Problem 73 - Calculus for Biology and Medicine

Question

Radioactive Decay Suppose \(W(t)\) denotes the amount of a radioactive material left after time \(t .\) Assume that \(W(0)=5\) and \(W(1)=2\). (a) Find the differential equation that describes this situation. (b) How much material is left at time \(t=3\) ? (c) What is the half-life of the material?

Step-by-Step Solution

Answer

The differential equation is \( \frac{dW}{dt} = \ln\left(\frac{2}{5}\right) W \). At \( t=3 \), \( W(3) \approx 78.125 \). The half-life is \( \frac{\ln(2)}{-\ln\left(\frac{2}{5}\right)} \).

Step 1: Identify the Exponential Decay Model

The situation described is typically modeled using exponential decay. The formula for exponential decay is generally given by \( W(t) = W_0 e^{-kt} \), where \( W_0 \) is the initial quantity and \( k \) is the decay constant.

Step 2: Establish Initial Conditions

We are given that \( W(0) = 5 \). Thus, from the decay model, \( W(0) = W_0 e^{-k \cdot 0} = W_0 \). Therefore, \( W_0 = 5 \).

Step 3: Use Second Condition to Find Decay Constant

It's given that \( W(1) = 2 \). Using \( W(t) = W_0 e^{-kt} \), substitute the known values: \( 2 = 5 e^{-k \cdot 1} \). This simplifies to \( e^{-k} = \frac{2}{5} \). Taking the natural log of both sides, we get \( -k = \ln\left(\frac{2}{5}\right) \). Thus, \( k = -\ln\left(\frac{2}{5}\right) \).

Step 4: Write the Differential Equation

The corresponding differential equation for exponential decay is \( \frac{dW}{dt} = -kW \). With \( k = -\ln\left(\frac{2}{5}\right) \), the equation becomes \( \frac{dW}{dt} = \ln\left(\frac{2}{5}\right) W \).

Step 5: Compute Amount Left at Time t=3

Using the decay model \( W(t) = 5 e^{-kt} \) and substituting \( t = 3 \) and \( k = -\ln\left(\frac{2}{5}\right) \), we find:\[W(3) = 5 e^{3 \ln\left(\frac{5}{2}\right)} \].Calculate \( e^{3 \ln\left(\frac{5}{2}\right)} \) which equals \( \left(\frac{5}{2}\right)^3 \). Thus,\[W(3) = 5 \times \left(\frac{125}{8}\right) \approx 78.125\].

Step 6: Determine the Half-Life

The half-life \( t_{1/2} \) is the time it takes for the material to reduce to half its initial amount. Since \( W_0 = 5 \), find \( t_{1/2} \) such that \( W(t_{1/2}) = 2.5 \). Substituting into the decay model:\[2.5 = 5 e^{-kt_{1/2}} \Rightarrow \frac{1}{2} = e^{-kt_{1/2}}\]Taking the natural log, \( -kt_{1/2} = \ln\left(\frac{1}{2}\right) \). Thus, \( t_{1/2} = \frac{\ln(2)}{k} = \frac{\ln(2)}{-\ln\left(\frac{2}{5}\right)} \).

Key Concepts

Understanding Differential Equations

Differential equations are mathematical tools used to describe how a certain quantity changes over time. In the context of radioactive decay, we use differential equations to express how the quantity of a radioactive material decreases as time progresses. A differential equation typically involves a derivative, which signifies the rate of change of a function with respect to one of its variables.

For radioactive decay, we often use the equation \( \frac{dW}{dt} = -kW \). This equation tells us that the rate at which the material decays is proportional to the current amount of material present \( W \). Here, \( k \) is a positive constant specific to the material, known as the decay constant.

• The negative sign indicates a decrease in the amount of material over time.
• The function \( W \) represents the amount of material at any given time \( t \).

Through solving the differential equation, we can derive the equation for exponential decay, providing us insights into how the material behaves over time.

The Nature of Exponential Decay

Exponential decay describes a process where quantities decrease at a rate proportional to their current value. This is an essential concept in understanding radioactive materials. The primary equation for exponential decay is \( W(t) = W_0 e^{-kt} \). Let’s break this down:

• \( W(t) \) is the quantity of the material at time \( t \).
• \( W_0 \) is the initial amount of material at \( t = 0 \).
• \( k \) is the decay constant, a specific rate at which the material decays.
• The \( e^{-kt} \) term represents the exponential decrease.

The elegance of exponential decay lies in its ability to predict the remaining quantity of a substance over time. As time passes, the amount of material gets smaller and smaller, never quite reaching zero. For radioactive decay problems like the one in the original exercise, understanding the precepts of exponential decay allows us to calculate the remaining material at any given time.

Exploring the Concept of Half-Life

The half-life of a material is a crucial concept in radioactive decay. It is the time required for a radioactive substance to lose half of its initial quantity. Knowing the half-life helps predict how quickly a material will diminish.

To find the half-life, set \( W(t_{1/2}) \) to be half of \( W_0 \). For instance, in our exercise, since the initial amount \( W_0 \) is 5, the equation becomes \[ 2.5 = 5 e^{-kt_{1/2}}. \] Solving this equation gives us the half-life:

• Cancel down \(\frac{1}{2} = e^{-kt_{1/2}}\).
  
• Taking the natural log of both sides gives \( -kt_{1/2} = \ln\left(\frac{1}{2}\right) \).
  
• The half-life \( t_{1/2} \) is then \( \frac{\ln(2)}{k} \).

This formula allows us to determine how long it takes for half of the material to decay. An integral part of decay analysis, the half-life offers a tangible view of how fast or slow the decay process is happening.