Problem 11 - Calculus and its applications

Question

Substance \(A\) decomposes at a rate proportional to the amount of \(A\) present. a) Write an equation that gives the amount \(A\) left of an initial amount \(A_{0}\) after time \(t\). b) It is found that \(10 \mathrm{lb}\) of \(A\) will reduce to \(5 \mathrm{lb}\) in \(3.3 \mathrm{hr}\) After how long will there be only 1 lb left?

Step-by-Step Solution

Answer

The equation is \( A(t) = A_0 e^{-kt} \). It takes approximately 10.99 hours to reduce to 1 lb.

Step 1: Understanding the Problem

We are given that substance \(A\) decomposes at a rate proportional to its current amount. This is a classic example of exponential decay, which is modeled by the differential equation \( \frac{dA}{dt} = -kA \), where \(k\) is a positive constant.

Step 2: Write the General Solution

To solve \( \frac{dA}{dt} = -kA \), we integrate both sides. This results in \( A(t) = A_0 e^{-kt} \), where \( A_0 \) is the initial amount of substance \(A\). Thus, the equation that gives the amount \( A \) left after time \( t \) is \( A(t) = A_0 e^{-kt} \).

Step 3: Use Given Values to Find k

We are told that \( A_0 = 10 \) lb reduces to \( 5 \) lb in \( 3.3 \) hr. Using the equation \( 5 = 10 e^{-3.3k} \), we can solve for \( k \). This simplifies to \( e^{-3.3k} = 0.5 \). Taking the natural logarithm of both sides, \( -3.3k = \ln(0.5) \), so \( k = - \frac{\ln(0.5)}{3.3} \).

Step 4: Solve for Desired Time where A = 1 lb

Now, use \( k \) and the equation \( A(t) = 10 e^{-kt} \) to find when \( A(t) = 1 \) lb. Substitute and solve: \( 1 = 10 e^{-kt} \), simplifying to \( e^{-kt} = 0.1 \). Taking the natural logarithm of both sides, \( -kt = \ln(0.1) \). Substitute \( k \) and solve for \( t \): \( t = - \frac{\ln(0.1)}{k} = 10.99 \) hr approximately.

Key Concepts

Differential Equation

Differential equations are mathematical expressions that describe how a particular quantity changes over time. They show the relationship between a function and its derivatives, which represent rates of change. In the context of exponential decay, the differential equation is typically written as \( \frac{dA}{dt} = -kA \). This equation describes how the substance \( A \) decreases over time. Here, \( A \) represents the amount of substance, \( t \) is time, and \( k \) is a positive constant that determines the rate of decay.

• The negative sign in \( -kA \) indicates that the amount is decreasing.
• The rate is proportional to the amount of substance currently present, meaning that as \( A \) decreases, so does the rate of decay.
• This results in an exponential decrease over time.

To solve these equations, integration is usually required, allowing us to find a general solution that predicts the amount of the substance at any time \( t \).

Exponential Function

The concept of an exponential function is central to understanding exponential decay. In general, an exponential function is of the form \( f(x) = a \cdot e^{bx} \), where \( e \) is the base of the natural logarithm, approximately equal to 2.71828. In the case of decay, the function used is \( A(t) = A_0 e^{-kt} \), where \( A_0 \) is the initial amount and \( k \) is the decay constant. This function describes how the quantity decreases exponentially over time.

• As \( t \) increases, \( e^{-kt} \) approaches zero, indicating the substance decreases down to nearly nothing.
• The decay rate is "exponential," meaning it decreases faster over time, particularly in the beginning when the amount is larger.
• Understanding the shape of this function helps to predict how quickly a substance will decay over time, which is useful in many scientific and engineering contexts.

Natural Logarithm

Natural logarithms are the inverse operations of exponential functions, and their base is the number \( e \). They are used to solve equations involving exponential functions by converting multiplication into addition, making equations easier to handle. In exponential decay, the natural logarithm helps us solve for unknown variables like time or the decay constant \( k \). For example, if we know the initial and current amounts of substance and want to find how long it took to reach the present amount, we use the natural logarithm to rearrange the equation.

• Taking the natural logarithm of both sides of an equation will turn an exponential relationship like \( e^{-kt} = 0.5 \) into a linear one, \( -kt = \ln(0.5) \).
• This step allows us to solve for \( t \) or \( k \) easily.
• Natural logarithms are particularly useful when working with continuous growth or decay processes, as in many scientific disciplines.