Problem 40 - Calculus Concepts

Question

Postage Rates Between 1919 and \(1995,\) the rate of change in the rate of change of the postage required to mail a first-class, 1 -ounce letter was approximately 0.022 cent per year squared. The postage was 2 cents in \(1919,\) and it was increasing at the rate of approximately 0.393 cent per year in \(1958 .\) (Source: Based on data from the United States Postal Service) a. Write a differential equation for the rate of change in the rate of change of the first-class postage for a 1 -ounce letter in year \(t,\) where \(t\) is the number of years after 1900 . b. Find both a general and a particular solution to the differential equation in part \(a\). c. Use the previous results to estimate how rapidly the postage is changing in the current year and the current first-class postage for a 1 -ounce letter. Comment on the accuracy of the results. If they are not reasonable, give possible explanations.

Step-by-Step Solution

Answer

Differential: \( \frac{d^2P}{dt^2} = 0.022 \); Postage 2023: ~89.41 cents; Rate change in 2023: ~1.823 cents/year.

Step 1: Set Up the Differential Equation

Given that the rate of change of the rate of change of postage is a constant 0.022 cent per year squared, we model this as a second derivative. Let \( P(t) \) be the postage rate in year \( t \) (where \( t \) is the number of years after 1900). Then the rate of change of the rate of change of \( P(t) \) is \( \frac{d^2P}{dt^2} = 0.022 \).

Step 2: Find the General Solution

To solve \( \frac{d^2P}{dt^2} = 0.022 \), integrate with respect to \( t \) to find the first derivative: \( \frac{dP}{dt} = 0.022t + C_1 \), where \( C_1 \) is an integration constant. Integrate again to find \( P(t) \): \( P(t) = 0.011t^2 + C_1t + C_2 \), where \( C_2 \) is another integration constant.

Step 3: Apply Initial Conditions

Use the given initial conditions to find \( C_1 \) and \( C_2 \). In 1919, the postage \( P(19) = 2 \) cents. In 1958, the rate of change \( \frac{dP}{dt}|_{t=58} = 0.393 \). Substitute these into the equations: \( 2 = 0.011(19)^2 + 19C_1 + C_2 \) and \( 0.393 = 0.022(58) + C_1 \). Solve these equations to find \( C_1 \) and \( C_2 \).

Step 4: Solve for Constants

From \( 0.393 = 1.276 + C_1 \), solve for \( C_1 = -0.883 \). Substitute \( C_1 \) into the first equation to find \( C_2 \): \( 2 = 0.011(361) - 0.883(19) + C_2 \). Calculate to find \( C_2 \): \( C_2 \approx 18.737 \).

Step 5: Estimate Current Postage Rate and Change

Use the particular solution \( P(t) = 0.011t^2 - 0.883t + 18.737 \) to estimate the rate of change and postage. For the current year 2023, \( t = 123 \): the rate of change \( \frac{dP}{dt}|_{t=123} = 0.022(123) - 0.883 \approx 1.823 \) and \( P(123) = 0.011(123)^2 - 0.883(123) + 18.737 \approx 89.41 \) cents.

Step 6: Comment on the Accuracy

The current estimation for the rate and cost of postage might not be accurate. If the actual postage in recent years is different, it could be due to changes in postal policies, inflation, or technological changes that were not accounted for in the simple quadratic model.

Key Concepts

Postage Rate

Postage rate is a term that refers to the price of sending a letter or parcel through the postal service. In our exercise, we look at how these rates have changed over a lengthy period, from 1919 to 1995. During this time, the cost of mailing a 1-ounce first-class letter increased, with its rate of change itself having a further rate of change. This is a concept known in calculus as a **second derivative**. The postage rate is significant as it reflects not only the cost of mailing but can also indicate economic factors like inflation or changes in postal service efficiency. Here, we identify the rate of change in postage over time and use it for further calculations.

Second Derivative

The second derivative is an essential concept in calculus, measuring how the rate of change of a quantity is itself changing. In our exercise, the second derivative of the postage rate function, \( \frac{d^2P}{dt^2} \), is given as a constant 0.022 cent per year squared. This indicates that the acceleration, or the rate of change of the rate of change, of the postage cost is consistent over time.

Understanding second derivatives is fundamental because they tell us about the concavity of the graph of a function. If the second derivative is positive, as in the case here, the graph is concave up, and the rate of increase of postage is accelerating.

Quadratic Model

A quadratic model is used when the relationship between variables can be graphed as a parabola. It typically involves a second-degree polynomial of the form \( at^2 + bt + c \). In this problem, the quadratic model \( P(t) = 0.011t^2 + C_1t + C_2 \) explains how postage prices change over time. The inclusion of a quadratic term (\( t^2 \)) reflects the second derivative constant and enables us to capture the acceleration in postage rates.

Using a quadratic model is appropriate here because we are dealing with a constant second derivative. It helps us interpolate or predict past and future values based on historical data, though it may not account for external factors such as technological advancements or policy changes.

Integration Constants

Integration constants \( C_1 \) and \( C_2 \) emerge when finding the antiderivative in solving differential equations, like in our postage problem. These constants represent initial conditions or specific values of a function at given points in time, which help personalize a general solution to a particular situation.

For example, in our exercise, we use the initial condition that the postage was 2 cents in 1919 to help find \( C_2 \), and the rate of change of 0.393 cents per year in 1958 to solve for \( C_1 \). By substituting these into our derived equations, we tailor the general formula to the historical conditions given. These constants are crucial for refining mathematical models to more closely fit real-world data.