Problem 12 - 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 \(8 \mathrm{~g}\) of \(A\) will reduce to \(4 \mathrm{~g}\) in \(3 \mathrm{hr}\). After how long will there be only 1 g left?

Step-by-Step Solution

Answer

It will take approximately 9 hours for the substance to reduce to 1 g.

Step 1: Understand the Rate of Decomposition

The rate at which substance \(A\) decomposes can be described by the differential equation \( \frac{dA}{dt} = -kA \), where \(k\) is the positive proportionality constant. The negative sign indicates the decrease in \(A\) over time.

Step 2: Solve the Differential Equation

The differential equation \( \frac{dA}{dt} = -kA \) can be solved by separation of variables. Integrate both sides: \( \int \frac{1}{A} \, dA = - \int k \, dt \), resulting in \( \ln |A| = -kt + C \). Exponentiating both sides gives \( A = Ce^{-kt} \).

Step 3: Determine Initial Conditions

We know that at \( t = 0 \), \( A = A_0 \). Plugging these into the equation \( A = Ce^{-kt} \), we find \( A_0 = Ce^{0} \), which simplifies to \( C = A_0 \). Thus, \( A(t) = A_0 e^{-kt} \).

Step 4: Use Given Data to Find \(k\)

Given that \(A_0 = 8 \) g and the amount reduces to \(4 \) g in \(3 \) hrs, use the equation \(4 = 8 e^{-3k} \). Solving for \(k\), divide both sides by \(8\) to get \(0.5 = e^{-3k}\). Take the natural logarithm of both sides: \(-3k = \ln 0.5\). Hence, \(k = - \frac{\ln 0.5}{3}\).

Step 5: Solve for Time to Reduce to 1g

Use the equation \( A(t) = A_0 e^{-kt} \) with \( A = 1 \) g and \( A_0 = 8 \) g: \( 1 = 8 e^{-kt} \). Simplify to \( \frac{1}{8} = e^{-kt} \). Take the natural logarithm of both sides to get \( -kt = \ln \frac{1}{8} \). Substitute \(k = - \frac{\ln 0.5}{3}\) into the equation and solve for \(t\): \( t = \frac{3 \ln \frac{1}{8}}{\ln 0.5} \).

Step 6: Calculate Time Numerically

Calculate the numerical value of \( t \) using the earlier expression: \( t \approx \frac{3 \ln(0.125)}{\ln(0.5)} \). The calculation gives approximately \( t \approx 9\) hours.

Key Concepts

Rate of Decomposition

When we talk about the rate of decomposition of a substance, we mean how quickly the substance is breaking down into other substances. This rate is often described mathematically by a differential equation. For example, if substance \( A \) decomposes at a rate proportional to its current amount, we can express this by:

• \( \frac{dA}{dt} = -kA \)

Here, \( \frac{dA}{dt} \) is the rate of change of \( A \) over time, \( k \) is a positive constant of proportionality, and \( -kA \) shows that the amount \( A \) is decreasing.
The negative sign is key—it indicates that as time progresses, the amount of \( A \) becomes less. This tells us that the substance is decomposing rather than accumulating. Essentially, the larger the amount of \( A \), the faster it decreases, akin to how a bigger snowball rolls down the hill more quickly.

Exponential Decay

Exponential decay is a process where a quantity decreases at a rate proportional to its current value. In the context of decomposition, after integrating the differential equation, we arrive at the equation:

• \( A(t) = A_0 e^{-kt} \)

Now, \( A_0 \) is the initial amount of the substance at time zero, and \( e^{-kt} \) is the exponential factor accounting for the decay rate over time \( t \).
The base \( e \) is Euler's number, a mathematical constant approximately equal to 2.71828, and it's essential when calculating any growth or decay that naturally occurs in processes. With exponential decay, each moment, the substance reduces by a constant factor relative to its size—a feature characterized by the exponential function.
The decay result is significant. It means rapidly decreasing material that can be predicted accurately by this tidy little equation, assisting in real-world applications like predicting chemical reactions.

Separation of Variables

Separation of variables is a technique used to solve differential equations, clarifying how quantities change with one another. To solve our decomposition problem using separation of variables, we start with:

• \( \frac{dA}{dt} = -kA \)

This first-order differential equation involves moving all \( A \) terms to one side and all \( t \) terms to the other:

• \( \int \frac{1}{A} \, dA = - \int k \, dt \)

After integrating, we get a natural log on one side, \( \ln |A| \), which upon exponentiating leads us to the exponential decay formula. This elegant detour from the original differential equation allows us to derive the practical formula \( A(t) = A_0 e^{-kt} \).
This technique highlights that the decomposition rate is intertwined with the changing amount of \( A \); separating variables unveils the relationship between time and the reduction of the substance, making it an invaluable tool in calculus, especially when dealing with real-life problems like radioactive decay or interest rates.