Problem 38 - Calculus Early Transcendentals

Question

(a) What effect does increasing \(y_{0}\) and keeping \(k\) fixed have on the doubling time or half-life of an exponential model? Justify your answer. (b) What effect does increasing \(k\) and keeping \(y_{0}\) fixed have on the doubling time and half-life of an exponential model? Justify your answer.

Step-by-Step Solution

Answer

(a) Increasing \( y_0 \) does not affect the doubling time or half-life. (b) Increasing \( k \) decreases both the doubling time and half-life.

Step 1: Understanding the Exponential Model

The exponential model can be expressed as \( y(t) = y_0 e^{kt} \), where \( y_0 \) is the initial value, \( k \) is the growth or decay rate, and \( t \) is time. The doubling time (for growth) or half-life (for decay) depends on \( k \) and is calculated via specific formulas.

Step 2: Calculating Doubling Time and Half-Life

The doubling time \( T_d \) is given by \( T_d = \frac{ ext{ln}(2)}{k} \). Conversely, the half-life \( T_h \) is \( T_h = \frac{ ext{ln}(0.5)}{k} \). Both depend only on \( k \) and not on \( y_0 \).

Step 3: Effect of Increasing \( y_0 \) with Fixed \( k \)

Since doubling time and half-life formulas \( T_d = \frac{ ext{ln}(2)}{k} \) and \( T_h = \frac{ ext{ln}(0.5)}{k} \) do not include \( y_0 \), changing \( y_0 \) does not affect the doubling time or half-life. They remain unchanged.

Step 4: Effect of Increasing \( k \) with Fixed \( y_0 \)

Increasing \( k \) will decrease both \( T_d \) and \( T_h \) since \( T_d = \frac{ ext{ln}(2)}{k} \) and \( T_h = \frac{ ext{ln}(0.5)}{k} \). A higher \( k \) results in faster growth or decay, thus shortening the doubling time or half-life.

Key Concepts

Doubling Time

In the realm of exponential growth, doubling time is the period it takes for a quantity to double in size or value. This concept is crucial when dealing with population growth, investments, and any context where an exponential increase is observed. Doubling time is denoted by the formula \( T_d = \frac{\ln(2)}{k} \), where \( k \) is the growth rate.

Notice something interesting? The initial value, often labeled as \( y_0 \), doesn't come into play here. This ultimately means that doubling time is determined solely by the growth rate \( k \).

• Independent of \( y_0 \): Changes in \( y_0 \) do not influence the time it takes for doubling.
• Direct Link to \( k \): A larger value of \( k \) results in a shorter doubling time, as growth happens more swiftly.

This means when calculating or estimating the time for any exponential model to double, focus primarily on the rate \( k \), not the starting quantity.

Half-Life

Half-life is a term often used in the context of radioactive decay, but it broadly applies to any context where a value decreases exponentially. It represents the time required for any quantity to reduce to half its original amount. The formula for calculating half-life is \( T_h = \frac{\ln(0.5)}{k} \), similar to the doubling time but used for decay scenarios.

Just like doubling time, half-life is entirely governed by \( k \), the decay rate. What does this mean for the initial value \( y_0 \)?

• Effect of \( y_0 \): There is none. The initial quantity \( y_0 \) does not affect how long it takes for half of it to disappear.
• Influence of \( k \): A larger decay rate \( k \) means a shorter half-life, as substances decay faster.

Thus, when considering half-life, concentration should be on the rate \( k \), as it is the driving force in determining how quickly you reach half your starting amount.

Growth Rate

The growth rate, denoted as \( k \), is a pivotal factor in exponential models, influencing both the doubling time in growth scenarios and the half-life in decay situations. It encapsulates how quickly or slowly a process occurs, and it sits at the heart of the exponential equation \( y(t) = y_0 e^{kt} \). Let's break down its wide-reaching influences:

• Role in Doubling Time: As \( k \) increases, \( T_d = \frac{\ln(2)}{k} \) decreases, meaning systems grow quicker.
• Effect on Half-Life: Similarly, increasing \( k \) causes \( T_h = \frac{\ln(0.5)}{k} \) to shrink, leading to faster decay.
• Independent Variable: While \( y_0 \) sets the initial condition, \( k \) is the determining factor for the rate of change.

In essence, whether you are analyzing the speed of viral spreads, growth of an investment, or decay of a radioactive substance, \( k \) is your guiding star. Understanding and manipulating \( k \) allows you to predict and control outcomes in exponential models.