Problem 70 - Calculus Single Variable

Question

The amount \(y\) of an isotope satisfies$$\frac{d y}{d t}=-k \cdot y(0) \cdot e^{-k t} \frac{\mathrm{g}}{\mathrm{yr}}$$ when \(t\) is measured in years and \(y(0)\) is measured in grams. If, after one year, the amount \(y\) is one-third the initial amount \(y(0),\) then what is the value of \(k ?\)

Step-by-Step Solution

Answer

The value of \(k\) is \(\ln(3)\).

Step 1: Understand the Problem

We are given a differential equation describing the rate of decay of an isotope. The differential equation is \(\frac{dy}{dt} = -k \cdot y(0) \cdot e^{-kt}\). We know that after one year, the amount of isotope reduces to one-third of its initial amount.

Step 2: Set Up the Relationship after One Year

When \(t = 1\), the amount of the isotope is \(\frac{1}{3} y(0)\). Therefore, we have the equation \(y(1) = \frac{1}{3} y(0)\).

Step 3: Solve the Differential Equation

The given differential equation \(\frac{dy}{dt} = -k \cdot y(0) \cdot e^{-kt}\) can be integrated to solve for \(y(t)\). We integrate both sides with respect to \(t\) to find \(y(t) = y(0) e^{-kt}\).

Step 4: Substitute the Condition into the Solved Function

Substitute \(t = 1\) into the expression for \(y(t)\): \(y(1) = y(0) e^{-k}\). We know \(y(1) = \frac{1}{3} y(0)\), so the equation becomes \(\frac{1}{3} y(0) = y(0) e^{-k}\).

Step 5: Solve for \(k\)

From \(\frac{1}{3} = e^{-k}\), take the natural logarithm of both sides to solve for \(k\): \(-k = \ln(\frac{1}{3})\). Thus, \(k = -\ln(\frac{1}{3})\).

Step 6: Simplify the Expression for \(k\)

Since \(\ln(a)^{-1} = -\ln(a)\), \(k = \ln(3)\). This is the solution for the decay constant.

Key Concepts

Radioactive Decay

Radioactive decay is a fascinating process where unstable atomic nuclei lose energy by emitting radiation.
Over time, this natural phenomenon results in a decrease in the number of radioactive atoms. The rate at which a radioactive substance decays is measured and analyzed using a differential equation. This rate of decay can be defined by the Law of Radioactive Decay, represented through a first-order differential equation like \[\frac{dy}{dt} = -ky,\]where \(dy/dt\) is the rate of change of the isotope's amount, \(k\) is the decay constant, and \(y\) is the substance's amount at time \(t\).
When a specific amount of radioactive substance decreases over time, such as reducing to one-third of its initial quantity after a year, it can help us determine the decay constant \(k\). Understanding these principles enables us to predict how quickly a radioactive substance will diminish, which is crucial in areas like nuclear medicine, carbon dating, and nuclear power management.

Exponential Functions

Exponential functions play a critical role in describing processes that change proportionally over time.
Below are some important points about exponential functions:

• General Form: An exponential function is generally represented as \(y(t) = y(0) e^{-kt}\), where \(y(t)\) is the amount present at time \(t\), \(y(0)\) is the initial amount, \(e\) is the base of the natural logarithm, and \(k\) is the rate of growth or decay constant.
• Decay and Growth: In radioactive decay, \(k\) usually represents a decay constant, leading to a progressively smaller value of \(y(t)\) as \(t\) increases. Conversely, if \(k\) were a positive constant, it would represent exponential growth.

Knowing how exponential functions operate enables us to apply them to model diverse real-world scenarios, including population growth, financial investments, and, most importantly for this topic, radioactive decay.

Initial Value Problem

An Initial Value Problem (IVP) is a crucial concept in differential equations. It involves finding a function that satisfies a differential equation and fulfills an initial condition.
In our case, the initial value problem is to find \(y(t)\) solving the differential equation \(\frac{dy}{dt} = -k \cdot y(0) \cdot e^{-kt}\) with the condition \(y(1) = \frac{1}{3}y(0)\).
A step-by-step approach to solving an IVP involves:

• Understanding the Differential Equation: Analyze the given differential equation to understand its components and behavior.
• Solving the Equation: Integrate or apply other mathematical methods to find the general solution.
• Applying Initial Conditions: Use given initial conditions, such as \(y(1) = \frac{1}{3}y(0)\), to find particular solutions by substituting values back into the general solution.

This structured method allows us to find unique solutions that satisfy both the differential equation and initial conditions, providing complete insight into the problem at hand.