Problem 38 - Calculus Concepts

Question

Technetium-99 is a radioisotope that has been used in humans to help doctors locate possible malignant tumors. Radioisotopes decay (over time) at a rate described by the differential equation $$ \frac{d s}{d t}=k s $$ where \(s\) is the amount of the radioisotope and \(t\) is time. Technetium-99 has a half-life of 210,000 years. Assume that 0.1 milligram of technetium-99 is injected into a person's bloodstream. a. Write a differential equation for the rate at which the amount of technetium-99 decays. b. Find a particular solution for this differential equation.

Step-by-Step Solution

Answer

(a) \(\frac{d s}{d t} = -\frac{\ln(2)}{210,000} s\); (b) \(s(t) = 0.1 e^{-\frac{\ln(2)}{210,000} t}\)."

Step 1: Initial Setup of the Problem

We are given a differential equation \(\frac{d s}{d t}=k s\) where \(s\) is the amount of technetium-99 and \(k\) is the decay constant. Technetium-99 has a half-life of 210,000 years, which will be used to find the value of \(k\).

Step 2: Relate Half-Life to Decay Constant

The half-life of a substance is the time it takes for half of it to decay. Using the formula \(\frac{s_0}{2} = s_0 e^{k t_{1/2}}\), where \(s_0\) is the initial amount and \(t_{1/2}\) is the half-life, we have \(\frac{1}{2} = e^{k \cdot 210,000}\). Solving for \(k\), we take the natural logarithm on both sides: \(\ln\left(\frac{1}{2}\right) = k \cdot 210,000\).

Step 3: Solve for Decay Constant k

Using the equation from Step 2, we find \(k\) as follows: \[k = \frac{\ln\left(\frac{1}{2}\right)}{210,000} = -\frac{\ln(2)}{210,000}.\]

Step 4: Write the Differential Equation

Substituting the value of \(k\) from Step 3 into the original differential equation, we have: \[\frac{d s}{d t} = -\frac{\ln(2)}{210,000} s.\] This is the differential equation describing the rate of decay of technetium-99.

Step 5: Find the General Solution

The differential equation is separable and can be written as \(\frac{d s}{s} = -\frac{\ln(2)}{210,000} dt\). Integrating both sides gives \(\ln |s| = -\frac{\ln(2)}{210,000} t + C\), where \(C\) is a constant of integration.

Step 6: Find the Particular Solution

Exponentiate both sides to solve for \(s\): \(s = e^C e^{-\frac{\ln(2)}{210,000} t}\). Letting \(C_1 = e^C\), we write \(s = C_1 e^{-\frac{\ln(2)}{210,000} t}\). Given \(s = 0.1\) mg at \(t = 0\), we find \(C_1 = 0.1\). Therefore, the particular solution is: \[s(t) = 0.1 e^{-\frac{\ln(2)}{210,000} t}.\]

Step 7: Conclusion

The equation \(\frac{d s}{d t} = -\frac{\ln(2)}{210,000} s\) describes the decay process, and the particular solution is \(s(t) = 0.1 e^{-\frac{\ln(2)}{210,000} t}\), given the initial condition that 0.1 mg of technetium-99 is present at \(t = 0\).

Key Concepts

Radioactive Decay

Radioactive decay is a natural process by which a radioactive isotope, like technetium-99, loses its radioactivity over time. This occurs as unstable atomic nuclei release particles or electromagnetic radiation. In this context, the amount of an isotope decreases exponentially, and the rate of decline is governed by a differential equation. For example, technetium-99 decays according to the equation \( \frac{ds}{dt} = ks \), where \( s \) is the amount of the isotope, \( t \) is time, and \( k \) is a decay constant.

The decay constant \( k \) is negative, indicating a decrease in the amount over time. This constant determines how quickly the material will lose its radioactivity. Finding the exact value of \( k \) is crucial for predicting the decay behavior of any radioactive substance.

This process is predictable and constant, making it useful for applications like medical imaging, where technetium-99 is instrumental in diagnostics.

Half-Life Calculation

The half-life of a radioactive substance is the period it takes for half of the material to decay. For technetium-99, this half-life is 210,000 years. This concept is vital because it provides a timescale over which exponential decay can be quantitatively understood.

To find the half-life of a substance like technetium-99 in mathematical terms, we utilize the relationship \( \frac{s_0}{2} = s_0 e^{kt_{1/2}} \). By dividing both sides of this equation by \( s_0 \) and taking the natural logarithm, we can solve for the decay constant \( k \): \( \ln\left(\frac{1}{2}\right) = kt_{1/2} \). This simplifies to \( k = \frac{\ln\left(\frac{1}{2}\right)}{t_{1/2}} \), leading to \( k = -\frac{\ln(2)}{210,000} \) for our example.

This calculated \( k \) can be used in any related differential equations to predict the rate of decay over time accurately. The constant \( k \), along with the knowledge of the initial quantity, allows us to model how the isotope's quantity decreases.

Separable Differential Equations

A differential equation is separable if it can be rewritten so that all terms involving one variable are on one side and all terms involving the other variable are on the other. This approach is particularly useful when dealing with decay equations like \( \frac{ds}{dt} = ks \).

In the case of technetium-99, we rearrange this equation into \( \frac{ds}{s} = k \, dt \). By integrating both sides, we effectively separate the variables and solve the equation as two independent integrals:

• Left side: \( \int \frac{ds}{s} = \ln|s| + C \)
• Right side: \( \int k \, dt = kt \)

The constant \( C \) comes from the integration process.

Exponentiating both sides results in the general solution \( s = C_1 e^{kt} \). Here, \( C_1 \) is a constant that accounts for the initial conditions. By using the provided initial conditions, such as \( s = 0.1 \) mg at \( t = 0 \), we can solve for \( C_1 \), giving us a particular solution: \( s(t) = 0.1 e^{-\frac{\ln(2)}{210,000} t} \).

This separable approach is powerful for solving many forms of differential equations, especially in modeling decay processes.