Problem 91 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The average state cigarette tax per pack (in dollars) from 2001 through 2007 is approximated by the function $$ T(t)=0.43 t^{0.43} \quad(1 \leq t \leq 7) $$ where \(t\) is measured in years, with \(t=1\) corresponding to the beginning of 2001 . a. Show that the average state cigarette tax per pack was increasing throughout the period in question. b. What can you say about the rate at which the average state cigarette tax per pack was increasing over the period in question?

Step-by-Step Solution

Answer

In summary, the average state cigarette tax per pack was increasing throughout the period from 2001 through 2007, as the derivative of the function T(t) = 0.43 * t^0.43 was positive, with T'(t) = 0.1849t^{-0.57}. However, the rate at which it was increasing was slowing down, since the exponent of t in the derivative function is negative.

Step 1: Find the derivative of T(t)

We have the function T(t) = 0.43 * t^0.43. First, let's find the derivative of T(t) with respect to t. We'll use the power rule for differentiation, which states that the derivative of a function x^n is nx^(n-1). \(T'(t) = \frac{d(0.43t^{0.43})}{dt}\) Apply the power rule: \(T'(t) = 0.43(0.43)t^{(0.43-1)}\)

Step 2: Simplify the derivative

Now, let's simplify the expression we've got: \(T'(t) = 0.1849t^{-0.57}\)

Step 3: Determine the sign of the derivative

Next, we need to check the sign of the derivative (T'(t)) within the interval (1 ≤ t ≤ 7). To do that, we can analyze the exponent of the t term in the derivative function. In this case, since the exponent is -0.57, raising any positive number t to this exponent will result in a positive value. So, T'(t) is positive for all t in the given interval. Thus, the function T(t) is increasing throughout the period from 2001 through 2007 (1 ≤ t ≤ 7).

Step 4: Analyze the rate at which the tax is increasing

From the derivative expression, we know that the function T(t) is increasing at a rate proportional to t^-0.57. Since the exponent is negative, we can conclude that the rate of increase of T(t) becomes slower as t increases. That implies that although the average state cigarette tax per pack was increasing over the period in question, its rate of increase was slowing down.

Key Concepts

Derivative

In calculus, the concept of derivatives plays an integral role in understanding how functions change. Derivatives represent the rate of change of a function with respect to its variables.
A simple way to think of a derivative is as a way of identifying how a function is changing at any particular point. If you visualize a graph of a function, the derivative at any point is equivalent to the slope of the tangent to the curve at that point.
For the function given as the average state cigarette tax, denoted by \( T(t) \), the derivative \( T'(t) \) provides insight into how the tax changes over the years. In mathematical terms, it allows us to examine the pace at which this tax increases between 2001 to 2007.

Increasing Function

An increasing function is a function that consistently increases as the input values increase. This concept can be best understood when you compare it to climbing up a hill: as you move ahead, you're always going up.
For a function to be classified as increasing, its derivative must be positive over the entire interval in question. This positivity signals that as the input \( t \) (in the given example, representing the year) grows, the output \( T(t) \) (the tax amount) also grows.

• In the provided exercise, \( T'(t) \) was calculated and found to be positive across the entire interval from 1 to 7.
• Thus, the function is increasing during the period of 2001 to 2007, showing us that the average state cigarette tax per pack was indeed rising consistently each year.

Rate of Change

The rate of change in mathematics describes how a quantity changes concerning another quantity. It's essentially the core idea behind the concept of a derivative in calculus.
This rate is particularly crucial in understanding real-world scenarios where you want to know not just if something is increasing or decreasing, but precisely how fast that change is happening.

• In the context of the provided exercise, the derivative \( T'(t) = 0.1849t^{-0.57} \) expresses the rate of change of the cigarette tax over the years.
• The rate was found to decrease over time, due to the negative exponent in the derivative. This indicates that even though the tax was increasing, the speed at which it increased was slowing down with each passing year.

Power Rule of Differentiation

The power rule of differentiation is one of the simplest and most used rules in calculus. This rule allows us to find the derivative of functions of the form \( x^n \). According to the power rule, the derivative of \( x^n \) is \( nx^{n-1} \).
This rule was applied in the exercise to find the derivative of the function \( T(t) = 0.43t^{0.43} \).
By using the power rule, we calculated the derivative as \( T'(t) = 0.43(0.43)t^{-0.57} \), simplifying the process of finding the rate at which the tax was changing.
The power rule is extremely handy and allows for straightforward differentiation of polynomial expressions, saving time and preventing errors when calculating complex derivatives.