Problem 8 - Advanced Engineering Mathematics

Question

(a) The solution of \(d A / d t=k A\) is \(A(t)=A_{0} e^{k t} .\) Letting \(A=\frac{1}{2} A_{0}\) and solving for \(t\) we obtain the half-life \(T=-(\ln 2) / k\). (b) since \(k=-(\ln 2) / T\) we have $$A(t)=A_{0} e^{-(\ln 2) t / T}=A_{0} 2^{-t / T}$$. (c) Writing \(\frac{1}{8} A_{0}=A_{0} 2^{-t / T}\) as \(2^{-3}=2^{-t / T}\) and solving for \(t\) we get \(t=3 T .\) Thus, an initial amount \(A_{0}\) will decay to \(\frac{1}{8} A_{0}\) in three half-lives.

Step-by-Step Solution

Answer

An amount \( A_0 \) decays to \( \frac{1}{8} A_0 \) in three half-lives, \( t = 3T \).

Step 1: Understanding the Differential Equation Solution

We start with the differential equation \( \frac{dA}{dt} = kA \). The solution to this equation is given as \( A(t) = A_0 e^{kt} \), where \( A_0 \) is the initial amount of \( A \). This equation describes exponential growth or decay depending on the sign of \( k \).

Step 2: Finding the Half-Life Expression

Let \( A = \frac{1}{2} A_0 \) and substitute into the solution to find the time \( t \), which gives us the half-life. This yields \( \frac{1}{2} A_0 = A_0 e^{kt} \). By dividing both sides by \( A_0 \), we obtain \( \frac{1}{2} = e^{kt} \). Taking the natural logarithm of both sides, \( \ln \frac{1}{2} = kt \), which simplifies to \( t = \frac{-\ln 2}{k} \). We denote this time as the half-life, \( T \).

Step 3: Expressing k in terms of T

We already found \( T = \frac{-\ln 2}{k} \). Rearranging gives \( k = \frac{-\ln 2}{T} \). This expression for \( k \) is used to rewrite the original exponential expression in decay form: \( A(t) = A_0 e^{-(\ln 2)t/T} \).

Step 4: Reformulating the Decay Expression

Substituting \( k = \frac{-\ln 2}{T} \) into \( A(t) = A_0 e^{kt} \), we reformulate it to \( A(t) = A_0 2^{-t/T} \). This uses the property of exponents \( e^{x \ln a} = a^x \), where \( e^{-(\ln 2)t/T} = 2^{-t/T} \).

Step 5: Solving for Three Half-lives

To find when the quantity becomes \( \frac{1}{8} A_0 \), set \( \frac{A_0}{8} = A_0 2^{-t/T} \). This simplifies to \( 2^{-3} = 2^{-t/T} \) because \( \frac{1}{8} \) corresponds to \( 2^{-3} \). Equating the exponents, \(-3 = -t/T\), we solve for \( t \) to find \( t = 3T \). This means the initial amount \( A_0 \) decays to \( \frac{1}{8} A_0 \) in three half-lives.

Key Concepts

Differential Equations

Differential equations play a crucial role in understanding exponential decay in fields like physics, chemistry, and biology. In this context, a differential equation involves a function and its derivatives, which relate to rates of change. For exponential decay, consider the equation \( \frac{dA}{dt} = kA \). Here, \( A \) represents the quantity that is changing with time, \( t \), and \( k \) is a constant that determines the rate of change.
The general solution of this equation is \( A(t) = A_0 e^{kt} \), where \( A_0 \) is the initial quantity of \( A \). The exponential function \( e^{kt} \) describes how \( A \) grows or decays over time. If \( k \) is negative, the function represents decay because the quantity decreases as time passes. This equation helps us model real-world processes, like radioactive decay or even cooling of a hot object, where the rate of decline accelerates over time.
To solve such differential equations, one needs to integrate using initial conditions to find specific solutions that describe how a system changes.
Use differential equations to predict how systems evolve or decay in various scientific and engineering problems.

Half-Life

The concept of a half-life is pivotal in understanding exponential decay processes. Half-life refers to the time it takes for a substance to reduce to half its initial amount. It is commonly used in contexts like radioactive decay, where materials decrease steadily over time.
Given the equation \( A(t) = A_0 e^{kt} \), setting \( A = \frac{1}{2}A_0 \) allows us to solve for the half-life \( T \). By substituting, we find:

• \( \frac{1}{2}A_0 = A_0 e^{kT} \)
• \( \frac{1}{2} = e^{kT} \)
• Taking the natural logarithm, \( \ln \frac{1}{2} = kT \)
• Thus, \( T = \frac{-\ln 2}{k} \)

Knowing the half-life helps in predicting how long a quantity will take to diminish to a specific fraction of its original amount. If a material has a half-life of, for example, 3 years, it would take 6 years to reduce to a quarter of its initial amount. Understanding half-life concepts aids in fields like nuclear physics, pharmacology, and even finance.

Exponential Functions

Exponential functions are a fundamental mathematical tool used for modeling exponential decay. They take the form \( A(t) = A_0 e^{kt} \), where \( A_0 \) is the initial quantity, and \( k \) dictates growth or decay rate.
For decay, \( k \) is negative. The power of exponential functions lies in their ability to handle continuous growth or decay. They can model phenomena like population decline, radioactive decay, or cooling of objects. These functions follow certain pivotal properties:

• The rate of change is proportional to the current value.
• They show consistent behavior over time, unaffected by the initial size.
• The half-life remains constant over any given period.

Exponential functions simplify complex calculations using natural logarithms. For instance, rewriting \( A(t) = A_0 e^{-(\ln 2)t/T} \) helps relate it to base-2 decay form, \( A(t) = A_0 2^{-t/T} \).
These functions provide insights across various domains, helping students and professionals alike model decay processes effectively and predict future states.