Question

The population \(P\) (in thousands) of Orlando, Florida from 1980 through 2005 can be modeled by \(P=131 e^{0.019 r}\) where \(t=0\) corresponds to \(1980 .\) (Source: U.S. Census Bureau) (a) According to this model, what was the population of Orlando in 2005 ? (b) According to this model, in what year will Orlando have a population of \(300,000 ?\)

Short answer

The population of Orlando in 2005 was approximately 196,200. The population will reach 300,000 around the year 2019 according to the given model.

Step-by-step solution

Finding the population in 2005

For part (a), \(t = 2005 - 1980 = 25\) means that \(t = 25.\) Substitute \(t = 25\) into \(P=131 e^{0.019 t}\) to find the population in 2005. The population then \(P = 131 e^{0.019 * 25}\).

Calculation

Doing the calculation, \(P = 131*e^{0.475} = approx. 196.2\) (in thousands). So, according to this model, the population in Orlando in 2005, was approximately 196,200.

Solving for the year when the population will reach 300,000

For part (b), set \(P = 300\) (since the population is in thousands) and solve for \(t\). So, \(300 = 131 e^{0.019 t}\).

Solving the equation

Divide both sides by 131 to isolate the exponential term: \(e^{0.019 t} = 300/131 = approx. 2.29\). Now take the natural logarithm of both sides: \(\ln(e^{0.019t}) = \ln(2.29)\). From the properties of logarithms, \(0.019t*ln(e) = \ln(2.29)\ and\ since\ ln(e) = 1,\ so\ 0.019t = \ln(2.29)\).

Calculation

Solve for \(t= \ln(2.29)/0.019\). Doing the calculation, \(t= approx. 39.48\). Adding this to 1980 gives the year as approximately 2019. So, according to this model, the population of Orlando will reach 300,000 in around 2019.

Key concepts

Exponential Growth

Exponential growth occurs when a quantity increases at a rate that is proportional to its current value. This results in a rapid escalation of the quantity over time. The general formula often used to represent exponential growth is \( P(t) = P_0e^{rt} \), where \( P(t) \) is the quantity at time \( t \), \( P_0 \) is the initial quantity, \( r \) is the growth rate, and \( e \) is the base of the natural logarithm, approximately equal to 2.71828.

In the context of population modeling, exponential growth indicates that the population is growing by a certain percentage each year. For example, if the population of Orlando is growing exponentially according to the formula \( P = 131e^{0.019t} \), the growth rate is 0.019 per year, and the population is expected to increase more rapidly as time progresses. It's important to note that exponential growth models are idealized and may not always account for factors that could limit growth, such as resource depletion or environmental resistance.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is the inverse operation of taking the exponential of a number with the base \( e \). This means that if \( y = e^x \), then \( x = \ln(y) \). The natural logarithm is fundamental in solving equations where the variable is an exponent, as it allows us to 'undo' the exponential function.

For example, when solving the equation \( 300 = 131e^{0.019t} \), the natural logarithm is used to isolate \( t \). By taking \( \ln \) of both sides, the exponent \( 0.019t \) becomes the main focus, making it possible to solve for \( t \) directly. It is critical to remember that \( \ln(e) = 1 \), which simplifies the process since multiplying by 1 does not change the value.

Solving Exponential Equations

Solving exponential equations involves isolating the variable that is in an exponent. This is often achieved by utilizing logarithms. When these equations are in the form of \( a \cdot e^{rt} = b \), where \( a \) and \( b \) are known constants, \( r \) is the known rate, and \( t \) is the unknown time, the procedure typically involves dividing both sides by \( a \) to get \( e^{rt} = b/a \) and then applying the natural logarithm to both sides.

To continue our example, after dividing by 131, we get \( e^{0.019t} = \frac{300}{131} \) and taking the natural logarithm of both sides results in \( 0.019t = \ln(\frac{300}{131}) \). This equation can then be easily solved for \( t \) by dividing both sides by 0.019. Using natural logarithms is particularly handy because it helps us reduce the exponent part of an exponential equation to a linear form, which is much simpler to handle.

Mathematical Modeling

Mathematical modeling uses mathematical language to describe real-world situations. In population growth modeling, a mathematical model attempts to represent how a population changes over time. These models are constructed based on assumptions and observed data. The model provided for Orlando's population uses an exponential growth function to estimate population changes.

Creating a model involves identifying variables and constants that impact the situation. In our case, the constant 131 represents the population (in thousands) in the base year 1980, \( e \) signifies the base of natural logarithm indicating continuous growth, 0.019 is the rate of growth per year, and \( t \) represents the number of years since 1980. Such a model helps in predicting future populations, assessing past growth and informing planning decisions. However, it's essential to remember that models are simplifications and may not fully capture all the dynamics of real-life scenarios.