Problem 16 - Differential Equations and Linear Algebra

Question

Consider the phenomenon of exponential decay. This occurs when a population \(P(t)\) is governed by the differential equation $$\frac{d P}{d t}=k P$$ where \(k\) is a negative constant. Cobalt-60, an isotope used in cancer therapy, decays exponentially with a half-life of 5.2 years (i.e., half the original sample remains after 5.2 years). How long does it take for a sample of Cobalt-60 to disintegrate to the extent that only \(4 \%\) of the original amount remains?

Step-by-Step Solution

Answer

It takes approximately \(21.5\) years for a sample of Cobalt-60 to disintegrate to the extent that only \(4 \%\) of the original amount remains.

Step 1: Solve the differential equation and determine the decay function

To determine the decay function, we need to solve the given differential equation \(\frac{d P}{d t} = k P\). This is a first order, separable differential equation. Separate the variables and integrate: \[\frac{d P}{P} = k d t\] \[\int\frac{d P}{P} = \int k d t\] \[\ln |P| = k t + C\] Now, solve for P: \[P(t) = P_0 e^{k t}\] Here, \(P_0\) is the initial population, and \(P(t)\) represents the population at time t.

Step 2: Find the value of k using half-life

We know that the half-life of Cobalt-60 is \(5.2\) years, which means that after \(5.2\) years, only half the initial population remains. We can use this information to find the value of k in the decay function. At t = 5.2, P(5.2) = 0.5P_0: \[0.5 P_0 = P_0 e^{5.2k}\] Solve for k: \[\frac{1}{2} = e^{5.2k}\] Take the natural logarithm of both sides: \[\ln{\frac{1}{2}} = 5.2k\] Find the value of k: \[k = \frac{\ln{\frac{1}{2}}}{5.2}\]

Step 3: Find the time it takes for the population to reach 4% of the initial population

We want to find the time it takes for the population to reach \(4 \%\) of the initial population (0.04P_0). We can plug this into the decay function to find the time. \[0.04 P_0 = P_0 e^{kt}\] Solve for t: \[\frac{0.04}{1} = e^{kt}\] Take the natural logarithm of both sides: \[\ln{0.04} = k t\] Now, substitute the value of k from Step 2: \[t = \frac{\ln{0.04}}{k}\] \[t = \frac{\ln{0.04}}{\frac{\ln{\frac{1}{2}}}{5.2}}\] Calculate t: \[t \approx 21.5\, \text{years}\] #Conclusion#It takes approximately \(21.5\) years for a sample of Cobalt-60 to disintegrate to the extent that only \(4 \%\) of the original amount remains.

Key Concepts

First Order Separable Differential Equations

Understanding first-order separable differential equations is crucial for many applications in physics, biology, economics, and more. These equations can describe how things change over time, and in our case, they are used to model the decay of radioactive substances like Cobalt-60.

A first-order separable differential equation has the form \(\frac{dy}{dx} = g(x)h(y)\), meaning the variables can be separated on opposite sides of the equation. To solve it, you integrate both sides after separating the variables, which is what was done in the textbook solution by having \(\frac{dP}{P}\) on one side and \(kdt\) on the other.

The integration of these separated terms involves finding the antiderivative, which, in the case of \(\frac{dP}{P}\), leads to the natural logarithm function, symbolized as \(\ln\). The process of solving and understanding separable differential equations is a cornerstone of mathematical modeling and helps to predict and analyze real-world systems.

Half-Life Calculation

The concept of half-life is central to the study of radioactive decay and pharmacokinetics, providing a measure of how quickly a substance decreases to half of its initial amount. The half-life of a substance is not affected by its initial amount but rather is a property intrinsic to the substance.

To calculate the half-life, or in our exercise, to use the half-life to find another variable such as the decay constant \(k\), we set up an equation using the decay formula, \(P(t) = P_0e^{kt}\). By substituting the half-life period for \(t\) and \(0.5P_0\) for \(P(t)\), we create an equation that can be solved for \(k\). It's also useful to introduce the concept of natural logarithms to solve for \(k\) when we have \(e^{kt}\) isolated on one side of the equation.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is a mathematical function that is the inverse of the exponential function with base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. It's particularly useful in solving equations involving \(e\), which commonly appear in continuous growth or decay problems like radioactive decay.

In our case, the natural logarithm is used to find the decay constant \(k\) and the time \(t\) it takes for the radioactive substance to reach a certain percentage of its initial value.
For instance, \(e^{kt}\) becomes \(kt\) when we take the natural logarithm of both sides, enabling us to isolate and solve for the time \(t\). This function's ubiquity in science and economics stems from its ability to linearize exponential growth and decay, making it a powerful tool in analysis and calculation.