Problem 18 - Calculus

Question

Since \(1980,\) textbook prices have increased at \(6.7 \%\) per year while inflation has been \(3.3 \%\) per year. \(^{7}\) Assume both rates are continuous growth rates and let time, \(t,\) be in years since the start of \(1980 .\) (a) Write a differential equation satisfied by \(B(t),\) the price of a textbook at time \(t\) (b) Write a differential equation satisfied by \(P(t),\) the price at time \(t\) of an item growing at the inflation rate. (c) Solve both differential equations. (d) What is the doubling time of the price of a textbook? (e) What is the doubling time of the price of an item growing according to the inflation rate? (f) How is the ratio of the doubling times related to the ratio of the growth rates? Justify your answer.

Step-by-Step Solution

Answer

Doubling time for textbook is 10.35 years, and for inflation, it's 21.02 years. Doubling times are inversely related to growth rates.

Step 1: Differential Equation for Textbook Price

The price of a textbook at time \(t\), denoted as \(B(t)\), increases at a continuous rate of \(6.7\%\) per year. This can be represented by the differential equation: \[ \frac{dB}{dt} = 0.067B(t). \] The solution to this equation models exponential growth.

Step 2: Differential Equation for Inflation

The price of an item growing at the inflation rate, denoted as \(P(t)\), increases at a continuous rate of \(3.3\%\) per year. This is expressed by the differential equation: \[ \frac{dP}{dt} = 0.033P(t). \] This reflects continuous exponential growth at the inflation rate.

Step 3: Solve Textbook Price Equation

Solve the differential equation \(\frac{dB}{dt} = 0.067B(t)\) using separation of variables or recognizing it as a standard exponential growth model. Its solution is: \[ B(t) = B_0 e^{0.067t} \] where \(B_0\) is the initial textbook price at \(t = 0\).

Step 4: Solve Inflation Equation

Similarly, solve the equation \(\frac{dP}{dt} = 0.033P(t)\) to get: \[ P(t) = P_0 e^{0.033t} \] where \(P_0\) is the initial price of the item subject to inflation at \(t = 0\).

Step 5: Calculate Doubling Time for Textbook Price

To find the time it takes for \(B(t)\) to double, set \(2B_0 = B_0 e^{0.067t}\). Solving for \(t\), we get: \[ t = \frac{\ln(2)}{0.067} \approx 10.35 \text{ years}. \]

Step 6: Calculate Doubling Time for Inflation Price

Similarly, set up \(2P_0 = P_0 e^{0.033t}\) to find the doubling time for the price due to inflation. Solving for \(t\) yields: \[ t = \frac{\ln(2)}{0.033} \approx 21.02 \text{ years}. \]

Step 7: Relationship Between Doubling Times and Growth Rates

The doubling time is inversely proportional to the growth rate, implying that a higher growth rate results in a shorter doubling time. Specifically, the ratio of the doubling times \(\frac{10.35}{21.02}\) is the inverse of the ratio of the growth rates \(\frac{0.033}{0.067}\).

Key Concepts

Exponential Growth

Exponential growth is a fundamental concept in understanding how certain quantities increase over time. Unlike linear growth, where amounts rise by a fixed sum, exponential growth occurs when increases are proportional to the current value. This means that as the quantity gets larger, the increase per unit time becomes greater.

Mathematically, exponential growth is described by the equation \( \frac{dy}{dt} = ky \), where \( y \) represents the quantity, \( t \) is time, and \( k \) is the growth rate constant. This results in a solution of the form \( y(t) = y_0 e^{kt} \), where \( y_0 \) is the initial quantity. In our exercise, both the price of textbooks and inflation follow this model:

• For textbooks: \( \frac{dB}{dt} = 0.067B \)
• For inflation: \( \frac{dP}{dt} = 0.033P \)

Understanding exponential growth helps us model phenomena such as population growth, investment compounding, and price increases like those seen in textbook costs and inflation.
Recognize that every system following exponential growth will exhibit characteristics such as continuous increase without bound, as long as the growth rate remains positive.

Continuous Growth Rate

The continuous growth rate is key to understanding how an increase accumulates continuously over time. It allows us to account for the compounding effect of growth occurring at every instant rather than at discrete intervals. This is particularly useful in economics and finance.

Continuous growth rates are typically expressed as percentages. For instance, a continuous growth rate of 6.7% per year means that the quantity grows by approximately 6.7% of its current size every single moment.
In mathematical terms, the continuous growth rate \( k \) in an exponential formula \( y(t) = y_0 e^{kt} \) dictates how fast the quantity expands.

• If \( k = 0.067 \), as seen with textbook prices, the price increases rapidly.
• With \( k = 0.033 \), corresponding to inflation, the increase is slower.

Using continuous growth models, we can predict future values and understand ongoing changes, which is particularly important for planning and resource allocation.

Doubling Time

Doubling time is a vital metric in estimating how long it will take a quantity to grow to twice its initial size. It gives a clear picture of the rapidity of exponential growth.

To calculate doubling time, we use the formula \( t = \frac{\ln(2)}{k} \), where \( \ln(2) \) is the natural logarithm of 2, approximately equal to 0.693, and \( k \) is the growth rate.

• For textbook prices with a growth rate of 0.067, the doubling time is around 10.35 years.
• For an item subject to inflation at a rate of 0.033, the doubling time is approximately 21.02 years.

The relationship between doubling time and growth rate is inverse. A larger growth rate means a shorter doubling time, as seen with textbook prices compared to inflation. This principle helps us compare different growth scenarios easily and determine which situation changes more rapidly.