Problem 31 - Calculus Early Transcendentals

Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. A quantity that increases at \(6 \% /\) yr obeys the growth function $$ y(t)=y_{0} e^{0.06 t} $$ b. If a quantity increases by \(10 \% / \mathrm{yr},\) it increases by \(30 \%\) over 3 years. c. A quantity decreases by one-third every month. Therefore, it decreases exponentially. d. If the rate constant of an exponential growth function is increased, its doubling time is decreased. e. If a quantity increases exponentially, the time required to increase by a factor of 10 remains constant for all time.

Step-by-Step Solution

Answer

Question: Determine the truth values of the following statements and provide an explanation if true or a counterexample if false. a) A quantity grows exponentially at 6% per year, and thus its growth function is represented by the equation \(y(t) = y_0 e^{0.06t}\). b) If a quantity increases by 10% every year, its value over a 3-year period will increase by 30%. c) If a quantity decreases by one-third of its value every month, it undergoes exponential decay. d) The doubling time of a quantity decreases as the growth rate constant (rate of change) increases. e) In exponential growth, the time it takes for a quantity to increase by a given factor remains constant, regardless of its initial value. Answer: a) True. The growth function for this quantity is \(y(t) = y_0 e^{0.06t}\). b) False. The actual percentage increase over 3 years is approximately 33.1%, not 30%. c) True. The decay function can be expressed as \(y(t) = y_0 (\frac{2}{3})^t\). d) True. The doubling time decreases as the rate constant (growth rate) increases. e) True. The time required to increase by a given factor remains constant in exponential growth.

Step 1: Statement a

The statement is true. A growth function has the equation \(y(t) = y_0 e^{kt}\), where \(y_0\) is the initial value, \(k\) is the growth rate, and \(t\) is the time. In this case, the growth rate is \(6\%\) per year, or \(0.06\) in decimal form. Therefore, the growth function for this quantity is \(y(t) = y_0 e^{0.06t}\).

Step 2: Statement b

The statement is false. To find the actual percentage increase, we must consider the effect of compounding. If a quantity increases by \(10\%\) per year, it can be represented by the growth function \(y(t) = y_0 (1+0.10)^t\). Over 3 years, the function becomes \(y(3) = y_0(1.10)^3\). The total increase after 3 years is \((1.10)^3 - 1 \approx 0.331\), or \(33.1\%\), not \(30\%\).

Step 3: Statement c

The statement is true. When a quantity decreases by a constant factor (like one-third), it follows an exponential decay function. In this case, it can be expressed as \(y(t) = y_0 (\frac{2}{3})^t\), where \(y_0\) is the initial value and \(t\) is the time in months.

Step 4: Statement d

The statement is true. The doubling time of an exponential growth function can be obtained using the formula \(T_d = \frac{ln 2}{k}\), where \(T_d\) is the doubling time and \(k\) is the rate constant. Since the denominator, \(k\), is in the denominator, when the rate constant (growth rate) increases, the doubling time decreases.

Step 5: Statement e

The statement is true. When a quantity is growing exponentially, the time it takes to increase by a given factor remains constant. This can be verified with the growth function \(y(t) = y_0 e^{kt}\). Suppose the initial value is multiplied by a factor of \(10\), then we have \(10y_0 = y_0 e^{kt}\). Canceling out \(y_0\), we get \(10 = e^{kt}\). This implies that \(t = \frac{ln 10}{k}\), which is a constant, independent of the initial value.

Key Concepts

Exponential Growth

Exponential growth occurs when the rate of growth of a quantity is proportional to its current value. This means as the quantity grows, its increase rate also accelerates.

The formula that represents exponential growth is:

• \[ y(t) = y_0 e^{kt} \]
• Here, \( y(t) \) is the quantity at time \( t \), \( y_0 \) is the initial amount, \( e \) is the base of natural logarithms, and \( k \) is the growth rate.

This concept is evident in many natural and economic systems, such as populations, investments, and technology.

For a clear example, consider an investment that grows at 6% per year. Over time, not only the investment amount grows, but the rate at which it grows also increases. This results in a "snowball effect," with more substantial growth in later periods.

Exponential Decay

Exponential decay is similar to exponential growth, but instead of increasing, the quantity decreases over time. This means the rate of decline becomes smaller as the overall quantity reduces.

Such a process can be expressed by the decay function:

• \[ y(t) = y_0 (1 - r)^t \]
• Where \( y_0 \) is the initial quantity, \( r \) is the decay rate, and \( t \) is the time.

Exponential decay often describes processes such as radioactive decay, depreciation of assets, or reduction of drug concentration in the body.

For instance, if a quantity decreases by one-third every month, this situation is accurately modeled by exponential decay. Such consistent reduction over time aligns well with the principles of exponential decay.

Doubling Time

Doubling time refers to the time it takes for a growing quantity to double in size. This is a handy concept for understanding how quickly quantities grow exponentially.

You can calculate doubling time using:

• \[ T_d = \frac{\ln 2}{k} \]
• Where \( T_d \) is the doubling time and \( k \) is the rate constant of growth.

Because of its sensitivity to changes in the rate constant \( k \), a small adjustment in \( k \) can significantly alter the doubling time. For example, increasing \( k \) results in shorter doubling times because the quantity grows faster.

Understanding doubling time is crucial in fields like finance and population studies, where forecasting growth rates accurately can impact decision-making.

Compounding Effect

The compounding effect is a powerful phenomenon where the value of an investment grows because interest is being earned on both the initial principal and the accumulated interest from prior periods.

This can be captured by the compound interest formula:

• \[ A = P (1 + r/n)^{nt} \]
• Where \( A \) is the future value of the investment, \( P \) is the principal investment, \( r \) is the annual interest rate, \( n \) is the number of times interest is compounded per year, and \( t \) is the time in years.

Unlike simple interest where interest is only earned on the principal, compounding can significantly amplify growth over time.

Imagine investing in a scheme that offers 10% per year; after 3 years, the growth will be more than simply 30%. This factor highlights the importance of considering compounding in financial decisions, as it results in more gains than naive linear calculations would suggest.