Problem 43 - Calculus

Question

With time, \(t,\) in years since the start of \(1980,\) textbook prices have increased at \(6.7 \%\) per year while inflation has been \(3.3 \%\) per year. \(^{27}\) Assume both rates are continuous growth rates. (a) Find a formula for \(B(t),\) the price of a textbook in year \(t\) if it \(\operatorname{cost} \$ B_{0}\) in 1980 (b) Find a formula for \(P(t),\) the price of an item in year \(t\) if it cost \(\$ P_{0}\) in 1980 and its price rose according to inflation. (c) A textbook cost \(\$ 50\) in \(1980 .\) When is its price predicted to be double the price that would have resulted from inflation alone?

Step-by-Step Solution

Answer

In the year 2000, the textbook price is predicted to be double the inflation price.

Step 1: Identify the Problem

We need to find formulas for the price of a textbook, \( B(t) \), and the price of an inflation-adjusted item, \( P(t) \), over time \( t \) with given continuous growth rates.

Step 1: Derive the Formula for Textbook Price

Using continuous growth rate, the formula for the price \( B(t) \) of a textbook in year \( t \) is given by \( B(t) = B_0 e^{r_b t} \), where \( r_b = 0.067 \) and \( B_0 \) is the initial price in 1980.

Step 2: Derive the Formula for Inflation-adjusted Price

For inflation, the price \( P(t) \) of an item after \( t \) years is \( P(t) = P_0 e^{r_i t} \) where \( r_i = 0.033 \) and \( P_0 \) is the initial cost in 1980.

Step 3: Specific Initial Conditions for 1980 Cost

Given a textbook cost \( \$50 \) in 1980, we can rewrite the formulas as \( B(t) = 50 e^{0.067 t} \) and \( P(t) = 50 e^{0.033 t} \).

Step 4: Determine When Textbook Price Doubles Inflation Price

To find the time when the textbook price is double the inflation price, solve \( 50 e^{0.067 t} = 2 \times 50 e^{0.033 t} \). Simplify to get \( e^{0.067 t} = 2 e^{0.033 t} \).

Step 5: Solve the Exponential Equation

Divide both sides by \( e^{0.033 t} \) to get \( e^{0.067t - 0.033t} = 2 \), which simplifies to \( e^{0.034t} = 2 \). Take the natural logarithm of both sides to solve for \( t \): \( 0.034t = \ln 2 \), resulting in \( t = \frac{\ln 2}{0.034} \approx 20.39 \) years.

Step 7: Conclusion

The time when the textbook price will double the inflation-adjusted price is approximately 20.39 years from 1980, which corresponds to the year 2000.

Key Concepts

Inflation Rate

Inflation is the overall rise in prices over a period, which decreases the purchasing power of money. It is generally expressed as a percentage rate. Understanding inflation is essential for comparing prices over time, as it helps adjust those prices by accounting for the changing value of money.

In the context of continuous growth rate, the inflation rate tells us how quickly money's value erodes continuously over time. For example, an inflation rate of 3.3% means that each year, on average, the cost of goods and services rises by 3.3%.

To calculate the price of an item based on inflation, we use the formula:
\[ P(t) = P_0 e^{r_i t} \]
where:

• \( P(t) \) is the price at time \( t \).
• \( P_0 \) is the initial price.
• \( r_i \) is the continuous inflation rate as a decimal (0.033 in this case).
• \( t \) is the number of years since the initial time.

Inflation can significantly impact how we measure the value of our money over time, and it is important to factor in when planning long-term financial decisions.

Continuous Growth Rate

Continuous growth rate refers to a constant relative growth rate, often expressed as a percentage or decimal, that determines how quantities such as prices or populations increase exponentially over time.

This type of growth is particularly powerful as it assumes compounding at every possible instant, leading to exponential increases. The key is that the base of the natural logarithm \( e \) (approximately 2.71828) serves as the factor of continuous growth, enabled through the formula:
\[ A(t) = A_0 e^{rt} \]
where:

• \( A(t) \) represents the value at time \( t \).
• \( A_0 \) is the initial value at time zero.
• \( r \) is the continuous growth rate.
• \( t \) is time.

This formula allows us to track how a quantity evolves over time when it grows at a constant rate. For textbook prices, for instance, a continuous growth rate of 6.7% implies each year's price grows by this factor on top of the previous year's price, leading to rapidly increasing costs. Managing and understanding continuous growth rates are crucial in finance, environmental studies, and any field dealing with exponential processes.

Textbook Prices

The prices of textbooks, like other commodities, can be affected by inflation and specific market trends. In this example, textbooks have been increasing at a rate higher than inflation, at a continuous growth rate of 6.7% per year.

To predict the price of textbooks over time, given an initial price, we use the continuous growth formula:
\[ B(t) = B_0 e^{r_b t} \]
where:

• \( B(t) \) is the textbook price at time \( t \).
• \( B_0 \) is the textbook price at the initial time (e.g., in 1980).
• \( r_b \) is the continuous growth rate for textbook prices (0.067 in this case).
• \( t \) is the number of years since the initial time (1980).

Such precise forecasting is pivotal, as higher education costs can significantly impact students' budgets. Understanding these trends allows students and institutions to make informed financial decisions, potentially advocating for policy changes or budgetary allowances to accommodate inflation and specific market-driven price hikes. It highlights the importance of financial literacy and future planning among students and educators.