Problem 359 - Calculus Volume 1

Question

For the following exercises, use \(y=y_{0} e^{k t}\). The effect of advertising decays exponentially. If \(40 \%\) of the population remembers a new product after 3 days, how long will \(20 \%\) remember it?

Step-by-Step Solution

Answer

It takes approximately 6 more days for 20% to remember.

Step 1: Identify Given Values and Objective

We are given that initially, 40% of the population remembers the product after 3 days. This gives us two points: at \( t = 3 \), \( y = 0.4 \). We want to determine when 20% remembers the product, i.e., find \( t \) when \( y = 0.2 \).

Step 2: Express the First Scenario in the Exponential Formula

Use the equation \( y = y_{0} e^{k t} \) at \( t = 3 \) with \( y = 0.4 \). You get: \[ 0.4 = y_{0} e^{3k} \]

Step 3: Express the Second Scenario in the Exponential Formula

Similarly, use the equation for when the percentage is 20%, i.e., \( y = 0.2 \): \[ 0.2 = y_{0} e^{k t} \]

Step 4: Solve for the Decay Constant \( k \) and Initial Condition \( y_0 \)

From the first equation \[ y_{0} e^{3k} = 0.4 \], solve for the initial condition \( y_0 \) or use it to express \( e^{3k} = \frac{0.4}{y_{0}} \). Substitute this expression back into the second equation \[ y_{0} e^{kt} = 0.2 \], allowing both sides to manipulate and solve for \( k \).

Step 5: Simplify and Solve for Time \( t \)

Divide the second equation by the first to eliminate \( y_0 \): \[\frac{0.2}{0.4} = \frac{e^{k t}}{e^{3k}}\] Which simplifies to:\[ 0.5 = e^{k(t - 3)}\] Taking natural logarithms of both sides, we get: \[ t - 3 = \ln(0.5)/k\] Therefore, solve for \( t \) to find: \[ t = \frac{\ln(0.5)}{k} + 3\]

Step 6: Solve for \( k \) Using the Initial Condition

From Step 2 rearrangement: \[ k = \frac{\ln(0.4/y_{0})}{3}\]Assuming an initial memory \( y_0 \) of 100%, substitute this into our expression from Step 5 for a value of \( k \) and solve the time duration required for \( y = 0.2 \).

Step 7: Compute Final Value for Time \( t \)

Use the previously derived formula to find \( t = 3 + \ln(0.5)/\left(\frac{\ln(0.4)}{3}\right)\). Simplifying this gives the value for \( t \).

Key Concepts

Decay Constant

The decay constant, often denoted as "\( k \)", plays a crucial role in exponential decay problems. It quantifies the rate at which a process, such as memory retention or radioactive decay, diminishes over time.

The decay constant is a fundamental part of the exponential decay formula:

• In the formula \( y = y_{0} e^{kt} \), \( k \) indicates how quickly the quantity \( y \) reduces from its initial value \( y_0 \).
• When \( k \) is negative, the equation describes a decay process. The larger the magnitude of \( k \), the faster the rate of decay.

To determine \( k\), we often require at least one additional measurement aside from the initial condition. Once \( k \) is known, it can be used alongside current observations to predict future values or to calculate conditions such as the time needed to reach a certain state.

Understanding the decay constant is essential for interpreting how systems evolve over time, making it particularly useful in fields like physics, biology, and economics. Think of it as a gauge of time's impact on any process that shows exponential decay.

Exponential Function

An exponential function is a mathematical expression denoted by \( y = y_{0} e^{kt} \), where \( e \) represents Euler's number (approximately 2.718). This function is pivotal in modeling situations where something grows or decays at a rate proportional to its current value.

The characteristics of exponential functions include:

• When \( k \gt 0\), the function represents growth, meaning \( y \) increases over time.
• Conversely, if \( k \< 0 \), it signifies decay, indicating that \( y \) decreases as time progresses.
• The initial value \( y_{0} \) is the quantity when \( t = 0 \), providing a baseline from which changes are measured.

These functions are widely used due to their ability to model processes that change rapidly, such as population growth, radioactive decay, or even memory retention. By understanding how to manipulate and interpret exponential functions, you can predict future outcomes based on past and present data.

Mastery of exponential functions allows us to capture and analyze dynamic changes accurately across multiple domains.

Logarithmic Equations

Logarithmic equations are used to solve for variables that sit as exponents in exponential equations. Such solutions are essential when dealing with problems involving exponential growth or decay.

In the context of exponential decay, taking the logarithm of an equation can help isolate the decay constant or time variable.

• Given the equation \( 0.5 = e^{k(t-3)} \), taking the natural logarithm simplifies it to \( \ln(0.5) = k(t-3) \).
• This straightforward application of logarithms allows us to solve for one unknown in terms of another, highlighting their utility in these calculations.

Using logarithms turns multiplicative problems, common in exponentials, into additive ones, which are typically much easier to handle.

Becoming comfortable with manipulating logarithmic equations is key in solving real-world problems effectively, as they directly relate to questions asking for specific time points or rates in processes governed by exponential decay or growth.