Problem 41 - Applied Calculus

Question

In 2000 , there were about 213 million vehicles (cars and trucks) and about 281 million people in the US. The number of vehicles has been growing at \(4 \%\) a year, while the population has been growing at \(1 \%\) a year. If the growth rates remain constant, when is there, on average, one vehicle per person?

Step-by-Step Solution

Answer

The average vehicle per person occurs in the year 2010.

Step 1: Define Variables

To solve the problem, define the following variables: let \( V(t) \) represent the number of vehicles at year \( t \), and \( P(t) \) represent the population at year \( t \). Let \( t = 0 \) be the year 2000.

Step 2: Initial Values and Growth Formulas

In year 2000, \( V(0) = 213 \) million vehicles and \( P(0) = 281 \) million people. Use the formula for exponential growth: \( V(t) = V(0) \, \times \, (1 + g_V)^t \) and \( P(t) = P(0) \, \times \, (1 + g_P)^t \), where \( g_V = 0.04 \) and \( g_P = 0.01 \).

Step 3: Setup Equation for Equality

Find the year \( t \) when \( V(t) = P(t) \). Substitute the growth formulas into the equality: \( 213 \times (1.04)^t = 281 \times (1.01)^t \).

Step 4: Simplify the Equation

Divide both sides by 213, giving \( (1.04)^t = \frac{281}{213} \times (1.01)^t \). Simplify this further to obtain the equation: \( \left( \frac{1.04}{1.01} \right)^t = \frac{281}{213} \).

Step 5: Solve for \( t \) Using Logarithms

Apply the logarithm: \( t \cdot \log_{10}\left( \frac{1.04}{1.01} \right) = \log_{10}\left( \frac{281}{213} \right) \). Solve for \( t \): \( t = \frac{\log_{10}(\frac{281}{213})}{\log_{10}(\frac{1.04}{1.01})} \).

Step 6: Compute \( t \)

Calculate the values: \( \log_{10}(\frac{281}{213}) \approx 0.119676 \) and \( \log_{10}(\frac{1.04}{1.01}) \approx 0.012136 \). So, \( t \approx 0.119676 / 0.012136 \approx 9.86 \).

Step 7: Determine the Year

Since \( t \approx 9.86 \), round up to the next whole year, which results in \( t = 10 \). Therefore, adding \( t = 10 \) to the base year 2000 gives the year 2010.

Key Concepts

Exponential Growth

Exponential growth is a process where the quantity of something, such as vehicles or population, increases by a fixed percentage each year. This type of growth is characterized by its accelerating increase over time; it grows faster as it becomes larger. For any quantity growing exponentially, the formula used is typically:

• \[ x(t) = x(0) \times (1 + g)^t \]

where:

• \( x(t) \) is the amount at time \( t \)
• \( x(0) \) is the initial amount,
• \( g \) is the growth rate expressed as a decimal

Let's apply this understanding to the scenario in the exercise: the vehicles grow at a rate of 4% per year, and the population grows at a rate of 1% per year. Through regular application of this growth principle, both figures expand, illustrating the compounding effect over time.

Vehicle Growth Rate

The vehicle growth rate refers to how quickly the number of vehicles is increasing over a period, particularly from 2000 in this exercise. This factor is critical as it determines when the number of vehicles will match the population count. Considering a growth rate of 4% per year:

• The formula \( V(t) = 213 \times (1.04)^t \) accounts for the exponential increase in vehicles.
• This percentage growth means cars and trucks grow significantly faster than the population.

The difference in growth rates between vehicles and population is why the vehicle count can potentially exceed the population count rapidly if these rates remain unchanged.

Logarithmic Calculation

Logarithms help solve exponential growth equations when determining the time it takes for one growing entity to catch up with another. In this exercise, finding when vehicles equal the population involves several steps:

• First, we set up the equation \( 213 \times (1.04)^t = 281 \times (1.01)^t \) to solve for \( t \).
• We simplify and rearrange using logarithms, specifically turning it into \( t \cdot \log_{10}(\frac{1.04}{1.01}) = \log_{10}(\frac{281}{213}) \).
• Solving it breaks down to \( t = \frac{\log_{10}(\frac{281}{213})}{\log_{10}(\frac{1.04}{1.01})} \).

Logarithms essentially enable finding \( t \) without tedious trial and error. They transform multiplicative exponential growth problems into more manageable linear calculations. In this case, they show that by the year 2010, roughly 10 years from 2000, the vehicle count would match the population count if growth rates remained constant.