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) \(\frac{dW}{dt} = -kW\); (b) \(W(5) \approx 3.275\); (c) half-life \(\approx 3.106\)

Step 1: Identify the basic exponential decay model

In radioactive decay, the amount of substance at any time \(t\), denoted as \(W(t)\), can typically be modeled by the equation \(W(t) = W(0) e^{-kt}\), where \(k\) is the decay constant. We start with the known condition \(W(0) = 10\).

Step 2: Formulate the Initial Differential Equation

To find the differential equation, recognize that the rate of change of \(W(t)\) with respect to time is proportional to \(W(t)\) itself. This can be written as \(\frac{dW}{dt} = -kW\).

Step 3: Use Given Information to Find the Decay Constant \(k\)

We know \(W(1) = 8\). Substitute into the decay model: \(8 = 10 e^{-k \cdot 1}\). Solve for \(k\):\[ \ln\left(\frac{8}{10}\right) = -k \Rightarrow k = -\ln\left(\frac{4}{5}\right) \approx 0.2231 \]

Step 4: Calculate the Amount at \(t=5\)

Using the value of \(k\) found, substitute \(t = 5\) into the decay formula:\[ W(5) = 10 e^{-0.2231 \times 5} \approx 10 e^{-1.1155} \approx 3.275 \]

Step 5: Determine the Half-Life of the Material

The half-life \(T_{1/2}\) is the time when \(W(t) = \frac{W(0)}{2} = 5\). Set up the equation: \(5 = 10 e^{-0.2231T_{1/2}}\).\[ \ln\left(\frac{1}{2}\right) = -0.2231 T_{1/2} \Rightarrow T_{1/2} = \frac{\ln(2)}{0.2231} \approx 3.106 \]

Step 6: Conclusion

We found the differential equation \(\frac{dW}{dt} = -kW\), calculated \(W(5) \approx 3.275\), and the half-life \(T_{1/2} \approx 3.106\).

Key Concepts

Differential Equations in Radioactive Decay

In the study of radioactive decay, differential equations play a crucial role in describing how the quantity of radioactive material decreases over time. Specifically, the relationship between the rate of change of the material's amount, denoted as \( \frac{dW}{dt} \), and the remaining quantity of the material \( W(t) \) is expressed through a differential equation. This particular differential equation is:

• \( \frac{dW}{dt} = -kW \)

where:

• \( W(t) \) is the amount of radioactive material at time \( t \)
• \( k \) is the decay constant

The negative sign indicates a decrease over time, which aligns with the nature of decay. Essentially, this equation suggests that the rate at which a material decays is directly proportional to the amount present at any given time. With this model, we can predict the future amount of a substance given its initial quantity and the decay constant.

Understanding Exponential Decay

Exponential decay is a fundamental concept in describing how quantities decrease at a rate proportional to their current value. In the context of radioactive decay, this means the more substance you have, the faster it decays. The exponential decay model is expressed as:

• \( W(t) = W(0) e^{-kt} \)

where:

• \( W(0) \) is the initial amount of the material
• \( k \) is the decay constant
• \( t \) is the time elapsed

By substituting different values of \( t \) into this equation, you can determine how much material remains at that time. The key feature of exponential decay is its consistency; it ensures that the decay process continues steadily over time. For instance, if you know the amount of material at two points, such as \( W(0) \) and \( W(1) \), you can calculate the decay constant \( k \), which will enable you to predict the material's quantity at any other time using this model.

The Concept of Half-Life

The half-life of a radioactive material is the time it takes for half of the substance to decay. This is a useful measure as it provides a straightforward understanding of the lifecycle of the material. In mathematical terms, the half-life \( T_{1/2} \) is defined by the equation:

• \( W(T_{1/2}) = \frac{W(0)}{2} \)

Thus, substituting into the decay model:

• \( \frac{W(0)}{2} = W(0) e^{-kT_{1/2}} \)

Solving this yields:

• \( T_{1/2} = \frac{\ln(2)}{k} \)

This result shows that the half-life is inversely proportional to the decay constant \( k \). The larger the decay constant, the shorter the half-life, meaning the material decays faster. Understanding half-life is essential for practical applications, such as determining how long a radioactive material will last or planning safe storage and handling.