frac{dW}{dt} = -kW \), where \( k \) is the decay constant. This equation signifies that the rate of decay of the material is proportional to the amount of substance present.

Step 3: Solve the Differential Equation

The general solution of \( \frac{dW}{dt} = -kW \) is \( W(t) = W_0 e^{-kt} \), where \( W_0 \) is the initial amount. Given \( W(0) = 10 \), we have \( W(t) = 10e^{-kt} \).

Step 4: Determine the Decay Constant \( k \)

Using \( W(1) = 8 \), substitute into the formula: \( 8 = 10e^{-k} \). Solving for \( k \), we get \( e^{-k} = \frac{8}{10} = 0.8 \). Thus \( -k = \ln(0.8) \) or \( k = -\ln(0.8) \approx 0.223 \).

Step 5: Calculate Material Left at \( t = 5 \)

Insert \( t = 5 \) into \( W(t) = 10e^{-0.223t} \): \( W(5) = 10e^{-0.223 \times 5} \). Evaluating gives \( W(5) = 10 \times 0.3679 \approx 3.68 \).

Step 6: Calculate the Half-Life

The half-life \( t_{1/2} \) satisfies \( \frac{W_0}{2} = W_0 e^{-kt_{1/2}} \). Simplifying gives \( e^{-kt_{1/2}} = \frac{1}{2} \). Solving \( -kt_{1/2} = \ln(0.5) \), we find \( t_{1/2} = \frac{\ln(0.5)}{-\ln(0.8)} \approx \frac{0.693}{0.223} \approx 3.11 \).

Key Concepts

Understanding Differential Equations in Radioactive Decay

A differential equation is a mathematical equation that involves derivatives of a function. It describes how the quantity of something changes with respect to another variable, often time. In the context of radioactive decay, the differential equation describes how the amount of radioactive material decreases over time. The general form used for radioactive decay is \[ \frac{dW}{dt} = -kW \]where:

• \( W \) is the amount of radioactive material present at time \( t \)
• \( k \) is the decay constant, a positive number indicating how quickly the substance decays

This equation captures the principle that the rate of decay (how fast the material is reducing) is proportional to the current amount of material. Essentially, more material equals a faster decay rate. In our problem, solving the equation involves finding a function that models the amount of material, \( W(t) \), as it decreases over time.

Exploring Exponential Decay

Exponential decay refers to quantities decreasing at a rate proportional to their current value. The solution to the differential equation in radioactive decay takes the form:\[ W(t) = W_0 e^{-kt} \]where:

• \( W_0 \) is the initial amount of the substance
• \( e \) is the mathematical constant approximately equal to 2.71828
• \( k \) is the decay constant, which dictates the decay rate

This formula represents the amount of radioactive material at time \( t \). In our specific example, starting with \( W_0 = 10 \), the formula becomes \( W(t) = 10e^{-kt} \). When evaluating this expression, using given data points allows us to find the decay constant \( k \). Once \( k \) is known, we can predict the remaining material at any future time, such as \( t = 5 \) in the exercise.

The Concept of Half-Life

Half-life is a key characteristic of radioactive material, representing the time it takes for half of the substance to decay. It provides a straightforward way to understand how quickly a material undergoes radioactive decay. The mathematical definition involves rearranging the exponential decay formula to find when the initial amount has halved:\[ \frac{W_0}{2} = W_0 e^{-kt_{1/2}} \]After canceling \( W_0 \) and simplifying, we have:\[ e^{-kt_{1/2}} = \frac{1}{2} \]By solving \( -kt_{1/2} = \ln(0.5) \), we find the expression for the half-life:\[ t_{1/2} = \frac{\ln(0.5)}{-k} \]In our exercise with \( k \approx 0.223 \), the half-life calculation yields \( t_{1/2} \approx 3.11 \). This value offers insight into the speed of decay, making it tangible by measuring how long it takes for significant reduction in substance amount.