Question

The population of Toledo, Ohio, in 2000 was approximately 500,000 . Assume the population is increasing at a rate of \(5 \%\) per year. a. Write the exponential function that relates the total population as a function of \(t\). b. Use a. to determine the rate at which the population is increasing in \(t\) years. c. Use b. to determine the rate at which the population is increasing in 10 years.

Short answer

The population increases by about 81,444 people after 10 years.

Step-by-step solution

Understand the Exponential Growth Formula

The formula for exponential growth is given by \( P(t) = P_0 \cdot (1 + r)^t \) where \( P_0 \) is the initial population, \( r \) is the growth rate as a decimal, and \( t \) is the time in years.

Identify Given Values

From the problem, the initial population \( P_0 \) is 500,000, and the growth rate \( r \) is 5\%, which is equivalent to 0.05 as a decimal.

Write the Exponential Function

Substitute the given values into the exponential growth formula: \( P(t) = 500,000 \cdot (1 + 0.05)^t \). Simplify this to \( P(t) = 500,000 \cdot (1.05)^t \).

Differentiate to Find the Growth Rate Function

To find the rate of increase, differentiate \( P(t) \) with respect to \( t \). The derivative is \( \frac{dP}{dt} = 500,000 \cdot \ln(1.05) \cdot 1.05^t \).

Calculate the Rate at 10 Years

Substitute \( t = 10 \) into the derivative function to find the rate of increase after 10 years: \( \frac{dP}{dt} = 500,000 \cdot \ln(1.05) \cdot 1.05^{10} \). Calculate this value to get approximately 81,444.22.

Key concepts

population growth

Population growth is essentially an increase in the number of inhabitants in a region over a period of time. In the context of Toledo, Ohio, we are considering an exponential model for population growth.
This means that instead of growing by a constant number each year, the population grows by a constant percentage. For Toledo, we start with an initial population, which is the population count in the year 2000. In mathematical terms, this is our starting point or initial population denoted as \( P_0 = 500,000 \).
The growth rate is given as a percentage — in this case, 5% per year. However, when we use this in calculations, it is important to convert this percentage into a decimal by dividing by 100, resulting in \( r = 0.05 \).

• The exponential function we use to describe this growth is \( P(t) = P_0 \cdot (1 + r)^t \).
• Inserting Toledo's values, it becomes \( P(t) = 500,000 \cdot (1.05)^t \).

This function gives us a mathematical way to predict future populations by simply varying \( t \), the number of years after 2000.

derivative of exponential functions

When dealing with exponential functions that describe real-world problems like population growth, the derivative of these functions provides crucial insight. Specifically, it tells us how fast the population is changing at any given point in time.
In calculus, the process of finding a derivative of a function gives us a new function, called the rate of change function. For our exponential function, \( P(t) = 500,000 \cdot (1.05)^t \), the derivative \( \frac{dP}{dt} \) is calculated to find the rate of growth.

• The derivative is \( \frac{dP}{dt} = 500,000 \cdot \ln(1.05) \cdot 1.05^t \).
• Here, \( \ln(1.05) \) is the natural logarithm of 1.05, which is an important constant when differentiating exponential functions.

This expression shows how the rate of population increase is related to both the natural logarithm of the growth factor \( 1.05 \) and the exponential function itself \( 1.05^t \). Analyzing the derivative enables us to understand not just the population at any time \( t \), but the pace at which it is growing.

calculus applications

Calculus isn't just an abstract mathematical concept; it has real-life applications, such as predicting population changes with exponential growth models. To tie it into our example, we use calculus to determine both the future population and its rate of change.
Once the derivative we calculated in the previous section is at our disposal, we can apply it to find the rate of growth at a specific point in time — say, 10 years after 2000. By setting \( t = 10 \) in \( \frac{dP}{dt} = 500,000 \cdot \ln(1.05) \cdot 1.05^t \), you essentially find out how many new people are added to the population in that particular year.

• Using \( t = 10 \), calculate: \( \frac{dP}{dt} = 500,000 \cdot \ln(1.05) \cdot 1.05^{10} \).
• This computation provides \( \approx 81,444.22 \), meaning around 81,444 people were added to the population in 2010.

This is a classic application of calculus, providing meaningful insights and predictions about real-world scenarios through mathematical modeling.