Problem 23 - Calculus and its applications

Question

In \(2011,\) U.S. federal receipts (money taken in) totaled \(\$ 2.30\) trillion. In \(2013,\) total federal receipts were \(\$ 2.77\) trillion. Assume that the growth of total federal receipts, F, can be modeled by an exponential function and use 2011 as the base year \((t=0)\). a) Find the growth rate \(k\) to six decimal places, and write the exponential function \(F(t),\) for total receipts in trillions of dollars. b) Estimate total federal receipts in \(2015 .\) c) When will total federal receipts be \(\$ 10\) trillion?

Step-by-Step Solution

Answer

Growth rate \( k \approx 0.094614 \); receipts in 2015 estimated at $3.38 trillion; $10 trillion by 2026.

Step 1: Understand the Exponential Growth Model

The exponential growth function is generally modeled as \( F(t) = F_0 e^{kt} \), where \( F_0 \) is the initial value (receipts in 2011), \( k \) is the growth rate, and \( t \) is the number of years since the start year (2011 in this case). We have \( F_0 = 2.30 \) trillion.

Step 2: Set Up Equation for 2013

Given that in 2013 (\( t = 2 \)), \( F(t) = 2.77 \) trillion, substitute the known values into the exponential equation to find \( k \): \( 2.77 = 2.30 \, e^{2k} \).

Step 3: Solve for k

Rearrange the equation to solve for \( k \): \( e^{2k} = \frac{2.77}{2.30} \). Therefore, \( 2k = \ln\left(\frac{2.77}{2.30}\right) \) and \( k = \frac{1}{2} \ln\left(\frac{2.77}{2.30}\right) \). Calculate \( k \) to six decimal places: \( k \approx 0.094614 \).

Step 4: Write the Exponential Function

Using the value of \( k \), write the exponential function: \( F(t) = 2.30 \times e^{0.094614t} \).

Step 5: Estimate Receipts for 2015

For 2015, \( t = 4 \). Substitute \( t \) into the exponential function: \( F(4) = 2.30 \times e^{0.094614 \times 4} \). Calculate \( F(4) \) to estimate total receipts in 2015. \( F(4) \approx 3.38 \) trillion.

Step 6: Solve for When Receipts Reach $10 Trillion

Set \( F(t) = 10 \) and solve for \( t \): \( 10 = 2.30 \times e^{0.094614t} \). Rearrange to find \( t \): \( e^{0.094614t} = \frac{10}{2.30} \), thus \( 0.094614t = \ln\left(\frac{10}{2.30}\right) \), yielding \( t \approx \frac{\ln\left(\frac{10}{2.30}\right)}{0.094614} \). Calculate \( t \) to find the year when receipts will reach \$10 trillion. \( t \approx 15.65 \), meaning sometime in 2026.

Key Concepts

Exponential Function

An exponential function is a type of mathematical model where a quantity grows or decays at a constant relative rate over time.
In the context of the exercise, the total federal receipts grow exponentially. This means that as the years pass, the amount of receipts increases at a rate that is proportional to the current amount.

• The general form of an exponential function is expressed as: \( F(t) = F_0 e^{kt} \), where:
  • \( F(t) \) is the value of the function at time \( t \).
  • \( F_0 \) is the initial amount or starting value. Here, it is the receipts in 2011, which is \( 2.30 \) trillion dollars.
  • \( k \) is the growth rate.
  • \( t \) is the time that has passed since the beginning (measured in years, where 2011 is \( t = 0 \)).

The exponential function is powerful for modeling scenarios where growth accelerates over time, such as population increases, investment returns, or indeed, federal receipts.
By understanding the structure of such a function, you are better equipped to predict future values given an initial amount and a growth rate.

Growth Rate Calculation

Determining the growth rate \( k \) is crucial for predicting future values in an exponential function.
In practical scenarios, like the one given, it involves working backward from known data to calculate \( k \). Once you know \( k \), you can apply it to the exponential function formula to predict future outcomes.
For the exercise, we used the information:

• 2011: \( F_0 = 2.30 \) trillion.
• 2013: \( F(t=2) = 2.77 \) trillion.

To find \( k \), substitute these into the exponential model, yielding:\[ 2.77 = 2.30 \cdot e^{2k} \]Solve for \( k \) by isolating the exponential component and using the natural logarithm:

• First, divide both sides by \( 2.30 \):\[ e^{2k} = \frac{2.77}{2.30} \]
• Next, take the natural logarithm (\( \ln \)) to break the exponential:\[ 2k = \ln\left(\frac{2.77}{2.30}\right) \]
• Finally, solve for \( k \):\[ k = \frac{1}{2} \cdot \ln\left(\frac{2.77}{2.30}\right) \approx 0.094614 \]

This calculation shows how quickly or slowly the total federal receipts are growing each year, with \( k \) representing the annual growth rate.

Mathematical Modeling

Mathematical modeling uses mathematical equations to represent and analyze real-world phenomena.
In this exercise, we applied an exponential model to predict federal receipts. This approach allows the use of existing data to forecast trends accurately, provided the growth pattern remains consistent.
The exponential model for federal receipts is given as:\[ F(t) = 2.30 \times e^{0.094614t} \]This model involves several steps to predict yearly values and assess long-term outcomes:

• Using it for future prediction, such as estimating receipts in 2015, which involved plugging \( t = 4 \) into the function, giving \( F(4) \approx 3.38 \) trillion.
• Determining when a threshold, like \$10 trillion of receipts, will be achieved, which is found by setting \( F(t) = 10 \) and solving for \( t \).

Mathematical models simplify complex systems and provide insights into growth trends. They highlight the effects of continuous growth and serve as essential tools in fields like economics, biology, and engineering. By understanding and utilizing these models, we can make informed decisions based on both current and projected data.