Problem 28 - Calculus

Question

Put the functions in the form \(P=P_{0} e^{k t}\). $$P=4(0.55)^{t}$$

Step-by-Step Solution

Answer

Rewritten function: \( P = 4e^{t \ln(0.55)} \).

Step 1: Identify Key Components

We are given the function \( P = 4(0.55)^{t} \). We need to express this function in the form \( P = P_{0} e^{k t} \). Here, \( P_{0} \) is the initial amount and \( k \) is the growth rate that we have to find.

Step 2: Rewrite the Function

Notice that we need our base to be \( e \). The current form has a base of \( 0.55 \). We rewrite the expression using the identity \( a^{t} = e^{t \ln a} \). This gives us: \( P = 4 \cdot (0.55)^{t} = 4 \cdot e^{t \ln(0.55)} \).

Step 3: Extract Parameters

Now we can see that \( P_{0} = 4 \). The exponent in the new form is \( kt \) where \( k = \ln(0.55) \). So the function in the required form is \( P = 4e^{t \ln(0.55)} \) or simply \( P = 4e^{k t} \) where \( k = \ln(0.55) \).

Step 4: Final Form

Thus, the function in the form \( P = P_{0} e^{k t} \) is: \[ P = 4 e^{t \ln(0.55)}. \] Here, \( P_{0} = 4 \) and \( k = \ln(0.55) \).

Key Concepts

Growth Rate

In exponential functions, the growth rate is a crucial component that determines how quickly or slowly a function increases or decreases over time. This growth rate is usually denoted by the variable \( k \) in the expression \( P = P_{0} e^{k t} \). Let's break it down some more:

• Exponential growth occurs when the growth rate \( k \) is positive. This indicates that the quantity is increasing.
• Exponential decay happens when \( k \) is negative. This suggests that the quantity is decreasing.
• The rate of change in exponential functions is constant and proportional to the value of the function at any point. This characteristic is what makes exponential functions so powerful in modeling real-life phenomena such as population growth or radioactive decay.

For the function \( P = 4(0.55)^{t} \), the growth rate \( k \) is calculated as \( \ln(0.55) \). Since \( 0.55 \) is less than 1, its natural logarithm will be negative, indicating that this is a case of exponential decay.

Natural Logarithm

The natural logarithm, often denoted as \( \ln \), is a special mathematical function. It is the logarithm to the base \( e \), where \( e \) is approximately 2.718. The natural logarithm provides a way to unravel exponential expressions and solve for variables that serve as exponents.

• One of the essential properties of the natural logarithm is that \( \ln(e^x) = x \). This is important when converting exponentials with different bases to base \( e \).
• In the exercise, to change \( (0.55)^t \) to a base \( e \) expression, we used \( a^{t} = e^{t \ln a} \) resulting in \( (0.55)^t = e^{t \ln(0.55)} \).
• The natural logarithm appears in many scientific and engineering domains because it handles continuous growth or decay processes very flexibly.

Natural logarithms are thus instrumental when rewriting exponential expressions, specifically in this case, when converting \( (0.55)^t \) into a form that fits base \( e \).

Base e

The number \( e \) is a mathematical constant known for being the base of natural logarithms. It plays a central role in exponential functions of the form \( P = P_{0} e^{k t} \). Understanding the properties of \( e \) helps in simplifying computations involving continuous growth and decay.

• Base \( e \) naturally arises in calculus, particularly in problems dealing with growth rates and integrals involving exponential functions.
• It is an irrational number, meaning it cannot be exactly expressed as a simple fraction.
• The value of \( e \), approximately 2.718, is key while working with natural logarithms, as \( \ln(e) = 1 \).

In the problem solution, the expression was converted from a base of 0.55 to base \( e \) to align it with the form \( e^{k t} \). This conversion allows for a more universal and mathematically convenient representation of exponential growth or decay processes.