Problem 28 - Applied Calculus

Question

In Problems \(25-28,\) put the functions in the form \(P=P_{0} e^{k t}\). $$P=4(0.55)^{t}$$

Step-by-Step Solution

Answer

The function is \( P = 4e^{t \cdot \text{ln}(0.55)} \).

Step 1: Identify the Given Function

The given function is \( P = 4(0.55)^t \). This function is in the form \( P = a \cdot b^t \), where \( a = 4 \) and \( b = 0.55 \).

Step 2: Convert Base to Exponential Form

To convert to the form \( P = P_0 e^{kt} \), we need to express \( 0.55^t \) using base \( e \), the natural exponential base. We start by writing \( 0.55 \) as an exponential with base \( e \): \( 0.55 = e^{ ext{ln}(0.55)} \).

Step 3: Substitute Back into the Original Function

Use the expression from Step 2: \( 0.55 = e^{ ext{ln}(0.55)} \). Substitute it back into the function: \( P = 4(e^{ ext{ln}(0.55)})^t \).

Step 4: Simplify the Exponentiation

Simplify the exponentiation: \( (e^{ ext{ln}(0.55)})^t = e^{t \cdot ext{ln}(0.55)} \). Thus, the function becomes \( P = 4e^{t \cdot ext{ln}(0.55)} \).

Step 5: Identify Parameters in the New Function Form

Identify the parameters: \( P_0 = 4 \) and \( k = \text{ln}(0.55) \). Hence, the function is expressed as \( P = 4e^{k t} \), where \( k = \text{ln}(0.55) \).

Key Concepts

Natural Logarithm

The natural logarithm is an essential concept when dealing with exponential equations, particularly those involving the base \( e \). The natural logarithm, denoted as \( \text{ln} \), is used to bring the expressions involving exponential functions into a more manageable form.

• It is the inverse operation of exponentiation involving the base \( e \).
• If \( a = e^b \), then \( b = \text{ln}(a) \).
• The natural logarithm can solve for the exponent in exponential equations.

For example, in the expression \( 0.55 = e^{\text{ln}(0.55)} \), we are expressing the number 0.55 as an exponential with base \( e \), which allows us to then substitute it into exponential forms such as \( P = P_0 e^{kt} \). This conversion is crucial as it aligns the function with the natural exponential base \( e \), making the analysis of exponential functions more straightforward.

Exponential Growth and Decay

Exponential growth and decay describe processes that increase or decrease at rates proportional to their current value, which is often depicted by the function form \( P(t) = P_0 e^{kt} \). Understanding whether a function represents growth or decay depends on the sign of \( k \).

• When \( k > 0 \), the process describes exponential growth.
• When \( k < 0 \), the process indicates exponential decay.

In the problem, the original function \( P = 4(0.55)^t \) is transformed into \( P = 4e^{t \cdot \text{ln}(0.55)} \). Here, \( k = \text{ln}(0.55) \), and because \( 0.55 \) is less than 1, \( \text{ln}(0.55) \) will be a negative number, indicating exponential decay. This is useful for exploring real-world phenomena such as population decreases, radioactive decay, or cooling processes.

Function Transformation

Function transformation involves changing the form of a mathematical function to a different but equivalent form. This can involve altering the base or the parameters of the function.

• Transformations help to analyze functions under different forms and are vital in mathematical modeling.
• A common transformation is changing from one base to another, particularly to the natural exponential base \( e \).
• In our case, transforming the function form \( P = 4(0.55)^t \) to \( P = 4e^{kt} \) using the natural logarithm allows it to fit the exponential model.

This is done by recognizing that \( 0.55 \) can be rewritten as \( e^{\text{ln}(0.55)} \), which facilitates applying exponential functions that are more innate to calculus analysis. Transformations simplify comparing growth rates, determining half-lives, or understanding continuous growth or decay processes.