Problem 71 - Calculus for Biology and Medicine

Question

Radioactive Decay Suppose \(W(t)\) denotes the amount of a radioactive material left after time \(t .\) Assume that \(W(0)=6\) and that $$ \frac{d W}{d t}=-3 W(t) $$ (a) How much material is left at time \(t=4 ?\) (b) What is the half-life of the material?

Step-by-Step Solution

Answer

(a) Near zero, as \( 6e^{-12} \) is negligible (b) Approximately 0.231 time units.

Step 1: Understanding the Problem

We need to find the amount of radioactive material left at time \( t = 4 \) given the differential equation \( \frac{dW}{dt} = -3W(t) \) with initial condition \( W(0) = 6 \). Additionally, we need to compute the half-life of the material.

Step 2: Solving the Differential Equation

The differential equation \( \frac{dW}{dt} = -3W \) is separable. We can write \( \frac{dW}{W} = -3 dt \). Integrating both sides gives \( \ln|W| = -3t + C \). Exponentiating both sides, we have \( W = e^{C}e^{-3t} \). Let \( C' = e^{C} \), so \( W = C'e^{-3t} \).

Step 3: Applying the Initial Condition

Using the initial condition \( W(0) = 6 \), substitute \( t = 0 \) into \( W = C'e^{-3t} \) to find \( W(0) = C' = 6 \). Thus, the solution is \( W(t) = 6e^{-3t} \).

Step 4: Finding the Amount at \(t = 4\)

Substitute \( t = 4 \) into \( W(t) = 6e^{-3t} \): \( W(4) = 6e^{-12} \). We need a calculator to evaluate \( e^{-12} \), which is a very small value, so \( W(4) \) is close to 0.

Step 5: Determining the Half-life

The half-life is the time \( t_h \) when \( W(t_h) = \frac{W_0}{2} = 3 \). Solving \( 6e^{-3t_h} = 3 \) gives \( e^{-3t_h} = \frac{1}{2} \). Taking the natural logarithm, \( -3t_h = \ln(0.5) \) or \( t_h = -\frac{\ln(0.5)}{3} \). This equals approximately \( 0.231 \).

Key Concepts

Differential Equations

Differential equations are a fundamental tool used to describe a wide range of physical phenomena, such as radioactive decay in our exercise. They involve equations that contain the derivatives of functions, which help us understand the rate of change of one variable with respect to another.
In our example, we see the differential equation \( \frac{dW}{dt} = -3W(t) \), which indicates that the rate of change of the amount of radioactive material \( W(t) \) is proportional to its current amount. The negative sign signifies that the material is decreasing over time, a common feature in decay processes.

• **Separable Equations:** This is a type of differential equation that can be rewritten such that all terms involving one variable are on one side, and all terms involving another variable are on the opposite side.
• **Integration:** Once separated, we integrate both sides to find the general solution to the equation.

For radioactive decay, understanding these principles allows us to predict how the material diminishes over time based on its initial quantity and decay rate.

Half-life

The concept of half-life is central to understanding radioactive decay. It represents the time required for half of a given amount of radioactive substance to decay or reduce to half its initial value. In the context of our problem, we determined the half-life by setting the amount \( W(t) \) to be half of the initial amount \( W_0 \).

• **Equation Formation:** Start by setting \( W(t) = \frac{W_0}{2} \) and solve for \( t \) using the derived equation \( W(t) = 6e^{-3t} \).
• **Natural Logarithm:** When solving \( 6e^{-3t} = 3 \), use the natural logarithm to solve for the time \( t_h \).

In our solution, applying these steps yields \( t_h = -\frac{\ln(0.5)}{3} \). This tells us precisely when half of the material will remain, emphasizing the exponential nature of decay processes characteristic of radioactive substances.

Exponential Decay

Exponential decay is a key concept describing processes where quantities decrease at a rate proportional to their current value. This is evident in the solution derived for our exercise \( W(t) = 6e^{-3t} \). Here, the formula shows that as time increases, the amount of material decreases exponentially.
This happens through several important features naturally encapsulated by the exponential function:

• **Constant Rate:** The decay occurs at a constant percentage rate, not a constant amount. For this problem, the rate is 3 times the current amount of material.
• **The Exponential Factor:** The negative exponent \(-3t\) represents how quickly the material decreases over time. The larger the magnitude, the faster the decay.

Understanding exponential decay provides insights into how radioactive materials behave over time, making it easier to predict future quantities based on current measurements. This mathematical model is frequently used in many fields, including physics, chemistry, and finance, highlighting its versatility.