Problem 10 - Calculus Concepts

Question

The percentage of people in the United States who earn at least \(t\) thousand dollars, \(25 \leq t \leq 150,\) can be modeled as $$ p(t)=119.931\left(0.982^{t}\right) \text { percent. } $$ a. Is \(p\) increasing or decreasing on the interval \(25 \leq t \leq 150 ?\) b. What is the concavity of \(p\) on the interval \(25 \leq t \leq 150 ?\)

Step-by-Step Solution

Answer

a. Decreasing; b. Concave up on \(25 \leq t \leq 150\).

Step 1: Analyze the Function Nature

To determine if the function is increasing or decreasing, consider the base of the exponential component. The function given is \[ p(t) = 119.931(0.982)^{t} \]where the base, 0.982, is less than 1. For exponential functions, when the base is less than 1, the function decreases as \( t \) increases. Hence, \( p(t) \) is decreasing on \([25, 150].\)

Step 2: Determine the First Derivative

To mathematically confirm that \( p(t) \) is decreasing, find the first derivative \( p'(t) \). \[p(t) = 119.931(0.982)^{t}\]First, apply the chain rule to differentiate:\[p'(t) = 119.931 \ln(0.982) \, (0.982)^{t}\]Since \( \ln(0.982) < 0 \), \( p'(t) < 0 \) for \( 25 \leq t \leq 150 \), confirming the function is decreasing.

Step 3: Find the Second Derivative

Now, assess the concavity by finding the second derivative \( p''(t) \). Take the derivative of the first derivative:\[p'(t) = 119.931 \ln(0.982) \, (0.982)^{t}\]Differentiating again gives:\[p''(t) = 119.931 \ln(0.982)^2 \, (0.982)^{t}\]Since \( \ln(0.982) < 0 \), \( \ln(0.982)^2 > 0 \), indicating that \( p''(t) > 0 \). Thus, the function is concave up on \( 25 \leq t \leq 150 \).

Key Concepts

Exponential Function

Exponential functions are a crucial part of calculus, characterized by a constant base raised to the power of a variable. In the function \( p(t) = 119.931(0.982)^{t} \), the base is 0.982, which is less than one. This specific feature implies that output values decrease as the input \( t \) increases. This behavior is typical of exponential functions when the base is between zero and one, as each increment in \( t \) results in multiplying by the base, which effectively shrinks the value of the function.
Exponential functions in calculus often represent real-world scenarios where growth or decay must be analyzed. The decay, in this case, comes from the diminishing earnings percentage relative to increasing values of \( t \). Understanding whether an exponential function depicts increasing or decreasing behavior can be essential for interpreting data patterns and trends.
Some key properties of exponential functions include:

• Fast initial changes, either growing or shrinking rates depending on the base.
• Behavior heavily steep or flattened, illustrating rapid decrease or increase.
• Limited range with outputs strictly positive or negative, governed by the function's constants.

First Derivative

The first derivative of a function provides valuable information about its rate of change. For \( p(t) \), determining \( p'(t) \) gives insights into whether it is increasing or decreasing over its domain. The derivative can be calculated using the chain rule, which is essential when handling exponential elements in the function.
For \( p(t) = 119.931(0.982)^{t} \), the chain rule aids in differentiating the exponential component, resulting in:
\[ p'(t) = 119.931 \ln(0.982) \, (0.982)^{t} \] Here, \( \ln(0.982) \) is negative, indicating that \( p'(t) \) is negative across the defined interval \([25, 150]\). This confirms that the function is decreasing, as a negative first derivative implies that as \( t \) increases, \( p(t) \) decreases.
The first derivative's sign is a simple way to determine function direction:

• If \( p'(t) > 0 \), the function is increasing.
• If \( p'(t) < 0 \), the function is decreasing.

Second Derivative

Exploring the second derivative reveals the concavity of the function, helping to understand its curvature nature. For our function, the second derivative \( p''(t) \) can be calculated by differentiating \( p'(t) \) again. It assesses how the rate of change (first derivative) is itself changing. Here, this results in:
\[ p''(t) = 119.931 \ln(0.982)^2 \, (0.982)^{t} \]
Since \( \ln(0.982) \) is negative, its square \( \ln(0.982)^2 \) is positive. Thus, \( p''(t) > 0 \), signifying that the function is concave up throughout \([25, 150]\). In simpler terms, the graph of \( p(t) \) is shaped like a right-side up bowl, suggesting that while it decreases, it does so at a declining rate.
Concavity tells us a lot about the behavior of functions:

• If \( p''(t) > 0 \), the curve is concave up (bowl-shaped).
• If \( p''(t) < 0 \), the curve is concave down (upside down bowl-shaped).

Understanding concavity is important because it offers insights into optimization problems and helps us anticipate function trends over particular intervals.