Problem 70 - Calculus for Biology and Medicine

Question

Radioactive Decay Suppose \(W(t)\) denotes the amount of a radioactive material left after time \(t\). Assume that \(W(0)=15\) and that $$ \frac{d W}{d t}=-2 W(t) $$ (a) How much material is left at time \(t=2 ?\) (b) What is the half-life of this material?

Step-by-Step Solution

Answer

(a) \( W(2) = 15e^{-4} \); (b) Half-life is \( \frac{\ln(2)}{2} \) units of time.

Step 1: Review the Differential Equation

The given differential equation is \( \frac{dW}{dt} = -2W(t) \). This represents an exponential decay process, where the rate of change of the material amount \( W(t) \) is proportional to \( W(t) \) itself with a constant factor \(-2\).

Step 2: Solve the Differential Equation

To solve \( \frac{dW}{dt} = -2W(t) \), we can separate variables by rewriting it as \( \frac{1}{W}dW = -2dt \) and integrating both sides. The left side integrates to \( \ln |W| \) and the right side integrates to \( -2t + C \), where \( C \) is a constant of integration.

Step 3: Solve for W(t)

Exponentiate both sides to solve for \( W(t) \): \( e^{\ln |W|} = e^{-2t + C} \). This simplifies to \( W(t) = e^{C}e^{-2t} \), or \( W(t) = Ce^{-2t} \) where \( C \) is another constant.

Step 4: Apply Initial Condition

Use the initial condition \( W(0) = 15 \) to find \( C \). Substitute \( t = 0 \): \( W(0) = Ce^{0} = C = 15 \). Hence, the function becomes \( W(t) = 15e^{-2t} \).

Step 5: Calculate W(2)

Substitute \( t = 2 \) into \( W(t) = 15e^{-2t} \). This gives \( W(2) = 15e^{-4} \). Calculate \( e^{-4} \) to find the amount of material left at \( t = 2 \).

Step 6: Determine the Half-Life

The half-life is the time \( t_{1/2} \) when \( W(t_{1/2}) = \frac{15}{2} \). Substitute into \( 15e^{-2t_{1/2}} = \frac{15}{2} \), simplify to \( e^{-2t_{1/2}} = \frac{1}{2} \) and solve for \( t_{1/2} \) using logarithms: \( -2t_{1/2} = \ln \left( \frac{1}{2} \right) \), so \( t_{1/2} = \frac{\ln(2)}{2} \).

Key Concepts

Differential Equation

The concept of differential equations plays a crucial role in understanding many natural phenomena, including radioactive decay. A differential equation is a mathematical expression that relates a function with its derivatives. In simpler terms, it shows how one variable changes with respect to another.
In the context of radioactive decay, the differential equation given is \( \frac{dW}{dt} = -2W(t) \). This tells us how the quantity of radioactive material \( W(t) \) decreases over time.
This specific equation indicates that the rate of change of \( W(t) \) is proportional to its current amount, characterized by the negative constant

• \( dW/dt \) represents the change in material over time
• \(-2 \) is the rate constant, determining how quickly the material decays
• \( W(t) \) is the amount of radioactive material at time \( t \)

Understanding such differential equations helps us model real-world processes mathematically, predicting how substances decay or transform over time.

Exponential Decay

Exponential decay describes a process where the quantity decreases at a rate proportional to its current value. In simpler terms, the material reduces rapidly at first and then slows down over time.
Given the equation \( W(t) = Ce^{-2t} \), we identify this is an exponential decay pattern:

• \( C \) is an initial constant, representing the initial amount of material
• \( e^{-2t} \) shows how the material decreases over time due to decay

This pattern is characterized by a rapid decline followed by a gentler slope as time progresses. The exponential decay equation is essential for understanding half-life and how much material remains over time.
It's a fundamental concept in understanding why radioactive materials don't decay all at once but rather over many half-lives.

Half-Life

Half-life is a concept representing the time required for a quantity to reduce to half its initial amount through exponential decay. For radioactive substances, it's a crucial measure of how quickly they lose their radioactive properties.
With our previous function \( W(t) = 15e^{-2t} \), the half-life \(t_{1/2}\) is determined:

• Set \( W(t_{1/2}) = \frac{15}{2} \), halving the initial condition
• Solve by calculating when \( e^{-2t_{1/2}} = \frac{1}{2} \)

Rearranging gives \( t_{1/2} = \frac{\ln(2)}{2} \), meaning the time for the material to reach half its original amount depends on the natural logarithm. This measure helps us understand the longevity and hazard duration of radioactive substances.

Integration

Integration is a mathematical technique used to find the original function from its derivative or rate of change. It is the reverse process of differentiation. In the context of our equation, integrating helps solve differential equations.
For the decaying material, we separate variables and express the equation as \( \frac{1}{W}dW = -2dt \). Through integration:

• The left side becomes \( \ln |W| \)
• The right side becomes \( -2t + C \)

The constant \( C \) is of particular importance as it helps refine our model based on the initial conditions. Integration ensures that our predictions about how much radioactive material remains are accurate by expressing them in terms of time.

Initial Condition

Initial conditions are critical for solving differential equations as they provide an exact starting point for our calculations. They allow us to find the specific solution relevant to our problem context.
Given \( W(0) = 15 \), it identifies the amount of material at the beginning, allowing us to solve for the constant \( C \) in the equation \( W(t) = Ce^{-2t} \).
These constraints enable us to calculate future values accurately:

• Set \( W(0) \) as the known quantity, \( 15 \) in our example
• This makes \( C = 15 \), establishing the function \( W(t) = 15e^{-2t} \) which governs our decay process

Using initial conditions helps tailor general solutions of equations to specific scenarios, ensuring our mathematical predictions match reality.