Problem 31 - Calculus Early Transcendentals

Question

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

a. A quantity increasing at a rate of 6% per year obeys the growth function \(y(t)=y_{0} e^{0.06 t}\) b. If a quantity increases by 10% per year, it increases by 30% over 3 years. c. If a quantity is decreased by one-third every month, it is decreased exponentially. d. If the rate constant of an exponential growth function is increased, the doubling time decreases. e. For an exponentially increasing quantity, the time required to increase by a factor of 10 remains constant for all time. Answer: a. True - The given function represents a growth of 6% per year. b. False - The actual growth is 33.1% over 3 years, not 30%. c. True - The given scenario represents exponential decay with a decrease by one-third every month. d. True - As the rate constant increases, the doubling time decreases. e. True - The time required to increase by a factor of 10 is constant for an exponentially increasing quantity, regardless of the current time.

Step 1: a. Determine the truth of growth function statement

The statement asserts that a quantity increasing at a rate of 6% per year obeys the growth function \(y(t)=y_{0} e^{0.06 t}\). This is a classic example of exponential growth. To determine if it is true, we can rewrite the given expression as \(y(t) = y_0 (1 + 0.06)^t\). As we can see, it represents a growth of 6% per year. Therefore, the statement is true.

Step 2: b. Determine the truth of a 10% per year increase leading to a 30% increase over 3 years

The statement claims that if a quantity increases by 10% per year, it increases by 30% over 3 years. This is a linear growth statement, but the actual growth here is exponential. Instead, we should have the formula: \(y(3) = y_0(1 + 0.10)^3\). Here, y(3) is the quantity after 3 years. So the growth will be: \(y(3)/y_0 = (1.1)^3 = 1.331\) This means there is a 33.1% increase, not a 30% increase as stated. Therefore, the statement is false.

Step 3: c. Determine the truth of a quantity being decreased exponentially after decreasing by one-third every month

The statement says that a quantity decreases by one-third every month, implying an exponential decrease. If a quantity has exponential decay, then it can be expressed by the formula \(y(t) = y_0 (1 - R)^t\), where R is the decay rate. In this case, R = 1/3. The formula becomes \(y(t) = y_0 (2/3)^t\) representing exponential decay with a decrease by one-third every month. So, the statement is true.

Step 4: d. Determine the truth of doubling time decreasing after increasing the rate constant

The statement claims that when the rate constant of an exponential growth function is increased, the doubling time decreases. Doubling time is the time it takes a quantity to double in value. We can rewrite the exponential growth function as \(y(t) = y_0( 1 + r)^t\), where r is the rate constant. Now, double the quantity: \(2y_0 = y_0(1+r)^t\) This simplifies to \((1+r)^t=2\). Taking the natural logarithm of both sides, we have: \(t * \ln(1 + r) = \ln 2\) This gives us the doubling time, t: \(t = \frac{\ln 2}{\ln(1+r)}\) From the equation, we can see that if r increases, the denominator in the fraction also increases. This means that the doubling time, t, will decrease. Therefore, the statement is true.

Step 5: e. Determine the truth of constant time required to increase by a factor of 10 for an exponentially increasing quantity

The statement claims that for an exponentially increasing quantity, the time required to increase by a factor of 10 remains constant for all time. We start with the exponential growth formula: \(y(t) = y_0 (1 + r)^t\) To increase by a factor of 10: \(10y_0 = y_0 (1 + r)^t\) Simplifying, we have: \((1+r)^t = 10\) Similar to the previous step, we take the natural logarithm of both sides: \(t * \ln(1+r) = \ln 10\) The time to increase by a factor of 10 is: \(t = \frac{\ln 10}{\ln(1+r)}\) This equation shows that the time to increase by a factor of 10 depends only on the rate constant r, not on the current time or initial value \(y_0\). Thus, the statement is true, as the time required to increase by a factor of 10 is constant for an exponentially increasing quantity, regardless of the current time.

Key Concepts

Exponential Decay

Exponential decay describes how a quantity diminishes over time following a consistent percentage rate. It is expressed with the formula \(y(t) = y_0 (1 - R)^t\), where:

• \(y_0\) is the initial amount.
• \(R\) is the decay rate, a number between 0 and 1, representing the fraction of reduction per time period.
• \(t\) is the time elapsed.

This formula shows a steady reduction in amount by the same percentage over equal time intervals.
To understand this better, consider a substance that decays by one-third every month. The decay rate \(R\) would be 1/3. Therefore, the expression becomes \(y(t) = y_0 (2/3)^t\). This signifies that the substance retains two-thirds of its quantity each month.
Exponential decay can model various real-world processes such as radioactive decay, depreciation of asset values, or cooling of an object. In these situations, knowing how fast something is decaying—its rate—helps predict future quantities quite effectively.

Doubling Time

Doubling time refers to the duration required for a quantity experiencing exponential growth to become twice its size. A key point with exponential growth is that the doubling time remains constant regardless of when you start measuring. If you know the rate of growth, you can calculate the doubling time using the formula:\[t_{d} = \frac{\ln 2}{\ln(1 + r)}\]where:

• \(t_d\) is the doubling time.
• \(r\) is the growth rate expressed as a decimal (for example, 6% becomes 0.06).

So, if an investment grows by 6% annually, the doubling time is \(\frac{\ln 2}{\ln(1 + 0.06)}\) which calculates to about 11.90 years. This means regardless of whether you start in year 1 or year 20, it will always take approximately 11.90 years to double.
This characteristic of constant doubling time is essential in financial planning, population studies, and any scenario where exponential growth models fit.

Rate Constant

The rate constant, often symbolized by \(r\), is a key player in understanding exponential growth and decay. It quantifies how quickly or slowly a quantity is increasing or decreasing. The rate constant is expressed as a fraction or percentage of the initial value, consistent across time.
In exponential growth, the growth function is expressed as:\(y(t) = y_0 (1 + r)^t\)

• Here, \(r\) represents the growth rate per time period.

When analyzing exponential decay, \(y(t) = y_0 (1 - R)^t\), \(R\) stands for the decay constant.
Adjusting the rate constant impacts the speed of growth or decay. For instance, increasing the rate constant in a growth scenario means the quantity grows faster, which also reduces the time required to achieve milestones like doubling it.
Hence, understanding the rate constant is crucial for projecting future scenarios accurately, whether you're dealing with investments, population growth, or decay processes.

Growth Function

A growth function mathematically represents how something expands over time. Typically, in exponential growth, you encounter the function:\(y(t) = y_0 e^{kt}\)

• \(y_0\) is the initial amount.
• \(e\) is the base of the natural logarithm, approximately equal to 2.718.
• \(k\) is the constant rate of growth.
• \(t\) is the time elapsed.

Using a growth function epitomizes scenarios where growth accelerates rapidly rather than uniformly. For example, if a population grows by 6% per year, we use\(k = 0.06\).
This leads to exponential growth because the population increase isn't a simple addition, but rather a consistent incorporation of a percentage of the existing population. Thus, every year the amount added becomes larger.
This model can apply to finances, biology, and technology where understanding rapid or compound growth trends is crucial for future planning and decision-making.